Question

A golf ball strikes a hard, smooth floor at an angle of 39.8 ° and, as the drawing shows, rebounds at the same angle. The mass of the ball is 0.0285 kg, and its speed is 30.4 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

Answer:

   I = 1.1 N s

Explanation:

The momentum is equal to the change of the moment

       I = Δp = m $v_{f}$-m v₀

They indicate that the ball arrived at an angle of 39.8º, let's use trigonometry to find the velocity component

      sin 39.8 = $v_{y}$ / v

      cos39.8º = vₓ / v

      $v_{y}$ = V sin 39.8

      $v_{y}$ = 30.4 sin39.4

      $v_{y}$ = 19.3 m / s

Let's calculate the momentum on the y axis

      I =  m  $v_{oy}$ - m $v_{fy}$

      I = 0.0285 19.3 - 0.285 (-19.3)  

      I = 1.1 N s

Answer 2

Answer:

The magnitude of the impulse is 1.33 kg m/s

Explanation:

please look at the solution in the attached Word file