Space, time and motion A A.1 Kinematics
You
must know:
✔ the meaning of the terms distance, displacement, speed, velocity and acceleration
✔ the difference between instantaneous and average values for velocity, speed and acceleration
✔ the qualitative effect of fluid resistance on projectiles, including time of flight, trajectory, velocity, acceleration, range and terminal speed.
You should be able to:
✔ use the kinematic equations of motion (suvat) for situations involving uniform acceleration
✔ represent and interpret motion using distance–time, speed–time, displacement–time, velocity–time and acceleration–time graphs
✔ analyse quantitatively projectile motion in the absence of fluid resistance.
The motion of an object is described by the vector quantities displacement, velocity and acceleration. The scalar counterparts of displacement and velocity are distance and speed.
The length of the curved path in Figure 1 is the distance travelled. The displacement is the vector joining X and Y. Both the length of XY and its direction relative to north are required to specify the displacement completely.
Distance is a scalar quantity and is the length of a path between two points.
Displacement is the vector difference between the initial position and the final position.
Position means the point in space where an object is situated. This can be referred to numerically as the distance or displacement from an origin.
Average speed is the total distance travelled by an object total time taken It is a scalar quantity.
Average velocity is displacement time . It is a vector quantity.
Acceleration is change in velocity time taken for change . It is a vector quantity.
Typical units for these quantities are:
• distance and displacement: metre (m), kilometre (km), megametre (Mm), astronomical unit (AU), light year (ly)
• speed and velocity: m s−1, km s−1, km h−1
• acceleration: m s−2, km s−2
Motion can be represented using distance–time and speed–time graphs. Vector quantities can be shown as displacement–time, velocity–time and acceleration–time graphs. When vector graphs are used, the signs of the axes, gradients and areas indicate the direction, so take care with these quantities. Figure 2 shows how quantities can be derived from the motion graphs.
▴ Figure 1 The difference between distance (red curve) and displacement (straight black line)
Assessment tip
When a question asks for a vector quantity such as velocity or displacement, you must give the direction of the quantity and the magnitude to gain full marks.
Approaches to learning
You may find that linking graphs to equations and theory is a good way to reinforce your learning. Graph lines have gradients and there are areas associated with graphs. Use these ideas to enhance your understanding.
=
Symbols used for motion in the IB Diploma Programme physics course:
s displacement
u initial velocity/speed
v final velocity/speed
a acceleration
t change in time
These are connected by the four kinematic equations of motion for uniform acceleration, often known as the suvat equations:
v = u + at
v2 = u2 + 2as
s = ut + 1 2 at2
s = (v + u)t 2
The suvat equations apply for uniform acceleration only. “Uniform” means constant here. The speed–time graph must be a straight line for the acceleration to be constant.
Figure 2 The meaning of the gradient and area for motion graphs
On page 234, there is more advice for constructing graphs, how to determine the gradient of a graph at a point and how to find the area under the graph.
Kinematic equations
The kinematic equations are a powerful way to solve problems involving motion in both one and two dimensions. They can only be used when the acceleration is constant.
For non-uniform accelerations (e.g. motion down a slope with a changing angle) you may have to equate the energies in a system.
Example 1 shows a common type of motion where an object is dropped from rest (or thrown vertically up or down) close to the Earth’s surface. The object is then said to be in freefall
A ball is released, from rest, from a distance above the ground. The graph plots vertical speed of the ball v against time t
These kinematic equations for uniform acceleration can be used to determine the acceleration of free fall, g, for objects released near the Earth’s surface. The distance fallen from rest and the time taken to fall are required. The quantity g links to the concept of (gravitational) field strength which is a crucial quantity in field theory.
a) Estimate
b) Determine:
a) The distance fallen by the ball is equal to the area under the curve. To calculate this area, count the number of squares under the curve. Then multiply this number by the area of one square. In this example, there are approximately 28 squares under the curve. Each square has an area of 1 s × 5 m s 1, which is equivalent to a distance of 1 × 5 = 5 m. Total distance = 28 × 5 = 140 m.
b) (i) You can find the instantaneous acceleration of the ball when t = 4.0 s by drawing a tangent to the curve at t = 4.0 s. The gradient of this tangent is the acceleration. Make sure you read off values as far apart as possible (at least half the length of the line).
