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4-1.Newton's Laws of Motion
medium
A particle of mass $\mathrm{m}$ is moving in the $x y$-plane such that its velocity at a point $(x, y)$ is given as $\vec{v}=\alpha(y \hat{x}+2 x \hat{y})$, 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
$\vec{F}=m \alpha^2(x \hat{x}+2 y \hat{y})$
(IIT-2023)
Solution
$\vec{v}=\alpha(y \hat{x}+2 x \hat{y})$
$v_x=\alpha y$
$\frac{d v_x}{d t}=\alpha \frac{d y}{d t}=2 \alpha^2 x \quad \frac{d v_y}{d t}=2 \alpha v_x=2 \alpha^2 y$
$\therefore \vec{F}=m \vec{a}=2 m \alpha^2(x \hat{x}+y \hat{y})$
Standard 11
Physics
