Question

A shell of mass 2m fired with a speed u at an angle $\theta$ to the horizontal explodes at the highest point of its motion into two fragments of mass m each. If one fragment, falls vertically, the distance at which the other fragment falls from the gun is given by

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

The total momentum of the system must remain conserved after the explosion. Since, at its highest point, the projectile has only a horizontal velocity, we must have
$\left(2m\right)ucos\theta =m\times 0+m\times u^{{}^{\prime }}$
$\therefore u^{{}^{\prime }}=2ucos\theta$
Thus the horizontal velocity of the other fragment is $2ucos\theta$. The time taken to reach the highest point (as well as the time taken to fall down from this point) being $\frac{usin\theta }{g}$, we have
Total horizontal distance covered by the other fragment
$=ucos\theta \times \frac{usin\theta }{g}+\frac{2ucos\theta \times usin\theta }{g}=\frac{3}{2}\frac{u^{2}sin2\theta }{g}$
Hence the correct choice is (a).