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Two blocks of masses m(1)= 2 kg and m(2)...

Two blocks of masses `m_(1)= 2 kg` and `m_(2) = 4 kg` are moving in the same direction with speeds `v_(1) = 6 m//s` and `v_(2) = 3 m//s`, respectively on a frictionless surface as shown in the figure. An ideal spring with spring constant `k = 30000 N//m` is attached to the back side of `m_(2)`. Then the maximum compression of the spring after collision will be

A

`0.06m`

B

`0.04m`

C

`0.02m`

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
C

At the time of maximum compression, the speeds of blocks will be the same. Let the speed be `v` and maximum compression be `x`.
Applying conservation of momentum,
`(m_(1)+m_(2))v=m_(1)v_(1)+m_(2)v_(2)`
`impliesV=4m//s` Applying conservation of mechanical energy,
`1/2kx^(2)+1/2(m_(1)+m_(2))v^(2)=1/2m_(1)v_(2)^(2)+1/2m_(2)v_(2)^(2)`
solving we get `x=0.02m`
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