A golf ball strikes a hard, smooth floor at an angle of θ = 29.6° and, as the drawing shows, rebounds at the same angle. The mass of the ball is 0.041 kg, and its speed is 45 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.)
first find the component of velocity perpendicular to the wall,that is the componnet of veloicty that will be refersed.
The change in velocity will be have twice this component...the component parallel does not change.
impulse=mass*2*velocityperpendicular. direction is away from the wall.
To find the magnitude of the impulse applied to the golf ball by the floor, we need to determine the change in the vertical component of the ball's momentum during the impact.
Given:
- Initial angle of the ball's trajectory, θ = 29.6°
- Mass of the ball, m = 0.041 kg
- Initial and final speed of the ball, v = 45 m/s
Let's break down the steps to find the magnitude of the impulse:
Step 1: Find the vertical component of the initial velocity (v₀y) and final velocity (vfy) using trigonometry:
v₀y = v₀ * sin(θ)
vf_y = vf * sin(θ)
Step 2: Calculate the initial momentum in the vertical direction (p₀y) and final momentum (pfy):
p₀y = m * v₀y
pfy = m * vf_y
Step 3: Determine the change in momentum (Δp):
Δp = pfy - p₀y
Step 4: Finally, the magnitude of the impulse (J) is equal to the absolute value of the change in momentum:
J = |Δp|
Let's calculate it step by step:
Step 1:
v₀y = 45 m/s * sin(29.6°) ≈ 22.55 m/s
vf_y = 45 m/s * sin(29.6°) ≈ 22.55 m/s
Step 2:
p₀y = 0.041 kg * 22.55 m/s ≈ 0.925 kg·m/s
pfy = 0.041 kg * 22.55 m/s ≈ 0.925 kg·m/s
Step 3:
Δp = 0.925 kg·m/s - 0.925 kg·m/s = 0
Step 4:
J = |0| = 0
Therefore, the magnitude of the impulse applied to the golf ball by the floor is 0 N·s (or kg·m/s).