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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 total change in momentum of ball when it stops rebounding is (Here, e is coefficient of restitution between the ball and the floor)

A
m2gh(ee(1+e))
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B
2mgh(e1e)
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C
m2gh(2e1+e)
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D
m2gh(1+e1e)
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Solution

The correct option is D m2gh(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

Change in momentum of ball after 1st collision,
Δp1=mv1(mv0)=mv1+mv0
Change in momentum of ball after 2nd collision,
Δp2=mv2(mv1)=mv2+mv1
Δp3=mv3(mv2)=mv3+mv2
Similarly change in momentum of ball after nth collision,
Δpn=mvn(mvn1)=mvn+mvn1

Therefore, change in momentum of ball:
Δp=Δp1+Δp2+Δp3+......+Δpn ...(i)
=mv0+2mv1+2mv2+.....+2mvn1+mvn
=mv0+2m(ev0)+2m(e2v0)+.....+2m(en1v0)+m(env0)
where number of collisions with floor n
Δp=mv0+2mev0(1+e+e2+e3+...) ...(ii)
Applying formula for sum of infinite GP, where a=1, r=e
=a1r=11e ...(iii)

From Eq.(ii) & (iii):
Δp=mv0[1+2e(11e)]
Δp=mv0(1+e1e)
Δp=m2gh(1+e1e)

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