Question

A glass ball collides with a smooth horizontal surface with a velocity $a\hat{i}-b\hat{j}$. If the coefficient of restitution of collision be $e$, find the velocity of the ball just after the collision.

Answer 1

Collision takes place along the normal. Therefore the magnitude normal component $\left(v_y\right)$ of the velocity of the glass ball is changed to $v_y^{\prime }=e{v}_y$ just after the collision whereas the horizontal component $\left(v_x\right)$ of its velocity remains constant due to the absence of any horizontal force
The velocity of the ball just after the impact
$=\overrightarrow{v}+{\overrightarrow{v}}_x+{\overrightarrow{v}}_y\Rightarrow {\overrightarrow{v}}^{\prime }=v_x^{\prime }\hat{i}+v_y^{\prime }\hat{j}$
where $v_x^{\prime }=a;v_y^{\prime }=eb\Rightarrow {\overrightarrow{v}}^{\prime }=a\hat{i}+eb\hat{j}$
Therefore the magnitude of the velocity ${\overrightarrow{v}}^{\prime }=\left|{\overrightarrow{v}}^{\prime }\right|=\sqrt{a^2+e^2{b}^2}$ and the direction is given as
$\theta ={\tan }^{-1}\left(\frac{v_x}{v_y}\right)={\tan }^{-1}\left(\frac{a}{eb}\right)$ to the normal (vertical)