Q#29 (Past Exam Paper – June 2012 Paper 22 Q2)
(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]
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.
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