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 ) = (10 xx 14 + 4 xx 0)/(10 + 4) = 10 m//s`
since spring force is internal force, it cannot change the linear momentum of the (two mass + spring) system. Therefore `v_(c )` remains the same.