Objectives
Students will learn the meaning of impulse force and how to calculate both impulse and impulse force in various situations.
Vocabulary
- impulse
Introduction
Earlier, we solved a problem dealing with change in momentum. We emphasized that momentum is a vector quantity and that care must therefore be taken in dealing with the direction of velocity. Now, we will again discuss a change in momentum, but we will investigate it from the point of view of Newton’s Second Law (N2L). The connection between the change in momentum and N2L will help to explain some everyday occurrences involving collisions, for example, why it is a good idea to bend your knees when landing on the ground, why it is a good idea to bring your arm back as you catch a fast ball, or why it hurts more to land on concrete than on a mattress bed.
We can explain the reasons for these statements by defining a quantity called an impulse.
Newton’s Second Law Revisited
Recall that N2L can be stated as \begin{align*}\sum F=F_{net}=ma\end{align*}
\begin{align*}F=m \frac{(v_f-v_i)}{\Delta t} = \frac{mv_f-mv_i}{\Delta t} \rightarrow F \Delta t=mv_f-mv_i=\Delta p \rightarrow F \Delta t = \Delta p \end{align*}
We call \begin{align*}F \Delta t=\Delta p\end{align*}
In typical collisions there is a rapid buildup of force between the objects colliding and a rapid diminishing of the force as they either come to rest or separate from each other. In either case the force is not constant and the time of interaction is brief. Figure below shows a baseball as it compresses and expands as it’s being hit by a bat. Figure below shows a graph of the force on the baseball as function of time during the interaction with the bat. The times for such interactions are measured in milliseconds and the force is in kilo-newtons.
Illustrative Example 7.2.1
If a ball thrown by Aroldis Chapman remains in contact with a baseball bat for 0.70 milliseconds:
(a) What is the impulse the ball experiences?
(b) What is the magnitude of the average force the ball experiences?
Recall the mass of the ball is 0.145-kg and the velocity is 46.95 m/s.
Answers:
a. In an earlier example, we showed that the change in momentum \begin{align*}\Delta p\end{align*}
\begin{align*}\Delta p = p_f-p_i = -6.808 - (+6.808) = -13.62 \rightarrow -13.6 \frac{kg*m}{s}\end{align*}
(Note-the negative sign indicates that the direction of the force on the ball is opposite to the ball’s incoming direction.)
Since \begin{align*}\Delta p = F \Delta t\end{align*}
b. The magnitude of the average force is \begin{align*}\bar F = \frac{\Delta p}{\Delta t} = \frac{13.6 \ N-s}{0.70 \times 10^{-3}}=19428=1.9 \times 10^4 \ N\end{align*}
Illustrative Example 7.2.2
We continue with our baseball example with a question for the catcher. It is a very good idea when catching a fastball to move your arm back as you catch the ball. Let’s determine the force on the catcher’s mitt under two conditions:
(1) The catcher moves his arm back 1.00-cm in catching the ball and
(2) The catcher moves his arm back 10.00-cm in catching the ball.
We use the values given in the original problem: \begin{align*}m=0.145 - kg\end{align*}
Case 1
There are several ways to work the problem. We will find the time it takes the ball to come to rest in each case and then use the change in momentum to solve for the force.
To find the time we divide the displacement of the ball by its average velocity, that is,
\begin{align*}v_{avg} &= \frac{(0+46.95)}{2} = 23.475 \rightarrow 23.48 \frac{m}{s}\\
x &= v_{avg} \Delta t \rightarrow \Delta t = \frac{x}{v_{avg}} = \frac{0.010 \ m}{23.475 \frac{m}{s}} = 4.26 \times 10^{-4} \rightarrow 4.3 \times 10^{-4} \ s\end{align*}
Recall that the change in momentum was calculated already as \begin{align*}-6.81 \frac{kg*m}{s}\end{align*}
This is a very large force, equivalent to 3,600 pounds. In all likelihood, this would result in serious damage to the catcher’s wrist.
Case 2
Since the displacement is ten times greater in this case, the time to bring the ball to rest will be ten times longer \begin{align*}\rightarrow \Delta t = \frac{0.10 \ m}{23.475 \frac{m}{s}}=4.26 \times 10^{-3} \rightarrow 4.3 \times 10^{-3} \ s\end{align*}
And as before \begin{align*}F=\frac{\Delta p}{\Delta t} = \frac{-6.81 \frac{kg*m}{s}}{4.26 \times 10^{-3} \ s} = -1,599 \rightarrow -1.60 \times 10^3 \ N\end{align*}
Check Your Understanding
Why is it less painful to fall on a mattress bed than on concrete?
Answer: Let’s say you’re standing up and decide to let yourself fall over like a slightly tipped bowling pin. In one case, at floor level, there is a mattress and in another case there is concrete. We’ll assume in each case you have the same initial velocity (close to zero) as you begin your fall. The velocity \begin{align*}v_1\end{align*}