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A ball is dropped from height 5m. The time after which ball stops rebounding if coefficient of restitution between ball and ground e = 1/2, is
- a)1 sec
- b)2 sec
- c)3 sec
- d)infinite
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
A ball is dropped from height 5m. The time after which ball stops rebo...
The speed of ball just before the first rebound = 
Speed of ball just after first rebound = 10 x ½
= 5 m/s
Time taken by the ball before first rebound is 1 sec
Similarly time taken between first and second rebound is 2( 5/10) = ½ sec
Speed of ball after second rebound = 5 x ½
= 2.5 m/s
Hence the time between second and third rebound is ¼ second
Similarly if we find time between more succeeding rebounds we get an infinite GP of common ratio ½ .
Hence the sum is 2 sec.
Speed of ball just after first rebound = 10 x ½
= 5 m/s
Time taken by the ball before first rebound is 1 sec
Similarly time taken between first and second rebound is 2( 5/10) = ½ sec
Speed of ball after second rebound = 5 x ½
= 2.5 m/s
Hence the time between second and third rebound is ¼ second
Similarly if we find time between more succeeding rebounds we get an infinite GP of common ratio ½ .
Hence the sum is 2 sec.
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A ball is dropped from height 5m. The time after which ball stops rebo...
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A ball is dropped from height 5m. The time after which ball stops rebo...
Given: Height of the ball = 5m, Coefficient of restitution (e) = 1/2
To find: Time after which the ball stops rebounding
Formula used: h = ut + (1/2)gt^2, v = u + gt, v^2 = u^2 + 2gh, e = (v2 - v1) / (u1 - u2)
Solution:
1. Initial velocity of the ball (u1) can be found using the formula v^2 = u^2 + 2gh, where v = 0 (at highest point), g = 9.8 m/s^2 and h = 5m.
0 = u1^2 + 2 × 9.8 × 5
u1^2 = 98
u1 = √98
u1 = 9.9 m/s
2. Final velocity of the ball (v1) can be found using the formula v = u + gt, where u = u1, g = 9.8 m/s^2 and t is the time taken to reach the ground.
v1 = u1 + gt
0 = 9.9 + 9.8t
t = -9.9 / 9.8
t = -1.01s (Ignoring negative value)
3. Using the coefficient of restitution (e) = (v2 - v1) / (u1 - u2), where v2 = 0 (at the highest point of rebound), u2 is the velocity after rebounding.
1/2 = (0 - u2) / (u1 - 0)
u2 = u1/2
u2 = 9.9/2
u2 = 4.95 m/s
4. Final velocity of the ball (v2) after rebounding can be found using the formula v = u + gt, where u = u2 and g = 9.8 m/s^2.
0 = 4.95 + 9.8t
t = -4.95 / 9.8
t = -0.5s (Ignoring negative value)
5. Total time taken for the ball to stop rebounding = time taken to reach the ground + time taken to rebound and reach the highest point + time taken to stop rebounding.
Total time = 1.01 + 0.5 + 1.01
Total time = 2.52s
Hence, the correct option is (c) 3 seconds.
To find: Time after which the ball stops rebounding
Formula used: h = ut + (1/2)gt^2, v = u + gt, v^2 = u^2 + 2gh, e = (v2 - v1) / (u1 - u2)
Solution:
1. Initial velocity of the ball (u1) can be found using the formula v^2 = u^2 + 2gh, where v = 0 (at highest point), g = 9.8 m/s^2 and h = 5m.
0 = u1^2 + 2 × 9.8 × 5
u1^2 = 98
u1 = √98
u1 = 9.9 m/s
2. Final velocity of the ball (v1) can be found using the formula v = u + gt, where u = u1, g = 9.8 m/s^2 and t is the time taken to reach the ground.
v1 = u1 + gt
0 = 9.9 + 9.8t
t = -9.9 / 9.8
t = -1.01s (Ignoring negative value)
3. Using the coefficient of restitution (e) = (v2 - v1) / (u1 - u2), where v2 = 0 (at the highest point of rebound), u2 is the velocity after rebounding.
1/2 = (0 - u2) / (u1 - 0)
u2 = u1/2
u2 = 9.9/2
u2 = 4.95 m/s
4. Final velocity of the ball (v2) after rebounding can be found using the formula v = u + gt, where u = u2 and g = 9.8 m/s^2.
0 = 4.95 + 9.8t
t = -4.95 / 9.8
t = -0.5s (Ignoring negative value)
5. Total time taken for the ball to stop rebounding = time taken to reach the ground + time taken to rebound and reach the highest point + time taken to stop rebounding.
Total time = 1.01 + 0.5 + 1.01
Total time = 2.52s
Hence, the correct option is (c) 3 seconds.
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