Two identical blocks with mass 5.0 kg each are connected to the opposite ends of a compressed spring. The blocks initially slide together on a frictionless surface with velocity 2 m/s to the right. The spring is then released by remote control. At some later instant, the left block is moving at 1 m/s to the left, and the other block is moving to the right. What is the speed of the center of mass of the system at that instant

Given a block system connected by a spring.

The blocks are identical and they have equal mass

M1 = M2 = M = 5kg

The initial speed of the block are equal

V1 = V2 = V = 2m/s

What is the centre of mass of the block speed at an instance of 1m/s.

Speed of center of mass cacan be calculated using

V_cm = Σ Mi•Vi / ΣMi

V_cm = (M1•V1 + M2•V2) / (M1 + M2)

Since, there is no external force on the system, then the velocity of the center of mass is constant.

V_cm = (M1•V1 + M2•V2) / (M1 + M2)

V_cm = (5*2 + 5 * 2) / (5 + 5)

V_cm = (10 + 10) / 10

V_cm = 20 / 10

V_cm = 2m/s

So, the velocity of the centre of mass at any instant is 2m/s