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Fundamentals of mechanics
FOCUS AREA
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SYLLABUS OUTCOMES
A student:
• analyses motion and events using scientific laws and models PY-11-01
CHAPTERS
2. Quantities of motion and motion relationships
3. Forces and motion
4. Momentum and energy
Quantities of motion and motion relationships 2
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SYLLABUS
CONTENT DOT POINTS: QUANTITIES OF MOTION AND MOTION RELATIONSHIPS
At the end of this chapter, I will be able to:
2A.1 Classify distance, displacement, time, speed, velocity and acceleration as either scalar or vector quantities
2A.2 Use vector diagrams to represent the displacement, velocity and acceleration of an object
2A.3 Add and subtract vectors in 2 dimensions
2A.4 Resolve vectors into components
2A.5 Solve total distance and displacement problems
2A.6 Explain how instantaneous speed, average speed, instantaneous velocity and average velocity are measured and calculated
2A.7 Solve problems using v av = Δ s Δt
2A.8 Solve problems involving the calculation of velocity in 2 dimensions
2A.9 Solve problems to determine speed, velocity, distance, displacement and time
2A.10 Solve problems in one and 2 dimensions using a = Δv Δt
2B.1 Use data to construct displacement–time graphs, velocity–time graphs and acceleration–time graphs
2B.2 Use data to analyse motion graphs to describe the displacement, velocity and acceleration of an object
2B.3 Analyse motion graphs to determine the relationships between displacement, initial velocity, final velocity, acceleration and time
2C.1 Solve problems involving uniformly accelerated motion in one dimension using s = u t + 1 2 a t 2 , v = u + a t and v 2 = u 2 + 2 a s
2C.2 Conduct a laboratory experiment to compare 2 methods of graphically determining a value for acceleration due to gravity and assess the accuracy and reliability of the experimental results
2C.3 Analyse problems of relative velocity involving motion in two dimensions using vAB = vA vB
SECTION
2A Scalar and vector quantities of motion
2B Analysing motion through graphs
2C Analysing uniformly accelerated motion and relative velocity
GLOSSARY TERMS acceleration
average speed
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average velocity component vectors displacement distance free fall gradient instantaneous speed instantaneous velocity magnitude net force origin parabolic path quantitative rectangular components resultant vector scalar SI units speed vector vector resolution velocity
Concept map
Quantities of motion
Vectors and scalars are used to understand and quantify motion
2A Scalar and vector quantities of motion
2B Analysing motion through graphs
Graphing and analysing relationships between quantities of motion
2C Analysing uniformly accelerated motion and relative velocity
2A Scalar and vector quantities of motion
SYLLABUS CONTENT
At the end of this section, I will be able to:
2A.1 Classify distance, displacement, time, speed, velocity and acceleration as either scalar or vector quantities
Worksheet 2A.1
Concepts used to model motion
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quiz questions download word download
2A.2 Use vector diagrams to represent the displacement, velocity and acceleration of an object
2A.3 Add and subtract vectors in 2 dimensions
2A.4 Resolve vectors into components
2A.5 Solve total distance and displacement problems
2A.6 Explain how instantaneous speed, average speed, instantaneous velocity and average velocity are measured and calculated
2A.7 Solve problems using v av = Δ s Δt
2A.8 Solve problems involving the calculation of velocity in 2 dimensions
2A.9 Solve problems to determine speed, velocity, distance, displacement and time
2A.10 Solve problems in one and 2 dimensions using a =
Scalars and vectors
The branch of physics in which motion and the forces causing motion are studied is called mechanics.
A physical quantity is anything in nature that can be measured and expressed as a number (magnitude) along with its unit.
In science, all quantitative measurements are either scalars or vectors
A scalar quantity only requires a magnitude; no direction is needed. For example, if someone asks you to pour 500 mL of milk to make a cake, it will not make sense to give a direction for the 500 mL of milk.
A vector is a quantity that requires both a magnitude (size) and direction. For example, if you give someone directions to a shop, just saying it is 500 m away is not sufficient. In this scenario, you will include a direction and say, for example, it is 500 m to the north.
Positive and negative signs can be applied to a vector to indicate the direction in which it points. For example, if the direction right or up is defined as positive (which is the usual convention), then left or down is negative. So, a displacement of –1 m simply means 1 m in the opposite direction. The magnitude of the vector is the same, and a displacement of –1 m is not less than a displacement of 1 m, so they are both the same sizes but opposite in direction.
2A.1 When modelling the world in motion, size and direction are usually equally important.
magnitude a value that defines only the size of a quantity
quantitative data or components that can be expressed or measured numerically
scalar a quantity that is fully described by magnitude (it does not need a direction)
vector a quantity that is described by both magnitude (how big it is) and a direction
displacement the direct straight-line distance with a direction joining the initial to final position of an object (i.e. it is the change in position)
Video: Concepts used to model motion
Figure
Table 2A.1 shows a list of examples of vector and scalar quantities.
Table 2A.1 Examples of vector and scalar quantities
Vector
Displacement, s
Velocity, v Acceleration, a Force, F
Momentum, p
Impulse, J Torque, τ
Gravitational field strength, g
Electrical field strength, E
Magnetic field strength, B
Scalar
Distance, d Speed, v Mass, m Volume, V Area, A Current, I Voltage, V Resistance, R Energy, E Power, P
Temperature, T Time, t
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distance the actual path length traversed by a moving object
NOTE
The vector arrow notation is explained later in this section.
Analysis of motion
Understanding how to analyse motion in a straight line is the starting point for analysing all motion. The analysis of motion is extensively used in a range of different fields such as flight, racing car driving and even for video umpires in sport.
Figure 2A.2 When hiking, the distance travelled and displacement are usually very different.
Distance versus displacement
Imagine a hiker who walks from camp A to camp B. The camps are separated by a distance of 10 km when you travel directly east. This means the displacement (vector quantity) of camp B from camp A is 10 km east – that is, it has a magnitude and a direction. However, it is impossible for the hiker to go directly east, as a steep mountain blocks the path. Therefore, to get to camp B, the hiker must walk a distance of 11.7 km around the mountain. This is the distance (scalar quantity) of the journey; it has only a magnitude.
Figure 2A.3 Example of a hike with a displacement of 10 km and a distance of 11.7 km
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This example highlights the difference between distance and displacement.
Displacement is the direct straight-line distance with a direction from the initial to final position of an object. Displacement is a vector quantity and therefore requires both a magnitude and a direction. Displacement is represented by the symbol s. When the hiker has reached camp B, their displacement from the origin (starting position) is 10 km east.
However, the hiker actually travelled 11.7 km in order to reach camp B; which is a measure of distance and is a scalar quantity, so it requires a magnitude only. Distance is the actual path length traversed by a moving object. Distance is represented by the symbol d.
The SI unit of distance and displacement is the metre (m).
For a moving object:
• Distance is always greater than or equal to displacement, because displacement is measured in a straight line and that is always the shortest path between two points.
• Distance can never be negative because that implies direction and it does not have direction.
• Displacement can be positive, negative or 0, where negative means the opposite direction to positive.
• Magnitude of displacement would be equal to distance travelled if the motion is along a straight line with no change in direction.
ACTIVITY 2A.1
Calculating routes to school
Consider three siblings, Layla, Chase and Mia, who each take different routes from their house to their school. The school is 5 km from their house in a straight line. Layla decides to take the bus, which tends to go on the side streets. Chase has a friend’s parent pick him up, and they primarily drive on the main roads. Mia decides to walk, and she manages to cut through a park to take the most direct route. Each of their routes are displayed in the diagram shown.
1. Compare the distance travelled by each sibling.
2. Compare the displacement travelled by each sibling.
3. Contrast distance and displacement using your findings in questions 1 and 2.
origin the initial position from where an object moves
SI units the International System of Units for measurement in science and engineering, based on seven basic physical quantities, within a decimal prefix structure, so units used to measure the same thing increase or decrease in multiples of 10
Camp B Camp A
Layla Chase Mia School
House
Video: Worked example 2A.1
WORKED EXAMPLE 2A.1
Distance and displacement
A track runner is running around a circular track that has a circumference of 500 m. The runner does exactly four laps, finishing where they started.
a. Calculate the distance travelled by the runner.
b. Calculate the displacement of the runner.
Solution
a. 500 × 4 = 2000 m
b. Displacement = 0 m
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Explanation
The distance is calculated by multiplying the length of the track by the number of laps.
Since the runner finishes where they started, they will have zero overall displacement because they have not moved relative to the origin.
Speed versus velocity
Velocity is the rate of change of displacement, meaning the change in displacement divided by the time taken. It is a vector quantity because it requires both a magnitude and direction. The direction of the velocity is the same as the direction of the change in displacement. Velocity can be either positive, when it is in the positive direction, or negative, when it is in the negative direction. For example, a velocity of 10 metres per second north means that every second you are moving 10 m further north. In this case a velocity of 10 metres per second means every second you are moving 10 metres further north. If we continue to define north as the positive direction, then a velocity of 10 metres per second in the negative direction (or a velocity of –10 m s –1) means that every second you are moving 10 m further south. The greater the velocity, the faster you are moving.
In general, when an object maintains its speed and travels with uniform velocity, also referred to as constant velocity, the symbol used to denote its magnitude is v This symbol is also used for the final velocity if either the magnitude or direction of velocity changes. The initial velocity in such a case is denoted by u.
Speed is the rate of change of distance (change in distance divided by change in time). This is a scalar quantity since it only involves a magnitude. The formula for speed is very similar to the formula for velocity; the two are shown side by side for comparison in Formula 2A.1.
For a moving object:
• Speed is always greater than or equal to velocity, since distance is always greater than or equal to displacement.
• Speed can never be negative, though it can be zero.
• Velocity can be negative, zero or positive. In fact, negative means the opposite direction to positive.
• Magnitude of velocity is equal to speed if the motion is along a straight line with no change in direction.
Instantaneous and average speed
Recall that speed is a scalar. Instantaneous speed is the speed of an object at any given instant of time. In a car, it is the speed the speedometer records at a given instant of time.
Average speed is the total distance travelled by an object divided by the total time taken. It does not matter whether the object speeds up, slows down or stops during its travel. Only the overall distance and the overall time are used.
If an object is moving with uniform speed, the instantaneous speed is equal to its average speed. To gauge the instantaneous speed of an object, the distance between two events is measured and the time between the two events is minimised as much as possible (actual instantaneous instantaneous speed is theoretically not achievable, since an instant of time cannot be measured). If the average speed is measured over a very short time interval, such as a millisecond or less, it can be taken as an approximation of instantaneous speed.
Instantaneous and average velocity
Similarly, the velocity can also be expressed at an instant or across a time interval. Recall that the velocity is a vector that depends on the displacement instead of the distance. Instantaneous velocity is the velocity of an object at any given instant of time. Average velocity is the change in displacement of an object divided by the total time taken.
Units for velocity and speed
In Stages 4 and 5 Science, the SI unit for speed and velocity, metres per second, is written as m/s, but in senior science the mathematical index notation m s –1 is used, where s –1 means “per second”.
instantaneous speed the speed of an object at any given instant of time
average speed the total distance travelled divided by the total time taken
instantaneous velocity the velocity of an object at any given instant of time
average velocity the change in displacement of an object divided by the time taken
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kmh–1
In everyday use and real-life applications, such as the speed of a vehicle or of a cricket ball, the non-SI unit “kilometres per hour” is most commonly used for speed, abbreviated to km/h (or km h–1 in a science context). Unless otherwise specified it should be converted to m s –1 for calculations in this course. The conversion factor to use is to divide the value in km h–1 by 3.6 (Figure 2A.4), so 60 km h–1 = 16.7 m s –1 .
Figure 2A.4 Conversion between kilometres per hour and metres per second
Note also that in everyday usage, without referring to direction, “velocity” is often used as a synonym for speed, but this should not be done in physics.
Representation of a vector symbolically
The symbol for a vector quantity is an arrow above the letter that represents the kind of vector, such as s for displacement and v for velocity. These will mostly be used from now on in this course. Note that the letter v is also used for speed, but without the arrow, meaning it is a scalar. The symbol d without an arrow is used for distance, which is also a scalar.
Worksheet 2A.2
FORMULA 2A.1
Average velocity and average speed
Analysing speed and velocity questions download word download pdf
Average velocity
Where:
v = Velocity (m s–1)
Δ s = Change in displacement (m)
Δ t = Change in time (s)
v = Speed (m s–1)
Δ d = Change in distance (m)
NOTE
Video: Worked example 2A.2
speed
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The triangle symbol Δ is the uppercase letter delta from the Greek alphabet. In mathematics and science, it is used to denote a “change in” a value. If velocity or speed is given in non-SI units such as kilometres per hour (km h–1), it should be changed to the SI units m s–1 in calculations, unless otherwise specified.
WORKED EXAMPLE 2A.2
Average speed and velocity
A bird flies east at 20 m s 1 for 200 m. It then turns around and flies west at 40 m s 1 for 30 seconds.
Neglecting the time taking to turn, calculate its average speed and average velocity. Solution Explanation
Calculate the time taken for the first part of the journey (towards east) t2 = 30 seconds d2 = 40 × 30 = 1200 m
The time taken for the second part of the journey (towards west) has already been given.
Calculate the distance travelled in the west direction. total time taken = 10 + 30 = 40 seconds total distance travelled = 200 + 1200 = 1400 m
Find the sum for the total time taken and distance travelled.
Calculate the average speed by dividing the total distance travelled by the time taken.
displacement = 1200 + 200 = 1000 m average velocity = displacement total time taken = −1000 40 = –25 m s 1
Calculate the displacement, then use this number to calculate the average velocity.
The negative sign indicates that the velocity is directed towards the west.
Acceleration
Acceleration is the rate of change of velocity, meaning the change in velocity divided by the time taken. It is represented by the symbol a , and it is a vector.
Imagine a car, starting from rest (0 m s–1) and travelling in a straight line in a particular direction, accelerating at 4 metres per second per second. By the end of the first second, that car will be travelling with a velocity of 4 m s –1 in that direction. When 2 seconds have passed, the car will have gained another 4 m s –1 to its velocity and will be travelling at 8 m s –1 in that direction.
The SI unit of acceleration is metres per second per second, also called metres per second squared. It is the unit of velocity (metres per second) divided by the unit of time (per second). You may see its symbol written as “m/s/s” or m/s2, but the official SI symbol is m s–2, and that is what you should use in this course. acceleration the rate of
FORMULA 2A.2
Acceleration
This can also be written as:
Where:
a = Acceleration (m s-2)
Δv = Change in velocity (m s–1)
v = Final velocity (m s–1)
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In everyday life, whenever an object increases its velocity, it is said to be accelerating and whenever its velocity decreases, it is said to be decelerating. Acceleration and deceleration may seem counterintuitive at times. For example, if an object accelerates from 8 m s 1 (or 8 m s 1 in the opposite direction to what was defined as positive) to 2 m s 1, its velocity has increased by 10 m s 1. Despite the increase in velocity, the object will be travelling more slowly. It is important to remember that what we witness as speed is actually the magnitude of the velocity.
• I f v > u, the final velocity of an object is greater than the initial velocity of the object, then the acceleration is positive. A positive acceleration indicates that the object is moving faster if the velocity is in the positive direction (or slowing down if the velocity is in the negative direction).
• For example, a car accelerates after passing the school zone. It increases its velocity from 10 m s–1 to 15 m s –1, a change in magnitude of +5 m s –1. If this takes 5 s, its acceleration is 1 m s-2 .
• If v < u, the final velocity of an object is less than the initial velocity of the object, then the acceleration is negative. A negative acceleration indicates that the acceleration is in the opposite direction to the positive. So, the object is slowing in velocity in the positive direction (decelerating) or increasing in (magnitude of) velocity in the opposite direction (accelerating in the reverse direction).
Worked example 2A.3
• For example, a car decelerates while approaching the school zone. It decreases its velocity from 15 m s–1 to 10 m s –1, a change in magnitude of –5 m s–1. If this takes 5 s, its acceleration is –1 m s–2. The negative sign indicates this is acceleration in the negative direction, in this case “decelerating” or slowing down.
WORKED EXAMPLE 2A.3
Acceleration multiple-choice question
A free-falling object increases its velocity from 4.9 m s 1 to 14.7 m s 1 downwards in one second. After a further second its velocity is 24.5 m s 1 downwards.
This data shows that the motion of a free-falling object is a case of A. uniform motion.
B. uniform acceleration.
C. no acceleration.
D. uniform velocity.
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Use the equation for average acceleration to determine the acceleration in the first second.
Determine the average acceleration in the next second.
B. This data shows uniform acceleration.Compare the acceleration in the first second against the acceleration in the next second. Both values are 9.8 m s–2, hence the acceleration is uniform.
Vector diagrams
A vector can be represented graphically or in a diagram, as well as in symbol form with an arrow above the letter, such as s, v or a
Figure 2A.5 As a raft moves into water that is moving more quickly, it will begin to accelerate in the direction of flow.
In a diagram, a vector is represented by a directed line segment (a straight line) with an arrowhead at one end, called the head, while the other end is called the tail, shown here. It should be labelled with the symbol of the vector, which is displacement in this case. The magnitude could be included in the label – for example “ F = 10 N”. The arrow is scaled; the length of the line segment gives the magnitude (size) of the vector, and its direction is represented by the arrow. Direction is usually measured as the angle in degrees of the arrow to a horizontal line (as in this diagram) or to a vertical line.
A vector could be shown on a set of compass axes as in Figure 2A.6, or just with a vertical or horizontal line if an angle measurement is included.
Conventions for vectors
Conventionally, a one-dimensional (1D) vector directed to:
• the east or to the right is positive
• the west or to the left is negative
• the north or up the page is positive
• the south or down the page is negative.
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Figure 2A.6 Representation of a force of 10 N acting 45° N of E, shown on a set of compass-direction coordinate axes. Alternatively, x and y axes could be used, or just positive x and y axes, or no axes at all.
• Two-dimensional (2D), three-dimensional (3D) and higher order vectors are commonly broken down into perpendicular one-dimensional (1D) components. Each component can be either positive or negative.
• For example, a 2D vector can be broken down into x and y components. A vector pointing northwest - using conventional positive directions - can be described as negative in the x-direction (west) and positive in the y-direction (north).
The direction of the vector can be described in a number of ways:
• In this course, directions are primarily given by the common physics-style notation that uses compass cardinal directions (north, east, south and west). Directions are described by an angle north/east/south/west of north/east/south/ west. For example, the vector in Figure 2A.6 would be read as 45 degrees north of east (or 45° N of E) and is commonly written as E45°N, where the reference compass direction is written first (in this example, that is E for east), the angle is written second (in this example, that is 45°), and the compass direction the angle is towards is written at the end (in this example, that is N for north).
• In maths, it is often stated as an angle in the anticlockwise direction from the positive x-axis. This is typically used on Cartesian planes (the two-dimensional, flat grid with a horizontal x-axis and a vertical y-axis that is used to locate points using pairs of coordinates) and can be impractical for vectors.
• Compass bearings from north are typically used in orienteering, where the direction in stated as an angle from north in a clockwise direction– that is, north is 0°.
• T here are many other notations for directions, such as unit vector notation, that are used in different situations.
Video: Worked example 2A.4 video
WORKED EXAMPLE 2A.4
Drawing vector diagrams
Draw labelled vector diagrams of the following:
a. a displacement of 3.7 m 30° S of W
b. a velocity of 110 m s–1 45° W of N.
Solution
a. 30° s = 3.7 m →
Explanation
Draw the horizontal line to indicate the east–west directions. The vector arrow should be oriented 30° south (down) from west (left) b.
questions
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Draw the vertical line to indicate the north–south directions. The vector arrow should be oriented 45° west (left) from north (up)
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Additional vector notation and terminology
Each vector quantity must have both a magnitude and direction. To represent the magnitude of a vector, vertical bars are used on either side of the vector symbol ( ). Often, a shorthand notation is used for the magnitude, where the symbol is written without the vector arrow. The direction is commonly represented by a theta (θ) with any additional descriptors included in subscript.
For example, when describing a velocity of 5 m s 1 to the southwest, using north and east as the conventional positive directions:
• Vector: v = 5 m s 1 at 45° E of N
• Magnitude: |v | = v = 5 m s 1
• Direction: θ = θ v = 45° W of S
When dealing with more than one vector, additional terminology is useful. Some examples are given in Table 2A.2. Recall that “positive” and “negative” apply to direction not magnitude.
Table 2A.2 Common terms for relating vectors to each other
Term
Description Diagram
Parallel vectorsTwo or more vectors having the same direction are parallel to each other. They could differ in magnitude; if not, they are equal vectors.
Equal vectorsParallel vectors having the same magnitude and same direction
Term Description Diagram
Antiparallel vectorsTwo vectors pointing in exactly opposite directions; they may have different magnitudes but one will be positive and the other negative
Opposite vectorsAntiparallel vectors having the same magnitude
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Conventionally positive vector
Conventionally negative vector
A vector pointing in a direction defined as positive; typically up, right, north, east. On the Cartesian plane, the x-axis defines right as positive and the y-axis defines up as positive.
A vector pointing in the opposite direction to positive; the negative vector of a given positive vector has the same magnitude but is antiparallel to it
Perpendicular vectorsVectors at right angles. Note that the magnitude of the resultant vectors can be calculated using Pythagoras’ theorem. resultant vector a single vector that is equivalent to the sum of two or more vectors
Vector addition
When adding any vector, you must remember that only the same types of vectors can be added. For example, you can add two velocity vectors, but you cannot add a velocity vector to an acceleration vector.
Recall that in mathematics, subtraction is the addition of a negative. So, subtracting a vector involves adding its negative.
Vector addition in one dimension
When objects are moving in one dimension (i.e. along the same straight line), the direction of the vector can be represented by either a positive or negative sign. In Figure 2A.7, you can represent the vectors on the balls as 5 m s 1 and 8 m s 1, using the convention that the direction right is positive.
2A.7 Two balls moving towards each other. The arrows indicate the magnitude and direction of the balls’ velocities.
An alternative positive direction could be defined. For example, you could say that the ball on the right is at the origin and its motion to the left is positive, so the vectors on the balls are 5 m s 1 and 8 m s 1 respectively. But you must be consistent within the same convention.
Figure
Video: Worked example 2A.5
WORKED EXAMPLE 2A.5
Vector addition in one dimension
1. A person walks 200 m east and then turns around and walks 50 m west.
a. Draw a vector diagram for this situation.
b. Calculate how far the person is away from the origin at the end of their journey.
2. Find the resultant displacement of two vectors directed 4 m east and 3 m east.
3. Find the resultant displacement of two vectors directed 4 m east and 3 m west.
4. Find the resultant displacement of two vectors directed 4 m west and 3 m east.
Solution
1.
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Explanation
The 200 m east vector is drawn to the right. The 50 m west vector then starts at the end (tail) of the previous vector and points to the left indicating the direction, west.
Find the sum of the two displacement vectors. To do this, the vectors are added head-to-tail and the resultant vector runs from the start of the first vector to the head of the final vector, as shown. The magnitude of the resultant vector is found by 200 + ( 50) = 150 m, where the positive indicated the direction east. Therefore, the final answer is 150 m to the east.
When two vectors of the same type are directed in the same direction, the resultant vector is the sum of the two vectors in the same direction using the head-to-tail method. The resultant displacement is 7 m east.
When two vectors of the same type are directed in the opposite direction, the resultant vector is the difference of the two vectors in the direction of the greater vector using the head-to-tail method. The resultant displacement is 1 m east.
When two vectors of the same type are directed in the opposite direction, the resultant vector is the difference of the two vectors in the direction of the greater vector using the head-to-tail method. The resultant displacement is 1 m west.
Vector addition in two dimensions
Vectors in two dimensions lie in the same plane, like on a sheet of paper, but can point in different directions.
When an aircraft is flying, the pilot must consider the direction of the wind and which way to point the nose of the plane in order to fly a straight line on the ground. This is one of many two-dimensional examples where finding the resultant vector acting on an object is needed.
The same set of principles applies when adding vectors in one dimension to adding vectors in two dimensions. This means that vectors are added head-to-tail, and the resultant vector extends from the start of the first vector drawn to the head of the last vector drawn.
There are two primary methods for adding vectors: the graphical method and the analytical method.
• Graphical method: Vectors are drawn to scale and are arranged such that the arrows can be added using the head-to-tail method. The tail of the resultant vector starts at the tail of the first arrow and the head of the resultant vector ends at the head of the final arrow after following the path of arrows from the head of the previous arrow to tail of the following arrow. The order in which you arrange and add the vectors does not affect the resultant vector. The graphical method is shown first.
