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

initial velocity is 5 m/s at 60 degree with vertical

so its components of velocity is given as

$v_x = 5 sin60 = 4.33 m/s$

$v_y = 5 cos60 = 2.5 m/s$

now after the collision the speed in vertical direction will change while horizontal speed will be same

$v_y' = e v_y$

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

$v_x' = v_x = 4.33 m/s$

now the net speed is given as

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

$v_{net} = \sqrt{4.33^2 + 1.25^2} = 4.51 m/s$

now the angle is given by

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

$\theta = tan^{-1}\frac{4.33}{1.25} = 74 degree$

so it will rebound with speed 4.51 m/s at angle 74 degree