Question

A body of mass $m$ is projected with a velocity $u$ making an angle $\theta$ with horizontal. Its horizontal range is $R$. The body splits into two equal parts at the highest point and one of the particle comes to rest. Then, the horizontal distance travelled by the other particle from the point of projection is

• A. $\frac{3R}{2}$
• B. $\frac{R}{2}$
• C. $2R$
• D. $R$

Answer 1

The correct option is A $\frac{3R}{2}$

At highest point horizontal velocity of particle before splitting $=ucos\theta$
using linear momentum conservation

$mu\cos \theta =\frac{m}{2}v+\frac{m}{2}\left(0\right)$

$\Rightarrow v=2u\cos \theta$
Since the horizontal component $=2$ times the initial velocity so range will also become twice since time depends only on vertical component of velocity

$\Rightarrow$ distance travelled $=\frac{R}{2}+R=\frac{3R}{2}$