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A moving block having mass m , collides...

A moving block having mass m , collides with another stationary block having mass 4m , The lighter block comes to rest after collision . When the intial velocity of the lighter block is v, then the value of coefficient of restitution (e ) will be

A

`0.4`

B

`0.5`

C

`0.8`

D

`0.25`

Text Solution

Verified by Experts

The correct Answer is:
D
From the law conservation of velocity ,
` " P=p " mv =4mv_(2) " " :. V_(2) = (v_(1))/4`
` :. ` Restitutional value ,
`(e)= (v_(2)-v_(1))/(v_(1)-v_(2))=((v_(1))/4-0)/(v_(1)-0)=1/4 =0.25`.
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