Questions: A ball of mass m=0.1 kg is thrown horizontally with an initial velocity Vinitial=10.0 m / s towards a wall. The ball bounces off the wall and comes back with a final, horizontal velocity Vfinal=-8.0 m / s. The contact between the ball and the wall lasts for a time Δt=0.8 ms (milliseconds). What is the magnitude of the force applied by the wall on the ball? Round off your answer to 0 decimal places. Do not forget the units.

Solution Steps

We are given the following values:

• Mass of the ball, \( m = 0.1 \, \text{kg} \)
• Initial velocity, \( V_{\text{initial}} = 10.0 \, \text{m/s} \)
• Final velocity, \( V_{\text{final}} = -8.0 \, \text{m/s} \)
• Contact time, \( \Delta t = 0.8 \, \text{ms} = 0.8 \times 10^{-3} \, \text{s} \)

The change in velocity (\(\Delta V\)) is given by: \[ \Delta V = V_{\text{final}} - V_{\text{initial}} = -8.0 \, \text{m/s} - 10.0 \, \text{m/s} = -18.0 \, \text{m/s} \]

The change in momentum (\(\Delta p\)) is calculated using the formula: \[ \Delta p = m \times \Delta V = 0.1 \, \text{kg} \times (-18.0 \, \text{m/s}) = -1.8 \, \text{kg} \cdot \text{m/s} \]

The force (\(F\)) applied by the wall can be calculated using the impulse-momentum theorem: \[ F = \frac{\Delta p}{\Delta t} = \frac{-1.8 \, \text{kg} \cdot \text{m/s}}{0.8 \times 10^{-3} \, \text{s}} = -2250 \, \text{N} \] The magnitude of the force is \(2250 \, \text{N}\).

Final Answer