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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
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The correct answer is:
be at rest
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.
$$
\begin{aligned}
& 5(0)=M \times 2 \mathrm{v}-2 M \times \mathrm{v}+2 \mathrm{M} \times \mathrm{v}^{\prime} \\
\Rightarrow \quad & \mathrm{v}^{\prime}=0
\end{aligned}
$$
i.e., the third fragment will be at rest.
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