Question

A ball with velocity of 5m/s impinge at angle of 60 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

ball is moving with speed 5 m/s at an angle 60 degree with vertical so two components of the motion of ball is given by

$v_x = 5 sin60 = 4.33 m/s$

$v_y = 5 cos60 = 2.5 m/s$

after collision the x component of velocity will not change while Y component of velocity will change

So after collision the velocity is given as

$v_x' = 4.33 m/s$

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

so 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 motion of the ball is given as

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

$\theta = 74 degree$

so ball will rebound with speed 4.51 m/s at an angle of 74 degree with the vertical