A particle of mass m is moving in the xy-plane such that its velocity at a point (x, y) is given as v→=α(yx^+2xy^)where α is a non-zero constant. What is the force F→ acting on the particle?

F → = 2 mα 2 ( y x ^ + x y ^ )

A particle of mass m is moving in the xy-plane such that its velocity at a point (x, y) is given as v→=α(yx^+2xy^)where α is a non-zero constant. What is the force F→ acting on the particle?

F → = 2 mα 2 ( y x ^ + x y ^ )

F → = mα 2 ( x x ^ + 2 y y ^ )

F → = 2 mα 2 ( x x ^ + y y ^ )

F → = mα 2 ( y x ^ + 2 x y ^ )

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

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

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

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

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

answer is A.

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Detailed Solution

$\begin{array}{l} V = α (y \hat{x} + 2 x \hat{y}) \Rightarrow a = \frac{dv}{dt} = α (Vy \hat{x} + 2 Vx \hat{y}) \\ = α (2 αx \hat{x} + 2 αy \hat{y}) = 2 α^{2} x \hat{x} + 2 α^{2} y \hat{y} Hence F = 2 {mα}^{2} (x \hat{x} + y \hat{y}) \end{array}$

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