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3.2 , motion in plane — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysics3.2 , motion in plane1 Mark
MCQ
Two particles are moving along two long straight lines, in the same plane, with the same speed $= 20 \,\,cm/s$. The angle between the two lines is $60^o$, and their intersection point is $O$. At a certain moment, the two particles are located at distances $3\,m$ and $4\,m$ from $O$, and are moving towards $O$. Subsequently, the shortest distance between them will be
  • A
    $50 \,\,cm$
  • B
    $40\sqrt 2 \,\, cm$
  • C
    $50\sqrt 2 \,\,cm$
  • ✓
    $50\sqrt 3 \,\,cm$
✓

Answer

Correct option: D.
$50\sqrt 3 \,\,cm$
d
$v_{Q}=-20 \hat{i}$

$v_{P}=-20 \cos 60 \hat{i}-20 \sin 60^{\circ} \hat{j}$

$=-10 \hat{i}-10 \sqrt{3} \hat{j}$

Assuming $P$ to be at rest, $v_{Q P}=v_{Q}-v_{P}=-10 \hat{i}+10 \sqrt{3} \hat{j}$

Now, $\therefore \tan \theta=\frac{10 \sqrt{3}}{10}=\sqrt{3}$ or $\theta=60^{\circ}$

where, theta is the angle of $v_{Q P}$ from $\mathrm{x}$ $-axis$ towards positive $y-axis.$

Shortest distance $=P M=P N \sin 60^{\circ}$

$=(100) \frac{\sqrt{3}}{2}=50 \sqrt{3} \mathrm{cm}$

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