Problem 0001
A shell of mass m is at rest initially. It explodes into three fragments having masses in the ratio 2 : 2 : 1. The fragments having equal masses fly off along mutually perpendicular direction with speed v. What willl be the speed of the third (lighter) fragment?
Answer:
From conservation of momentum
[tex]m{v_3} + 2\sqrt 2 vm=0[/tex]
[tex]{v_3} = -2\sqrt 2 v[/tex]
Hence, Speed of third fragment is [tex]2\sqrt 2 v[/tex]
