Question

A projectile of mass $3m$ is projected from the ground with a velocity $30\text{m/s}$ at ${45}^{\circ }$ with the horizontal. At the highest point, it explodes into two pieces. One of pieces has mass $2m$ and the other has mass $m$. Both the pieces fly off after explosion. Mass $2m$ falls at a distance of $100\text{m}$ from the point of projection. Find the distance of second mass from the point of projection, where it strikes the ground. (Take $g=10{\text{m/s}}^2$)

• A. $90\text{m}$
• B. $60\text{m}$
• C. $80\text{m}$
• D. $70\text{m}$

Answer 1

The correct option is D $70\text{m}$

Range of projectile in the absence of explosion :-
$R=\frac{u^2\sin 2\theta }{g}$
$=\frac{(30{)}^2\times \sin {90}^{\circ }}{10}$
$=90\text{m}$


The path of the center of mass will not change. So $x_{COM}$ will be $90\text{m}$.
Say, $m_1=m$ and $m_2=2m$

Let $x_1$ be the distance of $m_1$ from the point of projection, where it strikes the ground.
Mass $m_2$ falls at a distance $x_2=100\text{m}$ from the projection point.
$x_{COM}=\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}$
$\Rightarrow 90=\frac{m\left(x_1\right)+2m\left(100\right)}{m+2m}$
$\Rightarrow x_1=70\text{m}$

So, the mass $m$ will fall at a distance $x_1=70\text{m}$ from the point of projection.