Question
A body of mass 2 m is projected with velocity 100 m/s at angle 60° with horizontal. At highest point, it explodes into two fragments of equal masses. If after explosion one fragment retraces its path, then find the velocity of second fragment.
Solution
Given,
Initial velocity, u = 100 m/s
angle of projection of the particle, 
Therefore the velocity of body at highest point is,
At the highest point one of the fragment retraces its path, therefore after explosion the direction of the velocity of fragment reverses without change in the magnitude of velocity.
Therefore the velocity of fragment that retraces its path after explosion becomes 50 m/s.
Let V be the velocity of second fragment.
Now according to the law of conservation of linear momentum,
V = 150 m/s, is the velocity of the second fragment.
