Question

Two particles $A$ & $B$ are projected as shown in fig in x-y plane under the effect of force which provide a constant acceleration $a=11{\text{m/s}}^2$ in negative y- direction. Then match situation in Column-I with the corresponding result in column-II (All positions are given in metre) (${\overrightarrow{V}}_{AB}$=velocity of $A$ with respect to $B$; ${\overrightarrow{r}}_{AB}$ position of $A$ with respect to $B$)

$\begin{array}{cc}\text{Column-I}& \text{Column-II}\\ \text{(A) Seperation between the two particles is minimum at ...sec}& \text{(p)}0\\ \text{(B) Minimum seperation between two particles is ...m}& \text{(q)}0.5\\ \text{(C) Time when velocities of both particles are perpendicular to each other is ....sec}& \text{(r)}0.9\\ \text{(D) At the time of minimum seperation}{\overrightarrow{V}}_{AB}.{\overrightarrow{r}}_{AB}& \text{(s)}2\\ & \text{(t)}2\sqrt{5}\end{array}$

• A. a-t, b-p, c-p,s, d-q
• B. a-p, b-t, c-s, d-q
• C. a-p, b-t, c-p,s, d-q
• D. a-r, b-t, c-p,s, d-p

Answer 1

The correct option is D a-r, b-t, c-p,s, d-p


$\tan \theta =\frac{5}{20}=\frac{1}{4}$
$\tan \varphi =\frac{10}{20}=\frac{1}{2}$
$A,B:$
$d_{min}=\sqrt{425}\sin \alpha =\sqrt{425}\sin (\varphi -\theta )$
$d_{min}=5\sqrt{17}[\sin \varphi \cos \theta -\cos \varphi \sin \theta ]d_{min}=5\sqrt{17}[\frac{1}{\sqrt{5}}\times \frac{4}{\sqrt{17}}-\frac{2}{\sqrt{5}}\times \frac{1}{\sqrt{17}}]d_{min}=\frac{5}{\sqrt{5}}\left[2\right]=2\sqrt{5}$

$t=\frac{\sqrt{425\cos \alpha }}{\sqrt{500}}=\frac{5\sqrt{17}}{10\sqrt{5}}[\frac{2}{\sqrt{5}}\times \frac{4}{\sqrt{17}}+\frac{1}{\sqrt{5}}\times \frac{1}{\sqrt{17}}]\Rightarrow t=\frac{1}{10}[8+1]=\frac{9}{10}\text{sec}$


$C:$
${\overrightarrow{V}}_A=(8\hat{i}+6\hat{j})-\left(11t\right)\hat{j}$
${\overrightarrow{V}}_B=(-12\hat{i}+16\hat{j})-\left(11t\right)\hat{j}$
${\overrightarrow{V}}_A=8\hat{i}+(6-11t)\hat{j}$
${\overrightarrow{V}}_B=-12\hat{i}+(16-11t)\hat{j}$
${\overrightarrow{V}}_A.{\overrightarrow{V}}_B=-96+96-66t-176t+121t^2$
$0=-242t+121t^2=0$
$t=0\text{and}t=\frac{242}{121}=2$

$D$:
At minimum separation relative velocity is perpendicular to displacement, hence dot product is zero.