JEE Advanced 2023 — Paper 2 — PHY-I solution

JEE Advanced2023Paper 2PHY-I

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?

A
$F = 2 m α^{2} (x \overset{x}{^} + y \overset{y}{^})$
B
$F = m α^{2} (y \overset{x}{^} + 2 x \overset{y}{^})$
C
$F = 2 m α^{2} (y \overset{x}{^} + x \overset{y}{^})$
D
$F = m α^{2} (x \overset{x}{^} + 2 y \overset{y}{^})$

Solution

From the given velocity field, $v_{x} = α y$ and $v_{y} = 2 α x$ .

Acceleration components follow from the chain rule (or convective derivative):

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

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

Newton's second law gives

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

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