Question

A shell of mass ‘2m’ fired with a speed ‘u’ at angle ${}^{\prime }{\theta }^{\prime }$ to the horizontal explodes at the highest point of its trajectory 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. $\frac{u^{2}sin2\theta }{g}$
• B. $\frac{3u^{2}sin2\theta }{2g}$
• C. $\frac{2u^{2}sin2\theta }{g}$
• D. $\frac{3u^{2}sin2\theta }{g}$

Answer 1

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

At height point, projectile has u cos $\theta$ speed along horizontal direction.
According to law of conservation of linear momentum,
$2m\left(ucos\theta \right)=m\left(0\right)+m{v}^{\prime }$
$\Rightarrow v^{\prime }=2ucos\theta$
$\therefore$ Horizontal distance travelled by the fragment from highest point = $\left(2ucos\theta \right)=\frac{usin\theta }{g}=u^2\frac{sin2\theta }{g}$
$\therefore$ Distance of the fragment form the gum
$=\frac{u^{2}sin2\theta }{2g}+\frac{u^{2}sin2\theta }{g}$
$=\frac{3}{2}\frac{u^{2}sin2\theta }{g}$