Question 118 of 130

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

A$3$B$4$C$5$D$2$

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Answer: (B) $4$

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

$\left(m+m\right){v}^{'}=\left(m+m+M\right){v}_{f}$

$∴$ final velocity ${v}_{f}=\dfrac{m}{2m+M}v$

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

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

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

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

$6m=2m+M$

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