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(𝑦𝑥̂ + 2𝑥𝑦̂), where 𝛼 is a non-zero constant. What is the force 𝐹 acting on the particle? (A) 𝐹 = 2𝑚𝛼 2 (𝑥𝑥̂ + 𝑦𝑦̂ ) (B) 𝐹 = 𝑚𝛼 2 (𝑦𝑥̂ + 2𝑥𝑦̂ ) (C) 𝐹 = 2𝑚𝛼 2 (𝑦𝑥̂ + 𝑥𝑦̂ ) (D) 𝐹 = 𝑚𝛼 2 (𝑥𝑥̂ + 2𝑦𝑦̂ )

To determine the force $\mathbf{F}$ acting on the particle, we'll use Newton's second law, which states that the force is equal to the mass $m$ times the acceleration $\mathbf{a}$:

$$\mathbf{F}=m\mathbf{a}$$

First, we need to find the acceleration $\mathbf{a}$. Acceleration is the time derivative of velocity $\mathbf{v}$:

$$\mathbf{a}=\frac{d\mathbf{v}}{dt}$$

Given the velocity vector:

$$\mathbf{v}=\alpha (y\widehat{x}+2x\widehat{y})$$

We can express the acceleration as:

$$\mathbf{a}=\frac{d}{dt}[\alpha (y\widehat{x}+2x\widehat{y})]$$

Since $\alpha$ is a constant, it can be factored out:

$$\mathbf{a}=\alpha (\frac{dy}{dt}\widehat{x}+2\frac{dx}{dt}\widehat{y})$$

Recognize that $\frac{dx}{dt}$ and $\frac{dy}{dt}$ are the components of the velocity vector $\mathbf{v}$:

$$\frac{dx}{dt}=\alpha y\text{and}\frac{dy}{dt}=2\alpha x$$

Substituting these into the expression for acceleration:

$$\mathbf{a}=\alpha ((2\alpha x)\widehat{x}+2(\alpha y)\widehat{y})=2{\alpha }^{2}x\widehat{x}+2{\alpha }^{2}y\widehat{y}$$

Simplifying, we get:

$$\mathbf{a}=2{\alpha }^2(x\widehat{x}+y\widehat{y})$$

Now, using Newton's second law to find the force:

$$\mathbf{F}=m\mathbf{a}=m\cdot 2{\alpha }^2(x\widehat{x}+y\widehat{y})=2m{\alpha }^2(x\widehat{x}+y\widehat{y})$$

Thus, the correct answer is:

$$\boxed{{\displaystyle A}}$$

Answer: (A) $\mathbf{F}=2m{\alpha }^2(x\widehat{x}+y\widehat{y})$