A particle of mass m is moving in the xy-plane such that its velocity

Question

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 F⃗ acting on the particle?

Accepted Answer

Step 1: Identify the velocity components. The velocity vector of the particle is given. v⃗ = α(yx̂ + 2xŷ) Extract the x and y components of the velocity. vₓ = α y v_y = 2α x Step 2: Calculate the acceleration components. Differentiate the velocity components with respect to time t. aₓ = dvₓ/dt = α dy/dt = α v_y a_y = (dv_y)/dt = 2α dx/dt = 2α vₓ Substitute the expressions for vₓ and v_y. aₓ = α(2α x) = 2α² x a_y = 2α(α y) = 2α² y Step 3: Determine the acceleration vector. Combine the components to write the acceleration vector a⃗. a⃗ = aₓx̂ + a_yŷ a⃗ = 2α²(xx̂ + yŷ) Step 4: Calculate the force acting on the particle. Apply Newton's second law of motion. F⃗ = ma⃗ F⃗ = 2mα²(xx̂ + yŷ) Answer: F⃗ = 2mα²(xx̂ + yŷ)

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

Written solution

More from Laws of Motion

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

Written solution

More from Laws of Motion

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