- JEE Mains 2020
- Hard
- Numerical
A particle of mass \(m\) is moving along the \(x\) -axis with initial velocity \(u \hat i\). It collides elastically with a particle of mass \(10\, m\) at rest and then moves with half its initial kinetic energy (see figure). If \(\sin \theta_{1}=\sqrt{n} \sin \theta_{2}\) then value of \(n\) is.....

Answer & Solution
(C) \(10\)
Step-by-step Solution
Detailed explanation
\(m _{1} u _{1} \sin \theta_{1}= m _{2} u _{2} \sin \theta_{2}\)
i.e. \(mu _{1} \sin \theta_{1}=10 mu _{2} \sin \theta_{2}\)
\(\Rightarrow u _{1} \sin \theta_{1}=10 u _{2} \sin \theta_{2}\) \(...(i)\)
\(kf _{ m _{ 1 }}=\frac{1}{2} ki _{ m _{ 1 }}\) i.e. \(\frac{1}{2} mu _{1}^{2}=\frac{1}{2} \times \frac{1}{2} mu ^{2}\)
i.e. \(\sqrt{ u _{1}=\frac{ u }{\sqrt{2}}}\) \(...(ii)\)
Also collision is elastic : \(k _{ i }= k _{ f }\)
\(\frac{1}{2} m u^{2}=\frac{1}{2} m u_{1}^{2}+\frac{1}{2} \cdot 10 m \cdot u_{2}^{2}\)
\(\frac{1}{2} mu ^{2}=\frac{1}{2} \times \frac{1}{2} mu ^{2}+\frac{1}{2} \times 10 m \cdot u _{2}^{2}\)
\(\frac{1}{4} mu ^{2}=\frac{1}{2} \times 10 \times mu _{2}^{2}\)
\(u _{2}=\frac{ u }{\sqrt{20}}\)\(...(iii)\)
Putting \((ii)\) \(\&\) \((iii)\) in \((i)\)
\(\frac{ u }{\sqrt{2}} \sin \theta_{1}=10 \cdot \frac{ u }{\sqrt{20}} \sin \theta_{2}\)
\(\sin \theta_{1}=\sqrt{10} \sin \theta_{2} \quad \rightarrow\) Hence \(n =10\)
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