Question

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

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

Change in momentum $=2\times$( $v_2$-$v_1$)$=i-j+2i+2j=3i+j$
Direction of change in momentum makes:
$tan\theta =\frac{y-component}{x-component}=\frac{1}{3}$
$\theta ={\tan }^{-1}\left(\frac{1}{3}\right)$ with positive $x$ axis.

Thus for collision to happen in such a manner the wall has to be oriented at $90+$${\tan }^{-1}\left(\frac{1}{3}\right)$ with positive x axis.