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 m   s - 1 ) at time t = π 3   s . List-I List-II (I) A and B are moving on a horizontal circle of radius 1  …

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 $m s^{- 1}$ ) 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)	$\frac{\sqrt{3} + 1}{2}$
(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 from $t = 0$ . 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?

Aicharya · Accepted Answer

(I)
Angular velocity of both the particles is same. Therefore, angle between them will remain $90 °$ .
$|v_{r e l}| = \sqrt{2} ω R = \sqrt{2} m s^{- 1}$
Hence, $I \to S$
(II)
The velocity of A at any time t can be written as
$v_{A} (t) = \frac{5 π}{\sqrt{2}} \cos 45 ° \hat{i} + [\frac{5 π}{\sqrt{2}} \times \sin 45 ° - g \times t] \hat{j}$
$= \frac{5 π}{2} \hat{i} + (\frac{5 π}{2} - g t) \hat{j}$
Velocity of B at any time t
$v_{B} (t) = \frac{- 5 π}{2} \hat{i} + (\frac{5 π}{2} - g (t - 0.1)) \hat{j}$
Relative velocity at time t will be,
${\vec{v}}_{r e l} = 5 π \hat{i} - g \times 0.1 \hat{j}$
$|{\vec{v}}_{r e l}| = \sqrt{25 π^{2} + 1} m s^{- 1}$
Hence, $II \to T$
(III)
Relative position of A wrt B can be written as
$x = x_{A} - x_{B}$ $= x_{0} \sin t - x_{0} \sin (t + \frac{π}{2})$
Using trigonometric relation
$x = \sqrt{2} x_{0} \sin (t - \frac{π}{4})$
Therefore,
$v_{rel} = \frac{d x}{d t} = \sqrt{2} x_{0} \cos (t - \frac{π}{4})$
at $t = \frac{π}{3}$
$v_{r e l} = \sqrt{2} \cos (\frac{π}{3} - \frac{π}{4})$ $= \sqrt{2} \times \frac{\sqrt{3} + 1}{2 \sqrt{2}} = \frac{\sqrt{3} + 1}{2} m s^{- 1}$
Hence, $III \to P$
(IV)
The velocity of A is in $x y$ plane and it is $|v_{A}| = ω r = 1 m s^{- 1}$ and the velocity of B is along $z$ axis and $|v_{B}| = 3 m s^{- 1}$ . Therefore,
$v_{rel} = \sqrt{3^{2} + 1^{2}} = \sqrt{10} m s^{- 1}$
Hence, $IV \to R$

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

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