A small smooth disc of mass m and radius r moving with an velocity v along the positive x-axis collide with a big disc of mass 2m and radius 2r which was initially at rest centre at origin as shown in the figure. If the coefficient of restitution is 0, then the velocity of the larger disc after the collision is

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$\frac{8}{27} v \hat{i} - \frac{2 \sqrt{2}}{27} v \hat{j}$

$\frac{v}{3} \hat{i}$

$\frac{2 \sqrt{2}}{27} v \hat{i} - \frac{8}{27} v \hat{j}$

$\frac{8}{27} v \hat{i} - \frac{\sqrt{2}}{27} v \hat{j}$

answer is A.

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Detailed Solution

The larger disc will move along line of impact
As e = 0, So velocity of larger disc
$v^{'} = \frac{mv \cos θ}{m + 2 m} = \frac{v cosθ}{3}$
Velocity of larger disc
$\begin{array}{l} = v^{'} \cos θ \hat{i} - v^{'} \sin θ \hat{j} \\ = \frac{v}{3} \cos^{2} θ \hat{i} - \frac{v}{3} \sin θcos θ \hat{j} \end{array}$
(from the figure we can calculate $cosθ = \frac{2 \sqrt{2}}{3}, sinθ = \frac{1}{3})$
$= \frac{8}{27} v i - \frac{2 \sqrt{2}}{27} v \hat{j}$

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