Question

show all you work, including units, on separate paper. follow the \problem solving method\
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question 38 of 62
1.6 points
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a 0.10 kg rubber ball is falling straight down, and just before it strikes the ground, it is traveling at 12 m/s. the ball rebounds upward at 8.0 m/s. the collision with the floor lasts 0.040 s. using the impulse - momentum theorem, what is the magnitude of the average force that the floor exerts on the ball?
10 n
120 n
80 n
50 n

Explanation:

Step1: Recall Impulse-Momentum Theorem

The impulse - momentum theorem is \(J=\Delta p = F_{avg}\Delta t\), where \(J\) is the impulse, \(\Delta p\) is the change in momentum, \(F_{avg}\) is the average force, and \(\Delta t\) is the time interval. The formula for momentum is \(p = mv\), so \(\Delta p=m(v_f - v_i)\).

Let's define the upward direction as positive. The initial velocity \(v_i=- 12\space m/s\) (negative because it is downward) and the final velocity \(v_f = 8.0\space m/s\) (positive because it is upward). The mass \(m = 0.10\space kg\) and \(\Delta t=0.040\space s\).

Step2: Calculate the change in momentum

First, calculate \(\Delta p\):

$$
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$$

Step3: Solve for the average force

From \(J = F_{avg}\Delta t=\Delta p\), we can solve for \(F_{avg}\):

$$ F_{avg}=\frac{\Delta p}{\Delta t} $$

Substitute \(\Delta p = 2\space kg\cdot m/s\) and \(\Delta t = 0.040\space s\) into the formula:

$$ F_{avg}=\frac{2\space kg\cdot m/s}{0.040\space s}=50\space N $$

Answer:

50 N (corresponding to the option "50 N")