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 brutal 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 whole process. What is value of $M/m$ ?

• A. 4
• B. 5
• C. 3
• D. 2

Answer 1

Answer: (A)

4

Answer 2

$k_{i} = \frac{1}{2} mv^{2}_{0}$ From linear momentum conservation $mv_{0} = \left(2m+M\right)v_{f}$ $\Rightarrow v_{f} = \frac{mv_{0}}{2m+M}$ $\frac{k_{i}}{k_{f}} = 6$ $\Rightarrow \frac{\frac{1}{2} mv_{0}^{2} }{\frac{1}{2} \left(2m+M\right) \left(\frac{mv_{0}}{2m+M}\right)^{2}}= 6$ $\Rightarrow \frac{2m+M}{m} = 6$ $\Rightarrow \frac{M}{m} = 4$