A 0.15-kg baseball traveling in a horizontal direction with a speed of $20 \mathrm{m} / \mathrm{s}$ hits a bat and is popped straight up with a speed of \(15 \mathrm{m} / \mathrm{s} .\) (a) What is the change in momentum (magnitude and direction) of the baseball? (b) If the bat was in contact with the ball for 50 ms, what was the average force of the bat on the ball?

First, we'll find the initial and final momenta of the baseball. The initial state of the baseball is horizontal with velocity \(20m/s\), and the final state is vertical with velocity \(15m/s\). Let the initial momentum be \(\vec{p_1}\), and the final momentum be \(\vec{p_2}\). The initial momentum is: \(\vec{p_1} = m\vec{v_1} = 0.15kg \cdot 20m/s = 3kg·m/s\) in the horizontal direction. The final momentum is: \(\vec{p_2} = m\vec{v_2} = 0.15kg \cdot 15m/s = 2.25kg·m/s\) in the vertical direction.

Now, we'll find the change in momentum, which is the difference between the final and initial momenta. Since the direction changes, we must compute the change in momentum as a vector difference: \(\Delta \vec{p} = \vec{p_2} - \vec{p_1}\) We need to find the magnitude and direction of this difference. The magnitudes are: \(\Delta p_x = |p_{2x} - p_{1x}| = |0 - 3| = 3kg·m/s\) \(\Delta p_y = |p_{2y} - p_{1y}| = |2.25 - 0| = 2.25kg·m/s\) Now we can calculate the magnitude of the total change in momentum: \(\Delta p = \sqrt{(\Delta p_x)^2+(\Delta p_y)^2} = \sqrt{(3)^2+(2.25)^2} = \sqrt{9+5.0625} = \sqrt{14.0625} = 3.75 kg·m/s\) The direction is found using the inverse tangent function: \(\theta = \arctan\left(\frac{\Delta p_y}{\Delta p_x}\right) = \arctan\left(\frac{2.25}{3}\right) = 36.87^{\circ}\) above the negative horizontal axis. Thus, the change in momentum of the baseball is 3.75 \(kg·m/s\) at an angle of \(36.87^{\circ}\) above the negative horizontal axis.

From the impulse-momentum theorem, we know that the impulse exerted on the ball is equal to the change in momentum: \(\vec{I} = \Delta \vec{p}\) Next, we'll find the time interval during which the force acted on the ball, which is given as \(50ms\) or \(0.05s\).

We can now use the impulse-momentum theorem to find the average force exerted on the ball during the given time interval: \(\vec{I} = \vec{F}_{avg} \cdot \Delta t\) Now let's solve for the average force: \(\vec{F}_{avg} = \frac{\vec{I}}{\Delta t} = \frac{3.75 kg·m/s}{0.05s} = 75 N\) The direction of the average force is the same as the direction of the impulse, which is \(36.87^{\circ}\) above the negative horizontal axis. So, the average force exerted on the baseball by the bat is 75 N at an angle of \(36.87^{\circ}\) above the negative horizontal axis.