Q. 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 ?

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Solution:

$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 $

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