• Analytical method: This method uses vector components to perform addition, shown later in this section.
Resultant vectors
Vector addition involves combining two or more vectors to obtain a resultant vector. When adding two or more of the same type of vector, the vectors must be added headto-tail; the vector being added should start where the previous vector finished. The resultant vector is then drawn from the tail of the first vector to the head of the final vector. This is shown in Worked example 2A.6. Remember that the order in which you arrange and add the vectors does not affect the resultant vector.
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Vector addition for acceleration in two dimensions
Find the resultant acceleration of two vectors directed 4 m s 2 east and 3 m s 2 north.
When two vectors of the same type are perpendicular to each other, the resultant vector is the hypotenuse of a right-angled triangle. Its magnitude is obtained by using Pythagoras’ theorem and the direction goes from the tail of the first vector to the head of the second vector.
WORKED EXAMPLE 2A.6
Video: Worked example 2A.7
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WORKED EXAMPLE 2A.7
Finding total displacement
A hiker walks 5 km north, then 8 km east and finally 9 km south. What is the displacement of the hiker from their starting point? Include a direction in your answer.
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Explanation
First, draw a vector diagram of the situation.
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s = 82 + 42 = 8.94 km
From the diagram, you should be able to see that the north vector will partially cancel out the south vector. This will leave a right-angled triangle that has a length of 8 km on one side, 4 km on the other, and the hypotenuse will be the resultant displacement as shown.
Rearrange Pythagoras’ theorem, a2 = b2 + c2 , to determine the magnitude of the displacement.
Determine the angle by using tan (opposite over adjacent)
Video: Worked example 2A.8
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The displacement is 8.94 km 26.6° S of E.
WORKED EXAMPLE 2A.8
Vector addition for velocity in one dimension
A jet plane applies a thrust that would cause it to move through the air with a constant velocity of 150 m s 1 south (known as the airspeed) when it encounters a headwind blowing 25 m s 1 north. What is the velocity of the jet plane relative to the ground (known as the ground speed)?
Solution
positive direction = south
v = 150 + ( 25) = 125 m s 1 south
scorcher
Explanation
Determine which direction is positive.
A headwind is always in the opposite direction to the motion of the vehicle. So, the vector of the headwind to the north, which is negative using the defined positive direction (south). Add the vectors accordingly.
Note that the positive answer gave the direction: south.
We could also let south be negative and the solution would still be the same.
v = 150 + 25 = 125 m s 1
The negative indicates the direction: south.
WORKED EXAMPLE 2A.9
Vector addition for velocity in two dimensions
A boat applies a thrust that would cause it to move north with a constant velocity of 10 m s 1 (this is called the speed through water, STW) through a lake that has a current of 3 m s 1 east. Calculate the resultant velocity, v, of the boat (this is called the speed over ground, SOG) Ensure that you include a direction for the velocity.
Solution
Explanation
First, draw a vector diagram of this situation.
v = 102 + 32 = 10.4 m s 1
θ = 3
Rearrange Pythagoras’ theorem, a2 = b2 + c2, to determine the magnitude of the velocity.
Determine the angle by using a trigonometric function (sin, cos or tan). Note that the angle is measured from the north, and the angle moved towards the east. Hence, the angle is written as 16.7° east of north. Therefore, the final velocity is 10.4 m s 1 16.7°E of N.
Vector subtraction in two dimensions
As noted previously, the subtraction of vectors is when the negative value of one vector is added to the positive value of another vector. The vector that is to be subtracted is graphically reversed and then added to another vector in a vector diagram.
Vector subtraction can be defined as adding one vector to the negative value of another vector.
The vector subtraction of two vectors A and B is represented by A – B . It involves adding the negative of vector B to vector A – that is, A + (–B ).
WORKED EXAMPLE 2A.10
Vector subtraction for displacement
Consider A having a displacement of 5 m north and another vector, B, having a displacement of 3 m east. Find A – B
Explanation
The directions of A and B are as shown.
Explanation
Reverse the direction of the vector B to be subtracted: it is now – B , and it points west.
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The vectors are now added head-to-tail. The resultant vector runs from the start of the first vector to the head of the final vector. This will give a right-angled triangle that has a length of 5 m on the vertical side and 3 m on the horizontal side.
The hypotenuse will be the magnitude of A – B using Pythagoras’ theorem.
Determine the direction of A – B by using the tan trigonometric function (tan = opposite adjacent ) to calculate angle θ. Note that the angle is measured as the angle west of north.
Summary of vector addition advice and rules
1. Determine which direction is positive and keep track of signs (for compass points, N and E are usually + and S and W are usually –).
2. Reverse the direction of a vector that is subtracted, and subtract by adding a negative.
3. Draw vectors head-to-tail, and the resultant vector from the start of the first vector to the arrowhead of the second vector (or last).
4. I f vectors are perpendicular, find the magnitude using Pythagoras’ theorem.
5. I f vectors are perpendicular, find the direction using trigonometry.
Resolving vectors into components
In many cases it is useful to break up (“resolve”) a vector into its components. For example, when a projectile (an object in the air that only experiences the force of gravity) is launched at an angle, it is useful to determine the horizontal and vertical components of the projectile’s velocity. This is because, if air resistance is ignored, the horizontal component of the velocity will not change but the vertical velocity will be constantly changing because it is acted upon by the force of gravity throughout its flight.
Figure 2A.8 When the javelin is released, its velocity can be broken up into horizontal and vertical components. The magnitude of these two components affects how far the javelin will travel. Therefore, a javelin thrower must consider the launch angle of their javelin to maximise the distance that the javelin travels.
Rectangular components of a vector
Vector resolution is the process of obtaining the component vectors that, when combined according to the laws of vector addition, produce the given vector as the resultant.
Component vectors are the individual vectors obtained when a vector is resolved along chosen axes or directions.
Although there is a variety of different kinds of axes used in engineering, the most common axes for studying motion and force in physics are perpendicular horizontal (or x-axis) and vertical (or y-axis), as used in the rectangular Cartesian coordinate system in mathematics. The axes can also be labelled as the compass points north, south, east and west. The component vectors produced by this kind of resolution are referred to as rectangular components of a vector. This involves splitting the vector into two perpendicular components, often along the horizontal and vertical axes. These rectangular components and the vector itself form a right-angled triangle. Consequently, Pythagoras’ theorem and trigonometric functions can be used to solve problems involving the magnitudes of the vector and its components, as well as the angle of the vector.
Horizontal and vertical components
Figure 2A.9 A vector resolved into rectangular components and shown on a right-angled x, y coordinate system (a rectangular coordinate system)
vector resolution the process of breaking a vector into its component vectors that, when combined according to the laws of vector addition, produce the given vector component vectors individual vectors obtained when a vector is resolved along chosen axes or directions
rectangular components the horizontal and vertical components of a vector often shown on a Cartesian coordinate system
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The x- and y-axes of the coordinate system and the rectangular grid do not have to be shown in vector diagrams, as in the following. Consider a vector R at an angle θ to the horizontal.
The horizontal component vector R x lies along the x-axis. We use the convention of a dashed arrow for vector components. The length of the horizontal component can be determined by dropping the head of the arrow down to the same height as the tail, while keeping it along the same vertical line.
The vertical component lies along the y-axis. The length of the vertical component can be determined by moving the tail of the arrow across to the same vertical line as the head, while keeping it along the same horizontal line.
Video: Worked example 2A.11
If the angle between the vector and the horizontal axis is θ, then:
horizontal component = magnitude of vector multiplied by the cosine of the angle
R x = R cos θ
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vertical component = magnitude of vector multiplied by the sine of the angle
R y = R sin θ
According to the laws of vector addition, Pythagoras’ theorem and trigonometric identities, the magnitudes of the components will sum to the magnitude of the original vector. resultant vector =
R
WORKED EXAMPLE 2A.11
Finding vector components
Find the x and y components of the velocity vector with magnitude 15 m s 1 280° counterclockwise from the x-axis.
Draw a vector diagram.
Use trigonometry to find the horizontal and vertical component of the vector.
The vertical component is negative ( ) because the vector component is in the negative y-direction (downwards).
WORKED EXAMPLE 2A.12
Finding the x and y components of vectors with addition and subtraction
Construct the x component and y component of vectors A, B, C and D and then find A + B and A – B.
Vector A :
x component 10 cos 65 = 4.23 m east
y component 10 sin 65 = 9.06 m north
Vector B :
x component 15 m east, y component 0 m
Vector C :
x component 0 m, y component 20 m south
Vector D :
x component 45 cos 25 = 40.78 m west
y component 45 sin 25 = 19.02 m south
Use trigonometry to find the horizontal and vertical components of each vector. A + B : Along the horizontal:
Add the horizontal and vertical components of vectors A and B
Use the head-to-tail method.
Video: Worked example 2A.12 video
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Use the tangent function to determine the angle of the
Explanation A – B
Note: Since B is to be subtracted, we have to reverse the direction of B.
Along the horizontal: A x – B x = 4.23 m – 15 m = 10.77 m = 10.77 m W
Along the vertical:
Use the head-to-tail method.
Use the tangent function to determine the angle of the resultant vector.
Solving for unknown vectors
There will be situations where you will be given the resultant vector and will have to solve for an unknown vector. In these situations, you can determine the unknown vector by either drawing a vector diagram or determining an equation to solve for the change in the vectors. The unknown vector will add to the initial vector to cause a change and produce the resultant vector. A general equation to solve for a change in vectors is shown in Formula 2A.3.
FORMULA 2A.3
Solving for a change in vectors
V = The change in the initial vector
WORKED EXAMPLE 2A.13
Finding the displacement
A farmer completes one full lap around the boundary of a square field of side length 10.0 m in 40.0 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds?
Solution
Total time:
ttotal = 2 min 20 s = 2 × 60 + 20 = 140 s
The length of each side is 10 m.
The distance covered in 1 lap equals the perimeter of the square ABCD. 4 × 10 = 40 m
Explanation
Convert all units to SI units. To do so for 2 minutes and 20 seconds, convert 2 minutes to seconds and add the remaining seconds.
Let us assume the farmer starts from A. Draw a diagram of the situation.
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The time taken by the farmer to cover 1 lap is 40 s.
The speed the of farmer is given by:
v = d1 lap t1 lap = 40 40 = 1 m s–1
The total distance covered by the farmer in 2 min 20 s can be found using:
Total number of laps:
Therefore, after 140 seconds, the farmer will be at position C
The magnitude of the displacement can be found using:
AC = AB2 + BC2 = 100 + 100 = 200 = 14.1 m
Speed and total distance can be determined using the distance-speed-time triangle or by rearranging the equation:
v = d t
The total number of laps the farmer covered can be determined by dividing the total distance by the distance of 1 lap. Each side of the square field corresponds to a quarter (or 0.25) of a lap.
Use Pythagoras’ theorem to calculate the diagonal of the square field using the side lengths.
Note: Displacement is a vector quantity that measures the shortest distance (straight line) between the starting point and ending point and does not take the actual path travelled into account.
WORKED EXAMPLE 2A.14
Change in velocity
A ball is thrown east against a wall with a velocity of 5 m s 1. The ball then hits the wall and rebounds at a velocity of 3 m s 1 west. Calculate the magnitude and direction of the change in velocity of the ball.
Initial: u = 5 ms–1
Final: v = 3 ms–1
Let east be positive. The negative sign indicates the direction: west.
This can also be solved by drawing a vector diagram. The vector 3 m s 1 west is the resultant vector, so it must be drawn from the tail of the first vector, as shown.
Initial: u = 5 ms–1
Change: ∆v = 8 ms–1
Final: v = 3 ms–1
Video: Exam practice 2A
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The unknown vector is then drawn from the head of the first vector to the head of the resultant vector, as shown.
This is the head-to-tail method, because the vectors are added by connecting the head of the blue arrow to the tail of the red dashed arrow to create the resultant purple vector arrow.
EXAM PRACTICE 2A
Example question
A school bus is travelling south at 90 km h–1 for 5.0 minutes, then travels east at 100 km h 1 for a further 5.0 minutes.
a. Construct a labelled vector diagram of this situation. (2 marks)
b. Determine the total displacement of the bus. (2 marks)
c. Determine the average velocity of the bus. (2 marks)
What is the question asking?
In any question that involves a vector quantity, the direction of the vector must be considered and incorporated into your equation. Direction can be incorporated into an equation with a positive and negative sign representing different directions. This means that whenever you encounter a problem involving vectors, one of the first things you must do is determine which direction will be positive and which direction will be negative. Once you have made this decision, write it down on the page as this will allow you to keep track of the meaning of a negative or positive answer.
Typical structure
For most Physics questions that involve calculations, best practice is to complete the following:
1. Read the question carefully.
2. Create a list of known variables.
3. Identify the unknown variable(s)
4. Identify a formula that incorporates these variables.
5. Check if any conversions to SI units need to be completed.
6. Draw a diagram (where applicable).
7. Substitute your variables into the chosen formula.
8. Solve for the unknown variable.
9. Ensure your answer has the correct units and is correct to any appropriate significant figures and/or in scientific notation if required.
Marking criteria
Part a.
• Creates a vector diagram with directions and appropriate magnitudes
• Provides some relevant information
Part b.
• Correctly calculates the total displacement (including correct units and direction)
• Provides some relevant information
Part c.
• Correctly calculates the velocity (including correct units and direction)
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Draw a labelled diagram. Standard compass notation should be used. South and east are both taken to be positive.
Go online to access the Interactive Textbook
Sample response
b. Total time taken: t = 300 s south and then 300 s east.
t = 300 × 2 = 600 s
Given velocities:
90 km h–1 = 90 ÷
South:
s1 = 25 × 300 = 7500 m
East:
s2 = 27.8 × 300 = 8340 m
Resultant displacement:
s = 75002 + 83402
= 1.12 × 104 m
Direction:
θ = tan 1 (7500 8340) = 42° S of E
c. From part b:
Δs = 1.12 × 104 m
Total time:
Δt = 600 s
v = 1.12 × 104 600 = 18.7 m s 1
Direction:
42° S of E
v = 18.7 m s 1, 42° S of E
SECTION 2A REVIEW
Quiz
Review questions
Explanation
Time must be in SI units (s), so multiply given times in minutes by 60 to convert to seconds:
(5)(60) = 300
Velocity must be in SI units (m s–1)
The shortcut conversion of km h–1 to m s–1 is to divide by 3.6.
Use the formula for displacement: s = vt
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Use Pythagoras’ theorem: a2 = b2 + c2
Use trigonometry (SohCahToa) to find direction.
As displacement is a vector, a direction must be given.
Use the formula for velocity: v = Δs Δt
Again, a direction must be included as velocity is a vector.
The direction of the velocity is the same as the direction of the displacement used in the formula.
Review questions Reporting (student and teacher)
1. What is the difference between a vector and a scalar? Your answer must include an example to demonstrate your point.
2. A ball is thrown with a velocity of 18.1 m s−1 at an angle of 15.0° to the horizontal. Calculate the magnitude of the horizontal and vertical components of the ball’s velocity.
3. A car is travelling along a straight road at a constant velocity of 80 km h−1 south.
a. The driver approaches a pedestrian crossing and slows down to 40 km h−1 south. What is the magnitude and direction of the change in the car’s velocity?
b. The driver then accelerates due to a thrust force of 2250 N that overcomes the frictional force of 1750 N. What is the net force on the car, given that the vertical forces acting on the body are balanced?
4. An aircraft is attempting to travel north towards its landing strip. The aircraft has an air speed of 42 m s−1 (speed when there is no wind present). While the aircraft is trying to land, a 5.55 m s−1 wind blows from the west towards the east.
a. What is the final resultant velocity of the aircraft?
b. In order to fly due north, the pilot angles the aircraft into the wind. Explain how this enables the aircraft to fly north. Draw a diagram to aid your answer.
5. During the Australian Open a tennis ball is travelling west towards a player at 25 m s−1. The player strikes the tennis ball and sends the ball back east towards their opponent at 19.5 m s−1 What was the change in velocity of the tennis ball? Include a direction in your answer.
6. A swimmer intends to cross a river that is 1 km wide by swimming due north. The river has a current of 0.25 m s−1 east and the swimmer swims with a speed of 1.43 m s−1 in still water.
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a. Sketch a copy of the diagram and draw the path that the swimmer would take.
b. Calculate how much extra distance the swimmer would travel due to the current.
c. Suggest a way that the swimmer could swim directly north. Justify your answer.
7. A bridge is 400 m long. A 150 m long train crosses the bridge at a speed of 50 m s−1 Calculate the time taken to cross it.
8. When two objects move uniformly towards each other, they cross each other with a relative speed of 10 m s−1. If both the objects move in the same direction, then they cross each other with a relative speed of 6 m s−1. Calculate the speeds of both the objects.
9. An object moves in a circle of radius r and covers 3 4 of the circle. Calculate the ratio of the distance to the magnitude of the displacement.
10. An object is moving vertically upwards. Its velocity changes at a constant rate from 50 m s−1 to 20 m s−1 in 3 s. Calculate its acceleration explaining its direction and sign and what this means for its motion.
11. A bus first travels for 30 minutes with a velocity 35 km h−1 and then for 40 minutes with a velocity 50 km h−1 in the same direction, Calculate the:
a. total distance travelled
b. average velocity of the bus.
12. A car travels a distance 40 km with a velocity 15 km h−1 and then 55 km with a velocity 25 km h−1 in the same direction. Calculate the:
a. total time of journey
b. average velocity of the car.
Worksheet 2B.1
2B Analysing motion through graphs
SYLLABUS CONTENT
At the end of this section, I will be able to:
Analysing motion through graphs questions download word download pdf
2B.1 Use data to construct displacement–time graphs, velocity–time graphs and acceleration–time graphs
2B.2 Use data to analyse motion graphs to describe the displacement, velocity and acceleration of an object
2B.3 Analyse motion graphs to determine the relationships between displacement, initial velocity, final velocity, acceleration and time
Displacement–time graphs
Displacement–time graphs display how far an object is in a specified direction from its origin at a given point in time. In a displacement–time graph, the x-axis represents the time, and the y-axis represents the displacement. This can be shown by the displacement–time graph of a person going for a jog in Figure 2B.1.
The jogger leaves their origin, their house, and runs away from their house for 20 s, covering a distance of 80 m. The jogger then stops and rests for 28 s before turning around and running the 80 m back towards their house in 12 s. The jogger then continues to run in this direction past their house, and covers another 80 m in 20 s.
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gradient the slope of a graph. The gradient of a straight line can be calculated as the rise divided by the run.
Gradient of a displacement–time graph
There are a few useful principles that can be applied to all graphs. Many questions will ask about the gradient of a line between certain points or intervals. The gradient of the graph is given by:
gradient = rise r un
The gradient gives useful information – for example, the gradient of a displacement–time graph gives the velocity of the object because, in this case, rise = change in displacement and run = change in time.
Figure 2B.1 The displacement–time graph of a jogger
The shape of a displacement–time graph gives information about motion and velocity as shown in Table 2B.1.
Table 2B.1 Shapes of displacement–time graphs (s–t graphs)
Graph
0 s (m) t (s)
Shape, gradient and information
• Graph is a horizontal straight line
• Line parallel to time axis represents that the object is at rest as the displacement doesn’t change
• The gradient of a horizontal straight line is zero, which indicates that the object has a velocity of zero (i.e. at rest)
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0 s (m) t (s)
• Line 1 is a straight line with a positive gradient
• Positive gradient represents that the object is moving with constant (non-zero) velocity in the positive direction
• Line 2 has the steeper slope (larger gradient), which leads to the conclusion that the velocity of the second object is greater than the velocity of the first object in line 1
• Graph is a straight line with a negative gradient
• Negative gradient represents that the object is moving with constant (non-zero) velocity in the negative direction 0 s (m)
• Graph is a curve up
• The slope or steepness of the graph increases with time, which implies that the velocity is not constant but increasing (i.e. it is accelerating) 0 s (m)
• Graph is a curve up but levelling out
• The slope is decreasing with time, which implies that the velocity is not constant but decreases with time (i.e. it is slowing down or decelerating)
Video: Worked example 2B.1
WORKED EXAMPLE 2B.1
Gradient of a displacement–time graph
A body moves from rest in a straight line. Its motion is displayed in the following displacement–time graph.
What is the gradient of the graph? Include a unit in your answer and state what the gradient represents.
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two points, such as (0, 0) and (4, 8)
As you are dividing the values on the y-axis (rise) by the values of the x-axis (run), the gradient represents the velocity as v = Δs Δt . The unit for this graph is m s = m s–1
The first step in determining the gradient of a graph is to take two clear points from the line, determine the rise and the run, and then divide the rise by the run.
Displacement–time and velocity–time graphs compared
The gradient of the graph can be thought of as the steepness of the graph, and it is often important to be able to quickly assess the gradient of any graph. Consider the motion of an object displayed on the displacement–time graph and the velocity–time graph in Figure 2B.2.
a. Displacement–time graph
1. From A to B, the steepness is positive and constant, giving a constant positive gradient and therefore velocity.
2. The section B to C is steeper than the previous section (A–B), but it is still constant and positive, giving a larger constant positive gradient (greater velocity, i.e. faster)
3. Section C to D is flat and as it has no incline, the gradient and therefore the velocity is zero during this time (i.e. stationary)
4. The steepness of the segment D to E is negative, quite steep and constant. This gives a large, constant, negative gradient (velocity) so it is moving quickly in reverse, back to the start 0 m.
b. Velocity–time graph
To graph the velocity–time graph of the body for the same journey, we graph the value of the gradient of the displacement–time graph at each point. The two graphs are positioned so the points A–E are vertically aligned in both graphs.
1. From A to B the gradient in the top graph is 1 m s 1, as shown in the bottom graph. The dashed vertical line in the bottom graph indicates that the velocity change was instant. With most physical objects this is unrealistic, but that can be ignored here.
2. From B to C the gradient in the top graph is 2 m s 1
3. From C to D the gradient in the top graph is 0 m s 1, so the body is stationary.
4. From D to E the gradient is –3 m s–1, and the negative sign indicates the body has changed to the opposite (negative) direction.
Figure 2B.2 (a.) The displacement–time graph and (b.) the velocity–time graph of the same body moving in a straight line with a variety of different velocities
Velocity–time graphs
Velocity–time (v–t) graphs display the velocity of an object at a given point in time. This is demonstrated by the velocity–time graph below for a car doing a U-turn at a set of traffic lights.
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Figure 2B.3 The velocity–time graph of a car doing a U-turn at a set of traffic lights on a straight stretch of road
The car takes 50 s to uniformly accelerate to a speed of 15 m s 1. The car then travels at a constant velocity of 15 m s 1 for 100 s. The driver, seeing the traffic lights on red, brakes and the car takes 20 s to decelerate to a velocity of zero. The car remains stationary at the lights for 30 s. Then the light turns green and the car does a U-turn and accelerates for 50 s in the opposite direction to which it was originally travelling. This is why the velocity has become negative. The car then maintains a constant velocity of 10 m s 1 for 50 s in the negative direction.
Gradient of a velocity–time graph
On a velocity–time graph, the velocity is on the y-axis and time is on the x-axis. The gradient of the line represents the acceleration because gradient = rise r un = Δv Δt = a
Dividing m s 1 by seconds, you get the unit m s 1 s = m s 2 .
The shape of a velocity–time graph gives information about motion and velocity as shown in Table 2B.2.
Table 2B.2 Shapes of velocity–time graphs (v–t graphs)
Graph
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v (ms–1)
Shape, gradient and information
• Graph is a horizontal straight line above the x-axis
• Line parallel to time axis represents that the object is moving with uniform velocity in a positive direction
• The gradient of a horizontal straight line is zero, which indicates that the object has a uniform velocity (zero acceleration)
)
(s)
• Graph is a straight line with positive gradient
• Positive gradient represents that the object is moving with constant (non-zero) acceleration in the positive direction
• Line 2 has the steeper slope, which leads to the conclusion that the acceleration of the second object is greater than the acceleration of the first object (1)
v (ms–1)
• Graph is a straight line with negative gradient
• Negative gradient represents that the object is moving with constant acceleration in the negative direction, in this case causing constant deceleration
)
)
• Graph is a straight line with negative gradient below the x-axis
• Negative gradient represents that the object is moving with constant acceleration in the negative direction
• It is increasing in velocity in the opposite direction
• Graph is a straight line with negative gradient
• Its initial velocity is positive
• In section 1, the negative gradient represents that the object is moving with constant acceleration in the negative direction. In this case it is decelerating to zero.
