Q.
Two blocks of masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on frictionless horizontal surface. An impulsive force gives a velocity of 14$ms^{- 1}$ to the heavier block in the direction of the lighter block. The velocity of centre of mass of the system at that very moment is
A) $30ms^{- 1}$
B) 20$\ ms^{- 1}$
C) 10$\ ms^{- 1}$
D) 5$\ ms^{- 1}$
Explanation
(c) At the time of applying the impulsive force block of 10 kg pushes the spring forward but 4 kg mass is at rest.
Hence,
$v_{CM} = \frac{m_{1}v_{1} + m_{2}v_{2}}{m_{1} + m_{2}} = \frac{10 \times 14 + 4 \times 0}{10 + 4} = \frac{140}{14} = 10ms^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.