JEE Mains · Physics · STD 11 - Work,energy,power and collision
- JEE Mains 2020
- Hard
- MCQ
A particle moving in the \(xy\) plane experiences a velocity dependent force \(\overrightarrow{ F }= k \left( v _{ y } \hat{ i }+ v _{ x } \hat{ j }\right),\) where \(v _{ x }\) and \(v _{ y }\) are the \(x\) and \(y\) components of its velocity \(\overrightarrow{ v } .\) If \(\overrightarrow{ a }\) be the acceleration of the particle, then which of the following statements is true for the particle\(?\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) quantity \(\overrightarrow{ v } \times \overrightarrow{ a }\) is constant in lime.
Step-by-step Solution
Detailed explanation
\(\frac{ dv _{ x }}{ dt }=\frac{ k }{ m } v _{ y }\)
\(\frac{ dv _{ y }}{ dt }=\frac{ k }{ m } v _{ x }\)
\(\frac{ d v_{ y }}{ dv _{ x }}=\frac{ v _{ x }}{ v _{ y }} \Rightarrow \int v _{ y } dv _{ y }=\int v _{ x } dv _{ x }\)
\(v_{y}^{2}=v_{x}^{2}+C\)
\(v _{ y }^{2}- v _{ x }^{2}= constant\)
Option \((3)\)
\(\overrightarrow{ v } \times \overrightarrow{ a }=\left( v _{ x } \hat{ i }+ v _{ y } \hat{ j }\right) \times \frac{ k }{ m }\left( v _{ y } \hat{ i }+ v _{ x } \hat{ j }\right)\)
\(=\left( v _{ x }^{2} \hat{ k }- v _{ y }^{2} \hat{ k }\right) \frac{ k }{ m }\)
\(=\left(v_{x}^{2}-v_{y}^{2}\right) \frac{k}{m} \hat{k}\)
\(=\) \(Constant\)
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