• For an instant it is stationary, v = 0
• In section 2, the negative gradient represents that the object is moving with constant acceleration in the negative direction and speeding up
• This graph is typical of a ball being thrown vertically upwards, slowing down to zero for an instant then falling down, speeding up as it goes. The acceleration (gradient) is the acceleration due to gravity, which is the same on the way up as it is on the way down. Hence the straight line through the x–axis.
WORKED EXAMPLE 2B.2
Gradient of a velocity–time graph
A ball is thrown vertically into the air, and it leaves the thrower’s hand at an initial velocity of 10 m s 1. The velocity–time graph of the ball from the moment it leaves the thrower’s hand is shown (air resistance is ignored).
Video: Worked example 2B.2
a. You are told that when the velocity of the ball is zero, the time is 1.02 s. In terms of the ball’s height, what is the significance of when the velocity is zero?
b. Calculate the gradient of the line and comment on the significance of this gradient.
Solution
a. When the velocity of the ball is zero, the ball will be at its maximum height.
b. gradient = rise
This gradient is the acceleration due to gravity.
Explanation
The velocity is zero when the ball changes direction from going up to going down.
To determine the gradient, take two clear points from the line, determine the rise and the run and then divide the rise by the run.
As you are dividing the values on the y-axis (rise) by the values of the x-axis (run), the gradient represents Δv Δt . The unit for this graph is m s–1 s = m s–2
Note that the acceleration is changing the velocity. As seen above, the velocity has gone from being positive to zero to negative. Note the gradient is constant because the acceleration due to gravity is constant whether the ball goes up or down.
The area under a velocity–time graph
The area under a velocity–time graph represents the displacement. Consider the velocity–time graph of a hiker, walking with a constant straight-line velocity of 1 m s 1 for 5 seconds (Figure 2B.4).
The change in displacement of the hiker after 5 seconds can be determined by finding the area under the graph:
= ∆s = v∆t = (1)(5) = 5 m
Figure 2B.4 The velocity–time graph of a hiker walking at 1 m s−1 for 5 seconds
For non-constant velocity, the displacement is still found using the area under the velocity–time graph, as shown in Figure 2B.5. The unit is given by multiplying the x-axis unit with the y-axis unit as shown, m s 1 × s = m.
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Figure 2B.5 The area under a velocity–time graph gives the displacement, even if the velocity changes with time.
WORKED EXAMPLE 2B.3
Area under a velocity–time graph
A cyclist rides north up a straight road, turns around and then rides in the opposite direction. A velocity–time graph of this situation is shown below.
a. Find the displacement of the cyclist once they have travelled for 120 s.
b. What is the displacement after 180 s?
c. What is the total distance covered after 180 s?
1 2 (20)(5) + (60)(5) + 1 2 (40)(5) = 450 m
The displacement is 450 m north.
The area for the blue triangle is: area = 1 2 (60)(–11) = –330 m
Explanation
The displacement is the area under the graph. To find the area under the graph, divide the graph into a series of triangles and rectangles as shown.
The area under the graph up to 120 s, and therefore the displacement, can then be found by adding the area of all the shapes.
As this question asked for displacement (a vector quantity), a direction is required.
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Therefore, the total displacement of the bike after 180 s is: 450 + 330 = 120 m north
c. In the first 120 s the cyclist travels 450 m, and from 120 s to 180 s the cyclist travels 330 m. This gives a total distance covered of 780 m.
Note that this graph has a negative value for velocity; this indicates that the cyclist is travelling south. To calculate the displacement after 180 s, the graph needs to be divided into a rectangle and triangles as shown.
Distance is a scalar quantity. Therefore, no direction is required. This means that any negative velocities must be turned into positive values.
Acceleration–time graphs
In an acceleration–time graph, the x-axis represents the time and the y-axis represents the acceleration.
Acceleration–time graphs display the acceleration of an object at a given point in time. This can be demonstrated by the acceleration–time graph of a car travelling east that brakes uniformly as it approaches a hazard, travels at a lower constant velocity, then accelerates away.
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As seen in the acceleration–time graph in Figure 2B.6, when the car brakes, it accelerates at a constant rate of 4 m s–2 for 6 s. The negative value means that the car is accelerating in the west direction and therefore the velocity of the car is being reduced. The car then remains at constant velocity (0 acceleration) for 4 s before accelerating uniformly away in the positive direction at 3 m s–2
The area under an acceleration–time graph
In an acceleration–time graph, acceleration, a, is on the y-axis and time, t, is on the x-axis. The area under an acceleration–time graph is the change in velocity as:
For non-constant acceleration, the change in velocity is still found using the area under the acceleration–time graph. The unit is given by multiplying the units on the x- and y-axis together to give m s–2 × s = m s 1
Area under an acceleration–time graph
A car, initially travelling at 72 km h 1, is travelling on a straight road when it approaches some roadworks. The car slows to the new speed limit and then accelerates again once it has passed the roadworks. The acceleration–time graph of the car’s journey is shown.
Figure 2B.6 Acceleration–time graph of a car braking as it approaches a hazard
a. Calculate the speed that the car travelled past the roadworks in kilometres per hour.
b. Calculate the speed, in km h 1, of the car at the end of the 50 s.
Solution
a. initial velocity of the car
u = 72 3.6
= 20 m s–1
The change in velocity between 0 and 8 seconds can be found by calculating the area under the graph in that interval.
Δv = (–1)(8)
= –8 m s–1
final velocity
convert m s–1 to km h–1
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Explanation
Convert all units to SI units. The car is accelerating at a rate of 1 m s–2 for 8 s as it approaches the roadworks. The car is passing through the roadworks at a constant velocity (as there is no acceleration) from 8 s to 42 s. After 42 s, the car has left the roadworks and accelerates.
v = 20 – 8
= 12 m s–1
v = (12)(3.6)
= 43.2 km h–1
b. The change in velocity between 8 and 42 seconds is zero.
The change in velocity between 42 and 50 seconds can be found by calculating the area under the graph in that interval.
Δv = (2)(8) = 16 m s–1
final velocity
convert m s–1 to km h–1
v = 12 + 16
= 28 m s–1
v = (28)(3.6)
= 101 km h–1
Since you know from part a. that the car will be travelling at 12 m s 1 after 42 s, you only need to consider the change in velocity after this point.
Instantaneous velocity versus average velocity using graphs
Speed and velocity can be either instantaneous or average. Imagine a truck is moving down a straight road. A displacement–time graph of the truck’s motion is shown in Figure 2B.7.
The gradient of this line will be the velocity because: gradient = rise
But the gradient changes after 6 seconds and the line becomes steeper, indicating an increase in velocity. The velocity can be considered the instantaneous velocity while the section of the graph is linear, i.e. the slope of the graph is constant. This means that the instantaneous velocity for the first 6 seconds is:
The instantaneous velocity for the next 4 seconds is:
The average velocity for the whole 10-second trip is:
The formula for average speed and velocity must be used with care. As you can see from Figure 2B.7, considering different parts of the graph will give different results as the velocity of the object changes. You should remember that this formula gives the average velocity of a body and should not be confused with a formula that gives the final velocity.
Measuring instantaneous velocity during acceleration is much more difficult. In fact, any real-world measurement of velocity will always be an average because it is calculated by dividing the distance travelled by the time that it took the body to travel that distance. As the time interval approaches zero, the velocity becomes closer to the instantaneous velocity.
2B.7 Displacement–time graph of a truck driving along a road
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For example, imagine a car that is accelerating and therefore its velocity is constantly increasing. The displacement–time graph for the accelerating car is shown in Figure 2B.8.
Figure 2B.8 Displacement–time graph of an accelerating car. The value of the instantaneous speed is given by the gradient of the tangent at that point in time. As time passes, the gradient of the tangent increases, showing that the instantaneous velocity is also increasing.
Figure
Video: Worked example 2B.5
As the gradient is constantly changing, the velocity is also constantly changing.
WORKED EXAMPLE 2B.5
Average velocity and instantaneous velocity
What is the difference between the average velocity and the instantaneous velocity? Use Figure 2B.8 to help answer your question.
Solution
In Figure 2B.8 the average velocity is:
The instantaneous velocity when t = 2 and when t = 6 is:
Explanation
The average velocity of an object is the change in the displacement of an object divided by the time.
The instantaneous velocity of an object is the velocity of an object at a specific point in time.
Summary of graphs of motion
GRAPHSOFMOTION
Displacement–time Graph (s–t)
Slope → Velocity
• Steeper slope = higher speed
• Positive slope = moving forward
• Negative slope = moving backward
• Zero slope = object at rest
Shape:
• Straight line = uniform velocity
• Curved line = non-uniform velocity (acceleration or deceleration)
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Velocity–time Graph (v–t)
Slope → Acceleration
• Positive slope = acceleration in positive direction
• Negative slope = acceleration in negative direction
• Zero slope = constant velocity
Shape:
• Straight line = uniform acceleration
• Curved line = non-uniform acceleration
Acceleration–time Graph (a–t)
Slope → Rate of change of acceleration
• Positive slope = change in acceleration in the positive direction
• Negative slope = change in acceleration in the negative direction
• Zero slope = constant acceleration
Shape:
• Straight line = uniform rate of change of acceleration
• Curved line = non-uniform rate of change of acceleration
Area under curve = Change in displacement
Figure 2B.9 A summary of three types of motion graphs
Area under curve = Change in velocity
EXAM PRACTICE 2B
Example question
Use the given graph showing a skateboarder’s velocity in a straight line over a given period of time to answer the following questions.
a. Determine the skateboarder’s acceleration after 12 seconds. (2 marks)
b. Identify the time(s) at which the skateboarder is travelling at a constant velocity. (1 mark)
c. Determine is the total displacement of the skateboarder during this time period. (2 marks)
d. Identify the time at which the skateboarder experiences the greatest acceleration and the magnitude of that acceleration. (3 marks)
What is the question asking?
• Graphing questions are fundamentally checking to see if students can take formulae they know and apply them to a visual medium.
• The best tip for graphing questions is to first identify the variables on each axis. Only then should you start to consider the shape of the line.
• For example, a straight (horizontal) line on a distance–time graph would indicate that distance is not changing over the time period.
Marking criteria
Part a.
• Correctly calculates the magnitude and direction of the initial velocity
• Provides some relevant information
Part b.
Criteria
• Identifies correct time(s) and give them as intervals (i.e. between 2 and 4 seconds)
Part c.
Criteria
Marks
• Correctly calculates the magnitude and direction of the displacement 2
• Provides the main steps for calculating the displacement 1
Part d.
Criteria
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Marks
• Correctly identifies the correct time interval at which acceleration was greatest (i.e. acceleration was greatest between 2 and 4 seconds) 3
• Correctly calculates the magnitude of the acceleration 2
• Provides the main steps for calculating the acceleration 1
Sample response
Sample response
a.
b. The skateboarder is travelling with constant velocity between 4–10 s and 14–18 s.
c. Positive areas: 0–4 s triangle: 1 2 × 4 × 8 = 16 m 4–10 s rectangle: 6 × 8 = 48 m
10–14 s triangle: 1 2 × 4 × 8 = 16 m
Negative areas: 18–24 s triangle: 1 2 × 6 × 5 = 15 m
24–26 s triangle: 1 2 × 2 × 5 = 5 m
Total displacement: 16 + 48 + 16 – 15 – 5 = 60 m east
d. Compare gradients m of each section:
0–4 s: m = 8 − 0 4 = 2 m s–2
The greatest magnitude is 2.5 m s–2 during 24–26 s.
Explanation
At 12 s, the skateboarder is on the sloping section from 10 s to 14 s, where velocity changes from 8 m s–1 to 0 m s–1
This decrease in velocity corresponds to a negative acceleration as the skateboarder is slowing down.
Alternative answer can be given as 2 m s–2 west as east has been denoted as positive, therefore negative acceleration would occur in the opposite direction(i.e. west)
Points on the graph where the line is horizontal indicate no change in velocity.
Displacement is the area under the velocity–time graph.
Using the formula for the area of a triangle, 1 2 base × height, and a rectangle, base × height, we can solve for the areas of each section.
Remember that negative areas need to be subtracted from the total.
Direction must be included as displacement is a vector quantity.
You can cut down on some of the working by eliminating gradients that are clearly shallower than others and working with the steepest ones.
Using a = Δv Δt , solve for acceleration at each point and then compare.
Note that the question only asks for magnitude, so no direction is required despite acceleration being a vector.
SECTION 2B REVIEW
Quiz
Review questions
Review questions Reporting (student and teacher)
1. A car is travelling initially at 13 m s−1 in heavy traffic. An acceleration–time graph of part of the car’s journey is shown below. Assume that the positive acceleration is in the direction of travel.
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a. List all of the times that the car is braking.
b. What is the maximum velocity of the car?
c. Calculate the final velocity of the car.
2. A cyclist is initially travelling north with a velocity of 10 m s−1. The cyclist travels for 10 minutes and a velocity–time graph of the journey is shown.
a. Describe how the velocity and acceleration of the cyclist changes with time throughout the 10-minute journey. No exact values for acceleration are required.
b. What is the cyclist’s velocity after 7 minutes?
c. If the cyclist starts from the origin, how far from the origin are they after 10 minutes?
d. What is the total distance the cyclist covered in the journey?
e. Use the graph to calculate the cyclist’s acceleration between 4 and 8 minutes. Include a direction in your answer.
3. A banker accidently dropped a one-dollar coin, initially at rest, off a building. The building is 70.8 m tall and it takes the coin 3.80 s to reach the ground. The displacement–time graph of the coin’s velocity is shown below.
a. Describe how the velocity of the coin changes with time.
b. Calculate the average velocity for the coin for its 3.80-second fall and comment on how this would differ from the instantaneous velocity of the coin at the point of impact with the ground.
c. Which of the following is the velocity–time graph that matches the motion of the coin? Explain your choice.
d. Draw an acceleration–time graph of the motion of the coin. An appropriate scale, units, and x- and y-axis labels should be included.
e. What was the total distance covered by the coin?
4. A ball is thrown straight up in the air at 9.8 m s−1; once 2 seconds have passed the ball is caught at the same height that the ball was released from. Assume that the ball reaches a maximum height of 4.90 m above its origin.
a. Draw a displacement–time graph of the motion of the ball. An appropriate scale, units, and x- and y-axis labels should be included.
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b. Draw a velocity–time graph of the motion of the ball. An appropriate scale, units, and x- and y-axis labels should be included.
c. Draw an acceleration–time graph of the motion of the ball. An appropriate scale, units, and x- and y-axis labels should be included.
5. Some hikers start walking from their camp, which is considered the origin. They powerwalk north at a constant velocity for 200 m over 100 seconds. They then stop and rest for 20 seconds before taking 200 seconds to walk back to their camp. The hikers rest at camp for 80 seconds and then run 400 m south in 100 seconds.
a. Draw a displacement–time graph of the motion of the hikers. An appropriate scale, units, and x- and y-axis labels should be included.
b. During which time interval is the hikers’ speed the greatest?
c. What is the average speed in the first 110 s?
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2C Analysing uniformly accelerated motion and relative velocity
SYLLABUS CONTENT
Worksheet 2C.1
At the end of this section, I will be able to:
Analysing straight-line motion with uniform acceleration questions download word download pdf
2C.1 Solve problems involving uniformly accelerated motion in one dimension using s = u t + 1 2 a t 2 , v = u + a t and v 2 = u 2 + 2a s
2C.2 Conduct a laboratory experiment to compare 2 methods of graphically determining a value for acceleration due to gravity and assess the accuracy and reliability of the experimental results
2C.3 Analyse problems of relative velocity involving motion in two dimensions using vAB = vA vB
Formulas for uniform acceleration
There are three formulas for constant or uniform acceleration, also known as kinematic equations, that you need to know for this course. These formulas can then be applied to problems involving constant acceleration to quickly find a solution. Both vector quantities (initial/final velocity and acceleration, as denoted by the vector arrow) and a scalar quantity (time) are used in the formulas for uniform acceleration. When written without the vector arrows, the magnitudes of the vector quantities are used, often meaning that the problem has been broken down into one-dimensional components. Subscripts should be used to indicate which component is being considered, such as v x or v y for the velocity component in the x-direction and y-direction respectively.
FORMULA 2C.1
Formulas for uniform acceleration
Figure 2C.1 An object falling or rolling down a slope is the most familiar example we have in everyday life of uniformly accelerated motion.
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Where:
v = Final velocity (m s 1
u = Initial velocity (m s 1)
a = Acceleration (m s 2)
t = Time (s)
s = Displacement (m)
When analysing straight-line motion, graphs are extremely useful. From graphs, you can derive the formulas for uniform straight-line acceleration shown as follows.
An object moving with constant acceleration is represented by a diagonal line on a velocity–time graph that has a constant, non-zero, gradient. For example, a cyclist accelerates at a constant rate of 2 m s–2 for 5 seconds. The velocity–time graph of the cyclist’s motion is shown in Figure 2C.2.
Let’s say you wanted to find the average velocity of the cyclist. The distance travelled is the area under the graph, so d = 1 2 (5)(10) = 25 m. Now, to find the average velocity, the distance is divided by the time, so v = 25 5 = 5 m s –1 .
Figure 2C.3 shows that even though the cyclist has an ever-changing velocity over time, they will cover the same distance in 5 seconds as a cyclist moving at a constant velocity of 5 m s 1 in the same time. The average velocity for the accelerating cyclist would be less if the cyclist were to travel for less than 5 s and greater if travelling more than 5 s with the same acceleration.
Figure 2C.3 A geometric view of the area under the velocity–time graph for the uniformly accelerating cyclist. The black dashed line is the average velocity, and this diagram shows that the area under the plotted line in the velocity–time graph is the same as the area under the average velocity plotted on the same axes.
Figure 2C.2 Velocity–time graph for a cyclist accelerating at 2 m s–2 for 5 seconds
Deriving the formulas for constant acceleration
As you learned in Section 2B, the gradient of a velocity–time graph is the acceleration: gradient = a = rise
W hen the object starts at rest:
so, in this case:
This can be rearranged to give:
This is the first equation in Formula 2C.1.
Now consider the motion of another cyclist that is initially riding at 6 m s 1 and accelerates at a constant rate of 1 m s –2 for 5 s, as shown in Figure 2C.4. As you learned in Section 2B, the area under a velocity–time graph is the displacement. To find the displacement of the cyclist after their 5 s journey, the graph would have to be divided into a rectangle and a triangle.
From Figure 2C.4, the displacement is given by the area under the graph:
The displacement is also average velocity multiplied by time.
Figure 2C.4 Displacement of a cyclist initially riding at 6 m s−1 and accelerating at 1 m s−1 for 5 seconds, showing the calculation of the area under the curve
All the other formulas for uniform acceleration can be derived from this formula.
You have learned that the acceleration is given by:
You can rearrange this to:
= v – u
This means that at can be substituted for v – u giving:
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This is the second equation in Formula 2C.1.
The acceleration formula can then be rearranged to:
Then substituted into the equation as follows:
This is the third equation in Formula 2C.1.
This completes the formulas for uniform acceleration in this course. Note: You will not be required to derive these formulas.
Horizontal motion
Newton’s first law, which will be discussed in more detail in Chapter 3, states that an object in motion or at rest will remain in its state of motion unless acted upon by an unbalanced force. This means that when an object moves horizontally through space, its velocity will not change unless acted upon by an unbalanced force. Often objects will have a constant unbalanced force applied to them, like a car that is accelerating uniformly or a sprinter who is sprinting out of the blocks. In these situations of uniform acceleration, the equations of motion can be applied. When solving horizontal motion questions about objects with constant, one-dimensional acceleration, do the following:
• Draw a diagram of the situation and put all the information that you have on the diagram in SI units.
• C hoose an appropriate equation based on the data that you have and the data that you need.
• Remember that when vectors act in opposite directions, set one direction as positive; the opposite direction will then have a negative sign (except for t, which is a scalar quantity).
• E xpress your final answer to an appropriate number of significant figures. Include units in your answer and, if your answer is a vector, a direction as well.
Video: Worked example 2C.1
The SUVAT table method
Using the SUVAT table method to analyse motion with uniform acceleration is a structured way to solve problems with the relevant equations. It is named after the variables for which you construct a five-row table:
In the second column you list the values you need to find. You then choose the equation that has the variables you need. Follow the method in Worked example 2C.1.
WORKED EXAMPLE 2C.1
Using a SUVAT table
A cyclist slows from 12 m s 1 to rest in 6 seconds. Use the SUVAT method to find their acceleration and stopping distance. Solution Explanation s
Write the data given including relevant signs for direction. v
Substitute:
Rearrange for
You need to find s and a; you know u, v and t
Choose an equation which includes one unknown and variables we know, a, u, v, t; so v = u + at Rearrange the equation to solve for a
Figure 2C.5 Cheetahs are the fastest accelerating land animals on the planet; their acceleration is equal to most Lamborghini and Ferrari cars.
Solution
s = ut + 1 2 at2
Substitute:
s = 12(6) + 0.5( 2)(62)
= 72 36
= 36 m
The acceleration is –2 m s–2 and the stopping distance (displacement) is 36 m.
WORKED EXAMPLE 2C.2
Explanation
The stopping distance is the displacement. Choose another equation with the other unknown and variables you know, s, u, t, a; so s = ut + 0.5 at –2
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The negative sign in front of the acceleration indicates that the cyclist is accelerating in the negative direction (decelerating), causing the cyclist in this instance to slow down.
Using average velocity–time to find displacement
A ball travels forward with a velocity 4.65 m s 1. As it rolls over the ground, it slows to a stop in 1.75 s, due to a constant force acting on the ball, causing a constant deceleration (negative acceleration). Calculate the displacement.
Solution
t = 1.75 s
u = 4.65 m s 1
v = 0 m s 1
= 4.65 ms−1 a
s = v av × t
= u + v 2 × t
= 2.325 × 1.75
= 4.07 m
Vertical motion
When an object is launched vertically, and if air resistance is ignored, there is generally only one force acting on the object – the downwards force of gravity. Close to the surface of Earth, the force due to gravity will cause all free falling objects to accelerate at 9.8 m s –2 down to the surface of Earth, usually used as –9.8 m s–2 in calculations due to up being defined as positive and down as negative.
= ?
Video: Worked example 2C.2
Explanation
List the relevant data.
Draw a diagram of the situation.
= 0
Note: Although the problem involves acceleration, in this case we don’t need to calculate acceleration. The regular formula for displacement can be used to find the displacement of the ball.
free fall when an object is falling and the only force acting on the object is gravity (ignoring air resistance)
Figure 2C.6 A Peregrine falcon mid-dive. During such dives the peregrine falcon can reach speeds of 390 km h−1, which is about 30% faster than the top speed of a civilian helicopter.
For example, imagine a 1 kg mass and a 2 kg mass that are initially at rest are dropped and allowed to free fall. If air resistance is ignored, they will both reach a velocity of 9.8 m s 1 down after 1 second, and then reach a velocity of 19.6 m s 1 down after 2 seconds. If the 1 kg and 2 kg masses are thrown up with a velocity of 19.6 m s 1, the two masses will experience an acceleration downwards towards the surface of Earth (i.e. decelerate on the way up). After 1 second, the velocity of both masses would be reduced to 9.8 m s 1 up, and after 2 seconds the velocity of both masses would be instantaneously reduced to zero.
In the 1500s, Italian scientist Galileo Galilei conducted many experiments with falling objects to investigate the effect of gravitational acceleration on objects of varying mass. According to legend, it is said that he dropped two objects off the Leaning Tower of Pisa to see which landed first. Even though one of the objects was much lighter than the other, they both fell at the same rate, indicating that the acceleration on both masses was the same. In 1971, during the Apollo 15 Moon landing, astronaut Commander David Scott dropped a feather and a hammer at the same time. Since there was no air resistance on the Moon, both the feather and the hammer hit the ground at the same time.
These examples demonstrate that the acceleration due to gravity is unaffected by an object’s mass. Each pair of masses had the same rate of change in velocity for their respective scenarios. We say the acceleration due to gravity is independent of the masses of the objects.
The steps to solving a vertical motion problem are the same as the steps to solving a horizontal motion problem. However, in vertical motion problems (assuming air resistance is ignored) there are a number of key things to remember that will assist in solving them:
• T he constant acceleration acting on all free-falling bodies on Earth is 9.8 m s –2
• Gravity is usually given a negative sign in an equation of motion ( 9.8 m s–2).
Velocities and displacements directed upwards are given a positive sign.
• An object that is thrown upwards will slow down until it comes momentarily to a stop; gravity will then begin to bring the object back down to Earth.
• T he stationary point of an object that is projected up in the air is its maximum height. The time taken to reach the stationary is half of the total flight time if the object lands at the same level from which it was launched.
