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

(A) $\vec{F}=2 m \alpha^{2}(x \hat{x}+y \hat{y})$

(B) $\vec{F}=m \alpha^{2}(y \hat{x}+2 x \hat{y})$

(C) $\vec{F}=2 m \alpha^{2}(y \hat{x}+x \hat{y})$

(D) $\vec{F}=m \alpha^{2}(x \hat{x}+2 y \hat{y})$

JEE Advanced · 2023 · 04 Jun 2023 · Paper 2 · Kinematics

A particle of mass m is moving in the

x y

-plane such that its velocity at a point

(x, y)

is given as

v = α (y \overset{x}{^} + 2 x \overset{y}{^})

, where

α

is a non-zero constant. What is the force

F

acting on the particle?

Answer: (A)

Given

v = α (y \overset{x}{^} + 2 x \overset{y}{^})

, so

v_{x} = α y

and

v_{y} = 2 α x

.

The acceleration components are:

a_{x} = \frac{d v _{x}}{d t} = α \frac{d y}{d t} = α v_{y} = α (2 α x) = 2 α^{2} x

a_{y} = \frac{d v _{y}}{d t} = 2 α \frac{d x}{d t} = 2 α v_{x} = 2 α (α y) = 2 α^{2} y

Therefore,

F = m a = 2 m α^{2} (x \overset{x}{^} + y \overset{y}{^})

JEE Advanced · 2023 · 04 Jun 2023 · Paper 2 · Single correct option