Acceleration = 25 12 6.4 = 2.03 ≈ 2.0 m s 2 (to 2 s.f.)
(ii) The average acceleration is given by change in velocity time taken . The velocity of the ball increases from zero to about 24 m s−1 in a time of 8.0 s.
Average acceleration = 24 8.0 = 3.0 m s 2
The kinematic equations allow physicists to model the future behaviour of a system that involves movement. This model has important limitations. The moving object is taken to be a point in space. Acceleration is taken to be uniform. Nature of
A car changes its speed uniformly from 28 m s−1 to 12 m s−1 in 8.0 s.
a) C alculate the acceleration of the car.
b) Determine the distance travelled by the car during the 8.0 s of acceleration.
Solution
a) Use the suvat equations to calculate the acceleration. Start by writing down what you know, including symbols and units, and what you want to find out.
The equation you need is v = u + at.
Note that the units are consistent, if they were not (e.g. the speed was given in km h−1), you would need to convert them. Take care to substitute correctly into the equation:
12 = 28 + a × 8.0; therefore a = −2.0 m s−2
b) s = ut + 1 2 at2 = 28 × 8 + 1 2 × ( 2.0) × 82 = 224 64 = 160 m
Projectile motion
Samplecontent
u initial speed 28 m s−1 v final speed 12 m s−1 a acceleration ? m s−2 t change in time 8.0 s
The acceleration is negative. With this sign convention, this means the car is slowing down.
When the motion is in two dimensions, then it is known as projectile motion. The solutions still use the kinematic equations. Follow these steps to solve questions about projectile motion.
• Resolve the initial velocity into horizontal and vertical components.
• Recognize that the acceleration due to free-fall acts only on the vertical component. The symbol for the acceleration due to free-fall is g.
• As g is constant, the suvat equations are used in the vertical direction.
• Equations for constant velocity must be used in the horizontal direction. This is because the horizontal component of velocity does not change (assuming negligible air resistance).
• The horizontal and vertical motions can be combined at the end of the problem by adding the speeds vectorially.
In a uniform gravitational field (near the Earth’s surface, for example) the motion of the projectile is parabolic when there is no air resistance.
Acceleration in any uniform field leads to similar behaviour. You should link your understanding here to the movement of a charged particle in a uniform electric field. An electron travelling with an initial velocity that is at an angle to the electric field undergoes uniform acceleration and follows a parabolic path (Topic D.3).
Example 1
Nature of science
Example 2
Approaches to learning
Sometimes it is easier to use energy conservation (Topic A.3) for problems of projectile motion. This is particularly the case for problems that involve the velocity of a projectile rather than its time of flight. In a gravitational field, and in the absence of friction, the gravitational potential energy is transferred to the kinetic energy of the projectile (Topic A.3).
Nature of science
Example 3 involves a modelling process that is not always obvious. The solution involves assuming a value for g that is an average value quoted in the Physics data booklet. The assumption is that this value is constant for the whole motion (a very reasonable assumption). The soccer ball is assumed to be a point in space. Some of these assumptions are backed up by observations of the real-world behaviour of a soccer ball. Some are assumptions that any soccer player can tell you are not true.
Look for sentences such as “Assume that air resistance is negligible.” This makes your job easier. You will not have to carry out calculations involving air resistance for projectile motion, but you may have to answer qualitative (non-mathematical) questions about it. Assessment tip
Example 3
Idris kicks a soccer ball over a wall that is a horizontal distance of 32 m away. Assume that air resistance is negligible.
a)The ball takes 1.6 s to reach a point vertically above the wall. Calculate the horizontal component of the velocity of the ball.
b)The ball is at its maximum height as it passes above the wall. State the vertical component of its velocity at maximum height.
c)Determine the magnitude of the initial velocity of the ball when it was kicked.
d)Deduce the angle to the horizontal at which the ball was kicked.
Solution
a)The ball travels 32 m horizontally in a time of 1.6 s.
The horizontal component of velocity is 32 1.6 = 20 m s 1 and is constant during the motion.
b) At its maximum height, the ball only moves horizontally, and so the vertical component is zero.
c)Begin by finding the initial vertical component of the velocity, so you can combine it with the horizontal component. Use a suvat equation.
Note that, in projectile motion, typically the positive vertical direction is defined as upwards, so the initial vertical component u of the velocity is positive, but the acceleration a is negative.
