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Physics · Motion in Plane

JEE Advanced 2022 — Paper 1 — Question 32

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 ms−1\mathrm{m} \mathrm{s}^{-1} ) at time

t=π3 s\mathrm{t}=\frac{\pi}{3} \mathrm{~s}.

List -IList -II
(I)A and B are moving on a horizontal circle of radius 1 m with uniform angular speed ω=1rads−1\omega =1\text{rad}{{\text{s}}^{-1}}. The initial angular positions of A and B at time t=0\text{t}=0 are θ=0\theta =0 and θ=π2\theta =\frac{\pi }{2}, respectively. figure(P)3+12\frac{\sqrt{3}+1}{2}
(II)Projectiles A and B are fired (in the same vertical plane) at t =0=0 and t=0.1  ⁣ ⁣  ⁣ ⁣ s\text{t}=0.1\text{ }\!\!~\!\!\text{ s} respectively, with the same speed v=5π2\text{v}=\frac{5\pi }{\sqrt{2}} ms−1\text{m}{{\text{s}}^{-1}} and at 45∘{{45}^{\circ }} from the horizontal plane. The initial separation between AA and BB is large enough so that they do not collide. (g=10  ⁣ ⁣  ⁣ ⁣ m  ⁣ ⁣  ⁣ ⁣ s−2)\left( \text{g}=10\text{ }\!\!~\!\!\text{ m }\!\!~\!\!\text{ }{{\text{s}}^{-2}} \right). figure(Q)(3−1)2\frac{\left( \sqrt{3}-1 \right)}{\sqrt{2}}
(III)Two harmonic oscillators A and B moving in the x direction according to xA=x0sintt0{{\text{x}}_{\text{A}}}={{\text{x}}_{0}}\text{sin}\frac{\text{t}}{{{\text{t}}_{0}}} and xB=x0sin(tt0+π2){{\text{x}}_{\text{B}}}={{\text{x}}_{0}}\text{sin}\left( \frac{\text{t}}{{{\text{t}}_{0}}}+\frac{\pi }{2} \right) respectively, starting from t=0t=0. Take x0=1  ⁣ ⁣  ⁣ ⁣ m,t0=1  ⁣ ⁣  ⁣ ⁣ s{{x}_{0}}=1\text{ }\!\!~\!\!\text{ m},{{\text{t}}_{0}}=1\text{ }\!\!~\!\!\text{ s}. figure(R)10\sqrt{10}
Particle A is rotating in a horizontal circular path of radius 1 m on the xy plane, with constant angular speed w = 1 rad s–1. Particle B is moving up at a constant speed 3 m s–1 in the vertical direction as shown in the figure. (Ignore gravity.) figure(S)2\sqrt{2}
25π2+1\sqrt{25{{\pi }^{2}}+1}
  1. Option A:

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

  2. Option B:

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

  3. Option C:

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

    Correct
  4. Option D:

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

Answer: C

Step-by-step solution

ω\omega is same, therefore angle between velocity vectors remain same.

Vrel =12+12=2 m/s\mathrm{V}_{\text {rel }}=\sqrt{1^{2}+1^{2}}=\sqrt{2} \mathrm{~m} / \mathrm{s}

⇒v⃗1=5π2i^+5π3j^\Rightarrow \quad \vec{v}_{1}=\frac{5 \pi}{2} \hat{i}+\frac{5 \pi}{3} \hat{j}

v⃗1=−5π2i^+(5π3+1)j^\vec{v}_{1}=-\frac{5 \pi}{2} \hat{i}+\left(\frac{5 \pi}{3}+1\right) \hat{j}

vrel=∣∣v⃗2−v⃗1∣=25π2+1\mathrm{v}_{\mathrm{rel}}=\left|\left|\vec{v}_{2}-\vec{v}_{1}\right|=\sqrt{25 \pi^{2}+1}\right.

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2022
Paper
Paper 1
Subject
Physics
Chapter
Motion in Plane
Topic
Relative Motion in One Dimension

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