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 the vertical component of velocity is zero. If the heavier mass falls vertically down, how far on the ground from the 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 the 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 of the bomb in the vertical direction will be zero. i.e $v_y=0$


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