Use v = u + at. Substituting gives 0 = u + ( 9.8) × 1.6.
Rearranging gives u = 15.68 m s−1
You now have the information needed to calculate the vector velocity.
Magnitude = 202 + 15.682 = 25.4 ≈ 25 m s−1 (to 2 s.f.)
d)Angle to the horizontal = tan−1 15.68
Avoid early rounding of intermediate answers in multi-step calculations. Only round the final value to the appropriate number of significant figures.
A tennis ball is hit with a racket from a point 1.5 m above the floor. The ceiling is 8.0 m above the floor. The initial velocity of the ball is 15 m s−1 at an angle of 50° above the horizontal. Assume that air resistance is negligible.
ceiling
This answer could have achieved 2/3 marks:
Samplecontent
Determine whether the ball will hit the ceiling. [3]
▲ The component of speed is correct.
▼ The suvat equation used is not quoted and the substitution appears directly. Had there been an error here it would have been difficult to allow the error to be carried forward.
▲ The total height reached is calculated correctly.
This answer could have achieved 0/3 marks:
8.25 > 8 so the ball won’t hit the ceiling. A = A cos
Yes, 9.6 m > the 6.5 m required to hit the ceiling. Sample student answer
= 9.6
▼ The total height (8.25 m) is compared with the height of the ceiling (8.0 m) but the wrong conclusion is drawn.
▼ The component of speed is incorrect. This is the horizontal component.
▲ There is a stated comparison.
Effects of air (fluid) resistance
When an object moves through a fluid (gas or liquid), a viscous drag force acts on the object when energy is transferred from the moving object to the fluid. The faster the speed, the greater the retarding force that acts.
In the case of an object accelerating vertically downwards, the retarding force upwards eventually equals the weight downwards and the object reaches a terminal speed. In practice:
•the time to fall a particular distance increases
•the acceleration varies from g to zero (when terminal speed is reached)
•the terminal speed depends on the dimensions (size) of the object.
Figure 3 shows the difference between projectile motion in a vacuum and in air (drag). With air resistance:
•the overall range is shorter and speeds are slower
•the maximum height is lower
•the vertical acceleration is not constant and there is horizontal deceleration
•the trajectory is no longer a parabola.
▼ The answer assumes that the calculated speed is equal to the distance travelled.
▼ There is no addition of the 1.5 m from the position where the ball begins.
0 0 horizontal position drag vacuum vertical position 45°
▴ Figure 3 The effect of drag on an object projected at 45° to the horizontal
In badminton, a shuttlecock is hit from an initial height of 1.5 m. The graph shows how the vertical component v y of the velocity of the shuttlecock varies with time t. The positive vertical direction is upwards. The shuttlecock lands on the ground at t = 1.8 s.
a) Estimate the initial vertical acceleration of the shuttlecock. Comment on your answer.
b) Describe how the magnitude of the vertical acceleration of the shuttlecock varies with t.
c) Estimate the maximum height reached by the shuttlecock.
A steel ball is shot from the same initial height and with the same initial velocity as the shuttlecock. Air resistance on the steel ball can be ignored.
d) Determine the maximum height reached by the steel ball.
e) Draw, on the same axes, a graph to show how v y varies with t for the steel ball.
Solution
a) The tangent to the velocity–time graph drawn at t = 0 crosses the time axis at about t = 0.15 s. The initial vertical acceleration is equal to the gradient of this tangent, 0 16 0.15 0 = 110 m s 2. The magnitude of this acceleration is more than 10g, which means that the initial motion is very strongly affected by air resistance.
b) The magnitude of the vertical acceleration decreases with time, reaching 9.8 m s−2 at the maximum height (when t = 0.6 s) and decreasing further as the shuttlecock falls to the ground. The acceleration just before the impact with the ground (at t = 1.8 s) is still non-zero, which means that the shuttlecock has not yet reached its terminal speed.
c) The vertical displacement of the shuttlecock is equal to the area under the velocity–time graph from t = 0 to t = 0.6 s. The area covers about 7 grid squares, so the distance travelled in the vertical direction is approximately 7 × 0.2 × 2 = 2.8 m. The maximum height is obtained by adding this value to the initial height.
M aximum height = 1.5 + 2.8 = 4.3 m
d) The vertical distance moved by the ball can be found from suvat equations.
