Impulse and momentum problems, like energy problems, relate the situation before an interaction to the situation afterward. Consequently, the before-and-after pictorial representation remains our primary visualization tool. Let’s look at an example.

EXAMPLE 11.1 Hitting a baseball

A 150 g baseball is thrown with a speed of $20\text{m}\text{/}\text{s}$. It is hit straight back toward the pitcher at a speed of $40\text{m}\text{/}\text{s}$. The interaction force between the ball and the bat is shown in FIGURE 11.7. What maximum force $F_{\text{max}}$ does the bat exert on the ball? What is the average force of the bat on the ball?

FIGURE 11.7 The interaction force between the baseball and the bat.

FIGURE 11.7 Full Alternative Text

MODEL Model the baseball as an elastic object and the interaction as a collision. Assume the motion is along the x-axis.

VISUALIZE FIGURE 11.8 is a before-and-after pictorial representation. Because $F_x$ is positive (a force to the right), we know the ball was initially moving toward the left and is hit back toward the right. Thus we converted the statements about speeds into information about velocities, with $v_{\text{i}x}$ negative.

FIGURE 11.8 A before-and-after pictorial representation.

Full Alternative Text

Full Alternative Text

SOLVE The momentum principle is

$$\text{Δ}{p}_x=J_x=\text{area under the force curve}$$

We know the velocities before and after the collision, so we can calculate the ball’s momenta:

$$\begin{array}{l}{p}_{\text{i}x}=m{v}_{\text{i}x}=(0.15\text{ kg})\left(-20\text{m}\text{/}\text{s}\right)=-3.0\text{kg}\text{m}\text{/}\text{s}\\ p_{\text{f}x}=m{v}_{\text{f}x}=(0.15\text{ kg})\left(40\text{m}\text{/}\text{s}\right)=6.0\text{kg}\text{m}\text{/}\text{s}\end{array}$$

Thus the change in momentum is

$$\text{Δ}{p}_x=p_{\text{f}x}-p_{\text{i}x}=9.0\text{kg}\text{m}\text{/}\text{s}$$

The force curve is a triangle with height $F_{\text{max}}$ and width 3.0 ms. The area under the curve is

$$J_x=\text{area}=\frac{1}{2}(F_{\text{max}})(0.0030\text{s})=(F_{\text{max}})\left(0.0015\text{s}\right)$$

Using this information in the momentum principle, we have

$$9.0\text{kg}\text{m}\text{/}\text{s}=(F_{\text{max}})\left(0.0015\text{s}\right)$$

Thus the maximum force is

$$F_{\text{max}}={\displaystyle \frac{9.0\text{kg}\text{m}\text{/}\text{s}}{0\text{.0015 s}}}=6000\text{N}$$

The average force, which depends on the collision duration $\text{Δ}t=0.0030\text{s,}$ has the smaller value:

$$F_{\text{avg}}={\displaystyle \frac{J_x}{\text{Δ}t}}={\displaystyle \frac{\text{Δ}{p}_x}{\text{Δ}t}}={\displaystyle \frac{9.0\text{kg}\text{m}\text{/}\text{s}}{0\text{.0030 s}}}=3000\text{N}$$

REVIEW $F_{\text{max}}$ is a large force, but quite typical of the impulsive forces during collisions. The main thing to focus on is our new perspective: An impulse changes the momentum of an object.

Other forces often act on an object during a collision or other brief interaction. In Example 11.1, for instance, the baseball is also acted on by gravity. Usually these other forces are much smaller than the interaction forces. The 1.5 N weight of the ball is vastly less than the 3000 N average force of the bat on the ball. We can reasonably neglect these small forces during the brief time of the impulsive force; this is called the impulse approximation.

When we use the impulse approximation, $p_{\text{i}x}$ and $p_{\text{f}x}$ (and $v_{\text{i}x}$ and $v_{\text{f}x})$ are the momenta (and velocities) immediately before and immediately after the collision. For example, the velocities in Example 11.1 are those of the ball just before and after it collides with the bat. We could then do a follow-up problem, including gravity and drag, to find the ball’s speed a second later as the second baseman catches it. We’ll look at some two-part examples later in the chapter.