FOLLOW US ON TWITTER
SHARE THIS PAGE ON FACEBOOK, TWITTER, WHATSAPP ... USING THE BUTTONS ON THE LEFT


YOUR PARTICIPATION FOR THE GROWTH OF PHYSICS REFERENCE BLOG

Friday, September 20, 2019

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.

No comments:

Post a Comment

If it's a past exam question, do not include links to the paper. Only the reference.
Comments will only be published after moderation

Currently Viewing: Physics Reference | A ball is thrown vertically down towards the ground and rebounds as illustrated in Fig. 2.1.