We check, for each given trajectory, which quantities must remain conserved.

Given that the force is either zero or conservative:

• If force is zero, then $\vec p, \vec L, K, U, E$ are all conserved (with $U=$ constant).
• If force is conservative, then total energy $E=K+U$ is conserved.
• Angular momentum $\vec L$ is conserved only if torque about origin is zero, i.e. force is central: $\vec F \parallel \vec r$.
• Linear momentum $\vec p$ is conserved only if net force is zero.
• Kinetic energy $K$ is conserved if speed is constant.
• Potential energy $U$ is conserved if position-dependence gives constant value along the motion.

1. Path $P$

$\vec r(t)=\alpha t\,\hat i+\beta t\,\hat j$

Step 1: Velocity and acceleration

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

So net force is zero.

Step 2: Conserved quantities

• $\vec p$ conserved
• $\vec L$ conserved since $\vec \tau=\vec r\times \vec F=0$
• $K=\frac12 m v^2$ conserved
• $U$ constant (zero-force case)
• $E$ conserved

Thus, $P\to 1,2,3,4,5$

2. Path $Q$

$\vec r(t)=\alpha \cos \omega t\,\hat i+\beta \sin \omega t\,\hat j$

This is an ellipse.

Step 1: Velocity

$\vec v = -\alpha\omega\sin\omega t\,\hat i+\beta\omega\cos\omega t\,\hat j$

Speed squared: $v^2=\alpha^2\omega^2\sin^2\omega t+\beta^2\omega^2\cos^2\omega t$

Since $\alpha\neq \beta$, this is not constant in general. Hence $K$ is not conserved.

Step 2: Acceleration

$\vec a = -\alpha\omega^2\cos\omega t\,\hat i-\beta\omega^2\sin\omega t\,\hat j=-\omega^2\vec r$

So force is $\vec F = m\vec a=-m\omega^2\vec r$

This is a central conservative force (harmonic oscillator type).

Step 3: Conserved quantities

• $\vec p$ not conserved because $\vec F\neq 0$
• $\vec L$ conserved because $\vec \tau=\vec r\times \vec F=0$
• $K$ not conserved
• $U=\frac12 m\omega^2 r^2$; here $r^2=\alpha^2\cos^2\omega t+\beta^2\sin^2\omega t$ not constant, so $U$ not conserved
• $E$ conserved because force is conservative

Thus, $Q\to 2,5$

3. Path $R$

$\vec r(t)=\alpha(\cos\omega t\,\hat i+\sin\omega t\,\hat j)$

This is circular motion of radius $\alpha$.

Step 1: Velocity

$\vec v=\alpha\omega(-\sin\omega t\,\hat i+\cos\omega t\,\hat j)$

Hence $v^2=\alpha^2\omega^2$

So speed is constant, therefore $K$ is conserved.

Step 2: Acceleration

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

Again force is central conservative.

Step 3: Conserved quantities

• $\vec p$ not conserved because direction changes
• $\vec L$ conserved (central force)
• $K$ conserved (constant speed)
• $U=\frac12 m\omega^2 r^2=\frac12 m\omega^2\alpha^2=$ constant, since $r=\alpha$ constant
• $E$ conserved

Thus, $R\to 2,3,4,5$

4. Path $S$

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

Step 1: Velocity and acceleration

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

So force is constant: $\vec F=m\beta\hat j$

A constant force is conservative, with potential $U(y)=-m\beta y + C$

Step 2: Angular momentum

Check torque: $\vec \tau=\vec r\times \vec F=(\alpha t\,\hat i+\tfrac{\beta}{2}t^2\,\hat j)\times (m\beta\hat j)$ $\vec \tau = m\alpha\beta t\,\hat k \neq 0$

So $\vec L$ is not conserved.

Step 3: Other quantities

• $\vec p$ not conserved since force is nonzero
• $K$ not conserved since speed changes: $v^2=\alpha^2+\beta^2 t^2$
• $U$ not conserved since $y=\frac{\beta}{2}t^2$ changes with time
• $E$ conserved since force is conservative

Thus, $S\to 5$

Final matching

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

This matches Option A.

Comparison with stored answer

Stored correct answer: A

My derived answer: A

So they agree.