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Question

Two blocks A and B of mass m and 2m respectively are connected by a massless spring of force constant k. They are placed on a smooth horizontal plane. They are stretched by an amount x and then released. The relative velocity of the blocks when the spring comes to its natural length is

A
x3k2m
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B
x2k3m
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C
x2km
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D
x3km
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Solution

The correct option is A x3k2m
Let speed of block m be v1, speed of block 2m be v2,
Here v1,v2 are in opposite directions.

From conservation of momentum,
mv1=2mv2
v1=2v2
From conservation of energy,
12kx2=12mv21+12(2m)v22
12kx2=12m(2v2)2+12(2m)v22
12kx2=(3m)v22
v2=xk6m
v1=2xk6m,
Relative velocity
v1+v2=x3k2m
(a)

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