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{v}=\alpha(y \stackrel{\^}{x}+2 x \stackrel{\^}{y}), where \alpha is a non-zero constant. What is the force \overrightarrow{F} acting on the particle? [JEE Adv 2023 P2]

Correct Answer is : →F=2mα2(x^x+y^y)

\overrightarrow{F}=2 m \alpha^2(x \stackrel{\^}{x}+y \stackrel{\^}{y})

\overrightarrow{F}=m \alpha^2(y \stackrel{\^}{x}+2 x \stackrel{\^}{y})

\overrightarrow{F}=2 m \alpha^2(y \stackrel{\^}{x}+x \stackrel{\^}{y})

\overrightarrow{F}=m \alpha^2(x \stackrel{\^}{x}+2 y \stackrel{\^}{y})

JEE Advanced 2023 Paper 2 Solutions

Section: Physics

Question : 3 of 51

Marks: +1, -0

A particle of mass m is moving in the xy-plane such that its velocity at a point (x,y) is given as

→

=α(y

+2x

), where α is a non-zero constant. What is the force

→

acting on the particle?

[JEE Adv 2023 P2]

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