JEE Challenger
More from Motion in a Plane
PhyHardAdv 2022 • Paper 1

Match Relative Velocity Magnitudes for Different Particle Systems

List I describes four systems, each with two particles AA and BB in relative motion as shown in figures. List II gives possible magnitudes of their relative velocities (in m s−1\text{m s}^{-1}) at time t=π3 st = \frac{\pi}{3}\text{ s}.

Which one of the following options is correct?

Question Diagram 1

Options

A

I→R, II→T, III→P, IV→S\text{I} \rightarrow \text{R},\ \text{II} \rightarrow \text{T},\ \text{III} \rightarrow \text{P},\ \text{IV} \rightarrow \text{S}

B

I→S, II→P, III→Q, IV→R\text{I} \rightarrow \text{S},\ \text{II} \rightarrow \text{P},\ \text{III} \rightarrow \text{Q},\ \text{IV} \rightarrow \text{R}

C

I→S, II→T, III→P, IV→R\text{I} \rightarrow \text{S},\ \text{II} \rightarrow \text{T},\ \text{III} \rightarrow \text{P},\ \text{IV} \rightarrow \text{R}

Correct
D

I→T, II→P, III→R, IV→S\text{I} \rightarrow \text{T},\ \text{II} \rightarrow \text{P},\ \text{III} \rightarrow \text{R},\ \text{IV} \rightarrow \text{S}

Topics & Concepts

Motion in a PlaneRelative MotionKinematicsSHMProjectile Motion

Step-by-Step Solution

To determine the correct matching between List-I and List-II, we analyze each particle system individually at time t=π3 st = \frac{\pi}{3}\text{ s}.

System (I)

Particle AA and particle BB undergo uniform circular motion in the xyxy-plane with radius R=1 mR = 1\text{ m} and angular velocity ω=1 rad s−1\omega = 1\text{ rad s}^{-1}.

  • Initial angular positions at t=0t = 0: θA(0)=0,θB(0)=π2\theta_A(0) = 0, \quad \theta_B(0) = \frac{\pi}{2}

  • Angular positions at time tt: θA(t)=ωt,θB(t)=ωt+π2\theta_A(t) = \omega t, \quad \theta_B(t) = \omega t + \frac{\pi}{2}

  • The velocity vectors for AA and BB are given by: v⃗A=Rω(−sin⁡(ωt)i^+cos⁡(ωt)j^)\vec{v}_A = R\omega \left(-\sin(\omega t)\hat{i} + \cos(\omega t)\hat{j}\right) v⃗B=Rω(−sin⁡(ωt+π2)i^+cos⁡(ωt+π2)j^)=Rω(−cos⁡(ωt)i^−sin⁡(ωt)j^)\vec{v}_B = R\omega \left(-\sin\left(\omega t + \frac{\pi}{2}\right)\hat{i} + \cos\left(\omega t + \frac{\pi}{2}\right)\hat{j}\right) = R\omega \left(-\cos(\omega t)\hat{i} - \sin(\omega t)\hat{j}\right)

  • The relative velocity vector v⃗A/B=v⃗A−v⃗B\vec{v}_{A/B} = \vec{v}_A - \vec{v}_B is: v⃗A/B=Rω[(cos⁡(ωt)−sin⁡(ωt))i^+(cos⁡(ωt)+sin⁡(ωt))j^]\vec{v}_{A/B} = R\omega \left[\left(\cos(\omega t) - \sin(\omega t)\right)\hat{i} + \left(\cos(\omega t) + \sin(\omega t)\right)\hat{j}\right]

  • The magnitude of relative velocity is: ∣v⃗A/B∣=Rω(cos⁡(ωt)−sin⁡(ωt))2+(cos⁡(ωt)+sin⁡(ωt))2=Rω2|\vec{v}_{A/B}| = R\omega \sqrt{(\cos(\omega t) - \sin(\omega t))^2 + (\cos(\omega t) + \sin(\omega t))^2} = R\omega \sqrt{2}

Substituting R=1 mR = 1\text{ m} and ω=1 rad s−1\omega = 1\text{ rad s}^{-1}: ∣v⃗A/B∣=2 m s−1|\vec{v}_{A/B}| = \sqrt{2}\text{ m s}^{-1}

Thus, (I) →\rightarrow (S).

System (II)

Projectiles AA and BB are launched with speed v=5π2 m s−1v = \frac{5\pi}{\sqrt{2}}\text{ m s}^{-1} at an angle of 45∘45^\circ with the horizontal plane.

