JEE Advanced 2023 Paper 2 Q03 Physics Mechanics Friction & NLM Medium

JEE Advanced 2023 Paper 2 · Q03 · Friction & NLM

A particle of mass $m$ is moving in the $x$-$y$ plane such that its velocity at a point $(x, y)$ is given as $\vec{v} = \alpha(y\hat{x} + 2x\hat{y})$, where $\alpha$ is a non-zero constant. What is the force $\vec{F}$ acting on the particle?

  1. A. $\vec{F} = 2m\alpha^2(x\hat{x} + y\hat{y})$
  2. B. $\vec{F} = m\alpha^2(y\hat{x} + 2x\hat{y})$
  3. C. $\vec{F} = 2m\alpha^2(y\hat{x} + x\hat{y})$
  4. D. $\vec{F} = m\alpha^2(x\hat{x} + 2y\hat{y})$
Reveal answer + step-by-step solution

Correct answer:A

Solution

$v_x = \alpha y$, $v_y = 2\alpha x$. Acceleration: $a_x = dv_x/dt = \alpha\,dy/dt = \alpha v_y = \alpha\cdot 2\alpha x = 2\alpha^2 x$. $a_y = dv_y/dt = 2\alpha\,dx/dt = 2\alpha v_x = 2\alpha\cdot \alpha y = 2\alpha^2 y$. So $\vec{a} = 2\alpha^2(x\hat{x} + y\hat{y})$ and $\vec{F} = m\vec{a} = 2m\alpha^2(x\hat{x} + y\hat{y})$.

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