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

Correct Answer: The correct option is F⃗ = 2mα²(x x̂ + y ŷ).

Step-by-Step Explanation:

Step 1: Understand the given velocity vector
The velocity of the particle at any point $(x,y)$ is given by:
$\overrightarrow{v}=\alpha (y\hat{i}+2x\hat{j})$

From this velocity vector, we can identify the x-component and y-component of velocity as:
$v_x=\frac{dx}{dt}=\alpha y$
$v_y=\frac{dy}{dt}=2\alpha x$

Step 2: Find the acceleration components
Acceleration is the rate of change of velocity with respect to time. Using the chain rule for derivatives:

For the x-component of acceleration ($a_x$):
$a_x=\frac{d{v}_x}{dt}=\frac{d}{dt}\left(\alpha y\right)=\alpha \frac{dy}{dt}=\alpha v_y$

Substituting $v_y=2\alpha x$ into the expression:
$a_x=\alpha \left(2\alpha x\right)=2{\alpha }^{2}x$

For the y-component of acceleration ($a_y$):
$a_y=\frac{d{v}_y}{dt}=\frac{d}{dt}\left(2\alpha x\right)=2\alpha \frac{dx}{dt}=2\alpha v_x$

Substituting $v_x=\alpha y$ into the expression:
$a_y=2\alpha \left(\alpha y\right)=2{\alpha }^{2}y$

Step 3: Calculate the total acceleration vector
Combining the components gives the acceleration vector:
$\overrightarrow{a}=a_x\hat{x}+a_y\hat{y}=2{\alpha }^{2}x\hat{x}+2{\alpha }^{2}y\hat{y}=2{\alpha }^2(x\hat{x}+y\hat{y})$

Step 4: Calculate the force acting on the particle
According to Newton's Second Law of Motion, $\overrightarrow{F}=m\overrightarrow{a}$:
$\overrightarrow{F}=m·2{\alpha }^2(x\hat{x}+y\hat{y})=2m{\alpha }^2(x\hat{x}+y\hat{y})$

Thus, the force acting on the particle is F⃗ = 2mα²(x x̂ + y ŷ).