A particle moving in the xy-plane experiences a velocity dependent force F = k(v_y i + v_x j), where

Q: Why did we divide the two acceleration equations?

We divided them to eliminate the time variable $dt$. This is a standard mathematical trick for coupled differential equations. It allows us to find a direct relationship between the velocity components $v_x$ and $v_y$ without having to solve for them as explicit functions of time.

Q: Why is the magnetic force option wrong? Doesn't a magnetic force depend on velocity?

Yes, magnetic force depends on velocity, but it has a strict geometric constraint: it must always be perpendicular to the velocity vector ($\mathbf{F} = q(\mathbf{v} \times \mathbf{B})$). If you take the dot product of the given force and velocity, $\mathbf{F} \cdot \mathbf{v} = 2k v_x v_y$, which is not zero. Therefore, it cannot be a purely magnetic force.

Q: How do we know the kinetic energy isn't constant?

Kinetic energy is constant only if the net work done on the particle is zero. The rate of work done (Power) is $\mathbf{F} \cdot \mathbf{v}$. Since we just established that $\mathbf{F} \cdot \mathbf{v} = 2k v_x v_y \neq 0$, work is being done, and the kinetic energy must be changing.

Q: What are the rules for the cross product of unit vectors?

Remember the cyclic order: $\hat{\mathbf{i}} \to \hat{\mathbf{j}} \to \hat{\mathbf{k}} \to \hat{\mathbf{i}}$. Going forward gives a positive result (e.g., $\hat{\mathbf{i}} \times \hat{\mathbf{j}} = \hat{\mathbf{k}}$). Going backward gives a negative result (e.g., $\hat{\mathbf{j}} \times \hat{\mathbf{i}} = -\hat{\mathbf{k}}$). The cross product of a vector with itself is zero (e.g., $\hat{\mathbf{i}} \times \hat{\mathbf{i}} = 0$).

Q: Why is $\mathbf{v} \cdot \mathbf{a}$ not constant?

If you calculate the dot product, you get $\mathbf{v} \cdot \mathbf{a} = \frac{2k}{m} v_x v_y$. From our integration, we only proved that the difference of their squares ($v_x^2 - v_y^2$) is constant. The product $v_x v_y$ will generally change as the particle moves, so the dot product is not constant.

Question

A particle moving in the xy-plane experiences a velocity dependent force $\mathbf{F} = k(v_y \hat{\mathbf{i}} + v_x \hat{\mathbf{j}})$, where $v_x$ and $v_y$ are the x and y components of its velocity $\mathbf{v}$.
If $\mathbf{a}$ is the acceleration of the particle, then which of the following statements is true for the particle?

Options:

(A) Quantity $\mathbf{v} 	imes \mathbf{a}$ is constant in time.

(B) $\mathbf{F}$ arises due to a magnetic field.

(C) Kinetic energy of particle is constant in time.

(D) Quantity $\mathbf{v} \cdot \mathbf{a}$ is constant in time.

Sigma Percentile · Accepted Answer

### The Peculiar Force
Imagine a particle moving in the $xy$-plane. It isn't experiencing a standard constant force like gravity. Instead, it feels a highly peculiar, velocity-dependent force given by $\mathbf{F} = k(v_y \hat{\mathbf{i}} + v_x \hat{\mathbf{j}})$.

Notice the beautiful symmetry here: the force in the $x$-direction is proportional to the velocity in the $y$-direction, and the force in the $y$-direction is proportional to the velocity in the $x$-direction. This is a classic example of **cross-coupled kinematics**. By Newton's second law, the acceleration is simply the force divided by the mass $m$:

$$\mathbf{a} = \frac{k}{m}(v_y \hat{\mathbf{i}} + v_x \hat{\mathbf{j}})$$

### Uncoupling the Equations
To understand the trajectory and the invariants of this motion, we need to break the acceleration into its components:

$$a_x = \frac{dv_x}{dt} = \frac{k}{m}v_y$$
$$a_y = \frac{dv_y}{dt} = \frac{k}{m}v_x$$

