List I describes four systems, each with two particles A and B in relative motion, as shown in figures. List II gives possible magnitudes of their relative velocities (in ) at time [2022]

Question

List I describes four systems, each with two particles A and B in relative motion, as shown in figures. List II gives possible magnitudes of their relative velocities (in ${ms}^{- 1}$ ) at time $t = \frac{π}{3} s .$ [2022]

	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)	$\frac{\sqrt{3} + 1}{2}$
(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 $v = \frac{5 π}{\sqrt{2}} {m s}^{- 1}$ and at $45 °$ from the horizontal plane. The initial separation between A and B is large enough so that they do not collide. $(g = 10 {m s}^{- 2})$	(Q)	$\frac{\sqrt{3} - 1}{\sqrt{2}}$
(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 at $t = 0$ . Take $x_{0} = 1 m, t_{0} = 1 s .$	(R)	$\sqrt{10}$
(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 $3 {m s}^{- 1}$ in the vertical direction as shown in the figure. (Ignore gravity.)	(S)	$\sqrt{2}$
		(T)	$\sqrt{25 π^{2} + 1}$

Which one of the following options is correct?

Smart Achievers · Accepted Answer

(3)

$(I)$ $V_{B A}^{2} = V_{B}^{2} + V_{A}^{2} - 2 V_{B} V_{A} \cos θ$

As $ω_{A} = ω_{B}$ , $θ = 90 °$ remains constant.

Also, $V_{A} = V_{B} = 1 m/s$ $[∵ V = ω R]$

$V_{B A} = \sqrt{2} m/s$

So $I \to S .$

$(II)$ ${\vec{u}}_{A} = \frac{5 π}{2} \hat{i} + \frac{5 π}{2} \hat{j}$

${\vec{V}}_{A} |_{t = 0.1 sec} = \frac{5 π}{2} \hat{i} + (\frac{5 π}{2} - 10 \times 0.1) \hat{j} = \frac{5 π}{2} \hat{i} + (\frac{5 π}{2} - 1) \hat{j}$

${\vec{V}}_{B} |_{t = 0.1 sec} = - \frac{5 π}{2} \hat{i} + \frac{5 π}{2} \hat{j}$

After $ω = 1 {rad s}^{- 1} .$ sec, both projectile came in air. So the relative acceleration is zero. So relative velocity should not change after it.

$V_{rel} = V_{rel} (t = 0.1 sec) =$ $| 5 π \hat{i} - \hat{j} |$ $= \sqrt{25 π^{2} + 1}$

So $II \to T .$

$(III)$ $x = x_{A} - x_{B}$

$= x_{0} \sin t - x_{0} \sin (t + \frac{π}{2})$ $[∵ t_{0} = 1]$

$= \sin t - \cos t = \sqrt{2} (\frac{1}{\sqrt{2}} \sin t - \frac{1}{\sqrt{2}} \cos t)$

$= \sqrt{2} \sin (t - \frac{π}{4})$

$V_{rel} = \frac{d x}{d t} = \sqrt{2} \cos (t - \frac{π}{4})$ $= \sqrt{2} \cos (\frac{π}{3} - \frac{π}{4})$

$= \sqrt{2} \times \frac{\sqrt{3} + 1}{2 \sqrt{2}} = \frac{\sqrt{3} + 1}{2}$

So, $III \to P .$

$(IV)$ ${\vec{V}}_{A} and {\vec{V}}_{B} are always perpendicular$

So, $| {\vec{V}}_{B A} |$ $= \sqrt{V_{A}^{2} + V_{B}^{2}} = \sqrt{3^{2} + 1^{2}} = \sqrt{10} m/s$

So $IV \to R .$

Solution

1 (I) → (R); (II) → (T); (III) → (P); (IV) → (S)

2 (I) → (S); (II) → (P); (III) → (Q); (IV) → (R)

3 (I) → (S); (II) → (T); (III) → (P); (IV) → (R)

4 (I) → (T); (II) → (P); (III) → (R); (IV) → (S)

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