Three blocks A ,   B and C are lying on a smooth horizontal surface, as shown in the figure. A and B…

Three blocks $\mathrm{A},\mathrm{ }\mathrm{B}$ and $\mathrm{C}$ are lying on a smooth horizontal surface, as shown in the figure. $\mathrm{A}$ and $\mathrm{B}$ have equal masses, $m$ while $\mathrm{C}$ has mass $M.$ Block $\mathrm{A}$ is given an initial speed $v$ towards $\mathrm{B}$ due to which it collides with $\mathrm{B}$ perfectly inelastically. The combined mass collides with $C,$ also perfectly inelastically . $\frac{5}{6}\text{t}\text{h}$ of the initial kinetic energy is lost in the whole process. What is the value of $M/m$?

1. $3$
2. $4$
3. $5$
4. $2$

Solution

$mv=\left(m+m\right)v^{\prime }$

$\left(m+m\right)v^{\prime }=\left(m+m+M\right)v_f$

$\therefore$ final velocity $v_f=\frac{m}{2m+M}v$

Initial energy $=\frac{1}{2}m{v}^2$

Final energy $=\frac{1}{2}\left(2m+M\right){\left(\frac{mv}{2m+M}\right)}^2$

Given that $\frac{5}{6}\frac{1}{2}m{v}^2=\frac{1}{2}m{v}^2-\frac{1}{2}\left(2m+M\right){\left(\frac{mv}{2m+M}\right)}^2$

$\frac{1}{12}m{v}^2=\frac{1}{2}\frac{m^2{v}^2}{\left(2m+M\right)}$

$6m=2m+M$

$\frac{M}{m}=4$

Asked in: JEE Main 2019 (09 Jan Shift 1)

Practice more Center of Mass Momentum and Collision questions on Aicharya