Question

A shell is fired from a cannon with a velocity 'v' at an angle ′θ′ with the horizontal. At the highest point in its path it explodes into two pieces of equal masses. One of the pieces retraces its path and reaches the cannon. Then the velocity of the other piece immediately after collision is :

Answer 1

hope you find my ans helpful if yes tag as brain list answer

This is so straight forward answer.

just apply momentum conservation

m(v cosx) = m/2 (-ucosx ) + m/2 v1,   [ as it is divided into parts of equal mass n one part retraces to d cannon itself so its velocity will be vcosx but wid a negative.

we hav to find v1

here simply u get v1= 3v cosx

8

Let ‘2m’ be the mass of the cannon which is projected with velocity ‘v’ at an angle ‘?’ with the horizontal. The horizontal component of the velocity of the cannon is, vx = v cos?. This component remains constant. At the maximum height, the cannon has only this component as the vertical component is momentarily zero. So, momentum before exploding = 2mv cos? After exploding, one of the masses retraces the path of the cannon backwards. This is possible on if the part has velocity ‘-v cos?’. Therefore, momentum after explosion = -mv cos? + mu So, by conservation of momentum, 2mv cos? = -mv cos? + mu = > u = (3v cos?) m/s