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 $\alpha \ne \beta$. 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. $p$ is the linear momentum, $L$ 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 List-I with those quantities in List-II, which are conserved for that path. Column-I: (P) $r(t) = \alpha t i + \beta t j$; (Q) $r(t) = \alpha \cos(wt) i + \beta \sin(wt) j$; (R) $r(t) = \alpha(\cos(wt) i + \sin(wt) j)$; (S) $r(t) = \alpha t i + (\beta/2) t^2 j$. Column-II: (1) $p$; (2) $L$; (3) $K$; (4) $U$; (5) $E$.

Step 1: Define the conservation criteria for each physical quantity.

The force acting on the particle is either zero or conservative. The conservation conditions are:

• Linear momentum $\vec{p}$ is conserved if the net force is zero:

$$\vec{F} = 0$$

• Angular momentum $\vec{L}$ about the origin is conserved if the net torque about the origin is zero:

$$\vec{\tau} = \vec{r} \times \vec{F} = 0$$

• Kinetic energy $K$ is conserved if the speed $v$ is constant:

$$K = \frac{1}{2}mv^2 = \text{constant}$$

• Potential energy $U$ is conserved if the position-dependent potential energy remains constant.
• Total energy $E = K + U$ is conserved for any motion under a conservative force.

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Step 2: Analyze Path (P): $\vec{r}(t) = \alpha t\,\hat{i} + \beta t\,\hat{j}$

Find the velocity and acceleration by differentiating the position vector with respect to time:

$$\vec{v} = \frac{d\vec{r}}{dt} = \alpha\,\hat{i} + \beta\,\hat{j}$$ $$\vec{a} = \frac{d\vec{v}}{dt} = 0$$

Since the acceleration is zero, the net force is zero:

$$\vec{F} = 0$$

This leads to the following conservation results:

• $\vec{p} = m\vec{v}$ is constant because velocity is constant.
• $\vec{L} = \vec{r} \times m\vec{v} = (\alpha t\,\hat{i} + \beta t\,\hat{j}) \times m(\alpha\,\hat{i} + \beta\,\hat{j}) = 0$, which is constant.
• $K = \frac{1}{2}m(\alpha^2 + \beta^2)$ is constant.
• Since $\vec{F} = 0$, the potential energy $U$ is constant.
• $E = K + U$ is constant.

Thus, for path (P), the conserved quantities are (1), (2), (3), (4), and (5).

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Step 3: Analyze Path (Q): $\vec{r}(t) = \alpha \cos\omega t\,\hat{i} + \beta \sin\omega t\,\hat{j}$

Find the velocity and acceleration:

$$\vec{v} = -\alpha\omega\sin\omega t\,\hat{i} + \beta\omega\cos\omega t\,\hat{j}$$ $$\vec{a} = -\omega^2(\alpha\cos\omega t\,\hat{i} + \beta\sin\omega t\,\hat{j}) = -\omega^2\vec{r}$$

The force is a central force directed towards the origin:

$$\vec{F} = -m\omega^2\vec{r}$$

This leads to the following conservation results:

• $\vec{p}$ is not conserved because the velocity vector changes in both magnitude and direction.
• $\vec{L}$ is conserved because the torque about the origin is zero:

$$\vec{\tau} = \vec{r} \times \vec{F} = \vec{r} \times (-m\omega^2\vec{r}) = 0$$ $$\vec{L} = \vec{r} \times m\vec{v} = m\alpha\beta\omega\,\hat{k} = \text{constant}$$

• $K = \frac{1}{2}m\omega^2(\alpha^2\sin^2\omega t + \beta^2\cos^2\omega t)$ varies with time because $\alpha \neq \beta$.
• $U = \frac{1}{2}m\omega^2 r^2$ varies with time.
• $E = K + U = \frac{1}{2}m\omega^2(\alpha^2 + \beta^2)$ is constant.

Thus, for path (Q), the conserved quantities are (2) and (5).

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Step 4: Analyze Path (R): $\vec{r}(t) = \alpha(\cos\omega t\,\hat{i} + \sin\omega t\,\hat{j})$

This represents uniform circular motion of radius $\alpha$. Find the velocity and acceleration:

$$\vec{v} = \alpha\omega(-\sin\omega t\,\hat{i} + \cos\omega t\,\hat{j})$$ $$\vec{a} = -\omega^2\vec{r}$$

This leads to the following conservation results:

• $\vec{p}$ is not conserved because the direction of velocity changes continuously.
• $\vec{L} = m\alpha^2\omega\,\hat{k}$ is constant.
• $K = \frac{1}{2}m\alpha^2\omega^2$ is constant because the speed $v = \alpha\omega$ is constant.
• $U = \frac{1}{2}m\omega^2\alpha^2$ is constant because the distance from the origin is constant.
• $E = K + U$ is constant.

Thus, for path (R), the conserved quantities are (2), (3), (4), and (5).

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Step 5: Analyze Path (S): $\vec{r}(t) = \alpha t\,\hat{i} + \frac{\beta}{2}t^2\,\hat{j}$

Find the velocity and acceleration:

$$\vec{v} = \alpha\,\hat{i} + \beta t\,\hat{j}$$ $$\vec{a} = \beta\,\hat{j}$$

The force is constant:

$$\vec{F} = m\beta\,\hat{j}$$

This leads to the following conservation results:

• $\vec{p}$ is not conserved because velocity depends on time.
• $\vec{L}$ is not conserved because the torque about the origin is non-zero:

$$\vec{L} = \vec{r} \times m\vec{v} = m\alpha\beta\frac{t^2}{2}\,\hat{k}$$

• $K = \frac{1}{2}m(\alpha^2 + \beta^2 t^2)$ varies with time.
• $U = -m\beta y = -\frac{1}{2}m\beta^2 t^2$ varies with time.
• $E = K + U = \frac{1}{2}m\alpha^2$ is constant.

Thus, for path (S), the only conserved quantity is (5).

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Answer:

• (P) $\to$ (1, 2, 3, 4, 5)
• (Q) $\to$ (2, 5)
• (R) $\to$ (2, 3, 4, 5)
• (S) $\to$ (5)