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KINEMATICS

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♦ Kinematics is the branch of mechanics dealing with the motion of the bodies without reference to either mass or force causing it. ♦ A body is said to be at rest, When it is not changing its position with time, with respect to the surrounding. ♦ A body is said to be in motion. If it is changing its position with time, with respect to the surroundings. ♦ Uniform Velocity: If a body travels equal distances in equal intervals of time, however small these intervals may be along a straight line in a particular direction; then it is said to be moving with uniform velocity . ♦ Acceleration: Acceleration in the rate of change of velocity of a particle. Acceleration = Change velocity/Time.

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Uniform Acceleration: If equal changes in velocity take place in equal intervals of time however small these intervals of time may be then the body is said to be moving with uniform acceleration.

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♦ Equation of motion of uniformly accelerated body along a straight line. Let 'u' be its initial velocity i.e., the velocity at the start of the study of motion at t = 0 Let 'v' be its final velocity i.e., the velocity after time 't' has passed. 'a' its uniform acceleration and 's' is the distance travelled by it in time 't' ♦ Particle: A Particle is a small portion matter without extent. Mathematically, it is denoted by a point. ♦ Scalar: Physical Quantity which has only magnitude and direction is known as vector quantity. ex: Displacement, Velocity, Acceleration, Force, Weight etc. ♦ Distance and Displacement: The total length of the path of a particle. During its motions is called the distance covered by it. Suppose a particle has changed its position suppose a particle has changed from A (initial Position) to the position B (Final Position) form following diagram.

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♦ This may take place along any of the path shown. The distance travelled by the particle along each path in different. Distance in measured only in term of magnitude (length) and hence it is a scalar quantity. Here the references to direction is no necessary.

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Displacement: In the above figure, the straight path from A to B is called displacement. It is the shortest distance between the initial and final positions of the particle. Displacement is the change of position of a particle, measured along a straight line from the initial to final position. The change of position would have occurred by travelling along any of the paths. But the magnitude of the displacement is always the length of the straight line joining initial and final positions along which the particle may or may not have travelled and thus displacement is independent of the path or distance travelled. The Magnitude of displacement is length along a straight line. So the units of distance and displacement are the same. Speed and Velocity: Speed of a moving particle is the distance travelled in unit time. It gives us no idea of the direction in which the particle is moving. Hence speed is a scalar quantity. If the speed of a body is considered along a straight line in a particular direction, then it is called velocity. Instantaneous Speed and Instantaneous Velocity

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Instantaneous speed: If a particle travel a distance ∆s in a time ∆t, Where ∆s and ∆t are infinitesimally small the ratio ∆S/∆t in the average speed during the interval ∆t. If the ratio approaches a limit v as ∆t tends to zero. (represented as ∆t → 0) then this limit is the instantaneous speed of the particle v = Lt ∆s v = LT ∆t →0 ∆t Instantaneous Velocity: If a particle from A to B in the following figure, along the path ACB during a short intervals of time ∆t, the distance ∆s travelled by the particle in ∆r. As ∆t becomes smaller and smaller, the point B approaches coincides with the arc ACB. In the limit as ∆t → 0 A

. ∆r C

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∆s

A ∆B www.sakshieducation.com

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∆r The magnitude of ∆t approaches the instantaneous velocity v of the particle and the direction of the displacement coincides with the tangent to the path at A. ∆r The limit of ∆t as ∆t → 0 is called the instantaneous velocity v of the particle at A ∆r dr = ∆t →0 ∆t dt

v = LT

If the instantaneous velocity at every point is constant during the motion of a particle, the particle is serial to be moving with uniform velocity.

i) Motion of a stone thrown by a boy ii) A bullet fired from a pistol iii) A shot put thrown by an athlete

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ex:

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Projectiles: A Projectile is an object which is given an initial velocity and is then allowed to move under the action of gravity.

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A projectile is an object in flight through the atmosphere or space, which is no longer being propelled by any fuel. Thus a rocket is not a projectile central its propellant fuel burnt out. The science of the motion.

