A shell of mass $M$ is at rest initially. It explodes into three fragments having mass in the ratio

Question

A shell of mass $M$ is at rest initially. It explodes into three fragments having mass in the ratio $1 : 1 : 2$. If the fragments having equal mass fly off along mutually perpendicular directions with speed $V$,  the speed of the third (heavier) fragment is

Professor PC Thomas Classes · Answer

Masses are $M/4, M/4, M/2$. The two lighter fragments move at speed $V$ perpendicularly. Their momenta are $\frac{MV}{4}$ and $\frac{MV}{4}$. Resultant momentum is $\sqrt{(\frac{MV}{4})^2 + (\frac{MV}{4})^2} = \frac{\sqrt{2}MV}{4}$. The heavier fragment ($M/2$) must have equal and opposite momentum: $\frac{M}{2} V_3 = \frac{\sqrt{2}MV}{4} \implies V_3 = \frac{\sqrt{2}V}{2} = \frac{V}{\sqrt{2}}$.