The equation to use is v2 = u2 + 2as
s displacement ? m
u initial speed 16 m s−1
v final speed 0
a acceleration 9.8 m s−2
0 = 162 2 × 9.8 s, so s = 13.1 m. The final height will be 1.5 + 13.1 = 14.6 m, which is much greater than for the badminton shuttlecock!
e) The graph is a straight line with a negative gradient of 9.8 m s 2. The graph crosses the time axis at about t = 1.6 s, so the steel ball reaches its maximum height more than a second later than the shuttlecock. The total time of flight to the ground will also be longer for the steel ball.
Practice problems
Problem 1
An object is thrown vertically upwards from the edge of a vertical sea cliff. The initial vertical speed of the object is 16 m s−1 and it is released 95 m above the surface of the sea.
Air resistance is negligible.
a) C alculate the maximum height reached by the object above the sea.
b) Determine the time taken for the object to reach the surface of the sea.
Problem 2
An object is at rest at time t = 0. The object then accelerates for 12.0 s at 1.25 m s−2
Determine, for time t = 12.0 s:
a) the speed of the object
b) the distance travelled by the object from its rest position.
Problem 3
A cart moves up a ramp with an initial velocity of 4.0 m s−1. The graph shows the variation with time t of the velocity v of the cart.
a) C alculate the acceleration of the cart.
b) Determine:
(i) the maximum displacement of the c art from the starting position
(ii) the displacement of the c art at time t = 5.0 s.
Problem 4
A projectile is fired horizontally with an initial velocity of 12 m s−1 from the top of a building that is 8.0 m tall.
a) Ignoring air resistance, calculate:
(i) the time of flight of the projectile
(ii) the horizontal range.
b) The motion of the projectile is affected by air resistance. Explain what effect it has on your answers in part (a).
Example 4
A.2 Forces and momentum
You must know:
✔ real objects can be represented as points in space
✔ forces are interactions between bodies
✔ the nature of field forces (gravitational force, electric force and magnetic force)
✔ what is meant by translational equilibrium
✔ Newton’s three laws of motion
✔ what is meant by weight
✔ what is meant by linear momentum and impulse, and that an applied external impulse is equal to the change in momentum of a system
✔ what is meant by conservation of linear momentum, and how it applies to explosions and elastic and inelastic collisions
✔ what is meant by angular velocity, centripetal acceleration and centripetal force.
You should be able to:
✔ represent a force as a vector
✔ represent the forces acting on a body as a free-body diagram
✔ interpret the forces acting on a system in terms of Newton’s first law of motion
✔ use Newton’s second law of motion to relate the resultant force acting on an object to the object’s acceleration
✔ identify action–reaction pairs of forces using Newton’s third law of motion
✔ describe and solve problems involving contact forces (normal force, surface frictional force, elastic restoring force, viscous drag force and the buoyancy force)
✔ analyse the motion of an object moving around a circular path at a constant speed.
✔ solve problems involving conservation of linear momentum in one dimension
Additional higher level: solve problems involving conservation of linear momentum in two dimensions
Forces and their interaction
This topic is about contact forces.
The field forces tested in the IB Diploma Programme physics course are:
• gravitational force (Topic D.1)
• electric force (Topics D.2, D.3 and B.5)
• magnetic force (Topics D.2, D.3 and D.4).
The resultant force on an object is the (vector) sum of all the forces that act on it. It is important to take into account the directions when determining a resultant force.
ball
pull of Earth on ball
◂ Figure 1 The free-body diagram for a ball falling freely in the Earth’s gravitational field
In a lift (elevator) accelerating upwards, an object experiences a force on it acting upwards from the lift and a gravitational force acting downwards (Figure 2). The sum of the two vector lengths (drawn to scale) has a net upwards force. This is the resultant force acting on the object.
force exerted by li on object
force exerted by Earth on object object
Remember that “accelerating upwards” can mean two things.
• A lift starting from rest or moving upwards is accelerating upwards when it gets faster in the upwards direction.
Forces are said to be pushes or pulls. This implies a contact between one object and another. But forces also arise through field interactions, for example, when a mass interacts with a gravitational field.
A free-body diagram is used to represent the complete set of forces that act on an object. It allows you to analyse the effect of the forces on the object.
• The object is normally drawn as a single point (often representing the position of the centre of mass).
