Question

A mortar fires a shell of mass $M$ which explodes into two pieces of mass $\dfrac{M}{5}$ and $\dfrac{{4M}}{5}$ at the top of the trajectory. The smaller mass falls very close to the mortar. At 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. $\dfrac{{3R}}{2}$
B. $\dfrac{{4R}}{3}$
C. $\dfrac{{5R}}{4}$
D. None of these

Answer 1

The shell fragments into two pieces, one of which lands close to the mortar and the other at a distance of $D$. The centre of mass of the two pieces, on the other hand, will follow the original trajectory, and we can reasonably predict that the centre of mass will be at a distance $R$ from the mortar, which is the distance without explosion.

Complete step by step answer:
The CM will follow the same path as in the situation where there is no explosion. Assume the origin is the x-coordinate of the point of explosion. The coordinate of CM will be $R$ because the mortar would have fallen at a distance $R$ from the place of detonation if there had been no explosion.
${m_1} = \dfrac{M}{5}$
$\Rightarrow {m_2} = \dfrac{{4M}}{5}$
$\Rightarrow {X_{CM}} = \dfrac{{{m_1}{x_1} + {m_2}{x_2}}}{{{m_1} + {m_2}}}$
$\Rightarrow {x_1} = 0; \\\ \Rightarrow {x_2} = R; \\\$
Substituting ${x_1}$ and ${x_2}$ ;
${X_{CM}} = R = \dfrac{{\dfrac{M}{5} \times 0 + \dfrac{{4M}}{5} \times D}}{{\dfrac{M}{5} + \dfrac{{4M}}{5}}}$
$\therefore D = \dfrac{5}{4}R$

Therefore the value of $D$ is $\dfrac{{5R}}{4}$.

Note: The student must take care that the center of mass will remain at the same position as it was before the explosion because no external force is applied. In case if there was an external force then the method which we have applied in the given question would have led to the wrong answer.