List I describes four systems, each with two particles A and B in relative motion as shown in figure

Q: Why is the angle between velocities in System I always 90 degrees?

Since both particles move on the same circle with the same angular velocity, their relative angular position never changes. They started 90 degrees apart, so their position vectors and consequently their tangent velocity vectors remain 90 degrees apart.

Q: In System II, why is the time for particle B taken as $t - 0.1$?

Particle B is fired 0.1 seconds after particle A. Therefore, at any time $t$ (measured from A's launch), particle B has only been in the air for $t - 0.1$ seconds.

Q: Why do we add the magnitudes of velocities in System III?

At $t = \pi/3$, particle A has a positive velocity (moving right) and particle B has a negative velocity (moving left). Since they are moving in opposite directions, their relative speed is the sum of their individual speeds.

Q: How do we know the velocities in System IV are perpendicular?

Particle A is confined to the horizontal xy-plane, meaning its velocity vector only has x and y components. Particle B moves strictly along the vertical z-axis. Any vector in the xy-plane is orthogonal to the z-axis.

Q: Does the initial separation in System II affect the relative velocity?

No. Relative velocity depends only on the individual velocity vectors at that instant, not on their positions, as long as the acceleration (gravity) is uniform for both.

Question

List I describes four systems, each with two particles A and B in relative motion as shown in figure. List II gives possible magnitudes of their relative velocities (in $	ext{ms}^{-1}$) at time $t = \frac{\pi}{3}	ext{ s}$.

List-I:

P. A and B are moving on a horizontal circle of radius $1	ext{ m}$ with uniform angular speed $\omega = 1	ext{ rad s}^{-1}$. The initial angular positions of A and B at time $t = 0$ are $	heta = 0$ and $	heta = \frac{\pi}{2}$ respectively.

Q. Projectiles A and B are fired (in the same vertical plane) at $t = 0$ and $t = 0.1	ext{ s}$ respectively, with the same speed $v = \frac{5\pi}{\sqrt{2}}	ext{ m s}^{-1}$ and at $45^\circ$ from the horizontal plane. The initial separation between A and B is large enough so that they do not collide, ($g = 10	ext{ m s}^{-2}$).

R. Two harmonic oscillators A and B moving in the x direction according to $x_A = x_0 \sin \frac{t}{t_0}$ and $x_B = x_0 \sin \left(\frac{t}{t_0} + \frac{\pi}{2}ight)$ respectively, starting from $t = 0$. Take $x_0 = 1	ext{ m}, t_0 = 1	ext{ s}$.

S. Particle A is rotating in a horizontal circular path of radius $1	ext{ m}$ on the xy plane, with constant angular speed $\omega = 1	ext{ rad s}^{-1}$. Particle B is moving up at a constant speed $3	ext{ m s}^{-1}$ in the vertical direction as shown in the figure. (Ignore gravity.)

List-II:

1. $\frac{\sqrt{3} + 1}{2}$

2. $\frac{(\sqrt{3} - 1)}{\sqrt{2}}$

3. $\sqrt{10}$

4. $\sqrt{2}$

5. $\sqrt{25\pi^2 + 1}$

Sigma Percentile · Accepted Answer

The beauty of physics often lies in its ability to describe complex, intertwined motions through the elegant language of vectors. This problem is a masterclass in kinematics, presenting us with a quartet of distinct physical systems. Our mission? To determine the magnitude of the relative velocity between two particles, A and B, in each system at a specific instant: $t = \frac{\pi}{3}	ext{ s}$.

Let's embark on this journey and decode the motion, system by system.

### The Circular Chase (System I)

Imagine two particles, A and B, perpetually chasing each other on a horizontal circular track of radius $1	ext{ m}$. They both possess the same uniform angular speed, $\omega = 1	ext{ rad s}^{-1}$.

Because their angular speeds are identical, the angular separation between them is locked in time. They started with A at $	heta = 0$ and B at $	heta = \frac{\pi}{2}$. This $90^\circ$ phase difference means their position vectors are always perpendicular.

In uniform circular motion, the velocity vector is always tangential to the path, meaning it is perpendicular to the position vector. If the position vectors of A and B are at $90^\circ$ to each other, their velocity vectors must also be at $90^\circ$ to each other!

The speed of each particle is simply $v = \omega R = (1)(1) = 1	ext{ m s}^{-1}$. 
To find the magnitude of their relative velocity, we use vector subtraction. Since they are orthogonal, the Pythagorean theorem comes to our rescue:
$$|\vec{v}_{BA}| = \sqrt{v_A^2 + v_B^2} = \sqrt{1^2 + 1^2} = \sqrt{2}	ext{ m s}^{-1}$$
This elegantly matches with option **(S)**.

### The Delayed Duel (System II)

Now, the scene shifts to a vertical plane where two projectiles are fired towards each other. They both launch at a $45^\circ$ angle with a speed of $v = \frac{5\pi}{\sqrt{2}}	ext{ m s}^{-1}$. However, there is a catch: Particle B is fired $0.1	ext{ s}$ *after* Particle A.

Let's establish our initial velocity vectors. 
For A (fired at $t=0$):
$$\vec{u}_A = v\cos 45^\circ \hat{i} + v\sin 45^\circ \hat{j} = \frac{5\pi}{2}\hat{i} + \frac{5\pi}{2}\hat{j}$$

For B (fired at $t=0.1	ext{ s}$ in the opposite direction):
$$\vec{u}_B = -\frac{5\pi}{2}\hat{i} + \frac{5\pi}{2}\hat{j}$$

We need their velocities at $t = \frac{\pi}{3}	ext{ s}$. The horizontal components remain unchanged as there is no acceleration in the x-direction. For the vertical components, gravity ($g = 10	ext{ m s}^{-2}$) pulls them down.