WORKED EXAMPLE 2C.3
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Vertical motion for an object thrown upwards
A student throws a ball of mass 200 g into the air vertically upwards, with an initial velocity of 20 m s 1. The point at which the ball leaves the student’s hand is 1.22 m above the ground.
a. Calculate the maximum height above the ground reached by the ball.
b. Calculate the total flight time of the ball, assuming that the student catches the ball at a height of 1.22 m above the ground.
Figure 2C.7 Illustration of Galileo Galilei’s experiment demonstrating how gravitational acceleration is independent of the mass of the object.
Solution
a.
u = 20 ms−1
m = 0.200 kg
a = −9.8 ms−2
s = 1.22 m
v2 = u2 + 2as s = v2 – u2 2a = (0)2 – (20)2 (2)(–9.8) = 20.4 m
This is the displacement that the ball travelled before it reached its turning point
Explanation
Draw a diagram with all of the information given. Remember that when the maximum vertical height is reached, the velocity of the ball will be zero. This means if you consider the movement of the ball from when it is released to its maximum height, the final velocity of the ball will be zero.
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Apply an appropriate formula.
maximum height = 20.4 + 1.22 = 21.6 mTo get the maximum height above the ground, the initial height of the ball must be added.
b. v = u + at t = v – u a = (0) – (20) (–9.8) = 2.04 s total time = 2.04 × 2 = 4.08 s
Alternatively, use the symmetry of projectile motion at constant acceleration.
t = v – u a
= (–20) – (20) (–9.8) = 4.08 s
This problem can be solved with the knowledge that the time to reach the maximum height is half of the total flight time, as the ball lands at the same height that it was launched.
Use an appropriate formula with time.
There is often more than one way of reaching a solution. When a projectile is in free fall, there is symmetry in the projectile’s upwards and downwards movement. When the ball is travelling upwards, its upwards velocity is reducing by 9.8 m s 1 every second; in this case the ball will travel 20.4 m up before the ball no longer has any upwards velocity. After this momentary stationary point, the ball will then fall a distance of 20.4 m, which means that just before the ball is caught, it will be travelling down at –20 m s 1. This can then be used as the final velocity as shown, and the calculation is done in one step.
WORKED EXAMPLE 2C.4
Vertical motion for an object dropped from a height
Two balls are dropped to the ground from different heights. One ball is dropped 2 s after the other, but both strike the ground at the same time 5 s after the first ball is dropped.
a. From what height was the first ball dropped?
b. What is the difference in the heights from which they were dropped?
a. For the first ball, s = h1, u = 0, t = 5 s.
s = ut + 1 2 at 2
1 = (0) + 1 2 (–9.8)(5)2 = 122.5 m
The first ball was dropped from a height of 122.5 m.
b. For the second ball, s = h2, u = 0, t = 5 – 2 = 3 s.
s = ut + 1 2 at 2
h2 = (0) + 1 2 (–9.8)(3)2 = 44.1 m difference in heights = 122.5 – 44.1 = 78.4 m
WORKED EXAMPLE 2C.5
Write the information given for the first ball and use the relevant uniformly accelerated motion equation to determine the initial height.
Follow the method used for part a. to determine the initial height of the second ball. Subtract the initial height of the second ball from the first ball (as calculated in part a.)
Video: Worked example 2C.5
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parabolic path an approximately inverted U-shaped plane curve that is symmetrical. This is the path taken by a projectile in two dimensions that is only under the influence of gravity (or any constant acceleration) and air resistance is considered negligible.
Drop from a height
An initially stationary stone is dropped into a well and the sound of splash is heard after 5.36 s. If the water is at a depth of 122 m from the ground, what is the velocity of sound in air? Solution
s
Explanation
If t1 is the time taken by stone to reach the bottom of the well and t2 the time taken by sound to go up, then t1 + t2 = 5.36 s.
Parabolic paths
If you throw a ball at an angle, it will trace a parabolic path (parabola shape) as shown in Figure 2C.8, if the effect of air resistance is ignored.
Use the equation for velocity to determine the velocity of sound. Figure 2C.8 The parabolic path a ball follows when thrown with an
If you ignore air resistance, what are the forces acting on the ball throughout its flight?
Some students may say that since the ball is going up, the force on the ball must be up. This is not true. The ball may be going up, but the only force acting on the ball is the force due to gravity, which is downwards and constant.
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Figure 2C.9 When a body is in flight, the only force acting on it is the force due to gravity, which is downwards and constant. (Questions 1 or 2 on the HSC exam often ask a question involving this diagram.)
Objects will accelerate in the direction of the net force, so the constant downward force of gravity will create a constant acceleration on the ball of 9.8 m s–2 down or –9.8 m s –2 up. This means that whenever you are solving a question that involves an object in free fall, you will always know that the magnitude and direction of the acceleration acting on the object is 9.8 m s–2 down or –9.8 m s –2 up.
WORKED EXAMPLE 2C.6
Projectile motion
A student, Matt, is busy doing his Physics homework in class when suddenly his pen runs out of ink. Too lazy to stand up and walk to the rubbish bin to throw out the pen, he starts planning a way to throw out his pen from the back of the class with the least amount of effort. He figures that if he approximates the distance to the rubbish bin at the front of the classroom, he can work out how much velocity he needs to throw the pen with for it to land perfectly in the bin without hitting any other students. He has estimated the following:
• The distance to the centre of the opening of the rubbish bin from where he throws the pen is 9 m across and 1 m down.
• He needs to avoid the student in front of him, so will be throwing the pen at an angle of 30° from the horizontal.
net force the vector sum of all of the forces acting on a body; may also be referred to as the unbalanced force or the resultant force
Video: Worked example 2C.6
Calculate the velocity Matt needs to throw the pen at for him to execute this manoeuvre successfully.
Solution
Explanation
Use up as positive and towards the bin as positive. Define which direction is positive.
Horizontal (x) components:
u x = u cos 30 = 0.866u
s x = 9 m
a x = 0 m s–2
Vertical (y) components:
u y = u sin 30 = 1 2 u
s y = 1 m
a y = –9.8 m s–2
State all the variables given in the question and break them down into their horizontal and vertical components. Remember that since gravity has zero component horizontally, the horizontal acceleration will be zero.
To find the time taken, use the horizontal components. Since the horizontal velocity remains constant, the time can be found by rearranging s = ut + 1 2 at2 or using time = distance speed s y = u y t
Use the uniformly accelerated motion equations to determine the initial velocity from the vertical components.
So, the pen must be thrown with an initial velocity of 9.24 m s 1 at 30° for it to perfectly land in the bin.
PRACTICAL 2C.1
Comparing two methods of determining the acceleration due to gravity, g
Method 1: measuring the vertical motion of a falling ball using slow-motion video
Aim
To determine the acceleration due to gravity, g, by measuring the vertical motion of a falling ball using slow-motion video
Background information
If the object starts from rest and accelerates uniformly under gravity: s = ut + 1 2 at2 s = 1 2 at2 (u = 0)
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Since the object is falling from a height, its initial velocity u = 0 and a = g
Materials
• small dense sphere (steel ball, 10–20 mm)
• vertical white screen/board (≥1.5 m) with grid paper or pre-drawn scale (marks in cm)
• rigid vertical ruler (or tape) attached to the screen; clearly mark 0 cm at release level and marks at 1 cm, 4 cm, 9 cm, 16 cm and 25 cm below it
• tripod and clamp for camera/phone
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• smartphone or camera capable of slow motion (high frame rate: 120 fps or 240 fps)
• second smartphone running a digital stopwatch/clock with centisecond display
• release mechanism (trapdoor or pin release) that minimises push
• soft catcher (beanbag)
• marker, tape, and lighting to make the grid readable
• safety glasses, secure tripods and clamps
Experimental set-up
1. Fix the vertical screen/board flat and vertical. Attach grid paper or draw a vertical scale (1 cm squares) aligned with a vertical ruler. Mark the release level clearly and label it s = 0.
2. Attach the ruler/grid so that the plane of the grid and the path of the ball are the same plane (i.e. the ball falls directly in front of the grid, not offset to the side).
3. Place the digital clock display (phone showing stopwatch) in the same plane as the grid if possible so that the clock digits are readable in the same camera view. If the clock cannot be exactly in the same plane, place it directly adjacent to the grid and as close to the same distance from the camera as possible.
4. Mount the camera/phone on a tripod. The camera should be centred, levelled and sufficiently far away to include the whole falling path and the clock in one frame. Preferably use optical zoom (not digital).
5. Check lighting so the grid and clock digits are high contrast and clearly readable in slow-motion video.
6. Practise the release to be smooth and repeatable. Ensure the ball centre follows the plane marked by the grid.
Checklist before running the experiment
1. Is the camera perpendicular to the grid?
2. Are grid marks and clock digits sharp and readable in slow motion?
3. Does the path of the ball move towards or away from the grid? The ball should only move in two dimensions (horizontally and vertically), not in the third dimension (there should be no side-to-side movement when watching in line with the path of the ball).
4. Is release smooth and repeatable?
5. Have you tested one drop and checked frames to ensure you can identify t0 and the crossings clearly?
Procedure
1. Place the ball at the release position, with centre of ball aligned with the s = 0 mark. This point is your reference for t0
2. Start the digital stopwatch (visible on the stopwatch phone) simultaneously as you record and ensure the digits are clearly visible in the camera frame before you start recording. OR you can start the clock, start camera recording, then release.
3. Start the camera’s slow-motion recording mode. Immediately release the ball with the release mechanism. Allow the ball to fall and catch it with the beanbag.
4. Stop recording once the fall is complete. Repeat the drop at least 8 times to build statistics.
5. Transfer the video to a device where you can play frame-by-frame or at slower playback and clearly read the digital clock digits.
6. For each video:
a. Find the frame where the centre of the ball is exactly aligned with the s = 0 mark. Read the clock time on that frame and call this as t0.
b. Find frames where the ball’s centre aligns with the marks 1 cm, 4 cm, 9 cm, 16 cm and 25 cm below s = 0. For each mark read the clock time t
c. Compute the elapsed time for each mark: t n = t – t0.
7. For each t n compute t2 n
8. Convert distances from cm to metres before plotting.
Results
1. Copy the results table below to record your result. Give your table an appropriate heading. Time at s = 0, t0 = ……. s
3 9
4 16 5 25
2. Plot a graph of s (y-axis, in metres) against t2 n (x-axis, in s2).
3. Draw the line of best fit.
Discussion
1. Calculate the gradient of the line.
2. The slope, m, of the line equals 1 2 a
3. Using the data from the experiment, discuss:
a. validity
b. accuracy
c. reliability.
4. Identify at least three plausible error sources and suggest improvements. Suggested extensions/investigation questions
• Repeat with different drop heights and test whether percent error changes.
• Compare effect of air resistance by repeating with an object of different mass and same size (e.g. hollow vs. solid ball)
Avoiding parallax error
Parallax occurs when two objects (the falling ball and the scale/grid) are at different distances from the camera, so their apparent relative positions shift depending on the camera’s viewpoint. If the ball’s path is slightly in front of or behind the marked scale or grid, the measured vertical position of the ball relative to the scale will be offset and vary with small camera misalignment, leading to systematic error in distance measurement and therefore in computed a.
How parallax affects this experiment:
• If the ball falls in front of the scale or grid, marks appear lower than the ball’s true position; if behind, they appear higher. This causes systematic under/overestimates of s
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How to reduce parallax error:
1. Make the ball fall in the same plane as the scale. Do not let the ball deviate forward/back.
2. Place the camera perpendicular to the plane.
3. Place the camera further away and zoom in.
Method 2: using a simple pendulum
Aim
To determine the acceleration due to gravity, g, using a simple pendulum.
Background information
The formula for the period, T, of a simple pendulum is given by:
g
where l = length of pendulum.
Rearranging for g:
Materials
• retort stand and clamp
• cotton string (~1.2 m)
• small dense pendulum bob (metal ball)
• ruler or metre stick
• stopwatch
• protractor
• masking tape or marker
• graph paper
• split cork
Note: A split cork is a small cork with a slit cut halfway through it. It acts like a gentle clamp around the string of the pendulum. It allows you to secure the bob on the string and easily adjust the length.
Experimental set-up
Clamped split cork holding thread
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Figure 2C.10 Pendulum set-up for the experiment, with dashed lines showing the position of the bob at the limits of the oscillation, which should be no more than about 10° each side of the vertical.
Be careful
• Ensure the pendulum swings freely and does not strike anyone.
• Secure the clamp stand properly.
• Keep swing angles small to avoid instability and improve validity.
1. Attach the string to a fixed support (retort stand, clamp stand or overhead hook).
2. Thread the free end of the string through the split cork’s slit.
3. Tie the string to the pendulum bob (usually a metal sphere with a central hole or loop).
4. Because of the slit, the string sits inside the cork and can slide when loosened.
5. To change the length, slide the string through the cork.
6. Ensure the cork grips the string enough to hold the bob in place but allows controlled sliding.
7. Squeeze the cork lightly around the string so it holds the bob securely.
8. The friction from the slit should keep the bob from sliding during oscillation.
Procedure
1. Set up the pendulum and set the length, l (from the point of suspension to the centre of the bob), to 0.3 m.
2. Pull the bob slightly (≤10°) and release without pushing.
3. Time 20 oscillations (there and back) with a stopwatch. Record the total time, t, for 20 oscillations.
4. Repeat 3 times and calculate the average period T = t av 20 . Calculate T2
5. Increase the length, l, by 0.1 m and repeat steps 2–4. Continue increasing the length in increments of 0.1 m until the length of the pendulum reaches 1.0 m.
Thread l
Bob
Results
1. Copy this table adding rows for 8 trials. Give your table an appropriate heading.
Trial Length, l ( m ) Time for 20 oscillations, t 1 ( s ) Time for 20 oscillation, t 2 ( s ) Time for 20 oscillation, t 3 ( s ) Average time, t ( s ) Time for 1 oscillation, T = t 20 ( s )
2 ( s 2) 1 2 etc.
2. Plot a graph of T2 (y-axis) against l (x-axis)
3. Draw the line of best fit.
Discussion
1. Find the slope, m and calculate g = 4π2 m m s–2. Compare your result with accepted value 9.8 m s–2.
2. What assumptions are made in this experiment?
3. How does the length of the pendulum affect the period?
4. Why is it better to measure many oscillations instead of one?
5. Discuss validity, accuracy, reliability and precision of this experiment, with reference to the data you collected.
6. Identify at least three plausible sources of error and suggest improvements.
Conclusion
Write a conclusion comparing the two methods in terms of validity, accuracy and reliability.
Analysing the relative velocity
When we consider the motion of an object, we assume a fixed-point relative to which the given object is in motion. For example, if we say that water is flowing or wind is blowing, or a person is running with a speed “v”, we mean that all these are relative to Earth (which we have assumed to be fixed).
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Figure 2C.11
When a picture is taken from a moving vehicle, the relative velocity of the vehicle is zero, and any stationary objects surrounding the vehicle (like trees) move with a relative velocity in the opposite direction to the movement of the vehicle.
If v A and v B are velocities of two objects A and B relative to Earth, then the velocity of A relative to B is v AB = v A – v B
• A special case where the formula works using the magnitudes of the vectors occurs
• when two objects are moving along the same line in the same direction. The magnitude of velocity of A relative to B is then given by |v AB| = |v A| – |v B|. If |v AB| is positive, the direction of v AB is that of A (and B) and if it is negative, the direction of v AB is opposite to that of A (and opposite of B).
Case 1: >
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Case 2:
• I f two bodies are moving towards each other or away from each other with velocities v A and v B, then the magnitude of v B is made negative to account for the opposite direction. The magnitude of velocity of A relative to B is given by |v AB| = |v A| – (–|v B|) = |v A| + |v B| and is in the same direction as A.
Case 1: Away from each other
Case 2: Towards each other
• I f two bodies A and B are moving with velocities v A and v B in mutually perpendicular directions, then the magnitude of velocity of A relative to B is calculated using the Pythagoras’ theorem, |v AB| = |v A|2 + |v B|2
The direction of v AB with respect to v A is tan θ = –|v B| |v A| .
Velocity using relative motion of two objects in motion
Find the velocity of two objects if, when they move uniformly towards each other, they get 6.0 m closer each second, and when they move uniformly in the same direction with their original velocities, they get 6.0 m closer each 10 s.
Solution
Let the velocity of the two objects be v 1 and v 2 and let v 1 > v 2
Relative velocity v 1,2 = 6 m 1s = 6 m s 1 v 1,2 = v 1 + v 2 = 6 (1)
Explanation
In the first case, the objects are moving in opposite directions, so the relative velocity is the sum of their magnitudes (since one of the vectors will have a negative sign).
WORKED EXAMPLE 2C.7
Add equations (1) and (2):
In the second case, the objects are moving in the same direction, so the relative velocity is the difference between their magnitudes.
Solve the simulatenous equations by elimination.
Solving basic simultaneous equations is a skill that is required in Physics. To use the method of elimination, add or subtract multiples the equations such that one of the variables gets eliminated. Here, all the coefficients are one, so adding equations (1) and (2) eliminates v2 Alternatively, subtracting equation (2) from equation (1) would eliminate v1
Recall that there are multiple ways to solve simultaneous equations. The simultaneous equations can also be solved by substitution.
WORKED EXAMPLE 2C.8
Relative motion of two objects perpendicular to each other
Object A and B both have a speed of 10 m s 1. A is moving towards east while B is moving towards north starting from the same point. Find the velocity of A relative to B and the velocity of B relative to A.
Solution
Initial directions of A and B are as shown below.
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Explanation
Draw a diagram of the situation, clearly showing the magnitude and direction of each vector quantity.
Flip the direction of B and use the head-to-tail method to determine the resultant vector.
Find the magnitude by using Pythagoras’ theorem.
Use trigonometry, specifically the tangent function, to determine the orientation (or bearing) of the resultant vector.
Solution Explanation
The velocity of B relative to A is given by
To find the velocity of B relative to A, the same method can be used, except this time the equation is given by v BA = v B – v A.
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v BA = (–10)2 + 102
= 200
= 14.14 m s 1
θ = tan 1 10 10 = 45°
v BA = 14.14 m s 1 at 45° W of N
WORKED EXAMPLE 2C.9
Relative motion of a boat in a river
A boat applies a thrust that would cause it to move with a constant velocity of 20.0 km h 1 in the easterly direction (this is called the speed through water, STW). The current of the water is 4.0 km h 1 south. What is the velocity of the boat relative to:
a. an observer sitting on the shore (this is called the speed over ground, SOG)
b. a free-flowing log next to the boat?
Solution
a. Let the speed through water of the boat be v1 and the velocity of the water be v2
v1 = 20 km h–1 v2 = 4 km h–1
Explanation
Draw a vector diagram, clearly indicating the magnitude and direction of each vector quantity.
Use Pythagoras’ theorem to determine the magnitude of the resultant vector.
Use trigonometry, specifically the tangent function, to determine the orientation (or bearing) of the resultant vector.
Video: Worked example 2C.9 N
Velocity relative to the observer is 20.4 km h
b. The velocity relative to the free-flowing log = 20.0 km h 1 east.
The free-flowing log will be experiencing the same current as the boat. The only relative velocity between the boat and the log is the velocity of the boat.
Video: Worked example 2C.10
WORKED EXAMPLE 2C.10
Relative motion of two aircraft in level flight
An aircraft A flies due west at 250 m s 1 in level flight. At the same altitude and also in level flight, aircraft B flies at 200 m s 1 on heading 330° (30° W of N). What is the velocity of A relative to B?
Solution
Let the velocity of aircraft A be v A and the velocity of aircraft B be v B
Initial directions of A and B are as shown below.
Exam practice 2C
The velocity of A relative to B is given by v
Explanation
Draw a vector diagram, clearly indicating the magnitude and direction of each vector quantity.
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Flip the direction of B, calculate the horizontal and vertical components of the velocity, and use the head-to-tail method to determine the resultant vector.
Use Pythagoras’ theorem to determine the magnitude of the resultant vector.
Use trigonometry, specifically the tangent function, to determine the orientation (or bearing) of the resultant vector.
EXAM PRACTICE 2C
Example question
A group of students sets up an experiment to test the acceleration due to gravity. They drop a ball, initially at rest, from different heights and use an accurate device that can measure the instantaneous velocity of the ball. The students then record the final velocity at which the ball strikes the ground. The students aim to test the acceleration due to gravity by using the following equation:
v2 = u2 + 2as
Since the initial velocity is zero, the students rearrange the equation to: v2 = 2as
The students gain a set of results and then graph the height of the ball drop, s, on the x-axis and the velocity squared, v2, on the y-axis where they have defined down towards Earth to be positive.
a. Determine the gradient of the graph. Include the unit for the gradient in your answer.
(2 marks)
b. Use the gradient of the graph to determine the acceleration due to gravity, as measured by the students’ experiment.
(3 marks)
What is the question asking?
The gradient of a graph tells you how steep the graph is. It is calculated by the change in the y-value, ∆y, divided by the change in the x-value, ∆x gradient = ∆y ∆ x = rise run
This means that the unit for the gradient will always be the y-axis unit divided by the x-axis unit. Using this principle, we can derive the unit for the gradient of the displacement–time graph.
Figure 2C.12 Experiment results plotted as velocity squared versus height of drop
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For example, the gradient of a displacement–time graph, where the displacement is on the y-axis and the time is on the x-axis, is the velocity:
Understanding how to derive the unit for the gradient and how to interpret the meaning of the gradient is an essential skill in physics.
Marking criteria
Part a.
Criteria
• Chooses two points from the line of best fit and calculates the gradient (including SI units)
• Provides some relevant information
Part b.
• Transposes equation and substitutes in known values to solve for the acceleration due to gravity (including SI units)
• Identifies the transformation between graphed values and the linear equation y = mx + c
• Provides some relevant information
Sample response
Sample response
a. Two points on the line of best fit in Figure 2D.12 are (0.5, 10) and (2.5, 48.4).
The gradient can be calculated by:
= rise run = 48.4 – 10 2.5 – 0.5 = 19.2
Therefore, the gradient of the line of best fit is 19.2 m s–2
Explanation
Take two clear points on the line of best fit (not data points, since the line of best fit is a more reliable indicator of the experimental results). As it happens, the line of best fit passes through one of the data points, (0.5, 10)
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b.
This means the students’ data revealed that the acceleration due to gravity is 9.6 m s–2 down to Earth.
The second point used to calculate the gradient is on the line of best fit slightly above the rightmost data point, which can be approximately taken as (2.5, 48.4). Remember to determine and include the units for the gradient.
This question requires you to recognise that the gradient represents the following:
By defining y = v2 and x = s, the original equation v2 = 2as turns into a linear equation y = mx + c, where m = 2a is the gradient and c = 0 is the y-intercept.
Once you recognise the significance of the gradient, you can solve for the acceleration due to gravity using the gradient found in part a.
Note that this value is below the real acceleration due to gravity. This is not uncommon and you should not get discouraged if the value you have calculated does not match the accepted value. This is because this experiment may have had measurement uncertainties such as air resistance that decreased the final velocity of the ball and gave an acceleration due to gravity that was consistently lower than expected.
SECTION 2C REVIEW
Review questions
1. A passenger onboard a hot air balloon drops her camera. It takes the camera 7.88 s to fall from rest down to the ground. Ignore air resistance.
a. Show that the camera was dropped from a height of 304 m.
b. Calculate how much time it took for the camera to cover the first half (the first 152 m) of its journey to Earth.
c. Calculate the velocity of the camera just before it strikes the ground.
2. The minimum braking distance of a car is determined by the driver’s reaction time and the deceleration provided by the brakes. A car is travelling at a speed of 81 km h−1 when the driver sees a kangaroo jump on the road. The driver travels 9.44 m during the reaction time and a further 31 m while the car is braking with a constant (negative) acceleration.
a. What is the initial speed of the car in metres per second?
b. Calculate the driver’s reaction time.
c. What is the magnitude of the acceleration of the car during braking?
d. Calculate the total time that it took to stop the car.
3. On a construction site of a new skyscraper, Sophia takes an open elevator up to the top floor. The elevator moves at a constant speed of 5 m s−1. As Sophia passes the third floor, she drops a spanner outside the elevator and onto the third floor. Sami is standing on the third floor and sees Sophia drop the spanner. Sophia states that in her point of view the spanner is falling down as soon as she drops it so Sami must also see the spanner falling down as soon as Sophia drops it.
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a. Comment on the accuracy of Sophia’s statement.
b. At the exact moment that Sophia drops the spanner, Sami throws a bolt up in the air with a speed of 5 m s−1. Which would land on the floor of the third floor first? Explain your answer.
