First slide

Motion of centre of mass

Question

Two blocks of masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulse give a velocity of 14 m/s to the heavier block in the direction of the lighter block. The velocity of centre of mass is:

Easy

A

30 m/s
B

20 m/s
C

10 m/s
D

5 m/s

Solution

${Mv}_{cm} = m_{1} v_{1} + m_{2} v_{2}$
$v_{cm} = \frac{4 \times 0 + 10 \times 14}{14} \Rightarrow v_{cm} = 10 m / s$

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