Question

Calculate the velocity of the image with respect to a stationary observer in the following situation (Object moving at an angle with mirror and mirror moving at an angle with object).

• A. $5\sqrt{11}m/s$
• B. $5\sqrt{5}m/s$
• C. $5\sqrt{26}m/s$
• D. $5\sqrt{33}m/s$

Answer 1

The correct option is C $5\sqrt{26}m/s$

Lets resolve all the velocities as we discussed in the hint.

So we have four velocities . Let's discuss the effect of each velocity on the velocity of image one by one independently.
Horizontal velocity of object.
We know, horizontal speed of the image is same as that of the object and in opposite direction.
We have,
$\underset{v_{o\left(hor\right)}}{\overset{}{\to }}=5m/s\left(right\right)$
$\underset{v_{i\left(hor\right)}}{\overset{}{\to }}=5m/s\left(left\right)$

Vertical velocity of object.
We know, the vertical velocity of the image is equal to that of the object.
We have,
$\underset{v_{o\left(ver\right)}}{\overset{}{\to }}=5m/s\left(down\right)$
$\underset{v_{i\left(ver\right)}}{\overset{}{\to }}=5m/s\left(down\right)$

Horizontal velocity of mirror.
We know, horizontal speed of the image is two times that of the mirror and in the same direction as of mirror.
We have,
$\underset{v_{m\left(hor\right)}}{\overset{}{\to }}=10m/s\left(left\right)$
$\underset{v_{i\left(hor\right)}}{\overset{}{\to }}=20m/s\left(left\right)$
Vertical velocity of mirror
We know that the vertical velocity of the mirror has no effect on the velocity of image. So this component will add nothing to the velocity of image.
Now, adding in horizontal direction,
$\underset{v_{i\left(hor\right)}}{\overset{}{\to }}=5m/s\left(left\right)+20m/s\left(left\right)=25m/s\left(left\right)$
Adding in vertical direction,
$\underset{v_{i\left(ver\right)}}{\overset{}{\to }}=5m/s\left(down\right)$.

$\overrightarrow{V_{i\left(net\right)}}=\sqrt{(25{)}^2+(5{)}^2}=\sqrt{625+25}=\sqrt{650}=5\sqrt{26}m/s$.