Question

A particle of mass $3m$ is projected from the ground at some angle with horizontal. The horizontal range is $R$. 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 falls 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

The correct option is C $\frac{5R}{4}$

Centre of mass hits the ground at the position where original projectile would have landed.


$X_{COM}=\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}$

Here$X_{COM}=R,m_1=2m,m_2=m,x_2=\frac{R}{2}$

$R=\frac{2m{x}_1+m\frac{R}{2}}{2m+m}$
$R=\frac{2x_1+\frac{R}{2}}{3}$
$3R=2x_1+\frac{R}{2}$
$x_1=\frac{5R}{4}=x$