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) |
\[\Rightarrow \]  |
 |
  ( and represent 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. |
You need to login to perform this action.
You will be redirected in
3 sec