4. A truck is travelling at a constant velocity of 126 km h−1 when it passes a stationary police car. As soon as the two vehicles are level, the police car accelerates at 5 m s–2 to catch up to the speeding truck.
a. How much time does it take before the velocity of the police car is equal to the velocity of the truck?
b. If the police car continues to accelerate at the same rate and the velocity of the truck stays constant, how much time does it take before the police car is level with the truck?
5. Two students, Adam and Meg, are standing on a bridge that is 12.4 m above the water below. Adam has a 1 kg stone and Meg has a 0.5 kg stone. Adam throws his stone up with a velocity of 13.0 m s−1 and Meg throws her stone down with a velocity of 13.0 m s−1. Ignore air resistance.
a. What is the maximum height above the water achieved by the stone that Adam throws?
b. Adam says that both stones will hit the water with the same velocity, but Meg says that the stone that she throws straight down will hit the water below with the greater velocity. Who is correct? Justify your answer.
c. Whose stone hits the water first? Prove your answer by calculating the exact time for each person’s stone to hit the ground.
d. Adam’s stone is twice the mass of Meg’s stone. If air resistance is ignored and Meg threw her ball directly up, like Adam, what effect would the mass of the stone have on the stone’s motion? Explain your answer.
6. If a player throws a football at 30 m s−1 to a receiver who is running towards the throw at 10 m s−1, how fast is the football moving from the receiver’s point of view?
7. The air speed of a small plane is 215 km h−1 and the pilot has set a heading of 34° E of N. The wind is blowing at 57 km h−1 east. Determine the velocity of the plane relative to an observer on the ground.
Elevator speed 5 m s–1
Sophia
Sami
Chapter review
Revise and consolidate your learning with these online tools:
• Success criteria checklist (view online or download)
• Review questions (answer online or download)
• Suggested responses (view online with teacher enabled access)
• Depth study support (download)
Teachers can access these additional downloadable resources and Learning Management System features:
• Individual and whole class reports
• Assign tasks and track results
• Create (and print) custom practice exams
• Chapter tests (download)
• Chapter summaries (download)
Success criteria checklist
I
2A.1 Classify distance, displacement, time, speed, velocity and acceleration as either scalar or vector quantities 2, 12
2A.2 Use vector diagrams to represent the displacement, velocity and acceleration of an object 4, 19
2A.3 Add and subtract vectors in 2 dimensions 1, 4, 12
2A.4 Resolve vectors into components 3, 14
2A.5 Solve total distance and displacement problems 4
2A.6 Explain how instantaneous speed, average speed, instantaneous velocity and average velocity are measured and calculated 7, 21
2A.7
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2A.9 Solve problems to determine speed, velocity, distance, displacement and time 23
2A.10 Solve problems in one and 2 dimensions using a = Δ v
2B.1 Use data to construct displacement–time graphs, velocity–time graphs and acceleration–time graphs 5, 13, 16, 17, 18c
2B.2 Use data to analyse motion graphs to describe the displacement, velocity and acceleration of an object 5, 6, 7, 8, 9, 11, 13, 17
2B.3 Analyse motion graphs to determine the relationships between displacement, initial velocity, final velocity, acceleration and time 7, 8, 9, 12, 21
2C.1 Solve problems involving uniformly accelerated motion in one dimension using s = u t + 1 2 a t 2 , v = u + a t and v 2 = u 2 + 2a s 11, 14, 15, 16, 18
2C.2 Conduct a laboratory experiment to compare 2 methods of graphically determining a value for acceleration due to gravity and assess the accuracy and reliability of the experimental results 22
2C.3 Analyse problems of relative velocity involving motion in two dimensions using vAB = vA – vB 11, 19, 20
Multiple-choice questions
1.
The net force applied to the centre of the rope is
A. 350 N left.
B. 200 N right.
C. 50 N right.
D. 50 N left.
2. Which of the following is a not a scalar quantity?
A. mass
B. time
C. acceleration
D. speed
3. A towing cable is placed between two cars. The tension in the cable is 20 000 N and the cable makes an angle of 15° to the horizontal. The horizontal component of the tension force is:
A. 20 000 N
B. 19 320 N
C. 1456
D. 518 N
4. A basketball player is bouncing a basketball up and down. On one particular bounce, the ball hits the ground with a velocity of 7.80 m s 1 down and rebounds with a velocity of 5.30 m s 1 up. The change in the velocity is
A. 13.1 m s 1 up.
B. 13.1 m s 1 down.
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C. 2.50 m s 1 up.
D. 2.50 m s 1 down.
5. A ball is thrown vertically up into the air. Ignoring air resistance, which of the following is the best graph of the total distance travelled by the ball?
6. A person in a car travels past a bus stop at a constant velocity of 72 km h 1. While passing the pedestrians at the bus stop, the driver accidentally drops his sunglasses out of the window. Which one of the following statements correctly describes the motion of the sunglasses? Ignore air resistance.
A. The sunglasses will move back and down, as seen by the driver
B. The sunglasses will move vertically down and not move horizontally, as seen by the driver.
C. The sunglasses will move vertically down and back, as seen by the pedestrians at the bus stop.
D. Both A. and C. are correct.
Use the following information to answer Questions 7, 8 and 9.
On a vehicle-testing site, a new car is having its brakes tested. The car moves along a straight road before braking. Set up along the road are two infrared timing gates A and B that are used to measure the motion of the car midway through its brake test. A diagram of this situation is shown below.
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The velocity–time graph of the car’s motion once the car starts braking is shown below. At times A and B, the car crosses the timing gates A and B.
7. Calculate the average acceleration of the car between times A and B
3.66 m s–2
1.11 m s–2 C. 3.6 m s–2
8. What is the distance covered by the car between points A and B?
A. 144 m
B. 96 m
C. 48 m
D. 24 m
9. What is the average speed of the car measured by the infrared timing gates A and B?
A. 22 m s 1
B. 16 m s 1
C. 16 m s 1
D. 4 m s 1
10. A person can swim with a speed of 13 km h 1 in still water. If the speed of the current in a river is 4 km h 1, what will be the time taken by the person to go 68 km downstream?
A. 2.5 hours
B. 3 hours
C. 4 hours
D. 3.5 hours
Exam-style questions
11. A skydiver is falling vertically down with a constant velocity of 5 m s 1 downwards. When the skydiver is 100 m above the ground, a pen falls from their pocket. Assume that the pen falls with no air resistance and that the skydiver continues to fall with a constant velocity of 5 m s 1 down.
a. Calculate the speed with which the pen strikes the ground. (2 marks)
b. Determine how much time passes before the pen hits the ground. (2 marks)
c. Determine how much time passes once the pen hits the ground before the skydiver lands on the ground. (2 marks)
12. A runner places their water bottle on the ground and begins doing straight-line sprints away from, towards and past the water bottle. A displacement–time graph of the runner’s motion is shown below, where the a displacement of 0 m represents the position of the bottle.
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a. Identify what the total distance covered by the runner is. (2 marks)
b. Determine the longest distance that the runner completes without stopping for a break. (1 mark)
c. Identify what the maximum speed of the runner throughout their 300 s run is. (2 marks)
13. A track cyclist starts from rest and the first 5 s of their start is recorded in the acceleration–time graph shown.
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a. Construct a velocity–time graph of the cyclist’s motion in the first 5 s. (2 marks)
b. Another cyclist sets off, and their acceleration–time graph is shown. Calculate the final velocity of the cyclist after these 5 s have gone by. (2 marks)
14. A car is travelling with a velocity of 20 m s 1 north but slows down for a school zone to a velocity of 11 m s 1 north.
a. Identify what the change in velocity of the car is. Include a direction in your answer. (3 marks)
b. If the car travelled 88.7 m while braking, determine how much time the driver was braking for. (2 marks)
c. Determine what the acceleration of the car is. Include a direction in your answer. (3 marks)
15. A car is at rest at a set of traffic lights. When the lights change to green, the car accelerates to a velocity of 99 km h 1 in 7.91 s.
a. Calculate the magnitude of the acceleration of the car during this time. (3 marks)
b. Calculate the distance travelled by the car while accelerating. (2 marks)
16. A person is standing by a deep wishing well. They hold their coin 5 m above the surface of the water in the well and flick it directly up with a vertical velocity of 3.75 m s 1. The coin travels up and then falls down into the well. A diagram of this is shown below. 5 m
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a. Calculate the maximum distance above the water of the well that the coin reaches. (3 marks)
b. Calculate the speed with which the coin strikes the water. (2 marks)
c. Calculate the total flight time. (2 marks)
17. A truck is driving down a straight road when it passes a car at t = 0, the car then sets off and travels along the same straight road. A velocity–time graph of the two vehicles is shown below.
a. Describe the motion of the car. (3 marks)
b. Determine the car’s average acceleration, in metres per second, during the first 10 s. (2 marks)
c. Determine at what time the car overtakes the truck. (3 marks)
18. A person in a hot air balloon is filming a hawk. The person drops some of the hawk’s food, which is initially at rest, from the balloon. Seeing the food drop, the hawk flies down and passes the point where the food was initially dropped 0.730 s later with a velocity of 20 m s 1. Once the hawk is level with where the food was initially dropped from, it tucks into a dive position, which means only gravity is accelerating the hawk down.
a. Explore how much time passes after the hawk takes off when the food is caught by the hawk. (3 marks)
b. Determine how much distance the food travels before getting caught. (2 marks)
c. Use a velocity–time graph to construct a sketch of the motion of the hawk and its food. (2 marks)
19. Three friends go on a train ride. Rob is seated. The train is southbound at 15 m s 1 Relative to Rob, Ming walks towards the front of the train at 1.5 m s 1 and Aashvi runs towards the back of the train at 2.8 m s 1. There is a stationary observer outside at the train platform.
a. Identify what the velocity of the train relative to the stationary observer is.
b. Identify what Rob’s velocity relative to the stationary observer is.
c. Identify what Ming’s velocity relative to the stationary observer is.
d. Identify what Aashvi’s velocity relative to the stationary observer is.
e. Identify what Ming’s velocity relative to Aashvi is.
f. Identify what Aashvi’s velocity relative to Ming is.
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20. A boat wishes to cross a river that runs from north to south. The river is 39.2 m wide and the river’s current is 2.1 m s 1 S. The boat produces a velocity of 6.8 m s 1 Identify how long it takes the boat:
a. to cross the river perpendicular to the current
b. to land directly across from the starting point.
21. The relationship between the position and time of a moving object is shown on the graph. Use this graph to answer the following questions:
a. Determine what the instantaneous speed of the object is at 4 s.
b. Identify during which of the following times the object accelerates.
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22. The time period of oscillation of a simple pendulum is given by the following relation:
Where T = period of the simple pendulum.
The experiment is conducted using different lengths of the pendulum and measuring their time periods.
A graph is plotted for the square of time period T taken on the y-axis against the length l taken on the x-axis. The graph is a straight-line graph as shown.
a. Explore how you will calculate g from the formula and graph.
b. Determine whether the mass of the pendulum affects the value of g
c. Explain how the time period is affected if the length of the pendulum is doubled.
d. Determine the factors affecting the accuracy of the experiment.
23. An aircraft departs point O.
1. It flies 150.0 km on bearing 30° E of N in 30.0 minutes.
2. It then executes a steady right (clockwise) turn through 60.0°; while turning it follows a circular arc of length. (The arc connects the end of leg 1 to the start of leg 2.)
3. After completing the turn the aircraft continues on the new heading and flies 100.0 km on bearing due east (E) in 20.0 minutes.
Assume the speed is constant during each straight-leg (leg 1 and leg 2); treat the turn as motion along the circular arc (given arc length and time). Ignore the displacement between leg 1 and leg 2 of the journey. Calculate the displacement and velocity of the airplane.
24. A small car moves at a constant speed of 40 m s 1 along a quarter circular track (a 90° bend). It enters the curve moving due east and exits moving due north, completing the 90° turn in 5.00 s. Calculate its acceleration using vectors.
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Electricity and magnetism
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SYLLABUS OUTCOMES
A student:
• explains electric and magnetic field interactions PY-11-03
CHAPTERS
8. Electrostatics
9. Electriccircuits
10. Magnetism
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SYLLABUS
CONTENT DOT POINTS: ELECTROSTATICS
At the end of this chapter, I will be able to:
8A.1 Explain the role of the movement of electrons in charging particles and objects
8A.2 Explain how friction, induction and conduction can cause objects to become electrically charged
8A.3 Outline characteristics of an electric field surrounding a point charge
8A.4 Use the equation E = 1 4πε0 q r 2 to demonstrate that the electric field strength around a point charge follows the inverse square law
8A.5 Use field line diagrams to describe the electric field around point charges and between charged particles
8B.1 Explain the role of static electricity in producing a lightning strike*
8B.2 Solve problems involving charges in electric fields using F = qE
8B.3 Analyse the factors that affect the force between 2 charged particles using the electrostatic force law and F = 1
8B.4 Use field line diagrams to describe the electric field between 2 charged parallel plates
8B.5 Analyse the factors that affect the strength of the electric field between 2 charged parallel plates using E = V d
8B.6 Solve problems relating to the charged particles in electric fields using F = qE , E = V d , Fnet = m a and W =ΔKE
8C.1 Explain how the electric potential energy of a charged particle changes due to its position in an electric field
8C.2 Solve problems involving changes in electric potential energy in a uniform electric field using W = qV = qEd and W =ΔKE
8C.3 Account for the energy transformations that occur when a charge moves in a uniform electric field
8C.4 Solve problems involving the work done when moving a charge in an electric field using the relationships between W, ΔU and ΔKE
8C.5 Analyse the relationships between voltage, potential energy difference and work when changing the position of a charge in a uniform electric field using V = ΔU q
8C.6 Solve problems relating to changes in position of charges in electric fields using W = qV
*Syllabus dot point displayed out of syllabus order
SECTION
8A Charges and electric fields
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8B Forces due to electric fields
8C Energy associated with electric fields
GLOSSARY TERMS
charge
coulomb (C)
Coulomb’s law
electric field
electric field strength
electric potential energy
electric permittivity
electric permittivity of free space (ε0)
electron
field
field line
inverse square law
non-contact force
non-uniform electric field
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point charge
scalar field
uniform electric field
vector field
Concept map
How things become charged
The transfer of electrons results in charged objects.
8A Charges and electric fields
Electric fields depict the properties of the region of influence around charged objects.
8B Forces due to electric fields
In an electric field, a charged particle experiences a force.
8C Energy associated with electric fields
Utilising electric fields, components can store electric potential energy which can undergo a transformation to give this energy to the kinetic energy of charged objects.
Worksheet 8A.1
Title to come
8A Charges and electric fields
SYLLABUS CONTENT
At the end of this section, I will be able to:
8A.1 Explain the role of the movement of electrons in charging particles and objects
questions download word download pdf
8A.2 Explain how friction, induction and conduction can cause objects to become electrically charged
8A.3 Outline characteristics of an electric field surrounding a point charge
8A.4 Use the equation E = 1 4πε0 q r 2 to demonstrate that the electric field strength around a point charge follows the inverse square law
8A.5 Use field line diagrams to describe the electric field around point charges and between charged particles
A useful model of the atom
charge a property of matter that causes electric effects. The symbol for a point charge is the lower-case letter q. Like charges repel, opposite charges attract. The unit for charge is the coulomb, with the symbol C.
electron the lightest stable subatomic particle known. It carries a charge of –1.602 × 10−19 coulombs, which is considered the basic unit of electric charge.
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An atom can be modelled as a central nucleus composed of protons (positively charged) and neutrons (no charge ). Protons and neutrons in the nucleus of an atom are very tightly bound together, so any changes to the nucleus involve very large amounts of energy.
Orbiting the nucleus are electrons (negatively charged), as shown in Figure 8A.1. Each electron is so tiny that its position at any point in time is difficult to measure. The outer electrons are involved in chemical bonds (studied in chemistry). The electrons are also much lighter than protons. The less tightly bound outer electrons also allow objects to become charged due to having either excess electrons compared to protons (resulting in a negatively charged object) or having a lack of electrons compared to protons (resulting in a positively charged object).
NOTE
Many more subatomic particles have been discovered, leading to the development of the Standard Model. This explains matter and nuclear and electromagnetic forces, but the theory is covered in Year 12. Concepts such as quarks, leptons and bosons, including the recently detected Higgs particle, form the Standard Model. While it’s the best we currently have, this model cannot yet explain gravitational force nor dark matter or dark energy.
Video: Understanding electric fields
Figure 8A.1 A useful model (not to scale) of the atom showing a nucleus with protons (red), neutrons (blue) and orbiting electrons (green)
Electric charge
How objects become charged
There are three common ways in which an object can become charged and they all involve the transfer of electrons. These conditions that cause objects to become charged are known as charging by friction, conduction and induction.
1. Charging by friction
When two different materials are rubbed together electrons can be transferred from one to the other. This process is known as the triboelectric effect. For example, when an ebonite rod (a rod made of rubber that has tightly bound electrons) is rubbed with fur (which has electrons that are less tightly bound), electrons leave the fur and are transferred to the rod leaving the fur with a positive charge due to the loss of electrons and the ebonite rod with a negative charge due to the excess electrons.
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2. Charging by conduction
When a charged object comes into physical contact with a neutral (uncharged) conductor, there is an imbalance of electrons. The charged object has a surplus of electrons, which can be transferred to the neutral object to evenly distribute the charge across both objects, resulting in the neutral object becoming charged. Both objects end up having half of the initial charge.
3. Charging by induction
When an uncharged object experiences the presence of a charged object (without touching), the electrons in the uncharged object will relocate within the object due to electrostatic forces of attraction (between opposite charges) or repulsion (between like charges). On the side closest to the charged object, the uncharged object will acquire a charge opposite to that of the charged object. For example, if a positively charged ball is moved close to a neutral (uncharged) rod, the negative electrons will be attracted to the positive charge. The electrons relocate to the end of the rod nearest to the ball, causing a separation of charge. The end of the rod near the ball becomes negatively charged, while the opposite end becomes positively charged. The rod and ball experience a force of attraction between them.
Figure 8A.2 Example of charging by friction
Figure 8A.3 Example of charging by conduction
coulomb ( C ) the SI unit for charge. 1 C is equivalent to the combined charge of 6.242 × 1018 electrons.
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Describing and quantifying charge
In order to measure the charge of an object, a suitable unit is needed. As a positively charged object has lost electrons, and a negatively charged object has gained excess electrons one possibility is to use, as the basic unit, the charge carried. However, this charge is so tiny that it would be inconvenient for measuring the charges encountered in the laboratory or in everyday life.
Scientists have agreed to use the coulomb (C) as the unit of charge. A coulomb is the SI unit for charge, and is equal to an ampere-second, or 1 C = 1 A s. The magnitude of one coulomb is quite large, so a microcoulomb (μC, 10–6 C) is often used. Remember, one coulomb has one million microcoulomb in it, or 1 C = 106 μC.
How many electrons are needed to make up one coulomb of charge? To answer this, you need to know the size of the charge on an individual electron. This quantity of charge is known as the elementary or electronic charge, as it has long been regarded as the smallest charge that can be found in nature. Some experiments have indicated that smaller charges may exist, but the term “elementary charge” is still used to refer to the charge on an electron. While the electron carries negative charge, protons carry positive charge of the same magnitude.
Figure 8A.4 Example of charging by induction
Figure 8A.5 This photograph taken in Chicago in 1917, shows the apparatus used by Robert Millikan to measure the charge on the electron. He did this by observing the motion of tiny, charged drops of oil in an electrified field between two metal plates housed in the black metal tank, centre right. Millikan (1868–1953) was awarded
The total charge of an object, Q, can be calculated by multiplying the charge of one charged particle, q, by the number of charged particles, n Q = nq
For electrons with an elementary charge, qe, of 1.602 × 10 19 C, the number of electrons in 1 coulomb can be found by: n ×−1.602 × 10 19 =−1 C
Therefore: n = 1 1.602 × 10 19
= 6.242 × 1018
The number of electrons in 1 coulomb is 6.242 × 1018. This is a very large number!
Electric fields
If you charge objects (by friction, contact or induction) you will quickly notice some basic interactions between these charged objects. Put simply, like charges repel and unlike (or opposite) charges attract. That is, if two negatively charged objects are brought close together, they will experience a repulsive force trying to push them apart. Similarly, two positively charged objects will repel each other. However, a negatively charged object and positively charged object will experience an attractive force towards each other.
Attraction between masses occurs without the need for physical contact. This is called a non-contact force , or “action at a distance”. While working on understanding the nature of electromagnetism in 1849, the famous English scientist Michael Faraday proposed the concept of a field . Faraday’s “field” referred to “a region in the vicinity of a charged object affected by some force”.
A field is more precisely defined as a physical quantity that has a value at each point in space. For example, in various sports, “heat maps” are used as an indicator of the range and frequency of a player’s movement during a game (Figure 8A.6). They provide coaches with valuable analytical information. This is an example of a scalar field because it is a magnitude without direction at each point on the field.
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non-contact force when there is no physical contact between two objects, but acts via a field force field a region where an object feels a force, such as gravitational, electric or magnetic; more precisely defined as a physical quantity that has a value at each point in space scalar field a field with an assignment of a scalar to each point in a region in the space
Most
Least
Figure 8A.6 An impression of a heat map depicting the range and frequency of a soccer player’s movement during a game. Warm colours (red, orange, yellow) represent areas with higher frequency of occupation while the cool colours (green, blue) represent areas with lower frequency of occupation.
vector field a field with an assignment of a vector to each point in a region in the space
field line a line drawn to represent the direction of a field with arrows to indicate direction; field lines that are closer together indicate a stronger field electric field a physical field that creates a force on all charged particles within the field. Produced by charged particles and changing magnetic fields. A changing electric field also produces a changing magnetic field.
In contrast, gravitational, electric and magnetic fields are all vector fields ; they give both an indication of the value of the strength of the field at each point in space and also the direction of the field at that point.
The conventions for vector fields are that the field lines are drawn with arrows to indicate the direction of the field; field lines that are closer together indicate a stronger field and field lines further apart indicate a weaker field. Field lines never intersect or touch, if they did it would indicate a point where the field would be in two different directions simultaneously (which is not possible). Every charged object produces an electric field which can influence other charged objects.
Rules for drawing electric field lines
Electric fields are produced by electric charges and can be modelled by drawing electric field line diagrams. In an electric field, the direction of the field line at a particular point shows the direction of the force a positive charge would experience at that point.
When drawing two-dimensional electric field lines patterns around, and between, charged objects there are a few rules that need to be followed:
• electric field lines come out of positive charges and go into negative charges
• electric field lines start and end at 90° to conducting surfaces, with no gap between the lines and the surface
• field lines can never cross; if they did cross, it would mean that the force is acting in two different directions at that point, which is not possible.
Note that drawings are usually two-dimensional representations of three-dimensional fields.
Figure 8A.7a shows electric field lines around a positive charge. Figure 8A.7b shows electric field lines around a negative charge. As illustrated in the diagram, positively charged objects would be pushed away from the positive charge and would be pulled into the negative charge. Note that the charges would experience an equal and opposite electrostatic force.
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Figure 8A.7 (a.) Electric field around a fixed positive charge, +Q, and (b.) a fixed negative charge, Q These are known as radial fields (with lines arranged like the radii of a circle)
Figure 8A.8 Fields can be drawn between multiple charged particles. Since electric charges create vector fields, the field lines show the vector sums of the individual fields of the positive and negative charge.
Electric field strength
An electric field is the force field that exists in the space around every charge or group of charges. This field extends out in all directions, but the magnitude of the electric field decreases as the distance from the object increases. Because the electric field gets weaker the further out it is from the object, the field is described as a non-uniform electric field
The electric field strength , denoted by E, is measured in units of force per charge, newtons per coulomb (N C 1). Within a diagram, greater electric field strength is shown by the lines being closer together, and a weaker electric field strength is shown by the lines being farther apart.
A radial field indicates the relative strength of the field at varying distances from the charge. Close to the charge the field is relatively strong, which is represented by the field lines being close together (dense). Further away from the charge the field is relatively weak, which is represented by the field lines being further apart (less dense). The relative density of field lines can be found by choosing an area and measuring how much of the area is made up of field lines (or simply counting how many arrows pass through the area).
non-uniform electric field an electric field where the strength and direction of the field vary electric field strength is the strength of an electric field measured in newtons per coulomb (N C−1)
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Figure 8A.9 The electric field strength around a charged particle is shown by the density of the electric field lines. The blue box has four field lines passing through it, while the green box has only two field lines passing through it. This means that the electric field strength in the blue box is greater than the electric field strength in the green box. Essentially, if the field lines are closer together it is a stronger region within the field.
