Question

A shell is fired from a cannon with a velocity $v$ at an angle $\theta$ 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 reached the cannon. Then the velocity of the other piece immediately after explosion is :

• A. $3v\cos \theta$
• B. $2v\cos \theta$
• C. $v\cos \theta$
• D. $\frac{3}{2}v\cos \theta$

Answer 1

Answer: (A)

$3v\cos \theta$

Since the above piece retraces its path, its speed must be same as before in the opposite direction.

The speed at topmost point is: $vcos\theta$

Hence, by conserving momentum, since no horizontal force acts at the topmost point,

$mvcos\theta =-\frac{m}{2}vcos\theta +\frac{m}{2}{v}^{\prime }$, where $v^{\prime }$ is the velocity of the other piece

or, $v^{\prime }=3vcos\theta$