Q177 / 200 · WBJEE 2013 · Physics
A shell of mass $5 M$, acted upon by no external force and initially at rest, bursts into three fragments of masses $M, 2 M$ and $2 M$ respectively. The first two fragments move in opposite directions with velocities of magnitudes 2v and $v$ respectively. The third fragment will
- Amove with a velocily $v$ in a direction perpendicular To the other two
- Bmove with a velocity $2 v$ in the direction of velocity of the first fragment
- Correct answerCbe at rest
- Dmove with velocity $v$ in the direction of velocity of the second fragment
Solution
Let the velocity of third fragment is $v'$. Then by the conservation of linear momentum
$$
\begin{aligned}
& 5(0)=M \times 2 \mathrm{v}-2 M \times \mathrm{v}+2 \mathrm{M} \times \mathrm{v}^{\prime} \\
\Rightarrow \quad & \mathrm{v}^{\prime}=0
\end{aligned}
$$
i.e., the third fragment will be at rest.
Hence, the correct answer is option: 'C'.