A child sits stationary at one end of a long trolley moving uniformly with a speed V on a smooth horizontal floor. If the child gets up and runs about on the trolley in any manner, what is the speed of the CM of the (trolley + child) system?

Certainly! This is a classic problem in physics involving the conservation of momentum. Let me break it down for you.

Initially, when the child is sitting stationary at one end of the trolley, the center of mass (CM) of the system (trolley + child) is simply the center of the trolley, and it moves with a speed V.

When the child gets up and starts running about on the trolley, they exert a force on the trolley in the opposite direction to their motion. According to Newton's third law, the trolley exerts an equal and opposite force on the child.

Now, the total momentum of the system remains constant since there are no external forces acting on it. However, the distribution of mass within the system changes as the child moves.

Let M be the mass of the trolley and m be the mass of the child. Initially, the total momentum is M×VM×V.

When the child starts moving, to maintain the total momentum constant, the velocity of the trolley must decrease and the velocity of the child must increase.

Let VtVt be the final velocity of the trolley and VcVc be the final velocity of the child relative to the trolley.

The final momentum of the system is (M×Vt)+(m×Vc)(M×Vt)+(m×Vc).

Since momentum is conserved, we have:

M×V=(M×Vt)+(m×Vc)M×V=(M×Vt)+(m×Vc)

Now, to find the velocity of the center of mass of the system, we need to consider that the center of mass of the system moves with the same velocity as if all its mass were concentrated at that point.

The total mass of the system is M + m, and the velocity of the center of mass (V_cm) is given by:

Vcm=(M×Vt)+(m×Vc)M+mVcm=M+m(M×Vt)+(m×Vc)

Thus, to find the velocity of the center of mass of the system, we need to solve for VcmVcm using the equation derived above.