Question

The reflecting surface of a plane mirror is vertical. A particle is projected in a vertical plane which is also perpendicular to plane of the mirror. The initial velocity of the particle is $10 m / s$ and the angle of projection is $60^{\circ} .$ 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:

• A. $10 m / s$
• B. $5 m / s$
• C. $10 \sqrt{3} m / s$
• D. $5 \sqrt{3} m / s$

Answer 1

Answer: (A)

Horizontal component of the velocity is $10 \cos \left(60^{\circ}\right)=5$
And vertical component is $10 \sin 60^{\circ}=8.66$
The particle will touch the mirror after $=\frac{5}{5}=1 s$
In this time vertical velocity will be $v = u - gt$
$=8.66-9.8 \times 1=-1.14$
Now the relative velocity will be due to horizontal component of the particle.
Since the vertical velocity will be same
So, the relative velocity of the particle is $2 \times 5=10 m / s$.