Question

Find the velocity of image of a moving particle $\text{O}$ in the situation as shown in the figure.

• A. $\sqrt{164}\text{m/s}$, at ${\tan }^{-1}\left(4/5\right)$ with horizontal
• B. $10\text{m/s}$, at ${53}^{\circ }$ with horizontal
• C. $\sqrt{68}\text{m/s}$, at ${\tan }^{-1}\left(\frac{1}{2}\right)$ with vertical
• D. $\sqrt{104}\text{m/s}$, at ${\tan }^{-1}\left(\frac{1}{5}\right)$ with vertical.

Answer 1

The correct option is A $\sqrt{164}\text{m/s}$, at ${\tan }^{-1}\left(4/5\right)$ with horizontal

Let,
$V_o=$ velocity of object,
$V_i=$ velocity of image,
$V_m=$ velocity of mirror

Given:
$V_{ox}=10\cos {53}^{\circ }=6\text{m/s}$
$V_{oy}=10\sin {53}^{\circ }=8\text{m/s}$
$V_m=2\text{m/s}$

Using the formula of moving plane mirror and object along the optical axis

$V_{ix}=V_{ox}\pm 2V_m$

take positive sign when mirror moving towards object

$V_{ix}=V_{ox}+2V_m$

$V_{ix}=6+2\left(2\right)=10{\text{ms}}^{-1}$

And in vertical direction of moving mirror

$V_{iy}=V_{oy}$

$\Rightarrow V_{iy}=8{\text{ms}}^{-1}$

So, $V_i=\sqrt{V_{ix}^2+V_{iy}^2}=\sqrt{(10{)}^2+8^2}$

$\Rightarrow V_i=\sqrt{164}\text{m/s}$

And, $\tan \alpha =\frac{V_{iy}}{V_{ix}}=\frac{8}{10}=\frac{4}{5}$

$\Rightarrow \alpha ={\tan }^{-1}\left(\frac{4}{5}\right)$ with horizontal.

Hence, option $\left(b\right)$ is correct.