In the column-I below, four different paths of a particle are given as functions of time. In these f

Q: Why is potential energy conserved in Path A and C?

In Path A, the force is zero, so potential energy is constant. In Path C, the force is central and the particle moves in a circle, meaning its distance from the center is constant, so the potential energy $U(r)$ is also constant.

Q: Why is angular momentum conserved in Path B even though it's an ellipse?

The force in Path B is $\mathbf{F} = -m\omega^2 \mathbf{r}$, which is a central force directed towards the origin. The torque $\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F}$ is zero because $\mathbf{r}$ and $\mathbf{F}$ are anti-parallel. Hence, angular momentum is conserved.

Q: Why is linear momentum not conserved in Path C (circle) even though speed is constant?

Linear momentum is a vector quantity ($\mathbf{p} = m\mathbf{v}$). Even though the magnitude of velocity (speed) is constant, its direction is continuously changing as the particle moves along the circle. Therefore, linear momentum is not conserved.

Q: How do we know the force in Path D is conservative?

The acceleration in Path D is $\mathbf{a} = \beta \hat{\mathbf{j}}$, which is a constant vector. A constant force field (like uniform gravity) is always conservative, meaning the work done depends only on the initial and final positions, and total mechanical energy is conserved.

Q: Why is angular momentum not conserved in Path D?

In Path D, the force is constant and points in the $y$-direction, but the particle is moving in both $x$ and $y$ directions. The position vector $\mathbf{r}$ and the force $\mathbf{F}$ are not parallel, so the torque $\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F}$ is non-zero. A non-zero torque causes the angular momentum to change over time.

Question

In the column-I below, four different paths of a particle are given as functions of time. In these functions, $\alpha$ and $\beta$ are positive constants of appropriate dimensions and $\alpha 
eq \beta$. In each case, the force acting on the particle is either zero or conservative. In column-II, five physical quantities of the particle are mentioned: $\mathbf{p}$ is the linear momentum, $\mathbf{L}$ is the angular momentum about the origin, $K$ is the kinetic energy, $U$ is the potential energy and $E$ is the total energy. Match each path in Column-I with those quantities in Column-II, which are conserved for that path.

$\begin{array}{ll} 	ext{Column I} & 	ext{Column II} \ 	ext{A. } \mathbf{r}(t) = \alpha t \hat{\mathbf{i}} + \beta t \hat{\mathbf{j}} & 	ext{p. } \mathbf{p} \ 	ext{B. } \mathbf{r}(t) = \alpha \cos \omega t \hat{\mathbf{i}} + \beta \sin \omega t \hat{\mathbf{j}} & 	ext{q. } \mathbf{L} \ 	ext{C. } \mathbf{r}(t) = \alpha (\cos \omega t \hat{\mathbf{i}} + \sin \omega t \hat{\mathbf{j}}) & 	ext{r. } K \ 	ext{D. } \mathbf{r}(t) = \alpha t \hat{\mathbf{i}} + \frac{\beta}{2} t^2 \hat{\mathbf{j}} & 	ext{s. } U \ & 	ext{t. } E \end{array}$

Options:

(A) A $ightarrow$ p, q, r, s, t; B $ightarrow$ q, t; C $ightarrow$ q, r, s, t; D $ightarrow$ t

(B) A $ightarrow$ p, q, r, s, t; B $ightarrow$ r, t; C $ightarrow$ q, r, s, t; D $ightarrow$ q, t

(C) A $ightarrow$ q, r, s; B $ightarrow$ t; C $ightarrow$ p, q, s; D $ightarrow$ q, t

(D) A $ightarrow$ p, q, r, t; B $ightarrow$ q, t; C $ightarrow$ q, r, s, t; D $ightarrow$ q, t

Sigma Percentile · Accepted Answer

### The Symphony of Conservation Laws

In physics, conservation laws are the ultimate cheat codes. They allow us to bypass complex, time-dependent equations and jump straight to profound conclusions about a system's behavior. In this beautiful problem, we are presented with four distinct trajectories of a particle. Our mission is to act as cosmic detectives and determine which physical quantities—linear momentum ($\mathbf{p}$), angular momentum ($\mathbf{L}$), kinetic energy ($K$), potential energy ($U$), and total energy ($E$)—remain invariant as the particle dances through space.

