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Q.

An elastic spring is compressed between two blocks of masses 1 kg and 2 kg resting on a smooth horizontal table as shown. If the spring has 12J of energy and suddenly released, the velocity with which the larger block of 2 kg moves will be 

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a

2 m/s

b

4 m/s

c

1 m/s

d

8 m/s

answer is A.

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Detailed Solution

12Kx2=KEA+KEB and mAvA=mBvB

⇒vA=2vB...(i)

12 mAvA2+12 mBVB2=12

12 mA×4 VB2+12 mBVB2=12

on substituting the values of masses

we get vB=2 m/s

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