Question

A particle is moving with velocity $(\hat{i}+2\hat{j}+3\hat{k}){\text{ms}}^{-1}$. A plane mirror exists in the $xy-$ plane. The mirror has a velocity $(\hat{i}+2\hat{j}){\text{ms}}^{-1}$. The velocity of the image (in ${\text{ms}}^{-1}$) is-

• A. $\hat{j}+\hat{k}$
• B. $\hat{i}+2\hat{j}-3\hat{k}$
• C. $-\hat{j}-\hat{k}$
• D. None of these
  

Answer 1

The correct option is B $\hat{i}+2\hat{j}-3\hat{k}$

Given,

$\text{Velocity of object}{V}_o=(\hat{i}+2\hat{j}+3\hat{k}){\text{ms}}^{-1}\text{Velocity of the mirror}{V}_m=(\hat{i}+2\hat{j}){\text{ms}}^{-1}$

$\because$ For the plane mirror,

distance of object = distance of image

$\therefore \text{Velocity of object with respect to mirror}=-\left(\text{velocity of image with respect to mirror}\right)$

$\overrightarrow{V_{om}}=-\overrightarrow{V_{Im}}$

$\overrightarrow{V_{Im}}=-\overrightarrow{V_{om}}$

$\overrightarrow{V_I}-\overrightarrow{V_m}=-(\overrightarrow{V_o}-\overrightarrow{V_m})$

$\overrightarrow{V_I}=2\overrightarrow{V_m}-\overrightarrow{V_o}$

$=2(\hat{i}+2\hat{j})-(\hat{i}+2\hat{j}+3\widehat{k)}=\hat{i}+2\hat{j}-3\hat{k}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(B\right)$ is the correct answer.