question_answer

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 be moved, so as to keep the centre of mass at the same position?                                                                                [AIEEE 2006]

A)   \[\frac{{{m}_{2}}}{{{m}_{1}}}d\]     

B)        \[\frac{{{m}_{1}}}{{{m}_{1}}+{{m}_{2}}}d\]

C)   \[\frac{{{m}_{1}}}{{{m}_{2}}}d\]     

D)        \[d\]

Correct Answer: C

Solution :

[c] To keep the centre of mass at the same position, velocity of centre of mass is zero, so

\[\frac{{{m}_{1}}{{{\vec{v}}}_{1}}+{{m}_{2}}\,{{{\vec{v}}}_{2}}}{{{m}_{1}}+{{m}_{2}}}\,=0\]

(where, \[{{\vec{v}}_{1}}\] and \[{{\vec{v}}_{2}}\] are velocities of particles 1 and 2 respectively)

\[\Rightarrow \]

(andrepresent the small change in displacement so that andof particles)

Let 2nd particle has been displaced by distance then

Negative sign shows that both the particles have to move in opposite directions.

So,is the distance moved by 2nd particle to keep position of centre of mass unchanged.