A golf ball strikes a hard, smooth floor at an angle of 33.1 ° and, as the drawing shows, rebounds at the same angle. The mass of the ball is 0.0172 kg, and its speed is 59.2 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.)
3 answers
find the vertical component of momentum of the ball, double it...think out why..
I did 59.2 sin (33.1) for a new velocity. Then multiplied by my mass for the change in momentum (impulse). I don't think that's correct.
I get why it's twice. There's a rebound. But I don't understand how to do the vertical component. Is that correct how i showed above? but multiplied by 2?