A ball is thrown vertically down towards the ground and rebounds as illustrated in Fig. 2.1.

Question 29

(a) A ball is thrown vertically down towards the ground and rebounds as illustrated in

Fig. 2.1.

Fig. 2.1

As the ball passes A, it has a speed of 8.4 m s^-1. The height of A is 5.0 m above the

ground. The ball hits the ground and rebounds to B. Assume that air resistance is

negligible.

(i) Calculate the speed of the ball as it hits the ground. [2]

(ii) Show that the time taken for the ball to reach the ground is 0.47 s. [1]

(b) The ball rebounds vertically with a speed of 4.2 m s^-1 as it leaves the ground. The time the ball is in contact with the ground is 20 ms. The ball rebounds to a maximum height h.

The ball passes A at time t = 0. On Fig. 2.2, plot a graph to show the variation with time

t of the velocity v of the ball. Continue the graph until the ball has rebounded from the

ground and reaches B.

Fig. 2.2

[3]

(c) The ball has a mass of 0.050 kg. It moves from A and reaches B after rebounding.

(i) For this motion, calculate the change in

1. kinetic energy, [2]

2. gravitational potential energy. [3]

(ii) State and explain the total change in energy of the ball for this motion. [2]

Reference: Past Exam Paper – June 2012 Paper 22 Q2

Solution:

(a)

(i)

v² = u² +2as = 8.4² + (2×9.81×5) (= 168.66) 

v = 12.99 m s^-1 (allow 13 to 2sf but not 12.9)

(ii)

s = ut + ½ at²                           or t = (v-u) / a

{From the diagram the ball is thrown from a height of 5 m.}

5 = 8.4t + (½×9.81×t²)            or t = (12.99 – 8.4) / 9.81

t = 0.468 s                  

(b)

{We need to draw a graph to show the variation with time t of the velocity v of ball

At time t = 0 s, the ball is thrown with velocity 8.4 m s^-1 downwards. This corresponds to point (0, 8.4). The downward direction is taken as positive.

Due to gravity, the velocity of the ball increases until it becomes 12.99 m s^-1 when it touches the ground after 0.47 s. These were calculated in part (a) above. This corresponds to point (0.47, 12.99).

The gradient between these 2 points [(0,8.4) and (0.47,12.99)] represents the acceleration due to gravity. This is constant {graph is a straight line} and equal to 9.81 m s^-2.

The ball rebounds with a speed of 4.2 m s^-1. But since the ball is now moving upwards, v = - 4.2 m s^-1. The contact time of the ball with the ground is 20 ms = 0.02s [non-zero], the line representing the change from the maximum +ve speed to this –ve one is not vertical. The line is inclined.

As for the final stage, the velocity of the ball decreases to zero at the maximum height B due to the force of gravity opposing the motion.

a = v – u / t

Time taken for velocity to decrease to zero: t = (0 - -4.2) / (9.81) = 0.428 s.

 So, velocity is zero at t = 0.428 + 0.47 = 0.898 s.

The gradient is the same as for the first section (= 9.81 m s^-2) since the acceleration due to gravity is constant.}

(c) The ball has a mass of 0.050 kg. It moves from A and reaches B after rebounding.

(i)

1.

KE at B is zero, so ΔKE = ½ mv²   or ΔKE = ½ mu² – ½ mv²

ΔKE = ½ × 0.05 × 8.4² = (-)1.8 J

2.

{To find the maximum height h at B:

v² = u² +2as

{Here, the upward motion is taken as positive. So, acceleration due to gravity, which is downwards, is negative and opposes the motion. A negative sign is included in the equation.}

0 = (4.2)² + (2×-9.81×h)

Final maximum height, h = (4.2)² / (2×9.81) = (0.9m)

Change in PE = mgh – mgh₁ 

Change in PE = 0.05 × 9.81 × (0.9 – 5) = (-)2.0 J

(ii) The total change in energy is -3.8J. Energy is lost to ground (on impact) / energy of deformation of the ball / thermal energy in ball.