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 :
- v
- 2v
- 2 2v
- 3 2v
Answer
(3) 2 2v
Solution
Masses: 2m5, 2m5 and m5
Momentum of each equal fragment = 2m5v, at right angles to each other.
Resultant = 2 2m5v = 2 2 mv5
By conservation of momentum (initial momentum zero):
m5v' = 2 2 mv5
v' = 2 2v
Answer: (3)