Before we dive into the math, let's establish our ground rules:
- **Linear Momentum ($\mathbf{p}$)** is conserved if and only if the velocity vector $\mathbf{v}$ is strictly constant in both magnitude and direction.
- **Kinetic Energy ($K$)** is conserved if the *speed* $|\mathbf{v}|$ is constant.
- **Angular Momentum ($\mathbf{L}$)** is conserved if the net torque $\boldsymbol{	au} = \mathbf{r} 	imes \mathbf{F}$ is zero. This happens when the force is zero or when it is a central force (parallel or anti-parallel to $\mathbf{r}$).
- **Total Energy ($E$)** is conserved if the force acting on the particle is either zero or conservative.
- **Potential Energy ($U$)** is conserved if both $E$ and $K$ are conserved, since $E = K + U$.

Let's analyze each path one by one.

### Path A: The Straight Line (The Trivial Case)

The position vector is given by:

$$\mathbf{r}(t) = \alpha t \hat{\mathbf{i}} + \beta t \hat{\mathbf{j}}$$

Taking the first derivative with respect to time, we find the velocity:

$$\mathbf{v} = \frac{d\mathbf{r}}{dt} = \alpha \hat{\mathbf{i}} + \beta \hat{\mathbf{j}}$$

Notice that the velocity is a constant vector! Because $\mathbf{v}$ is constant, the linear momentum $\mathbf{p} = m\mathbf{v}$ is conserved. Since the velocity vector doesn't change, its magnitude (speed) is also constant, meaning the kinetic energy $K$ is conserved.

Differentiating velocity gives us acceleration:

$$\mathbf{a} = \frac{d\mathbf{v}}{dt} = 0$$

With zero acceleration, the net force $\mathbf{F}$ is zero. A zero force field is trivially conservative, so the total energy $E$ is conserved. Since both $E$ and $K$ are constant, the potential energy $U$ must also be constant.

Finally, let's check angular momentum:

$$\mathbf{L} = m(\mathbf{r} 	imes \mathbf{v}) = m(\alpha t \hat{\mathbf{i}} + \beta t \hat{\mathbf{j}}) 	imes (\alpha \hat{\mathbf{i}} + \beta \hat{\mathbf{j}}) = 0$$

The cross product of parallel vectors is zero, so $\mathbf{L}$ is constantly zero. Thus, for Path A, **all quantities (p, q, r, s, t) are conserved**.

### Path B: The Elliptical Dance

Here, the position vector traces an ellipse:

$$\mathbf{r}(t) = \alpha \cos \omega t \hat{\mathbf{i}} + \beta \sin \omega t \hat{\mathbf{j}}$$

The velocity is:

$$\mathbf{v} = -\alpha \omega \sin \omega t \hat{\mathbf{i}} + \beta \omega \cos \omega t \hat{\mathbf{j}}$$

The velocity vector is clearly changing with time, so linear momentum $\mathbf{p}$ is **not** conserved. What about speed? The square of the speed is:

$$|\mathbf{v}|^2 = \omega^2 (\alpha^2 \sin^2 \omega t + \beta^2 \cos^2 \omega t)$$

Because we are given that $\alpha 
eq \beta$, this expression fluctuates over time. Therefore, kinetic energy $K$ is **not** conserved.

Let's find the acceleration:

$$\mathbf{a} = -\alpha \omega^2 \cos \omega t \hat{\mathbf{i}} - \beta \omega^2 \sin \omega t \hat{\mathbf{j}} = -\omega^2 \mathbf{r}$$

This is a profound result! The force $\mathbf{F} = -m\omega^2 \mathbf{r}$ is a central force directed towards the origin. All central forces are conservative, so the total energy $E$ is conserved. However, since $K$ fluctuates, the potential energy $U$ must also fluctuate to keep $E$ constant.

