Final answer:
The impulse applied to the golf ball by the floor is calculated based on the change in momentum, which in this case is twice the product of the mass and the vertical component of the initial velocity. This results in an impulse of 1.90 kg·m/s.
Step-by-step explanation:
The student is asking about the magnitude of the impulse applied to a golf ball by the floor when it rebounds off the floor at the same angle it came in. To calculate the impulse (I), we use the equation I = Δp where Δp is the change in momentum of the golf ball. The momentum of an object is given by p = mv, where m is the mass and v is the velocity. Since the golf ball has the same speed just before and after the collision and the mass does not change, we are only interested in the change of direction.
Since the collision is elastic and the ball strikes and leaves the floor at the same angle, the horizontal component of the velocity (vx) doesn't change, only the vertical component (vy) changes direction. Thus, the total change in momentum is twice the vertical component of the initial velocity times the mass. Calculating the vertical component, vy = v × sin(ϴ), we get vy = 50.6 m/s × sin(30.0°) = 25.3 m/s. The total change in momentum is Δp = 2 × m × vy; substituting the values, Δp = 2 × 0.0377 kg × 25.3 m/s.
Finally, the magnitude of the impulse applied to the golf ball by the floor is just twice the product of the mass and the vertical component of the velocity, which yields I = 1.90 kg·m/s.