For A, the time in the air is exactly $\frac{\pi}{3}	ext{ s}$:
$$\vec{v}_A = \frac{5\pi}{2}\hat{i} + \left(\frac{5\pi}{2} - 10\left(\frac{\pi}{3}ight)ight)\hat{j} = \frac{5\pi}{2}\hat{i} - \frac{5\pi}{6}\hat{j}$$

For B, we must account for the delay. It has only been flying for $\left(\frac{\pi}{3} - 0.1ight)	ext{ s}$:
$$\vec{v}_B = -\frac{5\pi}{2}\hat{i} + \left(\frac{5\pi}{2} - 10\left(\frac{\pi}{3} - 0.1ight)ight)\hat{j} = -\frac{5\pi}{2}\hat{i} - \left(\frac{5\pi}{6} - 1ight)\hat{j}$$

The relative velocity $\vec{v}_{BA}$ is the vector difference $\vec{v}_B - \vec{v}_A$:
$$\vec{v}_{BA} = \left(-\frac{5\pi}{2} - \frac{5\pi}{2}ight)\hat{i} + \left(-\frac{5\pi}{6} + 1 - \left(-\frac{5\pi}{6}ight)ight)\hat{j} = -5\pi\hat{i} + \hat{j}$$

The magnitude is a straightforward calculation:
$$|\vec{v}_{BA}| = \sqrt{(-5\pi)^2 + 1^2} = \sqrt{25\pi^2 + 1}	ext{ m s}^{-1}$$
This perfectly aligns with option **(T)**.

### The Harmonic Dance (System III)

Next, we observe two particles executing Simple Harmonic Motion (SHM) along the x-axis. We are given their position functions:
$$x_A = \sin t$$
$$x_B = \sin\left(t + \frac{\pi}{2}ight) = \cos t$$

Velocity is the time derivative of position. Let's differentiate!
$$v_A = \frac{dx_A}{dt} = \cos t$$
$$v_B = \frac{dx_B}{dt} = -\sin t$$

At our target time $t = \frac{\pi}{3}	ext{ s}$:
$$v_A = \cos\left(\frac{\pi}{3}ight) = \frac{1}{2}	ext{ m s}^{-1}$$
$$v_B = -\sin\left(\frac{\pi}{3}ight) = -\frac{\sqrt{3}}{2}	ext{ m s}^{-1}$$

Notice the signs. Particle A is moving in the positive x-direction, while Particle B is moving in the negative x-direction. They are moving away from each other! Their relative speed is the sum of their individual speeds:
$$|\vec{v}_{BA}| = |v_B - v_A| = \left|-\frac{\sqrt{3}}{2} - \frac{1}{2}ight| = \frac{\sqrt{3} + 1}{2}	ext{ m s}^{-1}$$
This matches option **(P)**.

### The Orthogonal Ascent (System IV)

Finally, we step into three dimensions. Particle A is confined to a horizontal circular path in the xy-plane, moving with a constant speed $v_A = \omega R = 1	ext{ m s}^{-1}$.

Meanwhile, Particle B is ascending vertically along the z-axis with a constant speed $v_B = 3	ext{ m s}^{-1}$.

This setup is beautifully simple if you visualize it. Any vector lying entirely in the xy-plane is strictly orthogonal (perpendicular) to any vector pointing along the z-axis. Therefore, regardless of where Particle A is on its circular path, its velocity vector $\vec{v}_A$ will always be perpendicular to Particle B's velocity vector $\vec{v}_B$.

Once again, the Pythagorean theorem is our tool of choice for orthogonal vectors:
$$|\vec{v}_{BA}| = \sqrt{v_A^2 + v_B^2} = \sqrt{1^2 + 3^2} = \sqrt{10}	ext{ m s}^{-1}$$
This corresponds to option **(R)**.

### The Grand Finale

By systematically breaking down each physical scenario, applying the core principles of kinematics, and carefully managing our vector mathematics, we have successfully decoded the entire matrix. The final matching stands as:
**(I) $ightarrow$ (S), (II) $ightarrow$ (T), (III) $ightarrow$ (P), (IV) $ightarrow$ (R)**.

This problem is a fantastic reminder that whether particles are spinning, flying, oscillating, or ascending, the fundamental laws of relative motion remain universally powerful.

Animated Solution for Physics - Kinematics: List I describes four systems, each with two particles A and B in relative motion as shown in figure. List II gives possible magnitudes of their relative velocities (in $ms^{- 1}$ ) at time $t = \frac{π}{3} s$ .

List-I

List-II

Select Matching Pairs:

Visualized Solution

Analyzing the Four Systems

System I: Uniform Circular Motion

System I: Relative Velocity

System II: Projectile Motion Initial State

System II: Velocities at $t = \frac{π}{3} s$

System II: Relative Velocity

System III: Simple Harmonic Motion

System III: Relative Velocity at $t = \frac{π}{3} s$

System IV: 3D Kinematics

System IV: Relative Velocity

Final Matrix Match

The Sigma Insight: Relative Velocity

The Circular Chase (System I)

The Delayed Duel (System II)

The Harmonic Dance (System III)

The Orthogonal Ascent (System IV)

The Grand Finale

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