A basketball player drops a 0.60-kg basketball vertically so that it is traveling at 6.0 m/s when it reaches the floor. The ball rebounds upward at a speed of 4.7 m/s.
Determine the average net force that the floor exerts on the ball if the collision lasts 0.12 s.
I calculated delta p= 6.42 kg.m/s
I thought average net force was delta p / delta t but the that is not right.
Please help! Thank you
your analysis looks okay
your problem might be too many significant figures in your result
To determine the average net force exerted by the floor on the basketball, you can use the impulse-momentum principle.
Impulse (J) is defined as the change in momentum of an object and can be calculated using the equation:
J = Δp = m * Δv
Where:
J = impulse
Δp = change in momentum
m = mass of the basketball
Δv = change in velocity
In this case, the initial velocity (vi) of the ball is -6.0 m/s (negative direction because it is dropping downwards), and the final velocity (vf) is +4.7 m/s (positive direction because it is rebounding upward).
Δv = vf - vi = 4.7 m/s - (-6.0 m/s) = 10.7 m/s
Now, you can plug in the values into the equation:
J = Δp = m * Δv = (0.60 kg) * (10.7 m/s) = 6.42 kg·m/s
Next, you need to calculate the average net force (Favg) exerted over the 0.12 s collision.
Favg = J / Δt
Where:
Favg = average net force
J = impulse (change in momentum)
Δt = time interval (duration of the collision)
Plugging in the values:
Favg = 6.42 kg·m/s / 0.12 s
Favg = 53.5 N
Therefore, the average net force that the floor exerts on the basketball during the collision is 53.5 Newtons.
To determine the average net force that the floor exerts on the ball, you will need to use the principle of impulse-momentum.
Impulse is defined as the change in momentum of an object, which can be calculated using the equation:
Impulse = Change in Momentum = Delta p = m * (v2 - v1)
Where:
m = mass of the ball = 0.60 kg
v1 = initial velocity = 6.0 m/s (downward)
v2 = final velocity = 4.7 m/s (upward)
Substituting the given values, we have:
Delta p = 0.60 kg * (4.7 m/s - (-6.0 m/s))
Delta p = 0.60 kg * (4.7 m/s + 6.0 m/s)
Delta p = 0.60 kg * (10.7 m/s)
Delta p = 6.42 kg·m/s
As you've correctly calculated, the change in momentum (delta p) of the ball is 6.42 kg·m/s.
Now, to find the average net force (F_avg), you need to divide the change in momentum by the duration of the collision (delta t).
F_avg = Delta p / delta t
Given that the collision lasts for 0.12 s, substitute the values:
F_avg = 6.42 kg·m/s / 0.12 s
F_avg ≈ 53.5 N
Therefore, the average net force that the floor exerts on the ball during the collision is approximately 53.5 N.