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

Analyze the given condition and find the horizontal and vertical component of velocity and find the magnitude of the velocity after the collision by differentiating the obtained horizontal and vertical component of the velocity. Find the direction of velocity by using the formula of it.

Useful formula:

The formula of the direction of the velocity is given by

$\theta = {\tan ^{ - 1}}\left( {\dfrac{{{v_x}}}{{{v_y}}}} \right)$

Where $\theta$ is the direction(angle), ${v_x}$ is the horizontal component and ${v_y}$ is the vertical component of velocity.

Complete answer:

It is given that the,

Velocity at which the glass ball collides the smooth horizontal surface, $v = a\hat i - b\hat j$

The coefficient of the restitution collision is $e$

It is given that the glass ball collides with the horizontal surface, it means the collision takes place in the vertical direction normal to the plane. Hence when the velocity of the gall ball is separated into the horizontal and the vertical component, only the vertical component changes and the horizontal component remains the same. That is the normal component of the velocity ${v_y}$ is changed to the $e{v_y}$ and the horizontal component of the velocity ${v_x}$ remains the same.

The velocity of the ball at the time of the collision is considered as $\vec v$ .

$\vec v = {\vec v_x} + {\vec v_y}$

The velocity of the ball at the time of the collision is considered as $\vec v'$ .

$\vec v' = {\vec v'_x}\hat i + {\vec v'_y}\hat j$

From the given data, ${v'_x} = a$ and ${v'_y} = eb$ , since only the vertical component changes with coefficient of the restitution collision.

$\vec v' = a\hat i + eb\hat j$

Since the velocity is the vector quantity, it possesses both the magnitude and the direction. The magnitude of the velocity is found as follows.

$v = \left| {\vec v'} \right| = \sqrt {{a^2} + {{\left( {eb} \right)}^2}}$...... (1)

The direction of the velocity is calculated as follows.

$\theta = {\tan ^{ - 1}}\left( {\dfrac{{{v_x}}}{{{v_y}}}} \right)$

Substituting known values we get,

$\theta = {\tan ^{ - 1}}\left( {\dfrac{a}{{eb}}} \right)$....(2)

The above direction of the velocity is normal to the horizontal surface that is vertical.

Therefore, the velocity of the ball just after the collision will be $\sqrt {{a^2} + {{\left( {eb} \right)}^2}}$ at angle ${\tan ^{ - 1}}\left( {\dfrac{a}{{eb}}} \right)$ to the vertical.

Note: If any of the force or any other factor acts inclined to any plane it must be separated into the horizontal and the vertical component to solve the problem. In this problem it is given that the glass ball hit the smooth horizontal surface, so the direction must be vertical and since it is smooth the collision is elastic in nature.