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 $m{s}^{-1}$ ) at time $t=\frac{\pi }{3}s$.

List I		List II
I	$A$ and $B$ are moving on a horizontal circle of radius $1m$ with uniform angular speed $\omega =1rad{s}^{-1}$. 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 $t=0.1s$ respectively, with the same speed $v=\frac{5\pi }{\sqrt{2}}m{s}^{-1}$ and at $45^{\circ }$ from the horizontal plane. The initial separation between $A$ and $B$ is large enough so that they do not collide, $\left(g=10m{s}^{-2}\right)$.	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 $x_0=1m,t_0=1s$.	R	$\sqrt{10}$
IV	Particle $A$ is rotating in a horizontal circular path of radius $1m$ on the $xy$ plane, with constant angular speed $\omega =1rad{s}^{-1}$. Particle $B$ is moving up at a constant speed $3m{s}^{-1}$ 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. $I\to R,II\to T,III\to P,IV\to S$
• 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. $I\to T,II\to P,III\to R,IV\to S$

Answer 1

Answer: (C)

(I) $v_{BA}^2=v_A^2+v_B^2-2v_{AB}\cos \theta$
As ${\omega }_A={\omega }_B,\theta =90^{\circ }$ remains constant.
Also, $v_A=v_B=1m/s$
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) $x_A=\sin t$
$v_A=\cos t=\frac{1}{2}m/s$
$x_B=\mathrm{cost}$
$v_B=-\sin t=-\frac{\sqrt{3}}{2}m/s$
$v_{BA}=-\frac{\sqrt{3}}{2}-\frac{1}{2}$
(IV) ${\vec{v}}_A\&{\vec{v}}_{\underline{B}}$ are always perpendicular
So, $\left|{\overrightarrow{v}}_{BA}\right|=\sqrt{v_A^2+v_B^2}=\sqrt{10}m/s$