Question

# Two blocks of masses 5 kg and 2 kg are placed at rest on a frictionless surface and connected by a spring. An external hit gives a velocity of 21 m/s to the heavier block towards the lighter one. Velocities of both blocks (heavier and lighter one) in the centre of mass frame, just after the kick will be respectively: (Consider direction of motion of heavier block as +ve direction)

A
6 m/s, 10 m/s
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B
6 m/s,15 m/s
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C
15 m/s, 15 m/s
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D
3 m/s, 5 m/s
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Solution

## The correct option is B 6 m/s,−15 m/s vCM=m1v1+m2v2m1+m2 vCM=(5×21)+05+2=15 m/s From the frame of system's COM, centre of mass of system will remain at rest (no external force) ⇒ Velocity of the heavier block in COM frame, just after the kick is v′1=v1−vCM ∴v′=(21−15)=6 m/s & Velocity of the lighter block in COM frame ,just after the kick is v′2=v2−vCM ∴v′2=(0−15)=−15 m/s

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