Three blocks A, B and C are lying on a smooth horizontal surface,… - Vidyadip

MCQ

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. The combined mass collides with $C$, also perfectly inelastically if $\frac{5}{6}^{th}$ of the initial kinetic energy is lost in whole process. What is value of $M/m$ ?

A
$5$
B
$2$
✓
$4$
D
$3$

✓

Answer

Correct option: C.

$4$

c
Apply $LMC$ (Linear Momentum Conservation)

$mv = (2m + M) v'$

$v' = \frac{{mv}}{{2m + M}}$

Initial energy

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

Final energy

$\frac{1}{2}\,\left( {2m + M} \right){\left( {\frac{{mv}}{{2m + M}}} \right)^2}$

$Initial\, kinetic\, energy - Final\, Kinetic\, energy =$ $\frac{5}{6}$ of initial kinetic energy.

After solving, we get, $\frac{M}{m} = 4.$

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