Question

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?

Answer 1

This problem can be solved by using the concept that the state of motion of the centre of mass of a system can only be changed by an unbalanced external force acting on it. By determining whether external forces are acting on the system, we can solve this problem.

Formula used:
${{F}_{ext}}=m\dfrac{d{{v}_{cm}}}{dt}$

Complete step by step solution:
It is given that the (trolley + child) has been considered as a single system. Now, according to the question, the child gets up and runs about on the trolley. This can be considered as internal forces within the system. The child (who is a part of the system) is applying forces on the trolley (another part of the system). Hence, the forces applied by the child can be considered to be internal forces within the system.
Now, the state of motion of the centre of mass of a system can only be changed by unbalanced external forces acting on the system.
The instantaneous external force ${{F}_{ext}}$ on the system of mass $m$ is given by
${{F}_{ext}}=m\dfrac{d{{v}_{cm}}}{dt}$ --(1)
Where $\dfrac{d{{v}_{cm}}}{dt}$ is the rate of change of velocity of the centre of mass of the system.
Now, since using (1), we see that if there are only internal forces acting on the system, that is, if ${{F}_{ext}}=0$, we get,
$0=m\dfrac{d{{v}_{cm}}}{dt}$
$\therefore \dfrac{d{{v}_{cm}}}{dt}=0$ $\left( m\ne 0,\text{ since mass of the system is not zero} \right)$
$\therefore d{{v}_{cm}}=0$
$\therefore {{v}_{cm}}=\text{constant}$ $\left( \because dx=0,x=\text{constant} \right)$
Therefore, we see that if the external forces acting on a system are zero, the velocity of the centre of mass remains constant.
Therefore, going by this explanation, we get that since the child running on the trolley amounts to only internal forces within the system, the velocity of the centre of mass will not change.
The speed of the centre of mass will be nothing but the speed of the system, since it is a rigid system. Hence, the speed of the centre of mass of the trolley + child system will remain $V$.

Note: One important fact to be noticed in the question is that the trolley + child system moves upon a smooth floor. This means that the floor is frictionless and there are no forces of friction upon the system. If the question mentioned that the floor was not smooth, then the velocity of the centre of mass of the system would change since the force of friction would be a resistive external unbalanced force on the system.