Video: Worked example 8A.1
WORKED EXAMPLE 8A.1
Identifying charges from field patterns
The diagram shows the electric field pattern for four charges of equal magnitude arranged in a square.
video quiz questions download word download pdf widget reporting note taking scorcher
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a. Determine the polarity of each of the charges.
b. Identify the regions with stronger and weaker electric field strength.
Solution
a. Charges A and D are positive, B and C are negative.
b. The electric field strength is greatest near the charges, it is weaker between them and gets weaker still further away from the charges.
ACTIVITY 8A.1
Explanation
Electric field lines come out of positive charges and go into negative charges.
Electric field strength can be determined by identifying the areas of highest and lowest field line density.
Electric field between charged particles
Explore electric fields between charged particles using an online simulation such as the Charges and Fields simulation on the PhET website.
Generate electric fields by changing the variables. Record your observations.
1. Do you notice any pattern on how the electric field interacts with the charged particles?
2. The sensor illustrates the electric field strength at a given point. Place the sensor in the centre of the screen. Place a positively charged particle around the sensor. By only placing negatively charged particles in the field, can you find a way to make the red arrow:
a. disappear (indicating a zero field strength)
b. double in size?
3. Explore the interaction between multiple charged particles.
4. Write three pros and cons to using a digital interactive.
Electric fields around point charges
The strength of an electric field due to a point charge can be determined from Formula 8A.1. point charge an ideal situation in which all of the charge on an object is considered to be concentrated at a single point electric permittivity a property of a material that relates to its ability to become polarised and have an electric field produced or held in it electric permittivity of free space (ε0) the properties of a vacuum related to its ability to have an electric field produced or held in it.
FORMULA 8A.1
Electric field strength near a point charge
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Where:
E = Electric field strength due to a point charge (N C–1)
ε
0 = Electric permittivity of free space (8.854 × 10–12
q = Point electric charge generating the field (C)
r = Distance from the point charge (m)
From Formula 8A.1, there are three variables that determine the strength of an electric field near a point charge:
• T he magnitude of the point charge (the larger the value of charge in coulombs, the greater the electric field strength).
• T he physical properties of either the vacuum or matter around a charge, related to how an electrical field can be produced or stored in that region. This is known as the electric permittivity . Most commonly the electric permittivity of free space ( ε0) is used, which is true for a vacuum and almost identical in air.
• T he distance from the point charge (the larger the distance from the charge, the smaller the electric field because the field lines are more spread out).
Electric permittivity
Various substances respond differently when an electric field is applied to them. This variation is accounted for by including the term related to the electric permittivity, ε0, in Formula 8A.1. While most of the time the electric permittivity of free space can be used (it is approximately the same in air as in a vacuum), if calculating the electric field in other substances like glass or water, a different value will be needed.
A vacuum does not interact with an applied electric field, so the electric permittivity of free space can be used. Similarly, the permittivity of air is only 0.059% greater, so ε0 is again the appropriate value to use. Other substances like glass and water do interact with the field, increasing the value of the electric permittivity and therefore decreasing the strength of the electric field. Interestingly, as light is made up of oscillating electric and magnetic fields, this also accounts for why the speed of light is slower in water than a vacuum.
Table 8A.1 Comparison between electric permittivity of common materials and a vacuum
inverse square law the relationship between two variables where one is proportional to the reciprocal of the square of the other
The inverse square law
The relationship between electric field strength and distance, r, is described by the inverse square law. You may recognise the inverse square law from Chapter 6 (Section 6C) in calculations of the intensity of light waves at various distances. It occurs with electric fields for the same reason: as the distance increases the field lines for radii or an ever increasing spherical surface. The surface area of a sphere is 4
hence
appears in the electric field strength equation.
If the point charge is positive, the direction of the electric field is radially outwards. If the point charge is negative, the direction of the electric field is radially inwards. This is consistent with the known phenomenon that if you bring a positive charge close to another positive charge, the force will be repulsive, but if you bring a positive charge close to a negative charge, the force will be attractive.
Figure 8A.10 (a.) A positive electric charge and the direction of the electric field E away from it at point P, which is at distance rp (b.) The electric field–distance graph (E vs. r) for a positive charge. (c.) A negative electric charge and the direction of the electric field E towards it at point P, which is at distance rp. (d.) The electric field–distance graph for a negative charge, with negative values of E
Calculating the field strength of a point charge
Calculate the electric field strength and direction at a distance of 2.0 m from a charge of +2.0 × 10–4 C. Assume the electric permittivity is ε0 = 8.854 × 10–12 A2 s4 kg 1 m–3.
Use the formula for the electric field strength near a point charge (Formula 8A.1)
Electric field strength is a vector quantity so the direction must be specified.
= 4.5 × 105 N C 1 radially outwards from the positive point charge
WORKED EXAMPLE 8A.2
Video: Worked example 8A.2
Calculating the field produced by multiple charges
As an electric field is an example of a vector field, the electric fields at a point produced by multiple charges can be added using a vector sum to find the total field at that location. This means that at the point midway between two identical charges, the fields will cancel out and the net electric field will be equal to zero.
WORKED EXAMPLE 8A.3
Combining electric fields
In the diagram, positive charge A produces an electric field of 0.25 N C 1 at point P, pointing directly away from the charge. Charge B is also positive, but smaller in charge than charge A and therefore produces a smaller electric field of 0.15 N C 1, in a different direction.
Calculate the magnitude and direction of the net electric field at point P.
Use the head-to-tail method of vector addition to solve this problem.
Sketch a diagram of the problem. As the vectors are perpendicular, Pythagoras’ theorem can be used to calculate the magnitude of the resultant electric field.
Video: Worked example 8A.3
The angle above the horizontal can be calculated using a simple trigonometric function.
EXAM PRACTICE 8A
Example question
Explain the role of movement of electrons in charging objects. (6 marks)
What is the question asking?
This question is straight from the syllabus outcomes. It is easy to provide a simple answer to this question, but that is unlikely to get the full six marks on offer. You should find ways to demonstrate how the movement of electrons results in charged objects by outlining how objects become charged. In the answer, you should repeatedly connect the examples to the movement of electrons.
Exam practice
Marking criteria
Criteria
• Explains the role of electrons with diagrams showing at least two methods of how an object becomes charged 5
• Explains the role of electrons with a diagram showing one method of how an object becomes charged 4
• Describes how the electrons are the movable charge carriers in a conductor.
• Explains how an object becomes negatively charged when it possesses an excess of electrons, whereas a deficiency of electrons results in a positive charge
• Describes how the electrons are the movable charge carriers in a conductor OR
• Explains how an object becomes negatively charged when it possesses an excess of electrons, whereas a deficiency of electrons results in a positive charge
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3
2
• Provides some relevant information 1
Sample response
An object becomes charged when it has an uneven number of protons (positively charged) and electrons (negatively charged). Electrons can easily be transferred from object to object as they are light and mobile compared to protons which are bound to the nucleus. There are three common ways in which an object can become charged, and they all involve the transfer of electrons.
When two different materials are rubbed together, electrons can move by being transferred from one to the other. For example, when an ebonite rod is rubbed with fur, electrons leave the fur and are transferred to the rod, leaving the fur with a positive charge due to the loss of electrons and the ebonite rod with a negative charge due to the excess electrons.
When an uncharged object experiences the presence of a charged object (without touching), the electrons in the uncharged object will relocate within the object. On the side closest to the charged object, the uncharged object will acquire a charge opposite to that of the charged object. For example, if a positively charged ball is moved close to a neutral (uncharged) rod, the negative electrons will be attracted to the positive charge. The end of the rod near the ball becomes negatively charged.
SECTION 8A REVIEW
Quiz
Review questions
Review questions
1. Identify the three ways an object can become charged.
Reporting (student and teacher) Go online to access the Interactive Textbook
2. Identify an example from everyday life for each of the three ways that an object can become charged.
3. Compare the three ways an object can become charged.
4. Compare electric fields with gravitational fields and magnetic fields.
5. Describe how the inverse square law relates to the strength of an electric field.
6. The diagram shows a charge Q. Point X is 10 cm from the centre of Q and point Y is 20 cm from the centre of Q. The charge of Q = +2.0 × 10–8 C.
a. Calculate the magnitude and direction of the electric field at points X and Y.
b. How would your answers change if Q was a negative charge of twice the magnitude?
7. Two positively charged metal spheres (that act like point charges anywhere outside of the radius of the spheres) are at a distance of 30 cm apart as shown. Q1 = 2.0 × 10–8 C and Q2 = 8.0 × 10–8 C.
a. Determine how far from the middle of charge Q1, on the line joining Q1 to Q2, that the magnitude of the electric field is zero.
b. Sketch a diagram of how the net electric field Enet varies with the radius from Q1 on the line in the region between the two metal spheres.
8. Explain the difference between friction and conduction in causing an object to become electrically charged.
9. Calculate the magnitude and direction of the electric field 5.0 mm away from a negative point charge of 0.80 × 10–7 C.
10. Two positive point charges, each with a charge of 1.5 × 10–8 C, are arranged 3.0 cm from each other. A student wanted to determine the electric field at a location, point P, that would form an equilateral triangle with the two point charges as shown.
Calculatethe magnitude and direction of the electric field at point P.
11. Three charges are arranged in a line, as shown.
Draw an arrow at point X to show the direction of the resultant electric field at X.
Worksheet 8B.1
8B Forces due to electric fields
SYLLABUS CONTENT
At the end of this section, I will be able to:
Electric fields and forces questions download word download pdf
8B.1 Explain the role of static electricity in producing a lightning strike
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8B.2 Solve problems involving charges in electric fields using F = q E
8B.3 Analyse the factors that affect the force between 2 charged particles using the electrostatic force law and F =
8B.4 Use field line diagrams to describe the electric field between 2 charged parallel plates
8B.5 Analyse the factors that affect the strength of the electric field between 2 charged parallel plates using E = V d
8B.6 Solve problems relating to the charged particles in electric fields using F = q E , E
and W = ΔKE
Lightning and static electricity
Everyone is familiar with the lights (and sounds) of a thunderstorm and the east coast of Australia is well known for its storms. In particular, Sydney is highly susceptible to thunderstorms due to a combination of geographical factors. These include its coastal location with moist air and warm coastal waters, the presence of the “Sydney Basin” where the city is located in a low altitude area surrounded by mountains resulting in cold air flow into the basin to meet the warm and moist coastal air, and the occurrence of meteorological phenomenon known as an “East coastal Low” – a low pressure system known for rain and storms. This is why there are thunderstorms but what is it that causes the lightning?
The phenomenon of lightning is extremely complex and a topic for study at PhD level and beyond. Difficulties measuring in situ and in reproducing the conditions that create lightning in thunderclouds make it challenging to research. A simplified understanding is that the friction between molecules in a thunder cloud, as moist warm air currents rise rapidly while cooler air falls, causes loosely bound outer electrons to be separated from their atomic nuclei and attached to other molecules. This can be described as static electricity which creates an electric potential difference within the cloud.
Figure 8B.1 A lightning storm over Sydney, New South Wales, in 1991
Figure 8B.2 shows a photograph of a 3 × 108 V (300 million volts) lightning bolt discharging at a height of 10 km above the ground. Superimposed on the image are the negative charges at the bottom of the cloud that have induced a region of positive charge at the top of the surface of Earth. This charge distribution generates an electric field of the order of 30 000 V m 1. Lightning is a naturally occurring electrostatic discharge in which an electrical arc overcomes the insulating properties of air.
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Electric
Electric forces are an example of non-contact forces – forces that act on an object without coming physically in contact with it, relying upon a field instead. Unlike gravitational forces, which are always attractive, electric forces can be either attractive or repulsive.
Like charges repel and unlike charges attract. If the charges are the same size (magnitude) and the same distance apart, then these attractive and repulsive forces are the same size, F, and point in the directions as shown in Figure 8B.3.
Figure 8B.2 An electric field created by a thunderstorm
forces
Figure 8B.3 Like charges repel, unlike charges attract. The arrows indicate the direction of the force on each of the charges.
8B.4
Changing the charge on one object affects the force acting on both the objects. This is shown in the middle diagram in Figure 8B.4, where the left-hand charge is +2q and the right-hand charge is +q and the force on each charge is now 2F. Note that for each of the three diagrams in Figure 8B.4, the force on each of the paired objects remains equal and opposite to the force on the other paired object. This is a consequence of Newton’s third law: the force of the right-hand charge on the left-hand charge is equal and opposite to the force of the left-hand charge on the right-hand charge.
The force between any two charged objects (for example, q1 and q2) is directly proportional to the product of the charges on each object:
At a given fixed distance between two charges, the size of the force depends on the size of the charges on each object (Figure 8B.4). Video:
Force between two charges
The two positive charges shown produce an electric force, F, between them.
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a. Is the electric force produced an attractive or repulsive force?
b. Copy the diagram and draw four different ways a force of 8F can be produced between various positive charges if each charge is a whole number and the separation between the two charges remains the same.
Solution
a. Repulsive
b. The charges can be 8 : 1; 1 : 8; and 4 : 2; 2 : 4 as shown.
Explanation
Like charges repel.
Use the fact that the force is proportional to the product of the charges, F ∝ q1q2.
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What happens to the force when the distance between two charged objects is increased?
Newton demonstrated that the gravitational force between two masses varies with the product of the masses and the inverse square of the distance of separation:
In 1767, Benjamin Franklin established that electricity involves forces acting at a distance, similar to gravity. Knowing this, Joseph Priestley reasoned that the absence of an electric field inside a charged hollow conductor would only be possible if these forces decrease with distance according to an inverse square law, analogous to Newton’s law of gravitation.
However, it was not until 1788–1789, when the French scientist Charles-Augustin de Coulomb published a series of eight papers on different aspects of his carefully constructed experiments on electrically charged objects, that the scientific community accepted that the electric force satisfied the inverse square law.
Coulomb’s law
Combining the concepts of force being proportional to the product of the charges, F ∝ q1q2, and that force follows an inverse square law, F ∝ 1
, gives:
This is known as Coulomb’s law and describes the force between two charges at rest.
When the charges are in motion, additional forces are involved. These are discussed in later chapters. Coulomb’s law applies to situations where the charged objects are much smaller in size than the distance between them. Such objects are referred to as point charges.
If the charged object is a metal sphere, the entire charge can be considered to be concentrated at the centre of the sphere. This holds true provided the size of the objects is small compared to the distance of separation.
As shown in Figure 8B.5 and 8B.6, doubling the distance reduces the force, F, by a factor of four, so that it is now one-quarter as large (F 4 ); tripling the distance reduces the force by a factor of nine, so it is now one-ninth as large (F 9 ).
Coulomb’s law the force between two charges at rest is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them
8B.6 The inverse square law displayed as a graph. Coulomb’s law follows the inverse square law.
Coulomb’s law can be written as an equation rather than a proportionality. The proportionality constant turns out to be 1
which is sometimes simply written as k
FORMULA 8B.1
Coulomb’s law
Where:
F = Force on a charged particle in an electric field (N) ε0 = Electric permittivity of free space (8.854
q1, q2 = Charge of each particle (C)
r = Distance between charged particles (m)
Coulomb’s law dictates that the positive direction is the opposite direction to r, the distance between the particles (away from the other particle). A positive force represents a repulsive force, while a negative force represents an attractive force. It will be immediately obvious that this formula is very similar to Formula 8A.1, the electric field produced by one point charge. The only difference between calculating the strength of an electric field (in N C 1) and an electric force (in N) is the charge of the second point charge on the right-hand side.
We know that the electric field strength of q1 is:
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If we multiply the electric field strength of q
by charge q2:
Just like gravitational force is calculated by multiplying the strength of the gravitational field, g, by the mass of a particle, m, an electric force is calculated by multiplying the strength of an electric field, E, by the charge of a particle, q
Figure 8B.5 Coulomb’s law. The effect on the force between two charges as the distance is increased follows an inverse square law.
Figure
FORMULA 8B.2
Force on a charged particle in an electric field
F = qE
Where:
F = Force on a charged particle in an electric field (N)
q = Charge of the particle (C)
E = Strength of the electric field (N C 1)
By convention, the direction of the electric field is given by the direction of the force on a small positive unit test charge placed in the field.
WORKED EXAMPLE 8B.2
Using Coulomb’s law
One coulomb is a very large charge.
a. Calculate the force between two charges each of magnitude 1.0 C that are 3.0 m apart. Use ε0 = 8.854 × 10–12 A2 s4 kg 1 m–3
b. Calculate the mass that would experience the same magnitude of force due to the gravitational field near Earth’s surface. (Hint: force due to the gravitational field near Earth’s surface is the weight force)
c. What sort of an object with that mass could be hypothetically lifted off Earth’s surface with an electrostatic force of that size?
Solution
a.
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= 1.0 × 109 N
= 1.0 GN
b. FW = mg mg = 1.0 × 109 N m = 1.02 × 108 kg
c. The electrostatic force calculated would be more than enough, for example, to lift a small modern ocean cruise liner 3.0 m into the air!
WORKED EXAMPLE 8B.3
Explanation
Use Coulomb’s law. Since we are only calculating the magnitude, no direction is needed. The direction would depend on whether the charges are like or unlike.
Video: Worked example 8B.2
Use the equation for the weight force, FW = mg, with the gravitational field strength given in the question.
This is not expected knowledge, complete some research to determine what sort of objects have a mass of that size.
Finding the force on a charged particle from multiple charges
An electron is placed between two charges. A charge (+4.0 × 10 6 C) is located 0.30 m to the left of the electron, another charge ( 2.0 × 10 6 C) is located 0.40 m to the right of the electron.
a. Draw a diagram of the three charges.
b. Calculate the force on the electron from the positive charge.
c. Calculate the force on the electron from the negative charge.
d. Calculate the net force on the electron.
Video: Worked example 8B.3
uniform electric field an electric field in which the value of the field strength remains the same at all points
Solution
a.
+4.0 × 10–6 C
–2.0 × 10–6 C + –
b. Coulomb’s law dictates that positive is defined as the direction away from the other particle.
Explanation
The diagram should show the charges and distances labelled, and the electron closer to the positive charge to the left.
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Use Coulomb’s law and note that the negative force from the positive charge indicates an attractive force on the electron. It is important to add direction.
= 6.4 × 10 14
Therefore, the force on the electron from the positive charge is 6.4 × 10 14 N towards the positive charge (or to the left)
= 1.8 × 10 14 N
Therefore, the force on the electron from the negative charge is 1.8 × 10 14 N away from the negative charge (or to the left).
d. F net = 8.2 × 10 14 N to the left
Use Coulomb’s law and note that the positive force from the negative charge indicates a repulsive force on the electron. It is important to add direction.
Add the two forces as vectors. Both point to the left so the magnitudes can be added
The electric field and electric force between parallel charged plates
Between two parallel oppositely charged metal plates, a uniform electric field is established in the central region between the plates. The field strength is constant at all points within the region of the uniform electric field, so the field lines are parallel. The strength of the electric field is given by Formula 8B.3.
Electric field strength between parallel plates
E = V d
Where:
E = Electric field strength for a uniform electric field (V m 1 or N C 1)
V = Electric potential difference (voltage) between the plates (V)
d = Distance between the plates (m)
FORMULA 8B.3
NOTE
1. The letter V is used for both the electric potential difference, known as the voltage, V, and the unit of voltage, which is volts (V).
2. The unit V m 1 is equivalent to N C 1, since 1 V = 1 N m C 1
3. Bear in mind that the unit for the distance is metre. The distance between parallel plates is often given in centimetres or millimetres, so convert to metres before you start any calculations.
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Figure 8B.7 A uniform electric field created by two parallel oppositely charged metal plates
Figure 8B.8 shows a photograph of small seeds aligning in an almost-uniform electric field created between two parallel oppositely charged metal rods simulating plates using a common physics laboratory apparatus, in a carefully controlled demonstration by a teacher.
Figure 8B.8 Apparatus showing an almost-uniform electric field between two parallel oppositely charged plates. Note that this apparatus operates on a dangerous high voltage and should not be operated by untrained or unsupervised people.
WORKED EXAMPLE 8B.4
Calculating the strength of a uniform electric field
Calculate the magnitude of the electric field strength for the apparatus shown in the photograph in Figure 8B.8. The voltage used is 100 V and the distance between the two parallel plates is 2 cm.
Explanation
State the information given and convert all units to SI units.
Use Formula 8B.3 for electric field strength between parallel plates.
Video: Worked example 8B.5
WORKED EXAMPLE 8B.5
Equating forces in a uniform electric field
In Robert Millikan’s famous oil drop experiment, he relied on the physics described in the following situation. An oil drop with an excess charge of three electrons is balanced in a uniform electric field between two parallel charged metal plates as shown. The magnitude of the electric field is 10 kV m 1 .
Calculate the mass of the oil drop. The charge on an electron is 1.602 × 10–19 C.
Use g = 9.8 N kg 1
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E = 10 × 103 V m 1
qelectron = –1.602 × 10–19 C
qtotal = 3 × –1.602 × 10–19 C
FW = Fe mg = qE
m × 9.8 = 3 × 1.602 × 10–19 × 10 × 103 m = 4.9 × 10–16 kg
EXAM PRACTICE 8B
Video: Exam practice 8B
State the information given and convert all units to SI units. The total charge, qtotal, is based on the three excess electrons.
Since the oil drop was stationary, Newton’s laws state that the weight force due to gravity (FW = mg) is balanced by the equal and opposite electric force (Fe = qE )
Example question
Explain the role of static electricity in producing a lightning strike. (3 marks)
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What is the question asking?
The question is asking for an understanding of how objects become charged, how static electricity causes an electric field and can be discharged, and is asking for it to be applied to a real-life situation (lightning strikes). You should ensure you are factually correct about the location of positive and negative charges to demonstrate your understanding.
Marking criteria
Criteria Marks
• Explains how a spark can be caused, resulting in a lightning strike 3
• Connects lightning strikes to separation of charge or a large spark/discharge 2
• Provides some relevant information 1
Sample response
Friction between molecules in a thunder cloud causes loosely bound outer electrons to be separated from their atomic nuclei and attach to other molecules. This can be described as static electricity, which creates an electrical potential difference within the cloud. The bottom of the cloud becomes negative, and the top of the ground is positive by induction. As the charges build up, the electrostatic force between the clouds and the ground becomes so great that a discharge can occur as the electrical arc (or spark) overcomes the insulating properties of the air. This results in a lightning strike.
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SECTION 8B REVIEW
Review questions
1. Complete the following sentence using the terms: negative, positive, repel, attract.
Positivecharges repel ___________________ charges and attract ___________________ charges. It is said that “like charges ___________________ and opposite (or unlike) charges ___________________”.
2. Complete the following sentence using the terms: top, induction, discharge, static, bottom, friction.
During a thunderstorm, _______ between molecules in the clouds cause the ______ of clouds to become negatively charged and the _______ of clouds to become positively charged. The surface of the earth becomes positively charged by ___________. The ________ electricity may result in such a large electric potential difference that a powerful lightning bolt may cause the earth and clouds to undergo electrical __________.
3. Two positive charges are placed near each other; one is twice the charge of the other as shown.
Compare the magnitude and direction of the forces on the left and right charges.
4. Two positive charges on metal spheres are 30 cm apart as shown.
a. Determine how far from the middle of charge Q1, on the line joining Q1 to Q2, that the magnitude of the electric field is zero.
b. Sketch a diagram of how the electric field varies between Q1 and Q2 on the line in the region between the two metal spheres.
Quiz
5. The diagram shows two parallel charged plates in a vacuum.
a. Copy the diagram and sketch the electric field lines that exist between the two charged plates.
b. The best description of the field in the region between X and Y is that it is uniform/non-uniform and static/changing. Select the correct words.
c. i. Describe what happens if an electron is placed at point X.
ii. Describe what happens if an electron is placed at point Y.
d. i. Describe what happens if a proton is placed at point X.
ii. Describe what happens if a proton is placed at point Y.
e. If the potential difference and distance between the two plates are 20 V and 250 mm respectively, calculate the magnitude and direction of the electric field between the two plates.