• The forces acting on the object are drawn to scale and labelled clearly.
• These force vectors can be summed (by calculation or by drawing) to give the resultant force that acts on the object.
Assessment tip
You may be asked to draw a free-body diagram to show the relative magnitudes and directions of all the forces acting on a single body. The body is usually represented by a dot. Take care that:
• the vector lengths represent the relative magnitudes
• the directions are correct
• every force is labelled unambiguously
• only forces acting on the object are shown.
In the case of a ball falling freely close to the Earth’s surface, only a gravitational force acts on the object (Figure 1). The point is the centre of the sphere. The resultant force is downwards.
Samplecontent
• A lift that is already moving downwards and accelerating upwards is getting slower and will eventually come to rest.
◂ Figure 2 The free-body diagram for a ball in a lift that is accelerating upwards
Contact forces
Normal force FN
When an object rests on a surface in a gravitational field, a force acts at right angles to the surface. This force, called the normal force FN, has the same magnitude as the apparent weight of the object when the surface is horizontal. The magnitude of the normal force changes when the object accelerates. For example, suppose an object is in a lift:
• when the lift is at rest or moving at a constant velocity, the normal force is mg (the mass of the lift is m)
• when the lift is accelerating upwards with acceleration a, then the normal force will be mg + ma
• when the lift is accelerating downwards, then the normal force is mg – ma
Surface frictional force Ff
The magnitude of the frictional force acting between two surfaces depends on whether or not the surfaces are moving relative to each other.
• When there is no relative motion, the friction between the surfaces is called static friction.
• When there is relative movement between the surfaces, the friction is said to be dynamic.
In both cases, the magnitude of the frictional force F depends on the normal (perpendicular) force FN that acts between the surfaces.
Mathematically, this is written as:
Static friction
Dynamic friction F
where μs is the coefficient of static friction where μd is the coefficient of dynamic friction
In general, the maximum magnitude of static friction is greater than that for dynamic friction, and so the size of the frictional force between two surfaces decreases as the surfaces begin to slide.
Different language is used for fluid friction (the term “fluid” includes both liquids and gases). The terms air resistance and drag are used to describe the friction on a solid as it moves through a fluid.
Example 1
Nature of science
An empirical relationship is one that is determined by experiment, not by proof. There are other examples of empirical equations in the IB Diploma Programme physics course (e.g. the gas laws in Topic B.3). As an equation, FH = kx is not particularly useful as most materials deviate from it at large extensions. The energy stored by a stretched spring is discussed in Topic A.3.
The amount of frictional drag between the object and the fluid in which it moves (called the medium) depends on their relative speed. The greater this relative speed, the greater the drag force. As an object accelerates, frictional drag increases and the magnitude of the drag eventually equals the accelerating force. The two forces oppose each other, so the resultant force is zero and there is no further change in velocity. The speed at which this occurs is called the terminal speed. It depends on the size and shape of the object and the “stickiness” of the medium. In an automobile, when the maximum force of the engine equals the drag force, the vehicle travels at its terminal speed. Objects that are falling freely in a planet’s gravitational field also reach a terminal speed that depends on their size and shape.
Elastic restoring force FH
When solid objects are deformed, the forces between atoms lead to a force that attempts to restore the object to its original size and shape. This is known as the elastic restoring force FH (sometimes the “elastic force” for short).
An empirical equation is used to relate FH to the deformation of the object. In the case of a spring, the extension x of the spring is assumed to be directly proportional to FH:
FH = kx
where k is called the spring constant and has the SI units N m−1. The extension x is the change between the original length and the final length, and FH is the force needed to achieve this change. The equation was suggested by Robert Hooke following experiments in the 1660s.
Viscosity η is the resistance of a fluid to stress, where one plane of the liquid slides relative to another parallel plane. The unit of viscosity is the Pa s or N m−2 s. In fundamental SI units, this is kg m−1 s−1
Stokes’ law: when a sphere of radius r has a speed v, the viscous drag force on the sphere is Fd = 6πηrv Including the effects of buoyancy, when the density of the fluid is ρ and the density of the sphere is σ, the terminal speed
v t = 2r2g(σ ρ) 9η
Nature of science
Viscous drag in liquids and air resistance in gases both involve this empirical equation, which was suggested by the Irish scientist George Stokes.
