A particle of mass is moving in the -plane such that its velocity at a point is given as where is a non-zero constant. What is the force acting on the particle? [2023]

Question

A particle of mass $m$ is moving in the $x y$ -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? [2023]

Smart Achievers · Accepted Answer

(1)

$∵$ $\vec{v} = α (y \hat{x} + 2 x \hat{y})$

$∴$ $\vec{a} = \frac{d \vec{v}}{d t} = α (\frac{d y}{d t} \hat{x} + 2 \frac{d x}{d t} \hat{y})$

$= α (v_{y} \hat{x} + 2 v_{x} \hat{y})$

$= α (2 x α \hat{x} + 2 α y \hat{y})$

$= 2 α^{2} [x \hat{x} + y \hat{y}]$

Solution

Q.
A particle of mass $m$ is moving in the $x y$ -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 [2023]

1 $\vec{F} = 2 m α^{2} (x \hat{x} + y \hat{y})$

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

3 $\vec{F} = 2 m α^{2} (y \hat{x} + x \hat{y})$

4 $\vec{F} = m α^{2} (x \hat{x} + 2 y \hat{y})$

Ans.
(1)

$∵$ $\vec{v} = α (y \hat{x} + 2 x \hat{y})$

$∴$ $\vec{a} = \frac{d \vec{v}}{d t} = α (\frac{d y}{d t} \hat{x} + 2 \frac{d x}{d t} \hat{y})$

$= α (v_{y} \hat{x} + 2 v_{x} \hat{y})$

$= α (2 x α \hat{x} + 2 α y \hat{y})$

$= 2 α^{2} [x \hat{x} + y \hat{y}]$

Want Us to Email you About Special Offers & Updates?

Solution

Q. 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 [2023]

1 F→=2mα2(xx^+yy^)

2 F→=mα2(yx^+2xy^)

3 F→=2mα2(yx^+xy^)

4 F→=mα2(xx^+2yy^)

Ans. (1) ∵ v→=α(yx^+2xy^) ∴ a→=dv→dt=α(dydtx^+2dxdty^) =α(vyx^+2vxy^) =α(2xαx^+2αyy^) =2α2[xx^+yy^]