Questions: In the overhead view of the figure, a 450 g ball with a speed v of 5.8 m/s strikes a wall at an angle θ of 41° and then rebounds with the same speed and angle. It is in contact with the wall for 14 ms. In unit-vector notation, what are (a) the impulse on the ball from the wall and (b) the average force on the wall from the ball?

Solution Steps

The ball's initial velocity \( \vec{v_i} \) can be broken down into components: \[ \vec{v_i} = v \cos(\theta) \hat{i} + v \sin(\theta) \hat{j} \] Given \( v = 5.8 \, \text{m/s} \) and \( \theta = 41^\circ \): \[ \vec{v_i} = 5.8 \cos(41^\circ) \hat{i} + 5.8 \sin(41^\circ) \hat{j} \] \[ \vec{v_i} \approx 4.37 \hat{i} + 3.80 \hat{j} \]

After rebounding, the velocity \( \vec{v_f} \) will have the same magnitude but the x-component will be reversed: \[ \vec{v_f} = -v \cos(\theta) \hat{i} + v \sin(\theta) \hat{j} \] \[ \vec{v_f} = -4.37 \hat{i} + 3.80 \hat{j} \]

The mass of the ball \( m = 450 \, \text{g} = 0.45 \, \text{kg} \). The change in momentum \( \Delta \vec{p} \) is: \[ \Delta \vec{p} = m (\vec{v_f} - \vec{v_i}) \] \[ \Delta \vec{p} = 0.45 \left[ (-4.37 \hat{i} + 3.80 \hat{j}) - (4.37 \hat{i} + 3.80 \hat{j}) \right] \] \[ \Delta \vec{p} = 0.45 \left[ -8.74 \hat{i} \right] \] \[ \Delta \vec{p} = -3.93 \hat{i} \, \text{kg} \cdot \text{m/s} \]

Impulse \( \vec{J} \) is equal to the change in momentum: \[ \vec{J} = \Delta \vec{p} \] \[ \vec{J} = -3.93 \hat{i} \, \text{kg} \cdot \text{m/s} \]

The time of contact \( \Delta t = 14 \, \text{ms} = 0.014 \, \text{s} \). The average force \( \vec{F} \) is: \[ \vec{F} = \frac{\vec{J}}{\Delta t} \] \[ \vec{F} = \frac{-3.93 \hat{i}}{0.014} \] \[ \vec{F} \approx -280.71 \hat{i} \, \text{N} \]

Final Answer