MCQ
The reflection surface of a plane mirror is vertical. $A$ particle is projected in a vertical plane which is also perpendicular to the mirror. The initial velocity of the particle is $10\, m/s$ and the angle of projection is $60^o$. The point of projection is at a distance $5\, m$ from the mirror. The particle moves towards the mirror. Just before the particle touches the mirror the velocity of approach of the particle and its image is ......$m/s$
  • $10$
  • B
    $5$
  • C
    $10\sqrt 3 $
  • D
    $5\,\sqrt 3 $

Answer

Correct option: A.
$10$
a
$\left(V_{I, m}\right) x=-\left(V_{o, m}\right) x$

$\left(V_{1}-V_{m}\right) x=-\left(V_{0}-V_{m}\right) x$

$V_{i x}-O=V_{\otimes}$

$V_{i x}=-5 m / s$

$V_{a p p}=5-(-5)=10 m / s$

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