Practice portal › Laws of Motion › Impulse and Momentum Change
Asked in JEE Main 25th July 1st Shift 2021 · Bounces, catches and repeated impacts
Idea: the wall is smooth, so only the component of velocity perpendicular to the wall reverses. Resolve along X and compare.
Both balls arrive at the same speed u and leave at the same speed.
Ball a, head-on: its whole velocity is along X and it reverses, so
|Jₐ|=2mu.
Ball b, at 45°: only u cos 45° lies along X, and only that reverses, so
|J_b|=2mu cos 45°=(2mu)/(√2).
(|Jₐ|)/(|J_b|)=(2mu)/(2mu/√2)=√2, so the ratio is √2:1.
Neither the mass nor the 108 kmph is needed: both cancel in the ratio, and the answer would be the same for any common speed.
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