6. An oil drop of mass 8.0 x 10–16 kg is irradiated with X-rays giving it five excess electrons. It is then placed in between two charged parallel plates with an electric field pointing vertically downwards. The plates are 1.0 cm apart and use a voltage of 150 V.
a. Draw a free-body diagram labelling the direction of the electric force and gravitational force.
b. Calculate the magnitude and direction of the gravitational force (use g = 9.8 m s–1)
c. Calculate the magnitude and direction of the electrical force.
d. Calculate the net force on the particle.
e. Calculate the acceleration that the particle will experience (ignore air resistance)
7. a. Describe one difference between electrostatic forces acting on charges and gravitational forces acting on masses.
b. Describe one important similarity between electrostatic forces and gravitational forces. Hint: consider the effect of distance.
8. Four equal positive charges, Q, are placed at the corner of a square as shown. A positive test charge is placed at point X in the centre of the square.
a. If each point charge Q has an electric field strength of EX at point X, what is the net force acting on the positive test charge placed at point X?
b. Explain your answer to part a
8C Energy associated with electric fields
SYLLABUS CONTENT
At the end of this section, I will be able to:
8C.1 Explain how the electric potential energy of a charged particle changes due to its position in an electric field
Worksheet 8C.1
Title to come
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8C.2 Solve problems involving changes in electric potential energy in a uniform electric field using W = qV = qEd and W =ΔKE
8C.3 Account for the energy transformations that occur when a charge moves in a uniform electric field
8C.4 Solve problems involving the work done when moving a charge in an electric field using the relationships between W, ΔU and ΔKE
8C.5 Analyse the relationships between voltage, potential energy difference and work when changing the position of a charge in a uniform electric field using V = ΔU q
8C.6 Solve problems relating to changes in position of charges in electric fields using W = qV
Electric potential energy
In Chapter 4 it was demonstrated that gravitational potential energy can be stored in a gravitational field. Gravitational potential energy can be stored in a uniform gravitational field near the surface of Earth or in a changing gravitational field as we move away from the surface of Earth.
Similarly, electric potential energy can be stored in a uniform electric field between two oppositely charged parallel plates or in a changing electric field as we move away from a stationary charge. Electric potential energy is a form of energy that is stored in an electric field.
electric potential energy the energy a charge possesses due to its position in an electric field, representing the amount of work needed to move a unit charge from a reference point to a specific point in an electric field
Figure 8C.1 (a.) Gravitational potential energy is determined by the position of a mass in a uniform gravitational field, and (b.) electric potential energy is determined by the position of a charge in a uniform electric field.
Non-uniform gravitational field Gravitational potential energy increases
Non-uniform electric field
Figure 8C.2 (a.) Gravitational potential energy changes when a mass is moved in a non-uniform gravitational field, and (b.) electric potential energy changes when the positive charge is moved in a non-uniform electric field produced by the negative charge.
Electric potential
The amount of electric potential energy (ΔU) required to move a coulomb of charge in an electric field is known as the electric potential, potential difference or voltage, V, and is measured in volts. The voltage is the work done per unit charge; 1 volt = 1 joule per coulomb (J C 1). This is formally expressed in Formula 8C.1.
FORMULA 8C.1
Voltage resulting from a change of electric potential energy
Where:
= Voltage (V
ΔU = Change in electric potential energy
= Charge of the particle (C)
Formula 8C.1 shows that the greater the change in potential energy (per unit charge), the greater the voltage produced. Sometimes it is more useful to think about change in electric potential energy as work done on the particle. Formula 8C.1 can then be rearranged and combined with Formula 8B.3 to give the work done by a voltage or an electric field on a charged particle, expressed as Formula 8C.2.
FORMULA 8C.2
Work on a particle in an electric field between parallel plates
W = qV = qEd
Where:
W = Work done on a charged particle in a uniform electric field (J)
q = Charge of the particle (C)
V = Change in electric potential (potential difference) from the start to the end point of the particle in the electric field (V)
E = Electric field (N C 1)
d = Separation of parallel plates (m)
Figure 8C.3 shows a uniform electric field created between two charged metal plates, with plate A being positive and plate B being negative. The electric potential difference, ∆V, equals VAB. Usually, we simplify this to V
The positive charge, q, has a constant force acting down on it as it moves from A to B over a distance, d. The work done, W, on the charge is qV
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Figure 8C.3 Work, W, being done on a positive charge, q+, as it travels a distance, d, in a uniform electric field created between two parallel metal plates with a potential difference across the plates, V. Remember the electric field lines indicate the direction of the force a positive charge will feel at that point. The equal spacing and parallel nature of lines show that it is a uniform field.
Just as for a particle with mass having gravitational potential energy converted to kinetic energy in a gravitational field, calculating the work done is an effective way of determining the energy change of a charged particle in an electric field. This method can be used to calculate how electric potential energy changes into kinetic energy as the positive gains in velocity as it moves towards the negative plate (Formula 8C.3).
FORMULA 8C.3
Work and kinetic energy
Where:
W = Work done on a particle (J)
W = ΔKE
ΔKE = Change in kinetic energy of the particle (J)
Video: Worked example 8C.1 video
WORKED EXAMPLE 8C.1
Changes in energy for a charge at different electric potentials
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A charge of q = 2.5 × 10 6 C is located in an electric field near a negative point charge. Initially it has an electric potential of 12.0 V and is moved to a point closer to the point charge where the electric potential can be measured to be 5.0 V.
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a. Calculate the work done by the electrical field on the charge.
b. Determine the increase in kinetic energy if the particle was released from rest.
Solution
a. W = qV = q(V1 V0) = (2.5 × 10 7)(5.0 12.0) = 1.8 × 10 6 J
b. W = ∆KE = 1.8 × 10 6 J
Video: Worked example 8C.2
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Explanation
Work is equal to charge times change in electric potential. The work is negative because the negative charge did the work attracting the positive towards it (positive work is done against a force). Electric potential energy has decreased.
The particle is attracted to the negative point charge. The electric field does work on the particle turning electric potential energy into kinetic energy. The change in kinetic energy is equal to the work done. Kinetic energy has increased by the same amount as the electric potential energy decreased (law of conservation of energy)
WORKED EXAMPLE 8C.2
Changes in energy for a charge between charged parallel plates
Consider a set of charged parallel plates separated by 2.0 cm with an electric field of 4.0 ×
kg carrying a charge of 1.602 × 10 19 C is at rest next to the positive plate.
a. Calculate the initial electric potential energy of the proton at rest at the positive plate.
b. The proton accelerates towards the negative plate. What is the kinetic energy immediately before it reaches the negative plate?
c. Calculate the velocity of the proton immediately before it reaches the negative plate.
Potential energy can be calculated using Formula 8C.2.
Initially, all of the energy is electric potential energy, and kinetic energy is zero; through work done on the charge it is converted into kinetic energy (Formula 8C.3)
Using the definition of kinetic energy and the mass of a proton, the velocity can be calculated. Recall that the proton is initially at rest, so: Δv2 = v2 u2 = v2 0
Work done on a charged particle with horizontal velocity
What happens when a charged particle enters a uniform electric field at right angles to the field? Figure 8C.4 shows such an example. A positively charged particle, q, is ejected with a horizontal velocity, v, into a constant uniform electric field, E. The force acting on the charged particle in this case will always be constant and straight down; its magnitude is given by F = qE
This will cause the particle to follow a parabolic path while between the two horizontal charged plates. This is analogous to the parabolic path followed by a mass, m, projected horizontally in a uniform gravitational field, g. Ultimately, this means we could apply the uniformly accelerated motion equations from Chapter 2 to electric fields.
Energy changes in a uniform electric field
8C.4 The path of a positively charged particle, q+, entering a uniform electric field, E, at right angles is a parabola while in the field.
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Consider the energy changes of two positive charges, each with mass 4.5 × 10 3 kg and charge 6.0 × 10 7 C, in a region between two plates with a voltage of 60 V.
Figure
Figure 8C.5 The path of a horizontally projected mass, m, in a uniform gravitational field, g, is a parabola.
Table 8C.1 Comparison of energy changes where charge is at rest versus in movement
Diagram of motion
Video: Exam practice 8C
EXAM PRACTICE 8C
Example question
A uniform electric field is generated between two parallel pates 10.0 cm apart. A proton, initially at rest, moves from the positive plate to the negative plate. The voltage between the plates is 100 V. The mass of the proton is 1.673 × 10–27 kg.
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Calculate the:
a. magnitude of electric field between the two plates (1 mark)
b. magnitude of the electric force acting on the proton while in the uniform electric field (1 mark)
c. work done on the proton while travelling between the two plates (1 mark)
d. kinetic energy of the proton just before it arrives at the negative plate (1 mark)
e. speed of the proton just before it arrives at the negative plate. (2 marks)
Marking criteria
Criteria
a. Correct answer
b. Correct answer
c. Correct answer
d. Correct answer
e. Correct rearrangement or substitution of the formula for kinetic energy and correct answer
Sample
a. E = V d
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b. F = qE
= (1.602 × 10–19) × (1000)
= 1.6 × 10–16 N
c. W = qV
= (1.602 × 10–19) × (100)
= 1.6 × 10–17 J
d. W = ΔK
= 1.6 × 10–17 J
e. W = 1 2 mΔv2
= 1 2 m(v2 02)
v = 2W m
= 1.4 × 105 m s 1
Use the formula for the electric field strength between parallel plates.
Use the formula for the force on a charged particle in an electric field.
Note that the charge of the proton is not given in the original information. This is the same charge as carried by the electron, but positive not negative: +1.602 × 10–19 C.
To find the work done on the proton while travelling between the two plates, use the formula for work on a particle in an electric field between parallel plates.
The work done is also the change in kinetic energy (initial kinetic energy is zero since it started at rest), so the answer is the same as part c
To find the speed of the proton just before it arrives at the negative plate, use the formula for work:
W = ΔKE = 1 2 mΔv2
Rearrange the equation to make velocity, v, the subject. Again, since the proton is initially at rest: Δv2 = v2 u2 = v2 0
SECTION 8C REVIEW
Review questions
Review questions
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1. Describe what happens to the kinetic and potential energy of an electron moving toward the positive plate in a uniform electric field.
2. A +2.0 μC charge moves 0.15 m in a uniform electric field of 500 N C 1 in the same direction as the field lines. Calculate the work done on the charge and state what happens to its kinetic energy.
3. A 1.0 μC charge is moved 0.10 m through a uniform field, and the work done is 5 × 10–5 J. Calculate the magnitude of the electric field.
4. The diagram shows the force–distance graph for a positive charge. Estimate the change in electric potential energy if the distance between the points A and B is 20 cm.
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5. A particle with charge q = 2.0 × 10–6 C is moved in a uniform electric field. Its electric potential energy decreases by 4.0 × 10–4 J as it moves from point A to point B.
a. Calculate the potential difference, V, between points A and B.
b. State whether the particle gains or loses kinetic energy as it moves.
c. Explain the relationship between the work done by the electric field and the change in potential energy.
6. Two parallel plates are 3.0 cm apart. An electron starting from rest at the surface of the negative plate is accelerated between the plates through a potential difference of 90 V. The charge on the electron is –1.602 × 10–19 C.
Calculate the:
a. magnitude of electric field between the two plates
b. magnitude of the electric force acting on the electron while in the uniform electric field
c. work done on the electron while travelling between the two plates
d. kinetic energy of the electron just before it arrives at the positive plate.
7. Two point charges, +4.0 μC and –3.0 μC, are placed 0.25 m apart. A small particle with charge +1.5 μC and mass 2 × 10–5 kg is placed midway between them.
a. Calculate the magnitude of the electrostatic force on the particle due to each charge.
b. Determine the net force on the particle (magnitude and direction)
c. If the particle is released from rest, calculate the work done by the field as it moves 0.05 m toward the negative charge.
Quiz
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Linked questions
8A.1 Explain the role of the movement of electrons in charging particles and objects 1, 11
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8A.2 Explain how friction, induction and conduction can cause objects to become electrically charged 1, 12
8A.3 Outline characteristics of an electric field surrounding a point charge 2, 13
8A.4 Use the equation E = 1 4π ε0 q r 2 to demonstrate that the electric field strength around a point charge follows the inverse square law 3, 14
8A.5 Use field line diagrams to describe the electric field around point charges and between charged particles 5, 13
8B.1 Explain the role of static electricity in producing a lightning strike 4
8B.2 Solve problems involving charges in electric fields using F = qE 6, 16
8B.3 Analyse the factors that affect the force between 2 charged particles using the electrostatic force law and F = 1 4πε0 q1q2 r 2 7, 17
8B.4 Use field line diagrams to describe the electric field between 2 charged parallel plates 4, 15
8B.5 Analyse the factors that affect the strength of the electric field between 2 charged parallel plates using E = V d 8, 19
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8B.6 Solve problems relating to the charged particles in electric fields using F = qE, E = V d , Fnet = m a and W =ΔKE 6
I am now able to…
8C.1 Explain how the electric potential energy of a charged particle changes due to its position in an electric field
8C.2 Solve problems involving changes in electric potential energy in a uniform electric field using W = qV = qEd and W = ΔKE
8C.3 Account for the energy transformations that occur when a charge moves in a uniform electric field
8C.4 Solve problems involving the work done when moving a charge in an electric field using the relationships between W, ΔU and ΔKE
Linked questions
9, 18
10, 16, 18, 20
9, 18
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10, 16, 20
8C.5 Analyse the relationships between voltage, potential energy difference and work when changing the position of a charge in a uniform electric field using V = ΔU q 19
8C.6 Solve problems relating to changes in position of charges in electric fields using W = qV
Multiple-choice questions
10, 20
1. When a balloon is rubbed on a wool cloth, it becomes negatively charged. Which process best explains this?
A. conduction
B. induction
C. friction
D. magnetic effect
2. Two point charges, +2 μC and +3 μC, are separated by 0.5 m. The electric field is strongest
A. near the +2 μC charge.
B. at the midpoint.
C. near the +3 μC charge.
D. equally everywhere.
3. A point charge of +4.0 × 10 9 C is located in a vacuum. What is the magnitude of the electric field at a point 0.20 m from the charge?
A. 90 N C 1
B. 900 N C 1
C. 1800 N C 1
D. 0 as there are no electric fields in a vacuum
4. Which charging methods are associated with the formation and discharge of lightning strikes?
A. conduction only
B. conduction and friction
C. friction and induction
D. friction, induction and conduction
5. The diagram shows two positive charges of magnitude Q and 2Q.
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Which vector best represents the direction of the electric field at point P, which is equidistant from both of the charges?
6. The diagram shows a charge, Q. Point X is 10 cm from the centre of Q. The charge on Q = 8.0 × 10–8 C.
The electric field strength at point X is closest to
A. 7.2 N C 1
B. 7.2 × 101 N C 1
C. 7.2 × 103 N C 1
D. 7.2 × 104 N C 1
7. Two positive charges, 4Q and Q, are placed a distance, r, apart as shown. The force between them has a magnitude of F.
The distance between the two charges is doubled from r to 2r
Which one of the following is closest to the force between the two charges?
F 4
F 2
F
8. Two charged plates with equal and opposite charges are placed 7.5 mm apart. The electric field between the plates is equal to 200 N C 1. What is the voltage difference between the two plates?
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A. 1.5 V
B. 26.7 V
C. 1500 V
D. 26 700 V
9. A positive charge moves from a positive to a negative plate. Its kinetic energy increases because
A. its potential energy increases.
B. work is done by the electric field.
C. its mass decreases.
D. its voltage decreases.
10. A charge, which is initially at rest, is accelerated through a potential difference of 500 V and increases its kinetic energy by 4.0 × 10–5 J. Which one of the following is closest to the magnitude of the charge?
A. 4.0 × 10–8 C
B. 8.0 × 10–8 C
C. 4.0 × 10–7 C
D. 8.0 × 10–7 C
Exam-style questions
11. Explain why electrons, rather than protons, are primarily responsible for charging objects. (3 marks)
12. Explain, using a diagram, how induction can cause an object to become electrical charged. (5 marks)
13. Two equal negative charges are placed as shown.
a. Copy the diagram and construct a representation of the electric field lines between and surrounding the two charges. (2 marks)
b. One of the negative charges, –Q, is now replaced with the same magnitude positive charge, +Q, as shown. Copy the diagram and construct a representation of the electric field lines between and surrounding the two charges. (2 marks)
14. A point charge has an electric field strength of 200 N C 1 at 0.5 m. Calculate the expected field strength at 1.0 m and explain why it changes. (4 marks)
15. Copy the diagram and construct the electric field lines by drawing between the three parallel plates shown. (2 marks)
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16. The electron gun section of a particle accelerator accelerates electrons between two plates that are 10 cm apart and have a potential difference of 5000 V between them.
Mass of electron
9.109 × 10–31 kg
Charge on electron 1.602 × 10–19 C
a. Calculate the electric field between the plates. Include an appropriate unit. (2 marks)
b. Calculate the magnitude of the force on an electron between the plates. (2 marks)
c. Calculate the speed of the electrons as they exit the electron gun. Ignore any relativistic effects. Assume that the initial speed of the electrons is zero. (2 marks)
17. An electron is located at the origin of a coordinate system. A positive charge of +4.0 × 10 6 C is placed at position, (0.30 m, 0) and a negative charge of 2.0 × 10 6 C is placed at position (0, 0.40 m)
a. Construct a diagram showing the positions of the electrons and the two charges. Indicate the direction of the forces acting on the electron due to each charge. (3 marks)
b. Calculate the net force on the electron. (6 marks)
18. A particle with the charge 0.15 C freely moves from a potential of 20 V to 2 V. Explain how its electric potential energy and kinetic energy change. (3 marks)
19. Explain why increasing the distance between parallel plates affects the electric field strength, assuming the voltage is constant. (3 marks)
20. A charged particle carrying charge Q and travelling at speed v enters a uniform electric field, E. The charged particle is deflected vertically a distance d while travelling through the uniform electric field as shown.
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a. Is the charged particle positive or negative? Justify your answer. (2 marks)
b. Is the force acting on the charged particle constant or changing?
Justify your answer. (2 marks)
c. Is the speed of the charged particle as it leaves the uniform electric field less, the same or greater than when it entered the uniform electric field?
Justify your answer. (2 marks)
d. A particle carrying the same charge, Q, enters the field travelling at a speed 2v. Does the deflection distance, d, decrease, stay the same or increase? Justify your answer. (2 marks)
e. Another charged particle, carrying charge Q 2 , enters the uniform electric field travelling at speed v. Does the deflection distance, d, decrease, stay the same or increase? Justify your answer. (2 marks)
Magnetism 10
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SYLLABUS
CONTENT DOT POINTS: MAGNETISM
At the end of this chapter, I will be able to:
10A.1 Identify that magnetic field lines are continuous and used to represent the strength and direction of magnetic fields
10A.2 Use field line diagrams to analyse the magnetic field around a bar magnet, current-carrying wire and current-carrying loop
10B.1 Calculate the strength of the magnetic field produced by a currentcarrying wire using B = μ0I 2πr
10B.2 Describe the structure of a simple solenoid
10B.3 Account for the shape and direction of the magnetic field produced by a direct current passing through a solenoid
10B.4 Calculate the strength of the uniform magnetic field inside a solenoid using B = μ0NI L
10B.5 Conduct a secondary-source investigation to assess the choice of materials used in the core of electromagnets
10B.6 Conduct a practical investigation to examine the factors affecting the strength of an electromagnet
SECTION
10A Magnetic fields
10B Electromagnetism
GLOSSARY TERMS
electromagnet
electromagnetism magnetic magnetic dipole magnetic field
magnetic permeability constant (μ0) magnetise permanent magnet right-hand grip rule
solenoid static field temporary magnet uniform magnetic field
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Concept map
10A Magnetic fields
Are visualised and analysed using field line diagrams
The arrow convention uses dots and crosses to indicate a direction into the page and out of the page, respectively.
B out of the page B into the page
Current-carrying wires can produce magnetic fields
10B Electromagnetism
Wire wrapped into a coil produces a solenoid
The direction of the field can be determined using the right-hand grip rule
SNSN
SNNS SN
10A Magnetic fields
SYLLABUS CONTENT
At the end of this section, I will be able to:
10A.1 Identify that magnetic field lines are continuous and used to represent the strength and direction of magnetic fields
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quiz questions download word download
10A.2 Use field line diagrams to analyse the magnetic field around a bar magnet, current-carrying wire and current-carrying loop
Magnetism
You will be familiar with the fact that a permanent magnet attracts paper clips, nails, keys and other objects made of iron. This is because the permanent magnet is magnetising the other object and turning them into temporary magnets . All magnets, no matter what their shape (e.g. bar, horseshoe or cylindrical), have two ends or faces called poles, where the magnetic effect is strongest. By convention, these poles are called the north and south poles respectively. The technical term for this is a magnetic dipole
Whenever a magnet is broken, the individual pieces are each (weaker) magnets, also having two poles. This process of creating smaller and weaker magnets, by continually breaking up magnetised materials, can continue right down to subatomic particles. Even protons and electrons have magnetic dipoles.
Magnetic fields
Worksheet 10A.1 Dipoles and monopoles
Figure 10A.1 The neodymium–iron–boron magnet is the strongest permanent magnet material in wide-scale use today.
The region around a magnet, where magnetic effects can be experienced, is called the magnetic field and can be demonstrated using iron filings, as seen in Figure 10A.2a.
The direction of the field at a given point, is taken to be the direction of force that would be on a magnet’s north pole if it was placed at that point. As a result of this force, a compass will have its north arrow pointing in the same direction as the arrows on the magnetic field lines.
Arrows are used to indicate the direction of the force. Note that both of these figures show a two-dimensional representation of the magnetic field. The magnetic field strength (which is given the symbol B ) is a vector quantity, since it has both magnitude and direction. The SI unit of magnetic field is the tesla (T).
Video: Understanding magnetic fields
permanent magnet a material that naturally attracts or repels other magnets and certain metals and keeps that ability with negligible decline over time magnetise cause specific materials to behave like a magnet temporary magnet a material that only attracts or repels other magnets and certain metals while it is magnetised by external factors
magnetic having the property of attracting or repelling other magnets magnetic dipole any object that generates a magnetic field in which the field is considered to involve two opposite poles, emanating from the north pole and entering the south pole of a magnet magnetic field a region that contains a physical field that creates a force (perpendicular to the field) on moving charged particles and magnetic materials within the field. Produced by changing electric fields and magnetised materials. A changing magnetic field also produces a changing electric field, hence the self-propagation of electromagnetic waves.
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Figure 10A.2 All permanent magnets have a north and south pole. The magnetic field surrounding a magnet can be (a.) visualised using iron filings and (b.) represented using arrows. Magnetic field lines always move from the north pole to the south pole in continuous lines and never cross each other.
Measuring magnetic fields
The strength of Earth’s magnetic field at its surface is small, about one tenthousandth of a tesla (0.1 mT). A typical fridge magnet has a strength of approximately 30 mT. The strength of a magnetic field of a typical school bar magnet is about 10 mT, a typical MRI machine used in hospitals operates at 1.5 T to 3 T, and the strength of a very strong university physics laboratory magnet is about 5.0 T.
Physicists have developed very sensitive magnetic detectors and can measure magnetic fields as low as 8 × 10–15 T. The measurement of weak magnetic fields is not only for the development of scientific research into the mechanisms of magnetisation, but also allows us to trace and map geological mineral ore bodies from low-flying helicopters and enables medical diagnosticians to examine the very weak magnetic fields generated by the body, such as in the heart and brain.
Magnetic fields exist in three dimensions. A close-up photograph of the iron filings pattern between the north pole of a bar magnet and the south pole of a horseshoe magnet shows the three-dimensional nature of the magnetic field (Figure 10A.3).
Figure 10A.3 The three-dimensional nature of magnetic field lines can be seen when iron filings are gathered between the north pole of a bar magnet and the south pole of a horseshoe magnet.
In 1846, Michael Faraday introduced the concept of magnetic field lines to explain how magnets could attract and repel each other.
Faraday’s rules for magnetic field lines are listed below.
• Each magnetic field line is a continuous line that leaves the north end of the magnet, enters at a south end and passes through a magnet from the south end to the north end.
• Field lines do not intersect.
• The closeness of the magnetic field lines represents the strength of the magnetic field.
Magnetic field diagrams can be used to explain why magnets attract each other and repel each other, as shown in Figure 10A.4. The forces between magnets can be attractive, when unlike poles are placed near each other or repulsive, when like poles are placed near each other. Note that between the unlike poles, there is a section of almost uniform magnetic field
uniform magnetic field a magnetic field in which the value of the field strength remains the same at all points as demonstrated by parallel field lines
static field a field that is not changing
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U-shaped magnets (also known as horseshoe magnets) tend to have one side as the north pole and the other as the south pole. Between the two poles, field lines can be drawn from the north pole to the south pole. In this region, there often is an unchanging, uniform magnetic field present. The magnetic field is not changing and is therefore said to be a static field
Figure 10A.4 Magnetic field diagrams for (a.) attraction: north–south poles and (b.) repulsion: north–north poles
Figure 10A.5 Magnetic field lines shown by iron filings for (a.) attraction: north–south poles and (b.) repulsion: north–north poles
Figure 10A.6 A two-dimensional view of the magnetic field between the poles of a U-shaped magnet. Note the region of uniform magnetic field with the parallel field lines.
