Animated Solution for Physics - 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 ?

The journey to finding the force acting on a particle often begins with a deep dive into its kinematics. In this problem, we are presented with a particle of mass $m$ navigating the two-dimensional $xy$-plane.

However, its motion isn't simple; it is governed by a fascinating, position-dependent velocity field. Let's unravel this step-by-step.

The problem provides us with the velocity vector of the particle at any arbitrary point $(x,y)$:

To make sense of this, our first instinct should be to break this vector down into its fundamental Cartesian components. By comparing it to the standard form $\vec{v}=v_x\hat{x}+v_y\hat{y}$, we can immediately extract the individual velocities along the $x$ and $y$ axes.

Notice a beautiful symmetry here: the velocity in the $x$-direction depends entirely on the particle's $y$-coordinate, and the velocity in the $y$-direction depends on its $x$-coordinate. This cross-dependence is the hallmark of a dynamic, curving trajectory!

Newton's Second Law, $\vec{F}=m\vec{a}$, dictates that to find the force, we must first determine the acceleration vector $\vec{a}$. Acceleration is the rate of change of velocity with respect to time.

Let's start by calculating the $x$-component of acceleration, $a_x$. We do this by taking the time derivative of $v_x$:

Since $\alpha$ is a constant, it factors out of the derivative. We are left with $\alpha \frac{dy}{dt}$. But what is $\frac{dy}{dt}$? By definition, the rate of change of the $y$-coordinate is exactly the $y$-component of velocity, $v_y$!

Substituting our previously found expression for $v_y$, we get:

We have successfully isolated the $x$-component of acceleration. Now, we apply the exact same logical framework to the $y$-direction. We differentiate $v_y$ with respect to time to find $a_y$:

Again, the constant $2\alpha$ factors out, leaving us with $2\alpha \frac{dx}{dt}$. Recognizing that $\frac{dx}{dt}$ is simply $v_x$, we substitute its value:

With both components of acceleration calculated, we can now construct the complete acceleration vector $\vec{a}$:

Notice that both terms share a common factor of $2{\alpha }^2$. We can factor this out to reveal a very elegant structure. Finally, we invoke Newton's Second Law to find the net force $\vec{F}$:

Substituting our acceleration vector into the equation, we arrive at our final answer:

Physical Insight: Take a moment to look closely at the term $(x\hat{x}+y\hat{y})$. This is exactly the position vector $\vec{r}$ of the particle relative to the origin. Therefore, we can write the force as $\vec{F}=2m{\alpha }^2\vec{r}$. This tells us that the force acting on the particle is always directed radially outward from the origin, and its magnitude increases linearly with distance. This is a classic example of a repulsive central force!