Buoyancy Fb
When an object is submerged in a fluid and both are in a gravity field, then the object is subject to a buoyancy force Fb, which acts upwards on the object. This buoyancy is related to Archimedes’ principle.
Archimedes’ principle states that, for an object wholly or partly in a fluid, an upward buoyancy force acts on the object that is equal to the weight of fluid displaced by the object. The buoyancy force is sometimes known as the upthrust.
It is easy to show that Fb = ρVg, where ρ is the density of the fluid and V is the volume of the fluid displaced (this is, of course, the same as the volume of the object when the object is fully submerged).
The resultant force on a stationary object of density σ in a fluid of density ρ is ρVg (upwards) σVg (downwards) = (ρ σ)Vg. When:
• ρ > σ, the object floats on the surface
• ρ < σ, the object sinks to the bottom
• ρ = σ, the object has neutral buoyancy and neither rises nor sinks.
Viscous drag force Fd
Real fluids have viscosity that varies according to temperature. Some fluids are more viscous than others (e.g. at the same temperature, compare the flow of honey with water). George Stokes analysed the resistive (drag) force Fd acting on a sphere of radius r due to a viscous fluid while the sphere was falling through the fluid under conditions of streamline flow (flow that is not turbulent). When an object falls in a resistive medium, it will reach a terminal speed.
A sphere is falling at terminal speed through a fluid. The following data are available.
diameter of sphere = 3.0 mm
density of sphere = 2500 kg m−3
density of fluid = 875 kg m−3
terminal speed of sphere = 160 mm s−1
a) Draw a free-body diagram of the forces acting on the sphere.
b) Determine the viscosity of the fluid.
Solution
a) Drag force Fd and buoyancy Fb act upwards, and their combined magnitude is equal to the weight W of the sphere.
drag force,
buoyancy,
Samplecontent
b) Rearranging the Stokes’ law expression gives: η = 2r 2 g (σ ρ) 9 vt = 2 × (1.5 × 10 3 )2 × 9.8 × (2500 875) 9 × 0.16 = 0.050Pas
Newton’s laws of motion
Newton’s three laws of motion describe the interaction between an object and a force. Uniform velocity has both magnitude and direction, so both must be unchanging when the motion is uniform. For acceleration to occur, an object must be subject to a net external force.
When more than one external force acts on a body, you need to sum the vectors these must be added using the rules for vector addition (Topic T.3). When the forces cancel out to give zero resultant force or net force, the body is in translational equilibrium. When the forces do not cancel out to give zero resultant force, the forces are unbalanced.
Newton’s first law of motion (N1) states that an object will remain at rest or continue with uniform velocity unless a net external force acts on it.
Newton’s second law of motion (N2), in its simplest form, relates the resultant force F acting on an object of mass m to the acceleration of the object a by the equation F = ma. Both force and acceleration are vectors, whereas mass is a scalar. The direction of the net force and the direction of the acceleration that it produces are identical.
N2 defines the SI unit of force, the newton. One newton (1 N) is the force that gives a mass of 1 kg an acceleration of 1 m s−2. In fundamental units, this is kg m s−2
Newton’s third law (N3) states that action and reaction are equal and opposite.
Other interpretations of N2 and N3 are considered later in this topic.
The kinematics of motion when viscous drag is involved are discussed in Topic A.1.
Assessment tip
This book uses N1, N2 and N3 to abbreviate Newton’s first, second and third laws of motion. It is best not to use your own abbreviations in examinations unless you state clearly what they mean.
When you apply any law in an answer, it is good practice to state the law as well that way, the examiner will be clear that you understand how the physics arises from the law.
Assessment tip
When a question specifies that a speed or a velocity is constant, then remember the implications of this for Newton’s laws of motion. A constant velocity implies that neither the magnitude nor the direction is changing.
Example 2
A cyclist travels along a horizontal road at a constant velocity.
Discuss the forces acting on the system of the cyclist and the bicycle.
Solution
N1 states that, because the velocity of the system is constant, there is no net external force. The forward force generated by the cyclist pedalling the bicycle is equal in magnitude to the frictional forces acting on the system in the opposite direction.
The forces are also balanced in the vertical direction. The weight of the cyclist and the bicycle is equal in magnitude and has the opposite direction to the normal force from the road on the bicycle.