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An object can be projected either horizontally or obliquely (making an angle with the Horizontal)

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Horizontal Projection: When a rifle is kept horizontally and fired, the bullet leaving the rifle will have two velocities which are independent of each other. i. The Uniform horizontal Velocity ii. Vertical Velocity which is non-uniform due to acceleration due to gravity. The horizontal velocity will be uniform when the air resistance is neglected. The path taken by a projectile (called trajectory) is a parabola. This symmetric about y-axis. Oblique Projection: If a body is projected with a velocity u at an angle s with the horizontal as in the following diagram, the velocity can be resolved into two components i) A Horizontal component ux = ucosα ii) A vertical component uy = usinα

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vx=ux

v

vy

1∝

vx=ux

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vx=ux

uy vy

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The horizontal component ux is not affected by the acceleration due to gravity and remains constant, if the air resistance in neglected. ♦ Time taken to attain maximum height by the projectile t1 = u2 sin2 α ♦ Maximum height attained (h) h = 2g

♦ Time to flight (t) t =

2usin α g

♦ Horizontal Range: RANGE =

u2 sin 2α g

♦ Maximum horizontal range Maximumrange =

u2 g

♦ First Equation: Acceleration = Change of Velocity / time v −u t or at = v – u or v = u + at a=

♦ Second equation: S = ut + 1/2 at2 ♦ Third equation : v2 – u2 = 2as

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usin α g

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Distance Travelled in the nth second (Dn) = u + a(n – 1/2) ♦ The body moving under gravity will be subjected to uniform acceleration due to gravity 'g'. Thus 'a' in above equation is replaced by 'g'. The bodies are moving vertical distnace(s) is replaced by 'h'. Therefore the equations of motion a body moving under gravity are given by Bodies moving towards earth a) V = u + gt b. h = ut + 1/2 gt2 c. v2 – u2 = 2gh

Bodies moving away a. v = u – gt b. h = ut – 1/2 gt2 c. v2 – u2 = – 2gh

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♦ If bodies are at rest then initial velocity (u) = 0 ♦ Equation of motion for a freely falling body the initial velocity 'u' = 0 v = gt .....................(1) h = 1/2 gt2..............(2) v2 = 2gh .................(3)

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♦ Equation of motion when body goes upwards its final velocity (v) = 0 ♦ Maximum height reached by the body, (Final Velocity) v = 0

∴h =

u2 2g

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u2 =h 2g

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using, v2 – u2 = – 2gh (O)2 – u2 = – 2gh or + u2 = + 2gh

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Time of Ascent: The time taken by a body thrown up to reach maximum height 'h' is called its time of Ascent. t1 = u/g Time of Descent: The time taken for a freely falling body to touch the ground is called 'time of descent'' t2 = u/g Time of flight (t): Time of flight in the time for which a body remains in the air and is given by the sum of time of ascent and time of descent. t = t1 + t2 =

u u + g g

t=

2u g

♦ For a freely falling body initial velocity (u) is zero. ♦ For a freely falling body, the final velocity, (v) using www.sakshieducation.com

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v2

u2

– = 2gh v2 – (0)2 = 2gh or v2 = 2gh or

v = 2gh

Very short Answer Questions 1. Write the equations of motion for a freely falling body. - i) V = gt ii) h = 1/2 gt2 iii) v2 = 2gh

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2. Write the equation of motion for a body thrown upwards. - i) u = – gt ii) h = ut –1/2 gt2 iii) u2 = 2gh

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3. For a freely falling body 'g' is taken as positive. Why? - In case of freely falling body, the direction of motion of body and the direction of acceleration due to gravity (g) are same hence 'g' is taken as positive.

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4. For a body thrown up 'g' is taken as negative. why? - For a body thrown up, the direction of motion of body and direction of acceleration due to gravity (g) are opposite. Hence 'g' is taken as negative.

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5. Define the time of ascent. - The time taken by the body thrown up to reach its maximum height is known as time of ascent.

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6. What is meant by time of descent? - The time taken by the freely falling body to touch the ground is called time of descent. 7. Define time of flight? - The time for which the body remains in air is called time of flight. It is equal to the sum of time of ascent and time of descent. i.e. t = 2u/g Short Answers Questions 1. Define acceleration - Acceleration is the rate of change of velocity i.e. Acceleration = Change of Velocity/ Time 2. When the body is said to be in rest? - A body said to be at rest, when it is not changing its position with time with respect to the surroundings. www.sakshieducation.com

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1. a. b. c. d. e.

Essay Questions Obtain a formula to find the maximum height reached by a body when it is projected vertically upwards with a velocity 'u' A body is projected vertically upwards with an initial velocity 'u' as the body moving upwards a = – g as the body thrown up, final velocity (v) = 0 Considering v2 – u2 = –2gh Substituting the values, (O)2 – u2 = – 2gh (or) – u2 = – 2gh (or) u2 = 2gh (or) h = u2/2g

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Hence maximum height attained by the body is directly proportional to the square of the initial velocity.

