Question

A mortar fires a shell of mass $M$ which explodes into two pieces of mass $\frac{M}{5}$ and $\frac{4M}{5}$ at the top of the trajectory. The smaller mass falls very close to the mortar. In the same time the bigger piece lands a distance $D$ from the mortar. The shell would have fallen at a distance $R$ from the mortar if there was no explosion. The value of $D$ is
(Neglect air resistance)

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

Answer 1

The correct option is C $\frac{5R}{4}$

As we know explosion occurs due to internal forces only, thus COM will remains at the same place even after the explosion. However, the centre of mass of the two pieces will follow the original trajectory.

Take the coordinate of the point of projection as the origin.

Since the shell would have fallen at a distance $R$ from the point of projection if there were no explosion, hence the co-ordinate of COM will be $R$.

Given: $m_1=\frac{M}{5};m_2=\frac{4M}{5}$

Using the relation for COM,

$x_{com}=\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}[\because x_1=0,x_2=D]$

$R=\frac{(\frac{M}{5}\times 0)+(\frac{4M}{5}\times D)}{(\frac{M}{5}+\frac{4M}{5})}[\because x_{com}=R]$

$\therefore D=\frac{5}{4}R$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, option $\left(C\right)$ is the correct answer.