Question

A shell of mass $5M$, acted upon by no external force and initially at rest, bursts into three fragments of masses $M,2M$ and $2M$ respectively. The first two fragments move in opposite directions with velocities of magnitudes $2v$ and $v$ respectively. The third fragment will:

• A. move with a velocity $v$ in a direction perpendicular to the other two
• B. move with a velocity $2v$ in the direction of velocity of the first fragment
• C. be at rest
• D. move with velocity $v$ in the direction of velocity of the second fragment

Answer 1

The correct option is C be at rest

Let the velocity of third fragment is $v^{\prime }$. Then by the conservation of linear momentum:
$5M\times 0=M\times 2v-2M\times v+2M\times v^{\prime }$
or, $v^{\prime }=0$

So, the third fragment will be at rest.