Question

A ball with a velocity of 5ms-1 impings at angle 60° with the vertical on a smooth horizontal plane.if the cofficient of restitution is 0.5,find the velocity and direction after the impact?

Answer 1

velocity of the ball is 5 m/s just before it collide with the ground at an angle 60 degree with the vertical

now the component of velocity of ball is

$v_x = 5sin60 = 4.33m/s$

$v_y = 5cos60 = 2.5 m/s$

now after collision with the floor the X component of the velocity will remain same while Y component will change

so the velocity components after collision is given as

$v_x' = v_x = 4.33$

$v_y' = e v_y = 0.5 * 2.5 = 1.25 m/s$

now the net velocity is given as

$v_{net} = \sqrt{v_x'^2 + v_y'^2}$

$v_{net} = \sqrt{4.33^2 + 1.25^2}$

$v_{net} = 4.51 m/s$

now the direction of rebound of ball is

$\theta = tan^{-1}\frac{v_x'}{v_y'}$

$\theta = tan^{-1}\frac{4.33}{1.25}$

$\theta = 74 degree$

So the ball will rebound with speed 4.51 m/s at 74 degree with the vertical