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Q6.3:
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 ?

Solution :

Step 1: Identifying the System and External Forces

Define the system as the combination of the trolley of mass $M$ and the child of mass $m$. The motion occurs on a smooth horizontal floor, which implies that the coefficient of friction $\mu = 0$. Consequently, there are no external horizontal forces acting on the system.

[Per Newton's Second Law of Motion for a system of particles, $\vec{F}_{ext} = M_{total} \cdot \vec{a}_{CM}$]

Step 2: Analysis of External Force and Acceleration

Since the floor is smooth, the external horizontal force $\vec{F}_{ext} = 0$. Therefore:

$\vec{F}_{ext} = M_{total} \cdot \frac{d\vec{v}_{CM}}{dt} = 0$

This implies that the acceleration of the center of mass (CM) of the system is zero ($\vec{a}_{CM} = 0$). [Per the Law of Conservation of Momentum]

Step 3: Evaluating the Velocity of the Center of Mass

Given that $\vec{a}_{CM} = \frac{d\vec{v}_{CM}}{dt} = 0$, the velocity of the center of mass $\vec{v}_{CM}$ must be constant over time. The initial velocity of the system is dictated by the motion of the trolley moving uniformly with speed $V$.

Using the definition of the CM velocity:

$\vec{v}_{CM} = \frac{M\vec{V}_{trolley} + m\vec{V}_{child}}{M + m}$

Step 4: Application to the Specific Scenario

Initially, both the trolley and the child are moving at speed $V$. Thus, $\vec{V}_{trolley} = V\hat{i}$ and $\vec{V}_{child} = V\hat{i}$.

$\vec{v}_{CM} = \frac{M(V) + m(V)}{M + m} = \frac{(M + m)V}{M + m} = V$

When the child runs about on the trolley, these motions represent internal forces within the system. Internal forces cannot change the velocity of the center of mass. Therefore, regardless of the child's movement, the velocity of the CM remains constant.

Child (m) Trolley (M) moving at V V

Step 5: Summary of Conclusion

Because the internal actions of the child (running) do not exert any external force on the trolley-child system, the momentum remains conserved, and the center of mass continues to move with the initial uniform speed $V$ in the same direction.

Final Solution: The speed of the center of mass remains $V$ and it continues to move in the same direction as the trolley.


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