Question

Velocity of a particle of mass $2\text{kg}$ changes from $\overrightarrow{v_1}=-2\hat{i}-2\hat{j}\text{m/s}$ to $\overrightarrow{v_2}=(\hat{i}-\hat{j})\text{m/s}$ after colliding with a plane surface. Then

• A. The angle made by the plane surface with the positive $x-$axis is ${90}^{\circ }+{\tan }^{-1}\left(\frac{1}{3}\right)$
• B. The angle made by the plane surface with the positive $x-$axis is ${\tan }^{-1}\left(\frac{1}{3}\right)$
• C. The direction of change in momentum makes an angle ${\tan }^{-1}\left(\frac{1}{3}\right)$ with the positive $x-$ axis
• D. The direction of change in momentum makes an angle ${90}^{\circ }+{\tan }^{-1}\left(\frac{1}{3}\right)$ with the plane surface.

Answer 1

Answer: (A), (C)

The correct options are
A The angle made by the plane surface with the positive $x-$axis is ${90}^{\circ }+{\tan }^{-1}\left(\frac{1}{3}\right)$
C The direction of change in momentum makes an angle ${\tan }^{-1}\left(\frac{1}{3}\right)$ with the positive $x-$ axis

Impulse is change in momentum. Hence, impulse
$=2(\overrightarrow{v_2}-\overrightarrow{v_1})=2(3\hat{i}+\hat{j})$
As impulse is in the normal direction of colliding surface
$\tan \theta =\frac{1}{3}$
$\theta ={\tan }^{-1}\left(\frac{1}{3}\right)$
$\alpha ={90}^{\circ }+{\tan }^{-1}\left(\frac{1}{3}\right)$