Mechanics — Medium JEE physics MCQ
A particle of mass $m$ is moving in the $x y$-plane such that its velocity at a point $(x, y)$ is given as $\overrightarrow{\mathrm{v}}=\alpha(y \hat{x}+2 x \hat{y})$, 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. $\vec{F}=m \alpha^{2}(x \hat{x}+2 y \hat{y})$
Solution
The correct option is **A**.