Questions: Review A 2.5 kg mud ball drops from rest at a height of 14 m. Part A If the impact between the ball and the ground lasts 0.50 s , what is the average net force exerted by the ball on the ground? Express your at using two significant figures. F = □ N

Solution Steps

To find the velocity of the mud ball just before it hits the ground, we use the kinematic equation for an object in free fall: \[ v = \sqrt{2gh} \] where \( g = 9.81 \, \text{m/s}^2 \) (acceleration due to gravity) and \( h = 14 \, \text{m} \) (height).

\[ v = \sqrt{2 \times 9.81 \, \text{m/s}^2 \times 14 \, \text{m}} = \sqrt{274.68} \approx 16.57 \, \text{m/s} \]

The momentum just before impact is given by: \[ p = mv \] where \( m = 2.5 \, \text{kg} \) and \( v = 16.57 \, \text{m/s} \).

\[ p = 2.5 \, \text{kg} \times 16.57 \, \text{m/s} = 41.425 \, \text{kg} \cdot \text{m/s} \]

The average net force exerted by the ball on the ground can be found using the impulse-momentum theorem: \[ F_{\text{avg}} = \frac{\Delta p}{\Delta t} \] where \( \Delta p = 41.425 \, \text{kg} \cdot \text{m/s} \) (change in momentum) and \( \Delta t = 0.50 \, \text{s} \) (time duration of impact).

\[ F_{\text{avg}} = \frac{41.425 \, \text{kg} \cdot \text{m/s}}{0.50 \, \text{s}} = 82.85 \, \text{N} \]

Final Answer