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Exercises · 6.3

Q.A child sits stationary at one end of a long trolley moving uniformly with a speed VV 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?

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No external horizontal force acts on the (trolley + child) system, so momentum is conserved. The centre of mass continues moving at the original speed VV.

Why the centre of mass keeps moving at VV

When the child runs around on the trolley, you might think the system's speed changes. But here's the key insight: the child and trolley exert forces only on each other. The floor is smooth (frictionless), so there's no external horizontal force. Without an external force, the total momentum of the system cannot change.

The centre of mass velocity is directly tied to total momentum. If momentum is constant, the CM velocity stays constant too—regardless of how the child moves relative to the trolley.


Step-by-step reasoning

  1. Initial state of the system Both the trolley (mass MM, say) and the child (mass mm) move together at speed VV. The total momentum is:

Pinitial=(M+m)VP_{\text{initial}} = (M + m)V

The CM velocity is:

vCM=(M+m)VM+m=Vv_{\text{CM}} = \frac{(M + m)V}{M + m} = V

  1. The child starts running

    When the child runs forward, backward, or jumps, internal forces arise between the child and trolley. By Newton's third law, these forces are equal and opposite. The child pushes the trolley one way; the trolley pushes the child the other way.

  2. No external horizontal force

    The floor is smooth, so friction is absent. Gravity and the normal force act vertically and cancel out. Horizontally, the system is isolated.

  3. Conservation of momentum

    Since no external horizontal force acts, the total momentum remains:

Pfinal=Pinitial=(M+m)VP_{\text{final}} = P_{\text{initial}} = (M + m)V

at every instant, no matter what the child does.

  1. Centre of mass velocity The CM velocity is defined as:

vCM=PtotalM+m=(M+m)VM+m=Vv_{\text{CM}} = \frac{P_{\text{total}}}{M + m} = \frac{(M + m)V}{M + m} = V

This holds throughout the child's motion.

Tip

Internal forces (like the child's footsteps on the trolley) redistribute momentum within the system but cannot change the total. Only external forces can do that.

Watch out

A common mistake is to think that because the child or trolley individually speed up or slow down, the CM must also change speed. Remember: individual speeds can vary wildly, but the CM speed depends only on total momentum, which is conserved.


What actually happens?

If the child runs forward (in the direction of VV), the child speeds up and the trolley slows down. If the child runs backward, the trolley speeds up and the child slows down. But the weighted average—the CM velocity—remains VV throughout.

✓Final answer

The speed of the centre of mass of the (trolley + child) system remains V\boxed{V}.

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