Question

An object $O$ is just about to strike a perfectly reflecting inclined plane of inclination ${37}^o$. Its velocity is $5m/s$. Find the velocity of its image

• A. $3\hat{i}+4\hat{j}$
• B. $4\hat{i}+3\hat{j}$
• C. $4.8\hat{i}+1.4\hat{j}$
• D. $1.4\hat{i}+4.8\hat{j}$

Answer 1

The correct option is C $4.8\hat{i}+1.4\hat{j}$

$\begin{array}{c}\text{The image speed is equal to the object speed. Theretore,}\\ \text{the image speed would also be}5{ms}^{-1}\text{.}\end{array}$

Here, the object with a velocity $5{ms}^{-1}$ falls vertically, Whose image is formed at an agle same as the incident angle of the object i.e (37), which is considered with the normal of the inclined plane.

In Order to determme the speed of the image in terms of i and $j$, let us consider the $x$ -axis and $y$ -axis a we have,

$\begin{array}{cc}{37}^{\circ }+{37}^{\circ }+\theta & ={90}^{\circ }\\ \theta & =90-74={16}^{\circ }\end{array}$

$\begin{array}{c}\text{The velocity of the image along the}x\text{-axis is}\\ \text{inclined by}{16}^{\circ }\text{from the actual direction of the}\\ \text{image.}\\ \text{with regard i and}j\text{,}\\ \begin{array}{cc}{V}_I& =\sqrt{\cos {16}^{\circ }}+\sqrt{\sin }{16}^{\circ }\hat{j}\\ & =5\left(0.96\right)\hat{ı}+5\left(0.28\right)\hat{j}\\ & =(4.8\hat{i}+1.4\hat{ȷ})\cdot m{s}^{-1}\end{array}\end{array}$

$\text{Hence, option (c) is correct.}$