Question

Three block $A,B\text{and}C$ 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 ${\frac{5}{6}}^{th}$ of the initial kinetic energy is lost in the whole process. What is the value of $\frac{M}{m}?$

• A. $5$
• B. $4$
• C. $2$
• D. $3$

Answer 1

The correct option is B $4$

Initial Kinetic energy of block $\text{A}$

$K_1=\frac{1}{2}m{v}^2$

$\therefore$ From principle of linear momentum conservation

$mv=(2m+M)v_f$

$\Rightarrow v_f=\frac{mv}{2m+M}$

According to question, of ${\frac{5}{6}}^{th}$ the initial kinetic energy is lost in the whole process.
$\Rightarrow k_f=\frac{1}{6}{k}_i\Rightarrow \frac{k_i}{k_f}=6$

$\Rightarrow \frac{\frac{1}{2}m{v}^2}{\frac{1}{2}(2m+M){\left(\frac{mv}{2m+M}\right)}^2}=6$

$\Rightarrow \frac{2m+M}{m}=6$

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

Hence, option $\left(C\right)$ is correct.