Solve for magnitude and direction of ball's change in momentum and the average force that the floor exerts on the ball

Question

A basketball player drops a 0.5 kg basketball vertically so that it is traveling at 5.0 m/s when it reaches the floor. The ball rebounds upward at a speed of 3.2 m/s. The ball is in contact with the ground for a time of 10 ms.

Solve for (a) the magnitude and direction of the ball's change in momentum while in contact with the ground and (b) the average force that the floor exerts on the ball.

Evaluate your solution. Explain why the magnitude of your answer reasonable? How do you know what the sign of your answer should be? In what way is your answer consistent with your representations?

Answer 1

Hi Britney,

(a) Remember that Δp = p_f - p_i, where Δp is the change in momentum, p_f is final momentum, and p_i is the initial momentum. Also, p = mv, where m is mass and v is velocity, so we can rewrite the change in momentum equation to

Δp = p_f - p_i = mv_f - mv_i = m(v_f - v_i).

Letting the upwards direction be positive and downwards be negative, we have that v_i = -5.0 m/s (5.0 m/s downwards, hitting the ground) and v_f = +3.2 m/s (3.2 m/s bouncing upwards after contact), and the mass of the basketball is m = 0.5 kg. Plugging these values into the change in momentum equation gives us

Δp = (0.5 kg)(+3.2 m/s - (-5.0 m/s)) = (0.5 kg)(+8.2 m/s) = +4.1 kg m/s (4.1 kg m/s upwards).

(b) Recall that

Δp = F_avgt,

where Δp is the change in momentum, F_avg is the average force the floor exerts in the ball, and t is the duration of the collision. From part (a), we have that Δp = +4.1 kg m/s and the problem tells us that t = 10 ms = 0.01 s, so

F_avg = Δp / t = (+4.1 kg m/s) / (0.01 s) = +410 N (exerting 410 N of force upwards).

For the collision to take a very short amount of time (10 milliseconds is extremely short, an average human reaction time is around 150 milliseconds), a force that is much larger than the weight of the ball must act to push the ball upwards. The basketball's weight is m*g = 0.5 kg * 9.8 m/s = 4.9 N, so a force of 410 N exerted on the ball explains the very small collision time.

The "+" in the answer indicates that the force pushed the ball upwards. This makes sense, as the floor exerted a force that changed the ball's direction from down to up.

I hope this helps! Feel free to ask any questions.

Akshat Y.

Answer 2

A force acting for a period of time is called an impulse force and is equal to the change in momentum divided by the time interval F_avg = Δp / Δt.

Calling the momentum downward negative, and the momentum upward positive, the change in momentum is final momentum minus the initial momentum or 3.2 m/s x 0.5 kg - (- 5.0 m/s) x 0.5 kg = 4.1 mkg/s upward.

The change in momentum occurs over a time interval of 10 ms = 0.01 s. The average force is

F_avg = 4.1 mkg/s / 0.01 s = 410 N. The sign depends on your assignment of the signs of the initial and final momenta.

The force is large due to the short time interval involved.