Answer to Question #96848 in Classical Mechanics for Jake

Question

A pitcher throws a 0.145 so that it crosses home plate kg baseball horizontally with a speed of 40 m/s. It is hit straight back at the pitcher with a final speed of 50 m/s. a) What is the impulse delivered to the ball? b) Find the average force exerted by the bat on the ball if the two are in contact for 2.0 x 103 s. c) What is the acceleration experienced by the ball?

Answer 1

Part a): Let's choose the axis x along the trajectory of the ball and let positive direction be the same as the initial direction of the ball. Then, according to the impulse conservation

$$\vec p' = \vec p + \Delta \vec p$$

where $\vec p'$ -final impulse of the ball, ${\vec p}$ - initial impulse of the ball and $\Delta \vec p$ is the impulse delivered to the ball. Then

$$\Delta \vec p = \vec p' - \vec p$$

Let's use that $\vec p = m\vec v$ and project the velocities to the x-axis

$$\Delta {p_x} = m({v_x}' - {v_x})$$

And let's do the calculations

$$\Delta {p_x} = 0.145[{\text{kg}}]( - 50[\frac{{\text{m}}}{{\text{s}}}] - 40[\frac{{\text{m}}}{{\text{s}}}]) = - 13.05[\frac{{{\text{kg}} \cdot {\text{m}}}}{{\text{s}}}]$$

Minus sign in our reference system means that the direction of the $\Delta \vec p$ is "to the pitcher or in direction of the ball after hit".

Part b): We can use the Newton's second law

$$\vec F = \frac{{d\vec p}}{{dt}}$$

Let's approximate differentials by the finite differences, then for average force we can use

$$\left\langle {\vec F} \right\rangle = \frac{{\Delta \vec p}}{{\Delta t}}$$

Again, project to the x-axis and we get

$$\left\langle {{F_x}} \right\rangle = \frac{{\Delta {p_x}}}{{\Delta t}} = \frac{{ - 13.05[\frac{{{\text{kg}} \cdot {\text{m}}}}{{\text{s}}}]}}{{2 \cdot {{10}^{ - 3}}[{\text{s}}]}} = - 6525[{\text{N}}]$$

The meaning of the minus sign is the same.

Part c): Use Newton's second law again in the form

$$\vec F = \frac{{d\vec p}}{{dt}} = m\frac{{d\vec v}}{{dt}} = m\vec a$$

Averaging and expressing acceleration we get

$$\left\langle {\vec a} \right\rangle = \frac{{\left\langle {\vec F} \right\rangle }}{m}$$

Thus

$$\left\langle {{a_x}} \right\rangle = \frac{{\left\langle {{F_x}} \right\rangle }}{m} = \frac{{ - 6525[{\text{N}}]}}{{0.145[{\text{kg}}]}} = - 45000[\frac{{\text{m}}}{{{{\text{s}}^{\text{2}}}}}]$$