JEE Advanced 2023 · Paper 2 · Physics Q3

← JEE Advanced 2023 · Paper 2

PhysicsSingle correct+3 / -1Question 20 of 51

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

\vec F = 2m\alpha^2(x\hat x + y\hat y)

\vec F = m\alpha^2(y\hat x + 2x\hat y)

\vec F = 2m\alpha^2(y\hat x + x\hat y)

\vec F = m\alpha^2(x\hat x + 2y\hat y)

Answer and worked solution

Official answer: A

Acceleration along the path. With $v_x=\alpha y$ and $v_y=2\alpha x$ :
$a_x=\frac{dv_x}{dt}=\alpha\frac{dy}{dt}=\alpha v_y=2\alpha^2x,$
$a_y=\frac{dv_y}{dt}=2\alpha\frac{dx}{dt}=2\alpha v_x=2\alpha^2y.$

$\vec F=m\vec a=2m\alpha^2(x\hat x+y\hat y).$

Topics:Motion in a plane Newton's laws

← Previous question Next question →

JEE Advanced 2023 · Paper 2, Physics question 3