A particle is acted upon by a force of constant magnitude which is always perpendicular to the veloc

Q: Why does a perpendicular force do no work?

Work is defined as the dot product of force and displacement ($W = \vec{F} \cdot d\vec{r}$). When the force is perpendicular to the displacement (or velocity), the angle between them is $90^\circ$. Since $\cos(90^\circ) = 0$, the dot product is zero, meaning no work is done.

Q: If the kinetic energy is constant, why isn't the velocity constant?

Kinetic energy ($K = \frac{1}{2}mv^2$) is a scalar quantity that depends only on the *speed* $v$, not the direction. Velocity is a vector. The perpendicular force changes the direction of the velocity without changing its magnitude. Since the direction changes, the velocity vector is not constant.

Q: The problem says the force has a constant magnitude. Doesn't that mean acceleration is constant?

Acceleration is a vector ($\vec{a} = \frac{\vec{F}}{m}$). While its magnitude is constant, its direction must constantly shift to remain perpendicular to the changing velocity. A vector is only constant if *both* its magnitude and direction are constant.

Q: Could the particle move in a straight line?

No. If it moved in a straight line, the velocity direction wouldn't change, which would mean the perpendicular force is zero. But we are given a force of constant (non-zero) magnitude. This force will inevitably curve the path.

Q: Why must the path be a circle and not an ellipse or some other curve?

For an ellipse or other curves, the radius of curvature changes. Since the centripetal force is $F = \frac{mv^2}{R}$, and we know $F$, $m$, and $v$ are all constant, the radius of curvature $R$ must also be constant. A curve with a constant radius of curvature in a plane is a perfect circle.

Question

A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle. The motion of the particle takes place in a plane. It follows that

Options:

(A) its velocity is constant

(B) its acceleration is constant

(C) its kinetic energy is constant

(D) it moves in a circular path

Sigma Percentile · Accepted Answer

The problem presents us with a fascinating scenario: a particle moving in a plane, subjected to a force of constant magnitude that is always perpendicular to its velocity. This simple setup unlocks a cascade of physical consequences. Let's break it down step-by-step.

### The Work-Energy Connection

The most crucial piece of information is that the force $\vec{F}$ is always perpendicular to the velocity $\vec{v}$. In physics, when a force acts at a right angle to the direction of motion, it does no work.

We can see this mathematically through the definition of power, which is the rate at which work is done. Power is the dot product of force and velocity:

$$P = \vec{F} \cdot \vec{v} = |\vec{F}| |\vec{v}| \cos(90^\circ) = 0$$

Since the power is zero at all times, the total work done on the particle is zero.

According to the Work-Energy Theorem, the net work done on an object is equal to its change in kinetic energy:

$$W = \Delta K = 0$$

This implies that the kinetic energy $K$ of the particle remains strictly constant. Since $K = \frac{1}{2}mv^2$, and the mass $m$ is constant, the speed $v$ (the magnitude of velocity) must also be constant. This makes option (c) correct.

### Speed vs. Velocity: The Vector Trap

It is very tempting to think that if speed is constant, velocity must be constant too. However, velocity is a vector quantity; it has both magnitude (speed) and direction.

While the magnitude remains unchanged, the perpendicular force continuously pulls the particle sideways, altering its direction of motion. Because the direction is changing, the velocity vector $\vec{v}$ is not constant. Therefore, option (a) is incorrect.

### The Acceleration Dilemma

What about acceleration? By Newton's Second Law, acceleration is directly proportional to the net force:

$$\vec{a} = \frac{\vec{F}}{m}$$

We are given that the force has a constant magnitude, which means the acceleration also has a constant magnitude. But just like velocity, acceleration is a vector. Since the force must always remain perpendicular to the changing velocity, the direction of the force—and thus the direction of the acceleration—must continuously change.

Because its direction is not fixed, the acceleration vector $\vec{a}$ is not constant. This makes option (b) incorrect.

### The Inevitable Path: Uniform Circular Motion

Let's summarize what we know: the particle moves in a two-dimensional plane, it has a constant speed, and it experiences a force of constant magnitude that is always perpendicular to its motion.

This specific set of conditions is the exact geometric and physical definition of **Uniform Circular Motion**. The perpendicular force acts as the centripetal force, continuously turning the velocity vector without speeding it up or slowing it down, forcing the particle into a perfect circular trajectory.

Therefore, the particle moves in a circular path, making option (d) correct.

Animated Solution for Physics - Laws of Motion: A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle. The motion of the particle takes place in a plane. It follows that

Select Answer:

Visualized Solution

$F ⊥ v$

$P = F \cdot v$

$W = Δ K$

$K = \frac{1}{2} m v^{2}$

$v \neq = constant$

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

Uniform Circular Motion

The Sigma Insight: Dynamics of Circular Motion

The Work-Energy Connection

Speed vs

The Acceleration Dilemma

The Inevitable Path

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