Question

A particle of mass $4\text{m}$ 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 $3\text{m}$ 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 $\frac{7R}{6}$

Let the position of particle of mass 4m after the projectile be at the origin if the expolsion did not happened.

Let the larger mass($m_l$) falls at distance $x_l$ and smaller mass($m_s$) falls at distance $x_s$ from the origin.
Now the expression for the centre of mass of the whole system can be written as,
$0=\frac{m_l\left(x_l\right)+m_s\left(x_s\right)}{m_l+m_s}$
$0=\frac{3m(x-R)+m_s(-\frac{R}{2})}{3m+m}$
$x=\frac{7R}{6}$