We have a system of coupled differential equations. The most elegant way to solve this is to eliminate the time variable $dt$ entirely. By dividing the first equation by the second, we get:

$$\frac{dv_x}{dv_y} = \frac{v_y}{v_x}$$

Cross-multiplying yields a simple, separable differential equation:

$$v_x dv_x = v_y dv_y$$

Integrating both sides gives us a powerful invariant of the motion:

$$\int v_x dv_x = \int v_y dv_y \implies \frac{v_x^2}{2} = \frac{v_y^2}{2} + C'$$

$$v_x^2 - v_y^2 = C$$

This tells us that while the individual velocity components change, the difference of their squares remains strictly constant throughout the motion.

### The Cross Product Revelation
Now, let's evaluate the options. Option (a) asks us to check if the cross product $\mathbf{v} 	imes \mathbf{a}$ is constant. Let's compute it:

$$\mathbf{v} 	imes \mathbf{a} = (v_x \hat{\mathbf{i}} + v_y \hat{\mathbf{j}}) 	imes \frac{k}{m}(v_y \hat{\mathbf{i}} + v_x \hat{\mathbf{j}})$$

Using the distributive property of the cross product and remembering that $\hat{\mathbf{i}} 	imes \hat{\mathbf{i}} = 0$, $\hat{\mathbf{j}} 	imes \hat{\mathbf{j}} = 0$, $\hat{\mathbf{i}} 	imes \hat{\mathbf{j}} = \hat{\mathbf{k}}$, and $\hat{\mathbf{j}} 	imes \hat{\mathbf{i}} = -\hat{\mathbf{k}}$, we get:

$$\mathbf{v} 	imes \mathbf{a} = \frac{k}{m} (v_x^2 \hat{\mathbf{k}} - v_y^2 \hat{\mathbf{k}}) = \frac{k}{m} (v_x^2 - v_y^2) \hat{\mathbf{k}}$$

Substitute our invariant $v_x^2 - v_y^2 = C$ into this expression:

$$\mathbf{v} 	imes \mathbf{a} = \frac{k}{m} (C) \hat{\mathbf{k}}$$

Since $k$, $m$, and $C$ are all constants, the entire vector is a constant pointing in the $z$-direction. Thus, **Option (a) is perfectly correct**.

### Why the Other Options Fail
It is always good practice to understand why the incorrect options fail.
- **Option (b):** A magnetic force is always perpendicular to the velocity ($\mathbf{F}_m = q(\mathbf{v} 	imes \mathbf{B})$). Here, $\mathbf{F} \cdot \mathbf{v} = k(v_x v_y + v_y v_x) = 2k v_x v_y 
eq 0$. Since the dot product is not zero, the force is not purely magnetic.
- **Option (c):** Because $\mathbf{F} \cdot \mathbf{v} 
eq 0$, the force does work on the particle. By the work-energy theorem, the kinetic energy must change.
- **Option (d):** The dot product $\mathbf{v} \cdot \mathbf{a} = \frac{2k}{m} v_x v_y$. We only know that $v_x^2 - v_y^2$ is constant, not their product $v_x v_y$. Hence, this quantity varies with time.

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Visualized Solution

$Force and Acceleration$

$Component Analysis$

$Eliminating Time$

$Integration$

$Evaluating v \times a$

$Final Conclusion$

$Why other options fail?$

The Sigma Insight: Motion in a Plane

The Peculiar Force Imagine a particle moving in the $x y$ -plane

Uncoupling the Equations

The Cross Product Revelation Now, let's evaluate the options

Why the Other Options Fail It is always good practice to understand why the incorrect options fail. - Option (b): A magnetic force is always perpendicular to the velocity ( $F_{m} = q (v \times B)$ )

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Force and Acceleration

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Component Analysis

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Eliminating Time

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Integration

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Evaluating v×a

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Conclusion

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Why other options fail?