A shell is fired from a cannon with a velocity v m/s at an angle theta with the horizontal direction.At the highest point of this path , it explodes into two pieces of equal mass. One of the pieces retraces its path to the cannon.The speed in m/s of other piece immediately after the explosion will be ?

Let ‘2m’ be the mass of the cannon which is projected with velocity ‘v’ at an angle ‘θ’ with the horizontal. The horizontal component of the velocity of the cannon is, vx = v cosθ. 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 cosθ

After exploding, one of the masses retraces the path of the cannon backwards. This is possible on if the part has velocity ‘-v cosθ’.

Therefore, momentum after explosion = -mv cosθ + mu

So, by conservation of momentum,

2mv cosθ = -mv cosθ + mu

= > u = (3v cosθ) m/s

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