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

The Correct Option is D

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

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

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

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

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

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

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

Answer 2

\(F=m\frac{d\vec{V}}{dt}\)

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

\(\frac{d\vec{V}}{dt}=\alpha(\frac{dy}{dt}\hat{x}+2\frac{dx}{dt}\hat{y})\)

\(=\alpha[2x\alpha\hat{x}+2\alpha y\hat{y}]\)

\(=2\alpha^2[x\hat{x}+y\hat{y}]\)

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