A shell is fired from a cannon with a velocity `v (m//sec.)` at an angle `theta` with the horizontal direction. At the highest point in its path it explodes into two pieces of equal mass. One of the pieces retraces its path to the cannon and the speed (in `m//sec.`) of the other piece immediately after the explosion is

Options

A. `3v cos theta`
B. `2v cos theta`
C. `(3)/(2) cos theta`
D. `(sqrt(3))/(2)v costheta `

Answer

A

Explanation

Shell is fired with velocity v at an angle `theta` with the horizontal.
So its velocity at the highest point
= horizontal component of velocity `= v cos theta`
So momentum of shell before explosion `= m v cos theta`

When it breaks into two equal pieces and one piece retrace its path to the canon, then other part move with velocity V.

So momentum of two pieces after explosion
`= (m)/(2)(-v cos theta) + (m)/(2)V`
By the law of conservation of momentum
`m v cos theta = (-m)/(2) v cos theta + (m)/(2)V rArr V = 3 v cos theta`