Question

A plane mirror is placed along the vertical plane (yz-plane). If a particle is also projected in the vertical plane with speed of projection $10\text{m/s}$ and angle of projection as ${60}^{\circ }$ from the normal direction to mirror. The point of projection is at a distance of $5\text{m}$ from the mirror, and the particle is directed towards the mirror. The magnitude of velocity of approach of particle and image, just before the particle touches the mirror is

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

Answer 1

The correct option is A $10\text{m/s}$

The normal for plane mirror will be along x-direction. ($\therefore$ mirror is in yz plane)


For object, resolving the velocity and applying
$(v_{o,m}{)}_x=(v_{i,m}{)}_x$


The y-component of velocity of object and image will be same at all time.
Thus, relative velocity, $(v_{o,i}{)}_{xy}=0$
Velocity of approach for image and object will be,
$|v_{approach}|=|(v_{o,i}{)}_x|=|10\cos {60}^{\circ }-(-10\cos {60}^{\circ })|$
$\Rightarrow |v_{approach}|=10\cos {60}^{\circ }+10\cos {60}^{\circ }$
$\therefore |v_{approach}|=20\cos {60}^{\circ }=20\times \frac{1}{2}=10\text{m/s}$