Q 11-04-203NEETNEET 2022Top questionMedium
A shell of mass $m$ is at rest initially. It explodes into three fragments having mass in the ratio $2 : 2 : 1$. If the fragments having equal mass fly off along mutually perpendicular directions with speed $v$, the speed of the third (lighter) fragment is
Answer: (D) $2\sqrt{2}v$
Masses: $\dfrac{2m}{5}$, $\dfrac{2m}{5}$ and $\dfrac{m}{5}$. Total momentum stays zero.
The two equal fragments have perpendicular momenta of size $\dfrac{2m}{5}v$ each; their resultant is $\sqrt{2}\cdot\dfrac{2m}{5}v$.
The light fragment must carry an equal and opposite momentum:
$$\frac{m}{5}v' = \frac{2\sqrt{2}m}{5}v \;\Rightarrow\; v' = 2\sqrt{2}\,v$$
Solution by Sreeraj P, M.Sc Physics