Question

A child is standing at one end of a long trolley moving with a speed \[v\] on a smooth horizontal track. If the child starts running towards the other end of the trolley with a speed \[u\], the centre of mass of the system (trolley+child) will move with a speed.
A. zero
B. \[\left( {v + u} \right)\]
C. \[v\]
D. \[\left( {v - u} \right)\]

Answer 1

Hint: Use the formula for the speed of the centre of mass of the child and trolley. Determine the centre of mass of the child-trolley system before and after the child starts running.

Formula used:
The velocity \[{v_{CM}}\] of the centre of mass of the two objects is given by
\[{v_{CM}} = \dfrac{{{m_1}{v_1} + {m_2}{v_2}}}{{{m_1} + {m_2}}}\] …… (1)
Here, \[{m_1}\] and \[{m_2}\] are the masses of two objects and \[{v_1}\] and \[{v_2}\] are the velocities of the masses \[{m_1}\] and \[{m_2}\] respectively.
The equation for Newton’s law is
\[{F_{net}} = ma\] …… (2)
Here, \[{F_{net}}\] is the net force on the object, \[m\] is the mass of the object and \[a\] is the acceleration of the object.

Complete step by step answer:
Initially the trolley is moving with the speed \[v\] and the child is at the rest.
Hence, the speed of the child is the same as that of the trolley which is \[v\].
Let the mass of the child is \[m\] and the mass of the trolley is \[M\].
Calculate the speed \[{v_{CM}}\] of the centre of mass of the child-trolley system when the child is at rest.
Substitute \[m\] for \[{m_1}\], \[M\] for \[{m_2}\], \[v\] for \[{v_1}\]and \[v\] for \[{v_2}\] in equation (1).
\[{v_{CM}} = \dfrac{{mv + Mv}}{{m + M}}\]
\[ \Rightarrow {v_{CM}} = \dfrac{{\left( {m + M} \right)v}}{{m + M}}\]
\[ \Rightarrow {v_{CM}} = v\]
Hence, the speed of the centre of mass of the child-trolley system before the child starts running is \[v\].
The speed of the centre of mass of the child-trolley system will change only if there is change in the velocity of the system and there is a net force on the system.
But when the child starts running towards the other end of the trolley, the internal forces between the
child-trolley systems are balanced and hence there is zero net force on the system.
Therefore, according to equation (2), the net acceleration of the system will be zero which implies that the speed of the centre of mass is uniform (does not change).
Therefore, when the child starts running, then the speed of the centre of mass of the child-trolley system is \[v\].

So, the correct answer is “Option C”.

Note:
 The force on the child-trolley system is zero because their acceleration is zero and the acceleration is zero because the change in the speed of the child-trolley system is zero.
The speed of the centre of mass of the child-trolley system will change only if there is change in the velocity of the system and there is a net force on the system.