What about angular momentum? Since the force is anti-parallel to the position vector $\mathbf{r}$, the torque $\mathbf{r} 	imes \mathbf{F}$ is zero. We can verify this mathematically:

$$\mathbf{L} = m(\mathbf{r} 	imes \mathbf{v}) = m \alpha \beta \omega \hat{\mathbf{k}}$$

This is a constant vector! Thus, for Path B, **only angular momentum and total energy (q, t) are conserved**.

### Path C: The Perfect Circle

This path is a special case of Path B where the semi-major and semi-minor axes are equal:

$$\mathbf{r}(t) = \alpha (\cos \omega t \hat{\mathbf{i}} + \sin \omega t \hat{\mathbf{j}})$$

The velocity is:

$$\mathbf{v} = \alpha \omega (-\sin \omega t \hat{\mathbf{i}} + \cos \omega t \hat{\mathbf{j}})$$

The direction of velocity is constantly changing as the particle moves in a circle, so linear momentum $\mathbf{p}$ is **not** conserved. However, the speed is:

$$|\mathbf{v}| = \alpha \omega$$

This is a constant! Therefore, kinetic energy $K$ is conserved.

Just like in Path B, the acceleration is $\mathbf{a} = -\omega^2 \mathbf{r}$, which means the force is central and conservative. Total energy $E$ is conserved. Since both $E$ and $K$ are constant, the potential energy $U$ must also be constant.

Angular momentum is also conserved for the exact same reason as Path B:

$$\mathbf{L} = m \alpha^2 \omega \hat{\mathbf{k}}$$

Thus, for Path C, **angular momentum, kinetic energy, potential energy, and total energy (q, r, s, t) are conserved**.

### Path D: The Parabolic Arc

Finally, we have a parabolic trajectory:

$$\mathbf{r}(t) = \alpha t \hat{\mathbf{i}} + \frac{\beta}{2} t^2 \hat{\mathbf{j}}$$

The velocity is:

$$\mathbf{v} = \alpha \hat{\mathbf{i}} + \beta t \hat{\mathbf{j}}$$

The velocity vector changes with time, so linear momentum $\mathbf{p}$ is **not** conserved. The speed squared is $|\mathbf{v}|^2 = \alpha^2 + \beta^2 t^2$, which grows with time. Thus, kinetic energy $K$ is **not** conserved.

Differentiating velocity gives the acceleration:

$$\mathbf{a} = \beta \hat{\mathbf{j}}$$

The acceleration is a constant vector! A constant force field (like uniform gravity) is always conservative, meaning the total energy $E$ is conserved. Since $K$ is changing, $U$ must also be changing.

Let's check angular momentum:

$$\mathbf{L} = m(\mathbf{r} 	imes \mathbf{v}) = m\left(\alpha t \hat{\mathbf{i}} + \frac{\beta}{2} t^2 \hat{\mathbf{j}}ight) 	imes (\alpha \hat{\mathbf{i}} + \beta t \hat{\mathbf{j}}) = \frac{1}{2} m \alpha \beta t^2 \hat{\mathbf{k}}$$

The angular momentum depends on $t^2$, so it is **not** conserved. Thus, for Path D, **only total energy (t) is conserved**.

### The Grand Conclusion

By systematically applying the definitions of conservation laws, we have successfully decoded the physics of all four paths:
- **Path A:** p, q, r, s, t
- **Path B:** q, t
- **Path C:** q, r, s, t
- **Path D:** t

This perfectly matches option (a). This problem is a masterclass in understanding how mathematical trajectories dictate the physical realities of energy and momentum!

Select Answer:

Visualized Solution

$Conditions for Conservation$

$Path A: Straight Line$

$Path B: Elliptical Path$

$Path C: Circular Path$

$Path D: Parabolic Path$

$Final Matching$

The Sigma Insight: Torque and Angular Momentum

The Symphony of Conservation Laws

Path A

Path B

Path C

Path D

The Grand Conclusion

Similar Questions

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Select Answer:

Visualized Solution

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

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

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

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

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

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

The Sigma Insight: Torque and Angular Momentum

The Symphony of Conservation Laws

Path A

Path B

Path C

Path D

The Grand Conclusion

Similar Questions

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