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Question

A body falls from a height 'h' on a horizontal surface and rebounds. Then it falls again and rebounds and so on. If the coefficient of restitution is 12, then the total distance travelled by the ball before coming to rest is (neglect air resistance)

A
2h3
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B
3h2
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C
5h2
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D
5h3
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Solution

The correct option is D 5h3
When a body falls from height h it hits the ground with the velocity v=2gh

h=v22g

Now the ball undergoes collision, loses some energy and rebounds with a velocity ev where e is the coefficient of restitution. It will now rise to a height h1 and come back by covering the same distance h1

h1=(ev)22g

Now the ball will again strike the ground and undergo collision and will rebound with a velocity e2v. If it rises to a height h2

h2=(e2v)22g

Thus, the total distance is given as

d=h+2h1+2h2+ . . .

d=v22g+2e2v22g+2e4v22g+ . . .

d=v22g+2e2v22g(1+e2+e4+ . . .)

Here (1+e2+e4+ . . .) forms an infinite GP with first term a=1 and common rartio r=e2

So, sum S=a1r=11e2

d=v22g+2e2v22g11e2

d=v22g[1+2e21e2]

d=v22g[1e2+2e21e2]

d=h[1+e21e2]

d=h⎢ ⎢ ⎢ ⎢ ⎢1+(12)21(12)2⎥ ⎥ ⎥ ⎥ ⎥ (e=12)

d=h[5/43/4]

d=5h3

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