Question

A glass ball collides with a smooth horizontal surface (xz plane) with a velocity $V=ai-bj$. If the coefficient of restitution of collision be e, the velocity of the ball just after the collision will be :

• A. $\sqrt{e^2{a}^2+b^2}$ at angle $ta{n}^{-1}\left(\frac{a}{eb}\right)$ to the vertical
• B. $\sqrt{a^2+e^2{b}^2}$ at angle $ta{n}^{-1}\left(\frac{a}{eb}\right)$ to the vertical
• C. $\sqrt{a^2+\frac{b^2}{e^2}}$ at angle $ta{n}^{-1}\left(\frac{ea}{b}\right)$ to the vertical
• D. $\sqrt{\frac{a^2}{e^2}+b^2}$ at angle $ta{n}^{-1}\left(\frac{a}{eb}\right)$ to the vertical

Answer 1

The correct option is B $\sqrt{a^2+e^2{b}^2}$ at angle $ta{n}^{-1}\left(\frac{a}{eb}\right)$ to the vertical

Let the velocity of the ball after the collision be $v$ making an angle $\theta$ with the vertical

Horizontal component of velocity remains the same after the collision but its vertical component changes from $b\hat{j}$ to $v^{\prime }$

$\frac{v^{\prime }-0}{-b\hat{j}-0}=-e$

$⟹$ $v^{\prime }=be\hat{j}$

Thus velocity of the ball after the collision $\overrightarrow{v}=a\hat{i}+be\hat{j}$

$\therefore$ Magnitude of the velocity $|\overrightarrow{v}|=\sqrt{a^2+(be{)}^2}$

$tan\theta =\frac{a}{v^{\prime }}=\frac{a}{be}$

$⟹\theta =ta{n}^{-1}(\frac{a}{be})$