A golf ball strikes a hard, smooth floor at an angle of θ = 26.4° and, as the drawing shows, rebounds at the same angle. The mass of the ball is 0.047 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, and ignore the weight of the ball.)
To find the magnitude of the impulse applied to the golf ball by the floor, we need to calculate the change in momentum of the ball in the vertical direction during the impact.
Impulse (J) is defined as the change in momentum (ΔP) and can be calculated using the formula J = ΔP = mΔv, where m is the mass of the golf ball and Δv is the change in velocity.
In this case, we are only concerned with the vertical component of the ball's momentum, since the angle of impact and rebound is the same. Therefore, the change in velocity is equal to twice the vertical component of the ball's velocity, because the ball bounces back with the same angle and magnitude.
Given:
Mass of the golf ball (m) = 0.047 kg
Velocity of the golf ball (v) = 45 m/s
Angle of impact (θ) = 26.4°
First, we need to calculate the vertical component of the ball's velocity. Since the angle of impact and rebound is the same, the vertical component of velocity (v_y) is given by v_y = v * sin(θ) = 45 m/s * sin(26.4°).
v_y = 45 m/s * sin(26.4°)
v_y = 45 m/s * 0.438
v_y ≈ 19.71 m/s
Now, since the ball bounces back with the same angle and magnitude, the change in velocity in the vertical direction (Δv_y) is twice the vertical component of velocity, i.e., Δv_y = 2 * v_y.
Δv_y = 2 * 19.71 m/s
Δv_y = 39.42 m/s
Finally, we can calculate the magnitude of the impulse applied to the golf ball by the floor using the formula J = mΔv_y.
J = m * Δv_y
J = 0.047 kg * 39.42 m/s
J ≈ 1.854 Ns
Therefore, the magnitude of the impulse applied to the golf ball by the floor is approximately 1.854 Ns.