A ball of mass 50 g travelling at a speed of 2.0 m/s strikes a plane surface at an angle of incidence 45^ . The ball is reflected by the plane at an equal angle of reflection with the same speed. Determine the magnitude of the change in momentum of the ball.

Step-by-step solution

Mass of the ball, m = 50 g = 0.05 kg Speed of the ball, v = 2.0 m/s Angle of incidence, = 45^ The component of momentum parallel to the surface remains unchanged during the collision. The component of momentum perpendicular to the surface is reversed. Initial momentum perpendicular to the surface, p_ i = -mv Final momentum perpendicular to the surface, p_ f = mv Magnitude of the change in momentum, | p| = |p_ f - p_ i | = 2mv Substituting the given values: | p| = 2 0.05 2.0 45^ | p| = 0.2 1 2 | p| = 0.1 2 0.1 1.414