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A small steel ball bounces on a steel pl...

A small steel ball bounces on a steel plate held horizontally. On each bounce the speed of the ball arriving at the plate is reduced by a factor e (coefficient of restitution) in the rebound, so that `v_("upward")=eV_("downward.")` If the ball is initially dropped from a height of 0.4 m above the plate and if 10 seconds later the bouncing ceases, the value of e is

A

`sqrt(2/7)`

B

`3/4`

C

`13/18`

D

`17/18`

Text Solution

Verified by Experts

The correct Answer is:
D
A ball dropped from a height h and attaining height `h_(n)` after n rebounds. Then, `h_(n) = e^(2n) h` and total time taken by the ball is coming to rest is,
`t= ((1+e)/(1-e)) sqrt((2h)/g)`
`therefore t= 10 s, h=0.4 m`, putting these values, we get ,
`therefore 10 = (1+e)/(1-e) sqrt((2 xx 0.4)/(9.8))`
`rArr (1+e)/(1-e) = 35 rArr e= 17/18`
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