Question:
A particle of mass \(m\) with an initial velocity \(u\hat{i}\) collides perfectly elastically with a mass \(3m\) at rest. It moves with a velocity \(v\hat{j}\) after collision, then, \(v\) is given by :
A particle of mass \(m\) with an initial velocity \(u\hat{i}\) collides perfectly elastically with a mass \(3m\) at rest. It moves with a velocity \(v\hat{j}\) after collision, then, \(v\) is given by :
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Conserve vector momentum to get the velocity of the 3m mass in terms of \(u, v\), then conserve kinetic energy.
Updated On: Nov 5, 2026
- \(v = \sqrt{\frac{2}{3}}\,u\)
- \(v = \frac{u}{\sqrt{3}}\)
- \(v = \frac{u}{\sqrt{2}}\)
- \(v = \frac{1}{\sqrt{6}}\,u\)
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The Correct Option is C
Solution and Explanation
Momentum: \(m u\hat{i} = m v\hat{j} + 3m\vec{V}\), so \(\vec{V} = \frac{u\hat{i} - v\hat{j}}{3}\) and \(|\vec{V}|^2 = \frac{u^2 + v^2}{9}\).
Kinetic energy (elastic): \(\frac{1}{2}mu^2 = \frac{1}{2}mv^2 + \frac{1}{2}(3m)|\vec{V}|^2\).
\(u^2 = v^2 + 3 \cdot \frac{u^2 + v^2}{9} = v^2 + \frac{u^2 + v^2}{3}\).
\(\frac{2u^2}{3} = \frac{4v^2}{3} \Rightarrow v^2 = \frac{u^2}{2} \Rightarrow v = \frac{u}{\sqrt{2}}\).
Answer: (3) \(v = \frac{u}{\sqrt{2}}\)
Kinetic energy (elastic): \(\frac{1}{2}mu^2 = \frac{1}{2}mv^2 + \frac{1}{2}(3m)|\vec{V}|^2\).
\(u^2 = v^2 + 3 \cdot \frac{u^2 + v^2}{9} = v^2 + \frac{u^2 + v^2}{3}\).
\(\frac{2u^2}{3} = \frac{4v^2}{3} \Rightarrow v^2 = \frac{u^2}{2} \Rightarrow v = \frac{u}{\sqrt{2}}\).
Answer: (3) \(v = \frac{u}{\sqrt{2}}\)
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