Question

A golf ball strikes a hard, smooth floor at an angle of 27.0 ° and, as the drawing shows, rebounds at the same angle. The mass of the ball is 0.0200 kg, and its speed is 33.0 m/s just before and after striking the floor. What is the magnitude of the impulse applied to the golf ball by the floor? (Hint: Note that only the vertical component of the ball's momentum changes during impact with the floor, and ignore the weight of the ball.)

Answer 1

J = 3.564 N.s

Step-by-step explanation:

From the given information:

angle θ = 27°

mass = 0.0200 kg

speed = 33.0 m/s

To determine the impulse applied using the equation:

J = m(2V cos θ)

J = 0.0200 (2 × cos (27.0))

J = 0.0200 (2 × 0.8910)

J = 0.03564

J = 3.564 N.s