A car of mass 9.0 × 102 kg travels in a straight line away from traffic lights after they turn green. The graph shows how the speed of the car varies with time.
a) C alculate the acceleration of the car between t = 2.0 s and t = 5.0 s.
b) C alculate the force used to accelerate the car.
Solution
a) The change in speed is from 3.0 to 10.5 m s−1. It takes 3.0 s.
Acceleration = change in speed time taken for change = 7.5 3.0 = 2.5ms 2 = change in time taken for change = 7.5 3.0 = 2.5ms 2 = 2.5 m s−2
b) Force that accelerates the car = mass of car × acceleration
= 900 × 2.5 = 2.25 kN
Since the data used in the calculation were given to 2 significant figures, you should give the answer to 2 significant figures as well (2.3 kN).
Nature of science
Observations can sometimes be explained in more than one way. If you have seen a “Newton’s cradle” you can explain its motion using either Newton’s laws or energy ideas from Topic A.3.
Nature of science
Although the motion laws are credited to Newton, in fact, their development took place over many years by many scientists. Science was not then the shared endeavour that it is today. Nevertheless, Newton acknowledged his debt to earlier workers with his famous quotation where he described himself as standing on the shoulders of giants.
N3 implies that action–reaction pairs exist. These are pairs of forces that are:
• equal in magnitude and opposite in direction
• the same type of force (such as both gravitational, both electrostatic or both elastic restoring forces).
When a ball is released from rest above the Earth’s surface, a gravitational force acts downwards on the ball due to the attraction of the ball by the Earth. At the same time, the ball attracts the Earth with an upwards gravitational force that is equal in magnitude to the weight of the ball. This is an action–reaction pair.
When the ball sits at rest on a table, an additional force pair now acts. This pair is electrostatic in origin. It is:
• the upwards force of the table on the ball
• the equal downwards force of the ball on the table.
These forces arise because the table and the ball are both deformed by the other object. The electrostatic forces arise from the attempt by each to return to its original shape.
A car of mass 750 kg accelerates at 0.30 m s−2 in a straight line along a horizontal road. A resistive force of 550 N acts on the c ar.
a) C alculate the resultant force that acts on the car.
b) C alculate the force that the engine exerts on the car.
Solution
a) The accelerating force is 750 × 0.3 = 225 N. This is the net force acting. The answer is 230 N (to 2 s. f.).
b) The resistive force is 550 N. The engine must overcome this and provide the 225 N. So, the total engine force must be: 550 + 225 = 775 ≈ 780 N.
Note: the full number of significant figures is used until the final round-off and answer.
Momentum and impulse
Momentum is a conserved quantity. The momentum of a system is never lost or gained in a collision unless external forces act on it.
The law of conservation of momentum states that, when no external forces act on a system, the vector sum of the momenta before the collision is equal to the vector sum after the collision:
m1 × u1 +
where m1, m2, … are the masses of the objects, u1, u2, … are the initial velocities and v1, v2, … are the final velocities of the objects.
The equation can also be expressed as
550 + 225 = 775 ! 780N Δ ∑ ( m × v ) = 0, which means that the total change in momentum for a collision including every object that takes part must be zero. If you use this expression, you must use the signs of the velocities correctly.
When a girl throws a ball high into the air, she remains stationary but the force she exerts on the ball causes it to accelerate and gain vertical speed. The ball gains momentum because an external force has acted on it. When the system consists of the Earth plus the girl and ball, the ball still gains upward momentum. However, the girl also increases her downwards force on the ground as she throws. This gives momentum to the Earth, which recoils in the opposite direction to the ball’s velocity. The momentum change of the ball and the momentum change of the Earth are equal and opposite.
Newton’s second law of motion (N2) can be written in terms of change in velocity:
, where Δ means “change in”. This can be rewritten (provided mass is constant) as:
Samplecontent
Momentum = mass × velocity
Velocity is a vector quantity, mass is a scalar quantity, and so momentum is also a vector always specify both its magnitude and direction.
The fundamental unit of momentum is kg m s−1; this is equivalent to the newton second (N s).
Assessment tip
There are two parts to the conservation of momentum law:
• there is no change in momentum during a collision
• providing no external forces act, internal forces do not make any difference as they must act equally (N3) on all parts of the object.
