A particle of mass m is moving in the xy-plan | vedclass

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(A) Given velocity $\vec{v} = \alpha y \hat{i} + 2\alpha x \hat{j}$.The components of velocity are $v_x = \alpha y$ and $v_y = 2\alpha x$.The accelera

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given velocity $\vec{v} = \alpha y \hat{i} + 2\alpha x \hat{j}$.The components of velocity are $v_x = \alpha y$ and $v_y = 2\alpha x$.The acceleration components are $a_x = \frac{dv_x}{dt} = \alpha \frac{dy}{dt} = \alpha v_y = \alpha(2\alpha x) = 2\alpha^2 x$.Similarly,$a_y = \frac{dv_y}{dt} = 2\alpha \frac{dx}{dt} = 2\alpha v_x = 2\alpha(\alpha y) = 2\alpha^2 y$.Thus,the acceleration vector is $\vec{a} = a_x \hat{i} + a_y \hat{j} = 2\alpha^2 x \hat{i} + 2\alpha^2 y \hat{j} = 2\alpha^2(x \hat{i} + y \hat{j})$.Using Newton's second law,$\vec{F} = m\vec{a} = 2m\alpha^2(x \hat{i} + y \hat{j})$.

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

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