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Q7.3:
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

We define the system as the combination of the trolley and the child ($S = T + C$).

The system moves on a smooth horizontal floor. By definition, a "smooth" floor implies the absence of external friction ($f_{ext} = 0$). Furthermore, the normal force ($N$) exerted by the floor balances the weight ($Mg$) of the system, resulting in a net external vertical force of zero ($\sum \vec{F}_{ext, y} = 0$).

[Per Newton's Second Law for a system of particles: $\vec{F}_{ext} = M_{sys} \vec{a}_{CM}$].

Step 2: Analysis of Internal Forces

When the child stands up and runs on the trolley, all forces involved (the force exerted by the child's feet on the trolley, the frictional force between the child's shoes and the trolley surface, and the child's muscular forces) are internal forces to the system $(T + C)$.

[By Newton's Third Law, internal forces occur in equal and opposite pairs and cancel out: $\sum \vec{F}_{int} = 0$].

Step 3: Applying the Principle of Conservation of Momentum

Since $\vec{F}_{ext} = 0$, the net external force on the system is zero. According to the Law of Conservation of Linear Momentum, the momentum of the system must remain constant.

Initial Momentum ($\vec{P}_i$) = Final Momentum ($\vec{P}_f$)

The velocity of the Center of Mass (CM) is given by:

$\vec{V}_{CM} = \frac{\sum m_i \vec{v}_i}{M_{total}}$

Given the initial state, the child and trolley are moving together at a uniform speed $V$.

$\vec{V}_{CM(initial)} = \frac{(M_{trolley} + m_{child})V}{M_{trolley} + m_{child}} = V$

Step 4: Conclusion on CM Motion

Because there are no external horizontal forces acting on the (trolley + child) system during the child's movement, the acceleration of the Center of Mass is zero ($\vec{a}_{CM} = 0$). Consequently, the velocity of the Center of Mass remains constant regardless of the child's internal motion.

CM Velocity V is constant

The child's running motion constitutes internal redistribution of mass and velocity. While the individual velocities of the child and trolley change relative to the floor, the sum of their momenta remains invariant.

Final Solution: The speed of the CM of the (trolley + child) system remains $V$, as the internal forces do not change the total linear momentum of the system in the absence of external forces.


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