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JEE MainPhysicsModerateCentre of Mass

A small sphere of mass 1 kg is moving with a velocity
(6i^+j^)ms−1(6\widehat{i} + \widehat{j})ms^{- 1}. It hits a fixed smooth wall and
rebound with velocity (4i^+j^)ms−1(4\widehat{i} + \widehat{j})ms^{- 1}. The
coefficient of restitution between the sphere and the wall is

Show answer and solution
Impulse = Change in momentum
=1(4i^+j^)−1(6i^+j^)=−2i^= 1(4\widehat{i} + \widehat{j}) - 1(6\widehat{i} + \widehat{j}) = - 2\widehat{i}

Which is perpendicular to the wall. Componentof initial velocity along
i^=6i^\widehat{i} = 6\widehat{i}
⇒\Rightarrow Speed of approach =6
Therefore e=46=23e = \frac{4}{6} = \frac{2}{3}

Source: Grass-G / P-07 Physics paper 17 October.docx

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