Q. 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:
A) $3\sqrt{2}\,v$
B) v
C) $\sqrt{2}\,v$
D) $2\sqrt{2}\,v$
Explanation
[Solution] Masses are 2m/5, 2m/5, m/5. The resultant momentum of the two equal fragments is $\sqrt{2}(2m/5)v$, which must be balanced by $(m/5)v'$, so $v' = 2\sqrt{2}\,v$. Answer (4).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.