Question

Two particles $A$ and $B$ initially separated by distance $X$ are projected simultaneously in the directions shown in figure with speed $v_A=20\text{m/s}$ and $v_B=10\text{m/s}$ respectively. They collide in air after $\frac{1}{2}\text{s}.$ Find the distance $X$.

• A. $10\sqrt{5}\text{m}$
• B. $5\sqrt{3}\text{m}$
• C. $4\sqrt{6}\text{m}$
• D. $5\sqrt{2}\text{m}$

Answer 1

The correct option is B $5\sqrt{3}\text{m}$


Solving from frame of reference of B:-


$\overrightarrow{v}{}_{AB}=\overrightarrow{v_A}-\overrightarrow{v_B}$
$=(20\cos \theta \hat{i}+20\sin \theta \hat{j})-10\hat{j}$

Or, $\overrightarrow{v}{}_{AB}=20\cos \theta \hat{i}+(20\sin \theta -10)\hat{j}$

Condition for particles to collide is:
$y-$ component of relative velocity $=0$

$\therefore (v_{AB}{)}_y=0$
$\Rightarrow 20\sin \theta -10=0\Rightarrow \sin \theta =\frac{1}{2}$
$\therefore \theta ={30}^{\circ }$

Since, collision occurs at $t=\frac{1}{2}\text{s}$, applying kinematic equation:

$X=(v_{AB}{)}_x\times t$
$\Rightarrow X=20\cos {30}^{\circ }\times \frac{1}{2}$
$\therefore X=20\times \frac{\sqrt{3}}{2}\times \frac{1}{2}=5\sqrt{3}\text{m}.$