Question
Consider a two particle system with particles having masses $${m_1}$$ and $${m_2}.$$ If the first particle is pushed towards the centre of mass through a distance $$d,$$ by what distance should the second particle is moved, so as to keep the centre of mass at the same position?
A.
$$\frac{{{m_2}}}{{{m_1}}}d$$
B.
$$\frac{{{m_1}}}{{{m_1} + {m_2}}}d$$
C.
$$\frac{{{m_1}}}{{{m_2}}}d$$
D.
$$d$$
Answer :
$$\frac{{{m_1}}}{{{m_2}}}d$$
Solution :
Initially,
$$0 = \frac{{{m_1}\left( { - {x_1}} \right) + {m_2}{x_2}}}{{{m_1} + {m_2}}} \Rightarrow {m_1}{x_1} = {m_2}{x_2}\,\,.....(i)$$
Finally,
The centre of mass is at the origin
.PNG "Rotational Motion mcq solution image")
$$\eqalign{ & \therefore 0 = \frac{{{m_1}\left( {d - {x_1}} \right) + {m_2}\left( {{x_2} - d'} \right)}}{{{m_1} + {m_2}}} \cr & \Rightarrow 0 = {m_1}d - {m_1}{x_1} + {m_2}{x_2} - {m_2}d' \cr & \Rightarrow d' = \frac{{{m_1}}}{{{m_2}}}d\,\,\,\,\left[ {{\text{from (i)}}} \right] \cr} $$
Initially,
$$0 = \frac{{{m_1}\left( { - {x_1}} \right) + {m_2}{x_2}}}{{{m_1} + {m_2}}} \Rightarrow {m_1}{x_1} = {m_2}{x_2}\,\,.....(i)$$
Finally,
The centre of mass is at the origin
$$\eqalign{ & \therefore 0 = \frac{{{m_1}\left( {d - {x_1}} \right) + {m_2}\left( {{x_2} - d'} \right)}}{{{m_1} + {m_2}}} \cr & \Rightarrow 0 = {m_1}d - {m_1}{x_1} + {m_2}{x_2} - {m_2}d' \cr & \Rightarrow d' = \frac{{{m_1}}}{{{m_2}}}d\,\,\,\,\left[ {{\text{from (i)}}} \right] \cr} $$
