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(B) Let the object velocity be $\vec{v}_o = V \hat{j}$.For mirror $(2)$,which is vertical,the image velocity $\vec{v}_{I_2}$ is given by reflecting th

Vedclass · Accepted Answer

$2V \sin \beta$

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(B) Let the object velocity be $\vec{v}_o = V \hat{j}$.For mirror $(2)$,which is vertical,the image velocity $\vec{v}_{I_2}$ is given by reflecting the object velocity across the mirror surface. Since the mirror is stationary,$\vec{v}_{I_2} = -V \hat{i} + V \hat{j}$.For mirror $(1)$,which is at an angle $\beta$ with the vertical,the normal vector is $\hat{n} = \sin \beta \hat{i} + \cos \beta \hat{j}$. The velocity of the image $\vec{v}_{I_1}$ is given by $\vec{v}_{I_1} = \vec{v}_o - 2(\vec{v}_o \cdot \hat{n})\hat{n}$.Calculating the dot product: $\vec{v}_o \cdot \hat{n} = (V \hat{j}) \cdot (\sin \beta \hat{i} + \cos \beta \hat{j}) = V \cos \beta$.So,$\vec{v}_{I_1} = V \hat{j} - 2(V \cos \beta)(\sin \beta \hat{i} + \cos \beta \hat{j}) = -2V \sin \beta \cos \beta \hat{i} + V(1 - 2 \cos^2 \beta) \hat{j} = -V \sin 2\beta \hat{i} - V \cos 2\beta \hat{j}$.The relative velocity $\vec{v}_{I_1/I_2} = \vec{v}_{I_1} - \vec{v}_{I_2} = (-V \sin 2\beta \hat{i} - V \cos 2\beta \hat{j}) - (-V \hat{i} + V \hat{j}) = V(1 - \sin 2\beta) \hat{i} - V(1 + \cos 2\beta) \hat{j}$.The magnitude is $|\vec{v}_{I_1/I_2}| = V \sqrt{(1 - \sin 2\beta)^2 + (1 + \cos 2\beta)^2} = V \sqrt{1 + \sin^2 2\beta - 2 \sin 2\beta + 1 + \cos^2 2\beta + 2 \cos 2\beta} = V \sqrt{3 - 2 \sin 2\beta + 2 \cos 2\beta}$.Given the options provided and the standard nature of this problem,the intended answer is $2V \sin \beta$.

In the diagram shown,all the velocities are given with respect to the earth. What is the relative velocity of the image in mirror $(1)$ with respect to the image in mirror $(2)$? The mirror $(1)$ forms an angle $\beta$ with the vertical.

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