Question

A ball with a velocity of 5 m/s impinge at angle of 60 degree with the vertical on a smooth horizontal plane. if the coefficient of restitution is 0.5 find the velocity and direction after the impact

Answer 1

speed of the ball is 5 m/s at 60 degree with the vertical

now the components of the velocity of the ball in x and y directions are given as

$v_x = 5 sin60 = 4.33 m/s$

$v_y = 5 cos60 = 2.5 m/s$

so it collide with the floor and rebound inelastically

due to this inelastic collision its velocity will remain the same along x direction while in y direction it will rebound with other velocity

so it is given by

$v_x' = v_x = 4.33 m/s$

$v_y' = e*v_y = 0.5*2.5$

$v_y' = 1.25 m/s$

now as we know the net velocity is given as

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

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

$v_{net} = 4.51 m/s$

also in order to find the angle of rebound with vertical we can use

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

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

$\theta = 73.9 degree$

so it will rebound with speed 4.51 m/s at an angle 73.9 degree with vertical