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Question

A ball is dropped from a height h on to a floor. If in each collision its speed becomes e times of its striking value, then time taken by ball to stop rebounding is (Here, e is coefficient of restitution between the ball and the floor)

A
2hg(1+e1e)
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B
hg(2e2+e)
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C
2gh(2e2+e)
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D
hg(e1e)
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Solution

The correct option is A 2hg(1+e1e)
When the ball is dropped (u=0) from height h, time taken to reach the floor will be,
t0=2hg
Its speed while striking floor is
v0=2gh
After 1stcollision with floor, its rebound speed will be:
v1=ev0
Total time (t1) between 1st & 2nd collision is equal to time of flight of ball, after 1st collision.
t1=2v1g
Similarly, total time (t2) between 2nd & 3rd collision is equal to time of flight of ball, after 2nd collision.
t2=2v2g, where v2=ev1
(since ball will hit the ground at same velocity with which its starts)
Now, v2=ev1=e(ev0)=e2v0
Similarly, v3=e3v0

Thus, total time after which ball will stop rebounding is given by,
ΔT=t0+t1+t2+t3+.....
ΔT=t0+2v1g+2v2g+2v3g+....
ΔT=t0+2ev0g+2e2v0g+2e3v0g+....
Putting value of v0 in above equation,
ΔT=2hg+2ev0g(1+e+e2+e3+.....)
ΔT=2hg+2e2hg(1+e+e2+e3+.....)
ΔT=2hg[1+2e(1+e+e2+e3+.....)] ....(i)
Applying formula for sum of infinite GP, where a=1, r=e
=a1r=11e ...(ii)
From Eq (i) & (ii):

ΔT=2hg[1+2e(11e)]
ΔT=2hg(1+e1e)

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