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 then relative velocities (in $ms ^{-1}$ ) at time $t=\frac{\pi}{3} s$.

List I		List II
I	$A$ and $B$ are moving on a horizontal circle of radius $1 m$ with uniform angular speed $\omega=1 rad s ^{-1}$. The initial angular positions of $A$ and $B$ at time $t=0$ are $\theta=0$ and $\theta=\frac{\pi}{2}$ 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 $v=\frac{5 \pi}{\sqrt{2}} ms ^{-1}$ and at $45^{\circ}$ from the horizontal plane. The initial separation between $A$ and $B$ is large enough so that they do not collide, $\left(g=10 m s ^{-2}\right)$.	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 $x_B=x_0 \sin \left(\frac{t}{t_0}+\frac{\pi}{2}\right)$ 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 $\omega=1 rad s ^{-1}$. 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	$\sqrt{25 \pi^2+1}$

Which one of the following options is correct?

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

Answer 1

Answer: (C)

(I) $v_{B A}^2=v_A^2+v_B^2-2 v_{A B} \cos \theta$
As $\omega_{ A }=\omega_{ B }, \theta=90^{\circ}$ remains constant.
Also, $v _{ A }= v _{ B }=1 \,m / s$
So, $v _{ BA }=\sqrt{2} m / s$
(II) $\overrightarrow{ u }_{ A }=\frac{5 \pi}{2} \hat{ i }+\frac{5 \pi}{2} \hat{ j }$
$\overrightarrow{ v }_{ A }=\frac{5 \pi}{2} \hat{ i }+\left(\frac{5 \pi}{2}-10 \cdot \frac{\pi}{3}\right) \hat{ j }$
$=\frac{5 \pi}{2} \hat{ i }-\frac{5 \pi}{6} \hat{ j }$
$\overrightarrow{ u }_{ B }=-\frac{5 \pi}{2} \hat{ i }+\frac{5 \pi}{2} \hat{ j }$
$ \overrightarrow{ u }_{ B }=-\frac{5 \pi}{2} \hat{ i }-\left(\frac{5 \pi}{6}+1\right) \hat{ j }$
$ \overrightarrow{ v }_{ B , A }=-5 \pi \hat{ i }-\hat{ j } $
$ v _{ BA }=\sqrt{25 \pi^2+1}$
(III) $x _{ A }=\sin t$
$ v _{ A }=\cos t =\frac{1}{2} m / s $
$ x _{ B }=\operatorname{cost} $
$ v _{ B }=-\sin t =-\frac{\sqrt{3}}{2} m / s$
$ v _{ BA }=-\frac{\sqrt{3}}{2}-\frac{1}{2}$
(IV) $\vec{v}_{ A }\, \&\, \vec{v}_{\underline{B}}$ are always perpendicular
So, $\left|\overrightarrow{ v }_{ BA }\right|=\sqrt{ v _{ A }^2+ v _{ B }^2}=\sqrt{10}\, m / s$