Physics - Center of Mass Question with Solution | TestHub

Initially, the combined system of mass $(M+m)$ is at rest, so its total momentum is zero.

After the explosion, the two pieces of masses $M_1$ and $M_2$ move with velocities $V_1$ and $V_2$ respectively. By conservation of momentum, the total momentum after the explosion must also be zero: $M_1{V}_1+M_2{V}_2=0$. This implies $M_1{V}_1=-M_2{V}_2$. The negative sign indicates that $V_1$ and $V_2$ are in opposite directions.

The impulse exerted by the explosion on $M_1$ is equal to the change in momentum of $M_1$. Since $M_1$ was initially at rest (as part of the combined system), its initial momentum was zero.

Therefore, the magnitude of the impulse on $M_1$ is $|M_1{V}_1-0|=M_1{V}_1$. Due to Newton's third law, the impulse on $M_2$ will be equal in magnitude and opposite in direction, which is $|M_2{V}_2|$. Since $M_1{V}_1=M_2{V}_2$ in magnitude, both A and B represent the magnitude of the impulse. However, usually, the impulse on a specific mass is asked in terms of its own final momentum.