Magnetic dipoles
All magnetic field diagrams show a north pole and a corresponding south pole; in other words, magnetic fields are examples of dipole fields. This is different to electric and gravitational fields, where monopoles exist. To date, there is no evidence for the existence of magnetic monopoles.
The
magnetic monopole
To date, it has not been possible to isolate a single magnetic pole. However, since Paul Dirac’s 1931 theoretical prediction that they should exist, physicists have been searching for them continuously. The guiding principle for the prediction is the symmetry of nature. For example, the existence of the negative electron suggested the existence of the positive electron (positron). This positron was subsequently discovered and is now the basis of an invaluable medical diagnostic technique known as positron emission tomography (PET). So far, no one has been able to find an isolated magnetic pole (that is, without its corresponding opposite magnetic pole). The absence of experimental confirmation of the existence of magnetic monopoles creates a serious theoretical threat to the idea of an underlying symmetry between our understanding of electricity and of magnetism.
Representing a magnetic field in three dimensions
Representing a magnetic field often requires a three-dimensional view. To achieve this on a flat two-dimensional page, a convention is adopted called the arrow convention. The symbol of a circle with a dot in the middle is often used to represent a magnetic field coming out of a page and a circle with a diagonal cross is often used to represent a magnetic field going into the page. The cross represents the tail feather or fletching of an arrow moving away from you (into the page), while the dot represents the tip or arrowhead coming towards you (out of the page). Note that in some diagrams, just the arrowheads (dots) and tails (crosses) are used, without the surrounding circles.
Figure 10A.8 demonstrates how the arrow convention is used in practice. The eye on the left sees the magnetic field going into the page (arrow tails) and the eye on the right sees the magnetic field coming out of the page (arrow heads).
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Viewed from right Right
Figure 10A.7 Magnetic fields going into the page as viewed from the left and coming out of the page as viewed from the right, using the arrow convention.
Viewed from left Left NS
Electromagnetism
In 1820, the Danish physicist Hans Christian Ørsted noticed that the needle of a compass, placed next to an electrical wire carrying a current, created forces on the needle (Figure 10A.8a) that eventually turned the needle of the compass so that it was perpendicular to the wire (Figure 10A.8b).
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Figure 10A.8 In 1820, Hans Ørsted discovered that an electric current creates its own magnetic field that forces the compass needle to turn perpendicular to the current-carrying wire.
Ørsted had discovered a connection between electricity and magnetism –electromagnetism . He found that a wire carrying a current shows an associated magnetic field.
A straight current-carrying wire creates a circular magnetic field around it. The direction of the circular field is determined by using the right-hand grip rule (Figure 10A.9a). For this rule, the thumb indicates the direction of conventional current and the curled grip indicates the direction of the field lines. The strength of a magnetic field around a current-carrying wire is directly proportional to the magnitude of current flowing in the wire and inversely proportional to the distance from the wire.
Figure 10A.9 (a.) Ørsted’s discovery of the existence of a magnetic field, B, (purple lines) around a wire carrying an electric current, I (b.) The right-hand grip rule, where I is the current-carrying wire and B is the magnetic field. Recall the convention for current direction is from the positive to negative terminal.
electromagnetism the branch of physics that considers the connection between electricity and magnetism right-hand grip rule a mnemonic used to determine the direction of the magnetic field produced by a currentcarrying wire. The thumb indicates the direction of the current and the curled grip represents the field lines.
The arrow convention is used to indicate the direction of the current if the wires go into or come out of the page. Recall that a cross represents the tail of an arrow going into the page, while the dot represents the head of an arrow coming out of the page. Applying the right-hand grip rule creates magnetic fields surrounding the currents as shown in Figure 10A.10.
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Figure 10A.10 Magnetic field for current-carrying wires with (a.) current coming out of the page and (b.) current going into the page. This is also demonstrated with a side-view perspective, for (c.) out of the page and (d.) into the page.
Remember that the arrow convention is also used for the magnetic field going into or coming out of the page, as shown in Figure 10A.11.
B out of the page B into the page
Figure 10A.11 The direction of the magnetic field, B, in and out of the page, can be determined using the right-hand grip rule for a current-carrying wire, where I is the current and B is the magnetic field.
Two current-carrying wires arranged parallel to each other will each have their own magnetic field. The direction of the magnetic field around each wire is given by the right-hand grip rule. If the two wires are brought closer together, they interact; the respective field vectors add together to make a new field. The interaction can result in either an attraction or repulsion of the wires. When the currents are in the same direction, the wires attract (Figure 10A.13a). When the currents are in opposite directions, the wires repel (Figure 10A.13b).
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Figure 10A.12 Two current-carrying wires (a.) attract when current runs through them in the same direction and (b.) repel when current runs through them in opposite directions. a. b.
Figure 10A.13 Magnetic field between two current-carrying wires near each other. The fields of the two wires interact with each other: (a.) two wires with a current in the same direction (into the page) and (b.) two wires with a current in opposite directions.
A current-carrying wire can also be arranged into a loop. Figure 10A.15 shows a single loop of a current-carrying wire with current travelling in the anticlockwise direction. Using the right-hand grip rule, it can be determined that at all points around the outside of the loop of the current-carrying wire, there is a magnetic field that goes into the page, while on the inside of the loop, the magnetic field is going out of the page. The reverse is also true. When the current goes in a clockwise direction, all the points inside the loop will have a magnetic field going into the page and the points outside the loop will have a magnetic field going out of the page.
WORKED EXAMPLE 10A.1
Drawing magnetic fields
B into the page
Figure 10A.14 An anticlockwise loop of wire produces a magnetic field that comes out from the page inside the loop.
Consider the current-carrying wire shaped like an “S” shown. Determine the direction of the magnetic field at points A and B.
Solution
Around point A, as your thumb moves with the direction of the current, your fingers when on the A side of the wire will be pointing out of the page.
Around point B, as your thumb moves with the direction of the current, your fingers when on the B side of the wire will be pointing into the page.
Example question
Explanation
Use the right-hand grip rule at each point around the S-bend wire (refer to Skills 10B.2)
The S-bend closely resembles two loops. The bottom half is like a loop with the flow of current in an anticlockwise direction and the top half is like a loop with the flow of current in a clockwise direction.
A student is given a diagram showing magnetic fields around a bar magnet.
b. Explain how the spacing of the field lines represent the strength of the magnetic field. (4 marks) I I B out of the page
a. Describe what the magnetic field lines indicate about the direction of the magnetic field around the bar magnet. (3 marks)
EXAM PRACTICE 10A
Video: Worked example 10A.1
Video: Exam pratice 10A
What is the question asking?
While this is a simple question, it is important to focus on the command terms describe and explain. For part a. of the question the response doesn’t need to go into cause-and-effect but simply provide (describe) the features of the magnetic field lines and link it to the direction of the field. For part b., the command term is to explain. It has already been identified that the spacing is related to the strength but the question is asking for a deeper understanding of why.
Marking criteria
Part a.
Criteria
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Marks
• Correctly describes the direction including reference to north to south 3
• Correctly describes the direction 2
• Provides some relevant information 1
Part b.
Criteria
Marks
• Explicitly links closer spacing to increased strength with explanation 4
• Explicitly links closer spacing to increased strength with partial explanation 3
• Explicitly links closer spacing to increased strength 2
• Provides some relevant information 1
Sample response
a. The magnetic field lines show the direction of the magnetic field by pointing in the direction of the force of the north pole of a magnet, which is also pointing in the direction that a north arrow on a compass would point. The direction points from north to south of the bar magnet.
b. When the spacing of the field lines are closer together, the magnetic field is stronger. This is because the magnetic field lines are closer together near the poles of the magnet where the field is stronger and further away from the poles where the field is weaker.
Review questions
1. Explain the various features of a magnetic field that can be determined from viewing a magnetic field diagram.
2. Explain how the right-hand grip rule can be used to determine the direction of the magnetic field around a current-carrying wire.
3. Describe how the strength of the magnetic field changes with radius around a currentcarrying wire.
4. Two bar magnets are brought close together so their north poles are 1.0 cm apart. Draw the magnets and the magnetic field associated with this set-up.
5. Compare the magnetic fields around a current-carrying wire and a current-carrying loop.
Worksheet
10B Electromagnetism
SYLLABUS CONTENT
At the end of this section, I will be able to:
10B.1 Analysing magnetic fields and solenoids questions download word download pdf
10B.1 Calculate the strength of the magnetic field produced by a current-carrying wire using B = μ0I 2πr
10B.2 Describe the structure of a simple solenoid
10B.3 Account for the shape and direction of the magnetic field produced by a direct current passing through a solenoid
10B.4 Calculate the strength of the uniform magnetic field inside a solenoid using B = μ0NI L
10B.5 Conduct a secondary-source investigation to assess the choice of materials used in the core of electromagnets
10B.6 Conduct a practical investigation to examine the factors affecting the strength of an electromagnet
solenoid a coil of wire that creates a magnetic field when current is passed through it that is similar to a bar magnet magnetic permeability constant (μ0) how a vacuum exhibits magnetic fields. It takes the value of μ0 = 4π × 10–7 N A–2 and is used to calculate the strength of magnetic fields around wires and inside solenoids since the permeability of air is almost the same as a vacuum (1.00000037 μ0)
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Calculating the strength of the magnetic field produced by a current-carrying wire
Section 10A provided a qualitative model of the magnetic field produced by a current-carrying wire. Building on the qualitative model, a quantitative description of the magnetic field produced by a current-carrying wire further enhances our understanding of electromagnetic properties. The wire can be wrapped into a coil to produce a component known as a solenoid . This section describes the distinct electromagnetic properties of wires and solenoids.
Experimental observations of the magnetic force around a current-carrying wire show that there is a larger force on the wire:
• when closer (at a smaller radius) from the wire
• when the current is greater.
Combined, these findings can be expressed as Formula 10B.1.
FORMULA 10B.1
Strength of the magnetic field produced by a current-carrying wire
Where:
B = Magnetic field strength (T)
μ0 = Magnetic permeability constant (μ0 = 4π × 10–7 N A–2 or T m A–1)
I = Current in the wire (A)
r = Radial distance from the centre of the wire (m)
The magnetic permeability constant (μ0) represents how a vacuum conducts magnetic fields and has a value of μ0 = 4π × 10–7 N A –2. A larger permeability means a stronger magnetic field can be produced for a given current. In a vacuum, μ0 sets the baseline strength of magnetic interactions. It is comparable with the electric permittivity of free space, described in Chapter 8, which is the ability of a vacuum to exhibits an electric field.
SKILLS 10B.1
Understanding how variables relate to each other
Formula 10B.1, the strength of the magnetic field produced by a current-carrying wire, provides a helpful example for understanding how variables relate to each other:
The current, I, is part of the numerator, and therefore magnetic field, B, is directly proportional to current, I (for a constant radius). The radius, r, is part of the denominator, therefore magnetic field, B, is inversely proportional to radius, r (for a constant current). If we keep the remaining variables constant, the gradient gives us a lot of information. Consider an experiment that investigates the affect of current on the magnetic field, while keeping the radius constant. If we know the values for μ0 and π, then the gradient of the graph of B vs. I is m
. The radius can then be determined by r =
. Similarly, if the results for an experiment investigating the effect of the radius on the magnetic field (with constant current) are plotted with the magnetic field, B, on the y-axis and the inverse of the radius (r 1 or 1 r ) on the x-axis, then the gradient would be mB
. The current can be found using I =
Figure 10B.2 A graph of the magnetic field strength against the inverse of the radius. Can you determine the constant current that was used in this experiment?
Figure
Video: Worked example 10B.1
WORKED EXAMPLE 10B.1
Calculating the strength of the magnetic field produced by a current-carrying wire
Imagine a 2.0 m long, thin, straight wire connected to a circuit transmitting a current of 5.0 A.
Calculate the strength of the magnetic field 4.0 cm from the wire.
Solution
= 2.5 ×
Therefore, the magnetic field strength is 25 μT.
Solenoids
Video: Strength of a solenoid
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electromagnet a coil of wire through which a current is passed around a soft iron or low-carbon steel core to create a magnet
Explanation
Use Formula 10B.1 for the strength of the magnetic field produced by a current-carrying wire. Substitute values for I, r and μ0. Note that the length of the wire is irrelevant to the answer. After substituting, the common factor of 2π can be cancelled from both the numerator and denominator.
One tesla is very large, so a very small number is often expected in school laboratory settings.
Section 10A described what happens when a current-carrying wire is arranged in a loop. We will now explore what happens when there are many loops and the wire forms a coil. Recall how a single loop of wire concentrates the magnetic field in the centre of the loop (Figure 10B.3a). By making a coil of wire with lots of loops, it is possible to create a very intense magnetic field. Such a coil of wire is called a solenoid. When an electric current flows in the coil, a magnetic field is produced in and around the solenoid. Figure 10B.3b shows that the magnetic field created by the solenoid is similar to that of a bar magnet.
Figure 10B.3 (a.) A single loop of wire concentrates the magnetic field in the centre of the loop. (b.) A coil of wire with many loops (solenoid) creates an even stronger and more uniform internal magnetic field.
In 1823, an English electrical engineer, William Sturgeon, placed an iron rod inside a solenoid. This increased the strength of the magnetic field produced by the solenoid. The reason why the magnetic field strength increased with the iron rod has to do with the internal structure of the rod, which essentially turns into strong, temporary magnet (due to the magnetic field that is already present). The magnetic field strength of this temporary magnet reinforces and amplifies the field strength of the solenoid. Sturgeon had invented the electromagnet, which could be turned on and off, made stronger by increasing the current and constructed into large configurations. Modern electromagnets can produce magnetic fields up to a thousand times stronger than the magnetic field produced in a solenoid using the same current.
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Figure 10B.4 An electromagnet made by placing an iron rod inside a solenoid. The magnetic field surrounding the electromagnet is similar to that of the solenoid, but can be up to one thousand times stronger.
10B.5 500 kg of hot steel chips are transported into an industrial furnace by an electromagnet. The furnace then creates cast metal for wind turbines.
SKILLS 10B.2
Determining the direction of the magnetic field inside a solenoid Magnetic fields, like electric fields, are examples of vector fields. So, the magnetic field from a solenoid is equal to the sum of the magnetic fields from each individual part of the wire, producing a uniform magnetic field along the inside of the coil. Which direction does the field point? Just as the right-hand grip rule can show you the direction of the magnetic field around a single wire, the right hand can also show the direction of the magnetic field inside the solenoid. This time, curl the fingers of your right hand in the same direction as the current moving around the coil and your thumb will point in the direction of the magnetic field.
Figure 10B.6 The right-hand grip rule can be used to determine the direction of the magnetic field inside a solenoid. The curled fingers represent the flow of current in the coiled wires and the thumb is the resultant magnetic field direction.
Figure
Calculating the strength of the magnetic field inside a solenoid
Experimental observations of the magnetic force inside a solenoid show the strength of the magnetic field increases when the solenoid:
• has a greater current running through it
• has more turns of wire
• is shorter in length with the same number of turns.
Combined, these findings can be expressed as Formula 10B.2.
FORMULA 10B.2
Strength of the uniform magnetic field inside a solenoid
Where:
B = Magnetic field strength (T)
= Magnetic permeability
N = Number of turns of wire in the solenoid
I = Current in the wire (A)
L = Length of the solenoid (m)
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Calculating the strength of the uniform magnetic field inside a solenoid
WORKED EXAMPLE 10B.2
ACTIVITY 10B.1
Investigating the choice of materials used in the core of electromagnets
Various metals, each with different effects, can be used in the core of electromagnets. Identify and analyse at least three credible online sources and answer the following questions.
1. What are the properties of metals that make them more likely to produce a stronger electromagnet when used as the core of a solenoid?
2. Write a list of six metals in order from producing a stronger electromagnet to producing a weaker electromagnet.
3. For each of the metals listed in question 2., write the value of the magnetic permeability of the metal. What do you notice about the order of your list?
PRACTICAL 10B.1
Factors affecting the strength of an electromagnet
Aim
To investigate how the supplied current effects the strength of an electromagnet.
Materials
• 12 V DC powerpack
• rheostat (variable resistor)
• ammeter
• electrical leads with alligator clip attachments
• 0.50 m of thin, insulated copper wire
• 10 cm iron nail
• box of metal paperclips
Procedure
Be careful
• Electrical shocks may occur.
• Components may be hot after use and hot wires can burn if there are short circuits.
• Ensure voltage and current values are not exceeded.
• Turn off power supply when changing the circuit.
1. Wind the insulated copper wire along the length of the copper wire to create an electromagnet.
2. Count and record the number of turns of wire.
3. Connect a circuit to the powerpack with the following components in series: rheostat, ammeter and electromagnet.
4. Turn on the powerpack and adjust the rheostat until the current is 1.0 A.
5. Attempt to pick up as many paperclips from the bench at once and record the number of paperclips in a table.
6. Turn the powerpack off and allow any residual heat in the nail or wire to dissipate. You may notice that some of the paperclips stay stuck even after the power has been turned off. This is because regular steel can retain magnetisation. This can be eradicated by using heat, physical shock (such as hammer blows) or passing AC current through it (recommended).
7. Repeat steps 4–6 for currents of 2.0, 3.0, 4.0 and 5.0.
Results
Record all results in a table and construct a graph of number of paperclips vs. current.
Discussion
1. Describe in words the relationship between current and the electromagnet’s ability to pick up paperclips.
2. What do you think would be different if you had a coil that was twice as long around the nail with the same number of turns?
3. What do you think would be different if you had a coil that was the same length but contained twice as many turns of wire around the nail?
4. What do you think would be different if you removed the nail from the coil?
Conclusion
Write an appropriate conclusion relating to the aim of this investigation.
Extension
Optional extension activities include comparing the strength:
• of the electromagnet with and without the nail as the core
• with the number of turns per unit length of wire.
EXAM PRACTICE 10B
Example question
Two parallel wires are 10 cm apart, each carrying a current of 2.0 A down the page. Calculate the magnetic field at point X, 10 cm to the right of the right-hand wire (see diagram).
What is the question asking?
The question requires an understanding of how to calculate the magnetic field around a wire, the use of the right-hand rule to determine direction and the knowledge that magnetic fields are vector fields, therefore that magnetic fields from each wire can be added as vectors.
Marking criteria
• Correctly calculates the magnitude
Correctly identifies direction
Includes units (tesla
• Two out of three of the above criteria
• Calculates the correct magnitude of the magnetic field at X from
Sample response
Sample response
The distance to point X is 0.1 m from the right-hand wire and 0.2 m from the left-hand wire.
B = μ0I
2πr
= (4π × 10 7)×(2.0)
2π (0.2)
= 2.0 × 10–6 T
= 2.0 μT
B = μ0I
2πr
= (4π × 10 7)×(2.0)
2π (0.10)
= 4.0 × 10–6 T
= 4.0 μT
B = 2.0 × 10–6 + 4.0 × 10–6
= 6.0 × 10–6 T
Therefore, the magnetic field strength is 6 μT out of the page.
SECTION 10B REVIEW
Review questions
1. Describe the structure of a simple solenoid.
Explanation
State the relevant information.
Calculate the magnitude of the magnetic field at X from the left-hand wire.
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Calculate the magnitude of the magnetic field at X from the right-hand wire.
As both wires are going downwards, due to the right-hand grip rule, the fields at X will both be coming out of the page, so the two magnitudes add.
2. Calculate the strength of the magnetic field 6.0 cm away from a current-carrying wire with 2.0 A of current.
3. Calculate the strength of the magnetic field inside a 10 cm solenoid with 15 turns of wire containing 0.5 A.
4. Describe how the magnetic permeability of a metal relates to its effectiveness for use as the core of an electromagnet.
5. Explain why increasing the number of turns will increase the strength of the electromagnet (if all other properties remain the same)
Go
Chapter review
Revise and consolidate your learning with these online tools:
• Success criteria checklist (view online or download)
• Review questions (answer online or download)
• Suggested responses (view online with teacher enabled access)
• Depth study support (download)
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Teachers can access these additional downloadable resources and Learning Management System features:
• Individual and whole class reports
• Assign tasks and track results
• Create (and print) custom practice exams
• Chapter summaries (download)
• Chapter tests (download)
Success criteria checklist
I am now able to…
10A.1 Identify that magnetic field lines are continuous and used to represent the strength and direction of magnetic fields 1, 11, 12
10A.2 Use field line diagrams to analyse the magnetic field around a bar magnet, current-carrying wire and current-carrying loop 3, 4, 6, 14, 16
10B.1 Calculate the strength of the magnetic field produced by a current-carrying wire using B = μ0I 2πr 5, 13, 18
10B.2 Describe the structure of a simple solenoid 2
10B.3 Account for the shape and direction of the magnetic field produced by a direct current passing through a solenoid 6, 15
10B.4 Calculate the strength of the uniform magnetic field inside a solenoid using B = μ0NI L 10, 19
10B.5 Conduct a secondary-source investigation to assess the choice of materials used in the core of electromagnets 8, 17
10B.6 Conduct a practical investigation to examine the factors affecting the strength of an electromagnet 7, 9, 20
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Multiple-choice questions
1. A bar magnet is divided in two pieces. Which one of the following statements is true?
A The bar magnet is demagnetised.
B. The magnetic field of each separated piece becomes stronger.
C. The magnetic poles are separated.
D. Two new bar magnets are created.
2. A solenoid is best described as
A a loop of wire.
B. one straight wire carrying current.
C. two magnets aligned end to end.
D. a coil of wire wound into many loops.
3. A current carrying loop of wire produces a magnetic field coming out of the page inside the loop. What is the direction of the current in the wire?
A. into the page
B. out of the page
C. clockwise
D. anticlockwise
4. Which statement about magnetic field lines is correct?
A they can cross if the field is strong
B. the show both direction and relative strength
C. they only exist around magnets, not currents
D. they begin at the north pole and end in space
5. For a fixed current, doubling the distance r from a long straight wire changes the field strength B by a factor of:
A. 2
B. 0.5
C. 4
D. 0.25
6. The magnetic field inside an ideal solenoid is
A strongest at one end only
B. uniform and parallel to the solenoid axis.
C. zero everywhere.
D. circular around the solenoid axis.
7. An electromagnet’s strength can be measured in a school laboratory by
A. the colour of insulation of the wire.
B. the position of the leads.
C. the number of paper clips it can lift.
D. its effect on human hair
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8. A key characteristic of metals such as soft irons that are often used as an electromagnet core is that they
A. have very high electrical resistance.
B. are easily magnetised and demagnetised.
C. are non-magnetic.
D. stay permanently magnetised.
9. A fair test investigating electromagnet strength should change
A. multiple variables at once to speed up testing.
B. only one variable while keeping others constant.
C. the dependent variable and keep the independent constant.
D. the measuring method each trial.
10. A solenoid has 200 turns of wire, a length of 50 cm and carries a current of 2.0 A. What is the strength of the uniform magnetic field inside the solenoid?
A. 1.0 μT
B. 0.5 mT
C. 1.0 mT
D. 2.0 mT
Exam-style questions
11. Explain how magnetic field line diagrams represent both direction and relative strength of a magnetic field. (3 marks)
12. A compass is placed near the north pole of a bar magnet. Describe and explain the direction that it aligns. (2 marks)
13. A wire produces a magnetic field of 20 μT at a distance of 10 cm. Calculate the current in the wire. (3 marks)
14. Consider the magnetic field around the bar magnet in the diagram. Identify where there is the strongest magnetic field and explain how the diagram supports that answer. (2 marks)
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15. Explain why the magnetic field inside a solenoid is approximately uniform. (3 marks)
16. Compare the magnetic fields around a bar magnet and a solenoid. (4 marks)
17. Describe a secondary source that could be used to determine the relative benefits of different metals to use as the core of an electromagnet. (3 marks)
18. Calculate the magnetic field strength from a source 6.0 cm away from a long straight wire carrying 10 A. (3 marks)
19. The magnetic field inside a 6.0 cm long solenoid has a strength of 600 μT. Determine how many turns of a 1.5 A current-carrying wire would be needed to produce this magnetic field. (3 marks)
20. Outline an experiment that could be conducted in a school laboratory to investigate the effect of the number of turns on the strength of an electromagnet. (6 marks)
