Question

A particle of mass $3m$ is projected from the ground at some angle with horizontal . The horizontal range $R$. At the highest point of its path it breaks into two pieces $m$ and $2m$. The smaller mass comes to rest and larger mass finally at a distance $x$ from the point of projection where $x$ is equal to :

• A. $\frac{3R}{4}$
• B. $\frac{3R}{2}$
• C. $\frac{5R}{4}$
• D. $3R$

Answer 1

Answer: (C)

$\frac{5R}{4}$

As breaking of the particle doesn't involve any external force, motion of centre of mass won't be disturbed.
Final position of centre of mass will be at a distance R from the point of projection.
$2m(x-R)=m\ast R/2$
$x=5R/4$