A shell is fixed from a cannon with a speed of `100 m//s` at an angle `60^(@)` with the horizontal (positive x-direction). At the highest point of its trajectory, the shell explodes into two equal fragments. One of the fragments moves along the negative x-direction with a speed of `50 m//s`. What is the speed of the other fragment at the time of explosion.

Explanation

As we know in absence of external force the motion of centre of mass of a body remains unaffected. Thus, here the centre of mass of the two fragments will continue to follow the original projectile path. The velocity of the shell at the highest point of trajectory is `v_(M) = u cos theta = 100 x× cos 60^(@)` = 50 m/s Let `v_(1)` be the speed of the fragment which moves along the negative x-direction and the other fragment has speed `v_(2)`,. which must be along positive x-direction. Now from momentum conservation, we have
`mv=(-m)/2v_(1)+m/2v_(2)`
or `2v=v_(2)-v_(1)` or `v_(2)=2v+v_(1)`
`=(2xx50)+50=150` m/s