Question

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. $\frac{5}{6}\text{t}\text{h}$ of the initial kinetic energy is lost in the whole process. What is the value of $M/m$ ?

• A. $3$
• B. $4$
• C. $5$
• D. $2$

Answer 1

Answer: (B)

$mv=\left(m+m\right)v^{\prime }$
$\left(m+m\right)v^{\prime }=\left(m+m+M\right)v_f$
$\therefore$ 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 $\frac{5}{6}\frac{1}{2}m{v}^2=\frac{1}{2}m{v}^2-\frac{1}{2}\left(2m+M\right){\left(\frac{mv}{2m+M}\right)}^2$
$\frac{1}{12}m{v}^2=\frac{1}{2}\frac{m^2{v}^2}{\left(2m+M\right)}$
$6m=2m+M$
$\frac{M}{m}=4$