Question

A particle of mass $4m$ is projected from the ground at some angle with horizontal. Its horizontal range is $R$. At the highest point of its path it breaks into two pieces of masses $m$ and $3m,$ respectively, such that the smaller mass comes to rest. The larger mass finally falls at a distance $x$ from the point of projection, where $x$ is equal to :

• A. $\frac{2R}{3}$
• B. $\frac{7R}{6}$
• C. $\frac{5R}{4}$
• D. none of these

Answer 1

The correct option is B $\frac{7R}{6}$

Apply conservation of momentum at highest point:
$4mu\cos \theta =3mv\Rightarrow v=\frac{4}{3}u\cos \theta$ (i)
$R=\left(u\cos \theta \right)T$
$R_1=v\frac{T}{2}=\frac{4}{3}\frac{u\cos \theta T}{2}$
$=\frac{2R}{3}\Rightarrow x=\frac{R}{2}+R_1=\frac{R}{2}+\frac{2R}{3}=\frac{7R}{6}$.