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 gives a velocity of 14.m/s to the heavier-block in-the direction of the lighter block. The velocity of the centre of mass is

Options

A. 30 m/s
B. 20 m/s
C. 10 m/s
D. 5 m/s

Answer

C

Explanation

Just after collision
`v_c = (m_1 v_1 + m_2 v_2)/(m_1 + m_2) = (10 xx 14 + 4 xx 0)/(10 + 4) = 10 m//s`
Spring force is an internal force, it cannot cannot change the linear momentum of the (two mass+spring) system. Therefore `v_c` remains the same.