Question

A bomb is projected with initial velocity $9.8\text{m/s}$ making ${15}^{\circ }$ angle with the horizontal. Bomb explodes into two parts in the mass ratio 1 : 2 at the point where vertical component of velocity is zero. If the heavier mass falls vertically down, how far on the ground from projection point will the lighter mass fall?

• A. $4.9\text{m}$
• B. $19.6\text{m}$
• C. $9.8\text{m}$
• D. $2.45\text{m}$

Answer 1

The correct option is C $9.8\text{m}$

Let $R$ be the range of projected bomb and $m$ be the total mass of bomb.
Heavier mass $\left(m_1\right)=\frac{2m}{3}$
and lighter mass $\left(m_2\right)=\frac{m}{3}$


At the maximum height, velocity in the vertical direction will be zero. i.e $v_y=0$


Since, $F_{ext}=0$ in $x$- direction, so horizontal range attainer by COM will not change. According to question, heavier mass will fall vertically down i.e range is $R/2$. Let $x$ be the range of the lighter mass.
So, $\overline{x}=\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}$
$\Rightarrow R=\frac{(\frac{2m}{3}\times \frac{R}{2})+(\frac{m}{3}\times x)}{\frac{2m}{3}+\frac{m}{3}}$
$\Rightarrow R=\frac{R}{3}+\frac{x}{3}$
$\Rightarrow x=2R$
$\Rightarrow x=2\times \frac{u^2\text{sin}2\theta }{g}=\frac{2\times 9.8\times 9.8}{9.8}\times \frac{1}{2}=9.8\text{m}$
$\Rightarrow x=9.8\text{m}$