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A 2 kg block moving with 10m//s strikes ...

A `2 kg` block moving with `10m//s` strikes a spring of constant `pi^(2)N//m` attached to `2Kg` block at rest kept on a smooth floor. The velocity for which rear moving block remain in contact with spring will be-

A

`0 m//s`

B

`5 m//s`

C

`10 m//s`

D

`7.5 m//s`

Text Solution

Verified by Experts

The correct Answer is:
A

Let velocity of rear `2kg` be `v_(1)` and front `2kg` be `v_(2)` then `20 = 2v_(2) - 2v_(1) rArr v_(2) - v_(1) = 10`
Now by conservation of mechanical energy
`1/2(2)(10)^(2) = 1/2(2)v^(1^(2)) + 1/2(2)v_(2^(2))`
`rArr v_(1^(2)) + v_(2^(2)) = 100`
But `v_(2^(2)) + v_(1^(2)) - 2v_(1)v_(2) = 100`
`rArr v_(1)v_(2) = 0 rArr v_(1) = 0` as `v_(2) ne 0`
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