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:

Question

Aicharya · Accepted Answer

The speed of the third particle will be along angle bisector of the other two velocities. Applying the conservation of momentum, m v ' = 2 m v cos 45 ° + 2 m v cos 45 ° ⇒ v ' = 2 2 v

A shell of mass m is at rest initially. It explodes into three fragments having mass in the ratio 2 : 2 : 1 …

Solution