Q. Two blocks of masses 10 kg and 4 kg connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulse gives a velocity of $14\ ms^{- 1}$ to the heavier block in the direction of the lighter block. The velocity of the centre of mass is
A) $30\ ms^{- 1}$
B) $20\ ms^{- 1}$
C) $10\ ms^{- 1}$
D) $5\ ms^{- 1}$
Explanation
(c) Hence, $m_{1} = 10\ kg,\ m_{2} = 4\ kg$
$v_{1} = 14ms^{- 1},\ v_{2} = 0$
$v_{CM} = \frac{m_{1}v_{1} + m_{2}v_{2}}{m_{1} + m_{2}}$
$v_{CM} = \frac{10 \times 14 + 4 \times 0}{10 + 4} = 10ms^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.