A particle moves in the x-y plane under the influence of a force such that its linear momentum is p(

Q: Wait, why do we use the dot product to find the angle?

The dot product is defined as $\mathbf{A} \cdot \mathbf{B} = |\mathbf{A}||\mathbf{B}|\cos\theta$. It is the most direct algebraic way to extract the cosine of the angle between two vectors when they are given in Cartesian ($\hat{\mathbf{i}}$, $\hat{\mathbf{j}}$) coordinates.

Q: Why can't we just find the slopes of the vectors?

You absolutely can! The slope of $\mathbf{p}$ is $\frac{-\sin(kt)}{\cos(kt)} = -\tan(kt)$. The slope of $\mathbf{F}$ is $\frac{-\cos(kt)}{-\sin(kt)} = \cot(kt)$. The product of their slopes is $(-\tan(kt))(\cot(kt)) = -1$. Two lines with slopes multiplying to $-1$ are perfectly perpendicular!

Q: What physical motion does this mathematical setup represent?

This represents uniform circular motion. The momentum (and thus velocity) has a constant magnitude but a continuously changing direction. The force is always perpendicular to the velocity, acting purely as a centripetal force.

Q: Why did the constant $k$ come out during the differentiation step?

This is due to the chain rule in calculus. When differentiating a composite function like $\cos(kt)$ with respect to $t$, we first differentiate the outer cosine function to get $-\sin(kt)$, and then multiply by the derivative of the inner function $kt$, which is simply $k$.

Q: Does the angle between them ever change as time $t$ increases?

No! Notice how the time variable $t$ completely vanished when we calculated the final dot product. This means the dot product is zero, and the angle is exactly $90^\circ$, at every single instant in time.

Question

A particle moves in the $x$-$y$ plane under the influence of a force such that its linear momentum is $\mathbf{p}(t) = A[\hat{\mathbf{i}}\cos(kt) - \hat{\mathbf{j}}\sin(kt)]$, where, $A$ and $k$ are constants. The angle between the force and the momentum is

Options:

(A) $0^\circ$

(B) $30^\circ$

(C) $45^\circ$

(D) $90^\circ$

Sigma Percentile · Accepted Answer

The beauty of physics often lies in how abstract mathematical equations perfectly describe elegant physical realities. In this problem, we are given the linear momentum of a particle as a function of time and asked to find the angle between its momentum and the force acting on it.

At first glance, it might seem like we don't have enough information. We have the momentum, but where is the force? This is where the fundamental principles of classical mechanics come to our rescue.

### Analyzing the Setup

Let's take a close look at the given momentum vector:

$$ \mathbf{p}(t) = A\cos(kt)\hat{\mathbf{i}} - A\sin(kt)\hat{\mathbf{j}} $$

Notice the structure of this equation. The $x$-component and $y$-component are governed by cosine and sine functions, respectively. This tells us that the magnitude of the momentum is constant, but its direction is continuously rotating. The particle is undergoing uniform circular motion!

However, to find the angle between the momentum and the force, we need an explicit expression for the force vector.

### The Master Equation

This is where Newton's Second Law of Motion steps in. It states that the net force acting on a particle is equal to the rate of change of its linear momentum. Mathematically, this is expressed as:

$$ \mathbf{F} = \frac{d\mathbf{p}}{dt} $$

Our path forward is clear: we must differentiate the momentum vector with respect to time. Let's break it down component by component.

First, we differentiate the $x$-component. Using the chain rule, the derivative of $\cos(kt)$ is $-k\sin(kt)$.

$$ \frac{d}{dt}[A\cos(kt)] = -Ak\sin(kt) $$

Next, we differentiate the $y$-component. The derivative of $-\sin(kt)$ is $-k\cos(kt)$.

$$ \frac{d}{dt}[-A\sin(kt)] = -Ak\cos(kt) $$

Combining these results, we obtain the force vector:

$$ \mathbf{F}(t) = -Ak\sin(kt)\hat{\mathbf{i}} - Ak\cos(kt)\hat{\mathbf{j}} $$

### Final Calculation

Now we have both the momentum vector $\mathbf{p}$ and the force vector $\mathbf{F}$. To find the angle $	heta$ between them, the most powerful mathematical tool at our disposal is the dot product.

The dot product of two vectors is defined geometrically as:

$$ \mathbf{F} \cdot \mathbf{p} = |\mathbf{F}||\mathbf{p}|\cos	heta $$

It can also be calculated algebraically by multiplying the corresponding components and summing them up. Let's perform this algebraic calculation.

Multiplying the $\hat{\mathbf{i}}$ components:

$$ (A\cos(kt))(-Ak\sin(kt)) = -A^2k\sin(kt)\cos(kt) $$

Multiplying the $\hat{\mathbf{j}}$ components:

$$ (-A\sin(kt))(-Ak\cos(kt)) = A^2k\sin(kt)\cos(kt) $$

Now, we add these two products together to find the total dot product:

$$ \mathbf{F} \cdot \mathbf{p} = -A^2k\sin(kt)\cos(kt) + A^2k\sin(kt)\cos(kt) $$

Look at that! The two terms are exact opposites. They perfectly cancel each other out.

$$ \mathbf{F} \cdot \mathbf{p} = 0 $$

What does this mean physically? Since neither the force nor the momentum is a zero vector, their magnitudes are non-zero. Therefore, for the dot product to be zero, the cosine of the angle between them must be zero.

$$ \cos	heta = 0 \implies 	heta = 90^\circ $$

The force is always acting at a **$90^\circ$ angle** to the momentum. This perfectly aligns with our initial observation: in uniform circular motion, the centripetal force is always perpendicular to the instantaneous velocity (and thus momentum) of the particle!

Animated Solution for Physics - Laws of Motion: A particle moves in the $x$ - $y$ plane under the influence of a force such that its linear momentum is $p (t) = A [\hat{i} cos (k t) - \hat{j} sin (k t)]$ , where, $A$ and $k$ are constants. The angle between the force and the momentum is

Select Answer:

Visualized Solution

$p (t)$ Vector

Newton's Second Law

Differentiating $x$ -component

Differentiating $y$ -component

The Force Vector $F$

Dot Product Concept

Calculating $F \cdot p$ (Part 1)

Calculating $F \cdot p$ (Part 2)

Summing the Dot Product

Conclusion

The Sigma Insight: Newton's Laws of Motion

Analyzing the Setup

The Master Equation

Final Calculation

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