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A light spring of force constant k is at...

A light spring of force constant k is attached to a block of mass M placed on a horizontal firctionless surface from its one end and other end is fixed to a right support. The system is executing S.H.M. of amplitude `A_(1)` and time period `T_(1)`. At some instant, the block passed through the equilibrium position and a small object of mass m is placed on it. The new amplitude and time period of the system now becomes `A_(2)` and `T_(2)` respectively.
The velcities of block before and after the object is placed on it are `V_(1)` and `V_(2)` respectively. Then the ratio `V_(2)//V_(1)`

A

`((M+m)/(M))`

B

`((M)/(M+m))`

C

`((M-m)/(M)0^(1//2)`

D

`((M)/(M+m))(A_(2)/(A_(1))`

Text Solution

Verified by Experts

The correct Answer is:
B
Applying conservation of momentum, we have
`Mv_(1) = (M + m)v_(2)`
`therefore (v_(2))/(v_(1)) = (M)/(M + m)`
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