Q. A-6. A point object is placed at (0, 0) and a plane mirror 'M' is placed, inclined 30$°$ with the x axis.
(a) Find the position of image.
(b) If the object starts moving with velocity 1 $\hat{i}$ m/s and the mirror is fixed find the velocity of image.

Dear Student,

Please find below the solution to the asked query:

When a object moves in front of mirror, the component of velocity of the body, which is parallel to the mirror will remain same and the perpendicular component will be inverted. Here given that object is moving in X - direction with a velocity.
So,
$$\begin{gathered} V_{\left|\right|} = V \cos {30}^0 = 1 \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2} m/s \\ {V}_{perpendicular} = V \sin {30}^{30} = 1\times \frac{1}{2} = \frac{1}{2} m/s \end{gathered}$$
Now image will be having velocities as shown in the below diagram.

So, image will be having two components as shown in the figure. Therefore, the image velocity is the resultant of these two velocities.
$$\begin{gathered} \overrightarrow{V_{i,\left|\right|}} = \frac{\sqrt{3}}{2} m/s = \frac{\sqrt{3}}{2}\left(\cos {30}^0 \hat{i}+\sin {30}^0 \hat{j}\right) m/s = \left(\frac{3}{4} \hat{i}+\frac{\sqrt{3}}{4} \hat{j}\right) m/s \\ \overrightarrow{V_{i,per.}} = - \frac{1}{2} m/s = \frac{1}{2}\left(- cos {60}^0 \hat{i}+sin {60}^0 \hat{j}\right) m/s = \left(-\frac{1}{4} \hat{i}+\frac{\sqrt{3}}{4} \hat{j}\right) m/s \\ Therefore, \\ \overrightarrow{V_i } = \overrightarrow{V_{i,\left|\right|}} + \overrightarrow{V_{i,per.}} = \left(\frac{3}{4} \hat{i}+\frac{\sqrt{3}}{4} \hat{j}\right) m/s + \left(-\frac{1}{4} \hat{i}+\frac{\sqrt{3}}{4} \hat{j}\right) m/s = \left(\frac{1}{2} \hat{i}+\frac{\sqrt{3}}{2} \hat{j}\right) m/s \\ \Rightarrow \overrightarrow{V_i } = \left(\cos {60}^0 \hat{i}+\sin {60}^0 \hat{j}\right) m/s \end{gathered}$$

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