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. $\frac{3}{2}\frac{u^{2}sin2\theta }{g}$
• B. $\frac{2u^{2}sin2\theta }{g}$
• C. $\frac{u^{2}sin2\theta }{g}$
• D. $\frac{3u^{2}sin2\theta }{g}$

Answer 1

The correct option is A $\frac{3}{2}\frac{u^{2}sin2\theta }{g}$

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).