Remember to quote both parts when writing about this law.
changeinmomentum changeintime
changeinmomentum changeintime
This interpretation of N2 shows that the force acting on an object is its rate of change of momentum.
When N2 is written as F = ma, then the mass is assumed to be constant. When N2 is written as F = F
Δ (momentum)
t = changeinmomentum changeintime , then the mass can be changing (e.g. when a rocket burns fuel as it accelerates).
Example 3
Example 4
An apple is released from rest and falls towards the surface of Earth. Discuss how the conservation of momentum applies to the Earth–apple system.
Solution
The points to make are:
• The forces on the Earth and the apple are equal and opposite.
• No external force acts on this isolated system.
• Changes in the momentum of the Earth and the apple are equal and opposite.
• The momentum of the Earth–apple system stays the same and is conserved.
Assessment tip
When carrying out calculations, keep track of the direction in which objects move. Your work needs to be presented so that it is immediately clear what changes during a collision.
One way is to define a direction as positive (perhaps even drawing it on a diagram) velocities in the opposite direction are then negative. When the answer is negative, it indicates that the direction has changed.
In Topic A.3 you meet the concept of kinetic energy Ek. This is linked to momentum through 1 2 mv 2 = 1 2 × 1 m ( mv )2 = p 2 2m , where p is the momentum of a mass m.
A collision in which kinetic energy is removed from the system is known as inelastic. When energy is conserved in a collision, it is said to be elastic. In another type of collision, kinetic energy can appear during the collision. Examples of this include:
• explosions
• the release of a compressed spring (Example 7)
• when a stationary gun is fired.
A small steel ball is released from rest into a fluid. Ignore buoyancy effects in this question.
The graph shows how the speed v of the ball varies with time t
v
a) Draw a free-body diagram for the ball at:
(i) time t1
(ii) time t2
b) When t = t1, the gradient of the graph is a Deduce an expression in terms of the mass of the ball M, acceleration a and acceleration of free fall g for the magnitude of the drag force Fd acting on the ball at t = t1
c) Deduce, using your answer to part (a)(ii), the magnitude of the drag force acting on the ball when t = t2
Solution
a) (i) The ball is accelerating, and so there must be a net downward force on it. The weight force acts downwards and there is a smaller drag force upwards.
(ii) The vector lengths should be equal as the velocity is constant. The net force must be zero (N1). weight forceweight force ii. drag force i. drag force
b) The gradient of the graph is equal to the instantaneous acceleration of the ball at t = t1.
Net force Ftotal = Fweight Fd Fweight = Mg and Ftotal = Ma. So Fd = M(g a).
c) As the velocity is constant, N1 predicts that the upward and downward forces must be equal. The magnitude of the drag force must equal the magnitude of the weight force, Fd = Mg.
Two masses, m and M, on a frictionless horizontal table are connected by a compressed spring and released. Mass m moves with velocity v m
a) State the change in the momentum of mass m
b) Determine, in terms of m, M and v m :
(i) the velocity of mass M
(ii) the change in kinetic energy of the masses.
Solution
(a) The initial momentum of the system (both masses) is zero, and so the change in momentum of m is m × v m
(b) (i) initial momentum = final momentum. So 0 = mv m + MvM, where vM is the velocity of mass M. Therefore, vM = v M = m M v m v m . Mass M moves with a speed v M = m M v m v m, in the direction opposite to the velocity of mass m
(ii) The initial kinetic energy is zero.
Total final kinetic energy Totalfinalkineticenergy = 1 2 mvm 2 + 1 2 Mv M 2 = 1 2 mvm 2 + Mm2 vm 2 M
This is also the change in kinetic energy
Example 8
air-rifle
air-rifle pellet
The speed of the block with the embedded pellet immediately after the collision is 4.8 m s−1.
The mass of the pellet is 2.0 g.
The mass of the block is 56 g.
a) Determine the speed of impact of the pellet.
b) Compare the initial kinetic energy of the pellet with the kinetic energy of the pellet and block immediately after the collision.
Solution
a) Initial momentum of the system (pellet and block) = 0.056 × 0 + 0.002 × u1
The initial velocity of the pellet is u1 and movement to the right is taken to be positive.
Final momentum of the system = (0.056 + 0.002) × 4.8 kg m s−1
Example 5
Example 6
An
pellet is fired into a wooden block resting on a table. wooden block
Example 7