  • AA is fired at t=0 st = 0\text{ s} towards the right (+x+x direction): uAx=vcos⁡45∘=5π2 m s−1,uAy=vsin⁡45∘=5π2 m s−1u_{Ax} = v \cos 45^\circ = \frac{5\pi}{2}\text{ m s}^{-1}, \quad u_{Ay} = v \sin 45^\circ = \frac{5\pi}{2}\text{ m s}^{-1} v⃗A(t)=5π2i^+(5π2−gt)j^\vec{v}_A(t) = \frac{5\pi}{2}\hat{i} + \left(\frac{5\pi}{2} - gt\right)\hat{j}

  • BB is fired at t=0.1 st = 0.1\text{ s} towards the left (−x-x direction): uBx=−vcos⁡45∘=−5π2 m s−1,uBy=vsin⁡45∘=5π2 m s−1u_{Bx} = -v \cos 45^\circ = -\frac{5\pi}{2}\text{ m s}^{-1}, \quad u_{By} = v \sin 45^\circ = \frac{5\pi}{2}\text{ m s}^{-1} v⃗B(t)=−5π2i^+(5π2−g(t−0.1))j^\vec{v}_B(t) = -\frac{5\pi}{2}\hat{i} + \left(\frac{5\pi}{2} - g(t - 0.1)\right)\hat{j}

  • The relative velocity vector v⃗A/B=v⃗A−v⃗B\vec{v}_{A/B} = \vec{v}_A - \vec{v}_B is: v⃗A/B=(5π2−(−5π2))i^+(−gt+g(t−0.1))j^=5πi^−0.1gj^\vec{v}_{A/B} = \left(\frac{5\pi}{2} - \left(-\frac{5\pi}{2}\right)\right)\hat{i} + \left( -gt + g(t - 0.1) \right)\hat{j} = 5\pi \hat{i} - 0.1g \hat{j}

Given g=10 m s−2g = 10\text{ m s}^{-2}: v⃗A/B=5πi^−1j^\vec{v}_{A/B} = 5\pi \hat{i} - 1 \hat{j}

  • The magnitude of relative velocity is: ∣v⃗A/B∣=(5π)2+(−1)2=25π2+1 m s−1|\vec{v}_{A/B}| = \sqrt{(5\pi)^2 + (-1)^2} = \sqrt{25\pi^2 + 1}\text{ m s}^{-1}

Thus, (II) →\rightarrow (T).

System (III)

Harmonic oscillators AA and BB move along the xx-axis according to: xA(t)=x0sin⁡(tt0),xB(t)=x0sin⁡(tt0+π2)x_A(t) = x_0 \sin\left(\frac{t}{t_0}\right), \quad x_B(t) = x_0 \sin\left(\frac{t}{t_0} + \frac{\pi}{2}\right)

Given x0=1 mx_0 = 1\text{ m} and t0=1 st_0 = 1\text{ s}: xA(t)=sin⁡t,xB(t)=cos⁡tx_A(t) = \sin t, \quad x_B(t) = \cos t

  • Differentiating with respect to time to get velocities: vA(t)=dxAdt=cos⁡t,vB(t)=dxBdt=−sin⁡tv_A(t) = \frac{dx_A}{dt} = \cos t, \quad v_B(t) = \frac{dx_B}{dt} = -\sin t

  • Relative velocity vA/Bv_{A/B}: vA/B=vA−vB=cos⁡t+sin⁡tv_{A/B} = v_A - v_B = \cos t + \sin t

  • At time t=π3 st = \frac{\pi}{3}\text{ s}: vA/B=cos⁡π3+sin⁡π3=12+32=3+12 m s−1v_{A/B} = \cos\frac{\pi}{3} + \sin\frac{\pi}{3} = \frac{1}{2} + \frac{\sqrt{3}}{2} = \frac{\sqrt{3}+1}{2}\text{ m s}^{-1}

Thus, (III) →\rightarrow (P).

System (IV)

  • Particle AA moves in the xyxy-plane in a horizontal circle of radius R=1 mR = 1\text{ m} at constant angular speed ω=1 rad s−1\omega = 1\text{ rad s}^{-1}. Its velocity vector lies in the xyxy-plane with speed: vA=Rω=1⋅1=1 m s−1v_A = R\omega = 1\cdot 1 = 1\text{ m s}^{-1}

  • Particle BB moves upwards along the vertical zz-axis at constant speed: v⃗B=3k^ m s−1\vec{v}_B = 3\hat{k}\text{ m s}^{-1}

  • Since v⃗A\vec{v}_A is in the xyxy-plane and v⃗B\vec{v}_B is along the zz-axis, they are mutually perpendicular (v⃗A⊥v⃗B\vec{v}_A \perp \vec{v}_B). The magnitude of the relative velocity is: ∣v⃗A/B∣=vA2+vB2=12+32=10 m s−1|\vec{v}_{A/B}| = \sqrt{v_A^2 + v_B^2} = \sqrt{1^2 + 3^2} = \sqrt{10}\text{ m s}^{-1}

Thus, (IV) →\rightarrow (R).

Conclusion

The correct combination is: I→S, II→T, III→P, IV→R\text{I} \rightarrow \text{S},\ \text{II} \rightarrow \text{T},\ \text{III} \rightarrow \text{P},\ \text{IV} \rightarrow \text{R}

This corresponds to Option C.

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