Question

A plane mirror is moving with velocity $(-2\hat{i}+3\hat{j}+\hat{k})m/s$, which lies in $x-z$ plane. Find the velocity of image of a point object which is moving with velocity $(\hat{i}+2\hat{j}-5\hat{k})m/s$ (assume that object is located on front side of the mirror).

• A. $(-2\hat{i}+4\hat{j}-\hat{k})m/s$
• B. $(\hat{i}+4\hat{j}-5\hat{k})m/s$
• C. $(2\hat{i}-3\hat{j}+\hat{k})m/s$
• D. $(-\hat{i}+5\hat{j}-4\hat{k})m/s$

Answer 1

Answer: (B)

$(\hat{i}+4\hat{j}-5\hat{k})m/s$

$\begin{array}{c}{V}_m=-2\hat{i}+3\hat{j}+\hat{k}\\ \text{Position of plane mirror}=\frac{(\hat{i}+\hat{z})}{\sqrt{2}}\\ v_0=\hat{i}+2\hat{j}-5\hat{k}\\ \text{normal of plane mirror}=\hat{ȷ}\end{array}$

$\begin{array}{c}\text{The velocity of image remains same as}\\ \text{object in the directions which are not}\\ \text{parallel to normal of the mirror.}\\ \text{so velocity of object}=\hat{i}+x\hat{j}-5\hat{k}\\ \text{In the direction of normal to the mirror}\\ \begin{array}{c}{V}_{OM}=-V_{IM}\\ V_0-V_M=-(V_I-V_M)\end{array}\end{array}$

$\begin{array}{cc}{V}_M-V_0+V_M& =V_C\\ V_3& =2V_M-V_0\\ & =2\left(3\right)-2\\ & =4j\\ \text{So}x& =4\end{array}$

so velocity of image $=\hat{i}+4\hat{j}-5\hat{k}$

so option $B$ is correct