JEE Advanced 2022 — Paper 1 — PHY-III solution

JEE Advanced2022Paper 1PHY-III

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

List-I	List-II
(I) $A$ and $B$ are moving on a horizontal circle of radius 1 m with uniform angular speed $ω = 1$ rad s $^{- 1}$ . The initial angular positions of $A$ and $B$ at time $t = 0$ are $θ = 0$ and $θ = \frac{π}{2}$ , respectively.	(P) $\frac{3 + 1}{2}$
(II) Projectiles $A$ and $B$ are fired (in the same vertical plane) at $t = 0$ and $t = 0.1$ s respectively, with the same speed $v = \frac{5 π}{2}$ ms $^{- 1}$ and at $4 5^{\circ}$ from the horizontal plane. The initial separation between $A$ and $B$ is large enough so that they do not collide ( $g = 10$ ms $^{- 2}$ ).	(Q) $\frac{3 - 1}{2}$
(III) Two harmonic oscillators $A$ and $B$ moving in the $x$ direction according to $x_{A} = x_{0} sin \frac{t}{t _{0}}$ and $x_{B} = x_{0} sin (\frac{t}{t _{0}} + \frac{π}{2})$ respectively, starting from $t = 0$ . Take $x_{0} = 1$ m, $t_{0} = 1$ s.	(R) $10$
(IV) Particle $A$ is rotating in a horizontal circular path of radius 1 m on the $x y$ plane, with constant angular speed $ω = 1$ rad s $^{- 1}$ . Particle $B$ is moving up at a constant speed 3 ms $^{- 1}$ in the vertical direction as shown in the figure. (Ignore gravity.)	(S) $2$
	(T) $25 π^{2} + 1$

Which one of the following options is correct?

A
I $\to$ R, II $\to$ T, III $\to$ P, IV $\to$ S
B
I $\to$ S, II $\to$ P, III $\to$ Q, IV $\to$ R
C
I $\to$ S, II $\to$ T, III $\to$ P, IV $\to$ R
D
I $\to$ T, II $\to$ P, III $\to$ R, IV $\to$ S

Solution

(I) Both particles move on the same circle with the same speed $ω R = 1$ m/s but their velocity vectors are perpendicular (since their positions differ by $π /2$ always). Relative speed = $v^{2} + v^{2} = 2$ m/s. I $\to$ S.

(II) Both projectiles experience the same gravitational acceleration, so the relative velocity is constant in time once both have launched (for $t \geq 0.1$ s). At $t = 0.1$ s (just after $B$ launches), velocities are: $v_{A} = \frac{5 π}{2} cos 4 5^{\circ} \hat{i} + (\frac{5 π}{2} sin 4 5^{\circ} - g \cdot 0.1) \hat{j} = \frac{5 π}{2} \hat{i} + (\frac{5 π}{2} - 1) \hat{j}$ , and $v_{B} = - \frac{5 π}{2} \hat{i} + \frac{5 π}{2} \hat{j}$ (oppositely directed horizontal component). Relative velocity $v_{A} - v_{B} = 5 π \hat{i} - \hat{j}$ , magnitude $25 π^{2} + 1$ . II $\to$ T.

(III) $x_{A} - x_{B} = x_{0} [sin (t / t_{0}) - cos (t / t_{0})] = 2 x_{0} sin (t / t_{0} - π /4)$ . Relative velocity $= \frac{d}{d t} (x_{A} - x_{B}) = 2 cos (t - π /4)$ . At $t = π /3$ : $2 cos (π /3 - π /4) = 2 \cdot \frac{3 + 1}{2 2} = \frac{3 + 1}{2}$ . III $\to$ P.

(IV) $A$ 's speed is $ω R = 1$ m/s (horizontal); $B$ 's speed is 3 m/s (vertical). Velocities are perpendicular. Relative speed $= 1^{2} + 3^{2} = 10$ . IV $\to$ R.

The correct matching is option (C).

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