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

overrightarrow F =2 m α 2 ( x hat x + y hat y )

overrightarrow F =m α 2 (y hat x +2 x hat y )

overrightarrow F =2 m α 2 (y hat x +x hat y )

overrightarrow F =m α 2 (x hat x +2 y hat y )

Physics · Motion in one Dimension

JEE Advanced 2023 — Paper 2 — Question 16

A particle of mass $m$ is moving in the $x y$ -plane such that its velocity at a point $(x, y)$ is given as

$\overrightarrow{\mathrm{v}}=\alpha(\mathrm{y} \hat{\mathrm{x}}+2 \mathrm{x} \hat{\mathrm{y}})$ , where $\alpha$ is a non-zero constant. What is the force $\overrightarrow{\mathrm{F}}$ acting on the particle?

Option A:
$\overrightarrow{\mathrm{F}}=2 \mathrm{~m} \alpha^{2}(\mathrm{x} \hat{\mathrm{x}}+\mathrm{y} \hat{\mathrm{y}})$
Correct
Option B:
$\overrightarrow{\mathrm{F}}=m \alpha^{2}(y \hat{x}+2 x \hat{y})$
Option C:
$\overrightarrow{\mathrm{F}}=2 m \alpha^{2}(y \hat{x}+x \hat{y})$
Option D:
$\overrightarrow{\mathrm{F}}=m \alpha^{2}(x \hat{x}+2 y \hat{y})$

Answer: A

Step-by-step solution

$\vec{v}=\alpha(y \hat{x}+2 x \hat{y})$ $\vec{a}=\alpha\left(\frac{d y}{d t} \hat{x}+2 \frac{d x}{d t} \hat{y}\right)$

\vec{a}=\alpha \frac{d y}{d t} \hat{x}+2 \alpha \frac{d x}{d t} \hat{y} …(i) \end{gathered}$$ $$\begin{gathered} \frac{d x}{d t}=\alpha y …(ii) \end{gathered}$$ $$\begin{gathered} \frac{\mathrm{dy}}{\mathrm{dt}}=2 \alpha \mathrm{x} …(iii) \end{gathered}$$ From (i), (ii) and (iii) $\vec{a}=2 \alpha^{2} x \hat{x}+2 \alpha^{2} y \hat{y}$

\begin{aligned} & \overrightarrow{\mathrm{F}}=\mathrm{ma} \& \overrightarrow{\mathrm{~F}}=2 m \alpha^{2}(x \hat{x}+y \hat{y}) \end{aligned}

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2023
Paper: Paper 2
Subject: Physics
Chapter: Motion in one Dimension
Topic: Uniformly Accelerated Motion