Question

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

• A. F = mα$^2$(y$\hat{i}$+ 2x$\hat{j}$)
• B. F = mα$^2$ (x$\hat{i}$ + 2y$\hat{j}$)
• C. F = 2mα$^2$(y$\hat{i}$+ 2x$\hat{j}$)
• D. F = 2mα$^2$ (x$\hat{i}$ + y$\hat{j}$)

Answer 1

Answer: (D)

F = 2mα$^2$ (x$\hat{i}$ + y$\hat{j}$)

Answer 2

The correct answer is (D) F = 2mα$^2$ (x$\hat{i}$ + y$\hat{j}$)

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

$\Rightarrow V_x=\frac{dx}{dt}=\alpha y\,\,and\,\,V_y=\frac{dy}{dt}=2\alpha X$

$\vec{a}=\frac{d\vec{v}}{dt}=\alpha\frac{dy}{dt}\hat{i}+2\alpha\frac{dx}{dt}\hat{j}=[(\alpha)(2\alpha x)\hat{i}+2\alpha(\alpha y)\hat{j}]$

$\vec{a}=[(2\alpha^2x)\hat{i}+(2\alpha^2y)\hat{j}]$

$\vec{a}=2\alpha^2[x\hat{i}+y\hat{j}]$

$\vec{F_{net}}$ = $m\vec{a}=2m\alpha^2[x\hat{i}+y\hat{j}]$