A particle of mass 1 kg is subjected to a force which depends on the position as F = -k(x i + y j) k

Q: Why can't I just use equations of motion to find x and y at z = 0.5 m?

You absolutely can! The motion in the x and y directions is Simple Harmonic Motion (SHM). However, solving for the exact time when $z=0.5\text{ m}$ and substituting it back into the velocity equations is mathematically tedious and highly error-prone. Conservation laws offer a much faster, elegant route.

Q: Why is the torque not completely zero?

The force points towards the z-axis, not the origin. Because the particle has a non-zero z-coordinate, the force creates a torque that tries to rotate the particle's plane of motion. Only the torque *about the z-axis* is zero.

Q: What is the significance of z = 0.5 m in the question?

It's a classic distractor! Since $L_z$ is conserved, its value is the same at $z=0.5\text{ m}$ as it was at $t=0$. The exam setters put it there to tempt you into calculating the time $t$.

Q: How did you know $(x v_y - y v_x)$ is related to angular momentum?

By definition, $\vec{L} = \vec{r} \times \vec{p}$. If you expand the cross product $(x\hat{i} + y\hat{j} + z\hat{k}) \times m(v_x\hat{i} + v_y\hat{j} + v_z\hat{k})$, the $\hat{k}$ component is exactly $m(x v_y - y v_x)$.

Q: Does the mass value matter here?

Yes, but since $m = 1\text{ kg}$, the angular momentum $L_z$ is numerically equal to the given expression. If $m$ were $2\text{ kg}$, the expression would be equal to $L_z / 2$.

Question

A particle of mass 1 kg is subjected to a force which depends on the position as $\vec{F} = -k(x\hat{i} + y\hat{j})	ext{kgms}^{-2}$ with $k = 1	ext{ kgs}^{-2}$. At time $t = 0$, the particle's position $\vec{r} = \left(\frac{1}{\sqrt{2}}\hat{i} + \sqrt{2}\hat{j}ight)	ext{m}$ and its velocity $\vec{v} = \left(-\sqrt{2}\hat{i} + \sqrt{2}\hat{j} + \frac{2}{\pi}\hat{k}ight)	ext{ms}^{-1}$. Let $v_x$ and $v_y$ denote the x and the y components of the particle's velocity, respectively. Ignore gravity. When $z = 0.5	ext{ m}$, the value of $(x v_y - y v_x)$ is_________ $	ext{m}^2	ext{s}^{-1}$.

Sigma Percentile · Accepted Answer

### The Trap of Brute Force

When you first look at this problem, the instinct is to dive straight into Newton's Second Law. Since the force is given as $\vec{F} = -k(x\hat{i} + y\hat{j})$, you might write down the equations of motion for each axis.

For the x-axis, $m\ddot{x} = -kx$, which is the classic equation for Simple Harmonic Motion (SHM). The same goes for the y-axis. You could solve these differential equations to find $x(t)$ and $y(t)$, figure out the exact time $t$ when $z = 0.5	ext{ m}$, and then plug everything into the expression $(x v_y - y v_x)$.

While this brute-force method is physically correct, it is a mathematical nightmare under exam pressure. The condition $z = 0.5	ext{ m}$ is a brilliantly disguised distractor designed to waste your time. There is a much more elegant path.

### The Elegance of Angular Momentum

Let's take a closer look at the expression we need to evaluate: $(x v_y - y v_x)$. Does it ring a bell?

By definition, the angular momentum of a particle about the origin is $\vec{L} = \vec{r} 	imes \vec{p}$. If we expand this cross product, the z-component of the angular momentum is exactly $L_z = m(x v_y - y v_x)$.

Since the mass $m = 1	ext{ kg}$, the expression we are looking for is numerically equal to $L_z$. This is a massive hint! Instead of tracking the particle's exact coordinates, we just need to find out what happens to its angular momentum.

### Analyzing the Torque

To see if angular momentum is conserved, we must calculate the torque acting on the particle. Torque is given by $\vec{	au} = \vec{r} 	imes \vec{F}$.

Let's substitute the position and force vectors:

$$ \vec{	au} = (x\hat{i} + y\hat{j} + z\hat{k}) 	imes (-kx\hat{i} - ky\hat{j}) $$

Notice that the force vector has no z-component; it points directly towards the z-axis. When we compute the cross product, the z-component of the torque is:

$$ 	au_z = x F_y - y F_x = x(-ky) - y(-kx) = -kxy + kxy = 0 $$

This is the breakthrough! Because the z-component of the torque is strictly zero, **the z-component of the angular momentum must be conserved**. It will remain constant throughout the particle's entire journey.

### The Final Strike

Since $L_z$ is conserved, its value at the mysterious time when $z = 0.5	ext{ m}$ is exactly the same as its value at $t = 0$. We can completely ignore the z-coordinate condition!

Let's extract the initial values from the problem statement at $t = 0$:

$$ x = \frac{1}{\sqrt{2}}, \quad y = \sqrt{2} $$
$$ v_x = -\sqrt{2}, \quad v_y = \sqrt{2} $$

Now, we simply substitute these into our conserved expression:

$$ x v_y - y v_x = \left(\frac{1}{\sqrt{2}}ight)(\sqrt{2}) - (\sqrt{2})(-\sqrt{2}) $$

$$ x v_y - y v_x = 1 - (-2) = 3 $$

And just like that, without solving a single differential equation, we arrive at the final answer. **Always look for symmetries and conservation laws before resorting to brute force!**

Enter Numerical Value:

Visualized Solution

Visualizing the System

Angular Momentum $L_{z}$

Torque Setup $τ$

Calculating $τ_{z}$

Conservation of $L_{z}$

Initial Values at $t = 0$

Final Calculation

Symmetry and Conservation

The Sigma Insight: Torque and Angular Momentum

The Trap of Brute Force

The Elegance of Angular Momentum

Analyzing the Torque

The Final Strike

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Enter Numerical Value:

Visualized Solution

Visualizing the System

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

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

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

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

Initial Values at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t=0