JEE Mains · Physics · STD 11 - Work,energy,power and collision
- JEE Mains 2020
- Hard
- MCQ
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
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(v=\frac{u}{\sqrt{2}}\)
Step-by-step Solution
Detailed explanation
From momentum conservation
\(\overrightarrow{ P }_{ i }=\overrightarrow{ P }_{ f }\)
\(m ( ui )+3 m ( D )= mvj +3 m \overline{ v }_{2}\)
\(mui - mvj =3 m \overline{ v }_{1}\)
\(\bar{v}_{1}=\frac{u i-v j}{3}\)
or \(\left| v _{1}\right|=\frac{\sqrt{ u ^{2}+ v ^{2}}}{3}\)
or \(v_{1}^{2}=\frac{u^{2}+v^{2}}{9} \ldots .(1)\)
As collision is perfectely elastic hence
\(k_{i}=k_{j}\)
\(\frac{1}{2} m u^{2}+\frac{1}{2} 3 m 0^{2}=\frac{1}{2} m v^{2}+\frac{1}{2} 3 m v_{1}^{2}\)
\(\Rightarrow u ^{2}= v ^{2}+3 v _{1}^{2}\)
\(u ^{2}= v ^{2}+3 \frac{\left( u ^{2}+ v ^{2}\right)}{9}\)
\(\Rightarrow 3 u ^{2}=3 v ^{2}+ u ^{2}+ v ^{2}\)
\(\Rightarrow 2 u ^{2}=4 v ^{2}\)
\(v =\frac{ u }{\sqrt{2}}\)
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