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2. Show that the time of ascent in equal to the time of descent. a. The time taken by a body thrown up to reach maximum height 'h' is called its time of ascent (t1) b. At maximum height, its velocity (v) = 0 c. Using V = u – gt, d. Substituting, v = 0 we get u = gt (or) t = u/g i.e. t1 = u/g..........................................(1)

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e. The time taken by freely falling body to touch the ground is called the time descent (t2) f. At maximum height 'h', initial velocity (u) = 0 g. Considering h = ut + 1 / 2gt 22

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(or)

t 22 =

2h ⇒ t2 = g

2h g

................................(2)

u2 h. But we know that h = 2g

substituting this in (2) ∴ t2 =

2 u2 × g 2g

(or) u2 u h= 2 = g g

..........................................(3)

i. From (1) and (3) we observe that time of ascent is equal to time of descent. www.sakshieducation.com

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3. Derive a formula for the velocity on reaching the ground when a body is dropped from height h a. For a freely falling body, its initial velocity u = 0 b. Let the time taken to reach the ground be 't' and final velocity be v. c. Using equation of motion v2 – u2 = 2gh d. Substituting the values v2 – 0 = 2gh (or) v2 = 2gh (or)

v = 2gh

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ii) Time of Ascent (t1) = u/g = 10/10 = 1 sec

u2 102 = 5m (h) = = 2g 2 × 10

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maximum height reached

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Problems with Solutions 1. A stone is thrown vertically up with an initial velocity of 10 ms–1. Find maximum height reached and the time of ascent. (Take g = 10ms–2) - As the stone thrown vertically final velocity v = 0 v=0 u = 10ms–1 i) g = 10ms–2

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2. A ball is thrown up and attains maximum height of 80 m. Find its initial speed? (Take g = 10ms–1) - Maximum height reached = 80m g = 10ms–1 consider formula for maximum height reached h = u2/2g 80 = u2/2 × 10 (or) u2 = 80 × 2 × 10 u = 1600

u = 40 ms–1 3. A stone is dropped from the top of a building and is found to reach ground in 1 second. Find the height of the building. (Take g = 10ms –2) - For a freely falling body, initial velocity (u) = 0 Time of descent (t) = 1 sec using h = ut + 1/2 gt2 h = 0(1) + 1/2 (10) (1)2 (or) h = 1/2 × 10 h = 5m Hence height of building = 5m www.sakshieducation.com

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I. Choose the correct Answers 1. When a body thrown up, its final velocity is b) – 9.8 m/s2 c) 9.8 m/s a) 9.8 m/s2

d) 0

2. Maximum height reached a) t =

2h g

u2 b) h = 2g

c) h =

u2 b) 2g

c)

2u d) h = g

u g

3. Time of flight 2g u

d)

u g

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2u a) g

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4. When a body is dropped from a height its final velocity is b) v = gh

5. Unit of acceleration is a) ms–1 b) ms–2

c) v= 2gh

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a) v = 2gh

c) ms2

d) v= gh

d) ms1

5. 6. 7.

III. 1. 1. 2. 3. 4.

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Fill in the blanks When a body moves towards the earth acceleration due to gravity is taken as ____ When a body is projected upwards the acceleration due to gravity is taken as ______ When a body is dropped from a height 'h', the velocity of the body on reaching the earth is ______ Maximum height reached by a body when it is projected upwards with a velocity 'u' is _____ Time of ascent is directly proportional to ______ For a body moving under the influence of gravity, time of ascent is equal to ______ A body projected vertically upwards with a initial velocity of 40 m/s. The maximum height reached by the body is _____ (g = 10ms –2)

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II. 1. 2. 3.

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6. A body is projected vertically upwards with a velocity 20 m/s. The maximum height reached by the body is (Take g = 10) a) 20m b) 200 m c) 40 m d) 10 m

Match the following Group A Time of Ascent () Time of descent () Time of flight () Maximum height ()

5. Final velocity

()

Group B A. h = u2/2g B. 2u/g C. u/g D. gt E. v = 2gh F. 2h / g www.sakshieducation.com

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Answers I. 1. 2. 3. 4. 5. 6.

d b a a b a

II. 1. Positive 2. Negative

u2 2g

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4. h =

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3. v = 2gh

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C F B A E

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III. 1. 2. 3. 4. 5.

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5. initial Velocity 6. Time of descent 7. 800 m/s

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