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# Impulse

• 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 $\sum F=F_{net}=ma$ . We will rewrite this equation dropping the “net” subscript. We will also assume that $F$ represents the average net force (we will see why in a moment) and express the acceleration as $a=\frac{\Delta v}{\Delta t}=\frac{(v_f-v_i)}{\Delta t}$ , giving us:

$F=m \frac{(v_f-v_i)}{\Delta t} = \frac{mv_f-mv_i}{\Delta t} \rightarrow F \Delta t=mv_f-mv_i=\Delta p \rightarrow F \Delta t = \Delta p$

We call $F \Delta t=\Delta p$ the impulse-momentum equation. The left-hand side is referred to as the impulse and the right-hand side is referred to as the change in momentum. The equation suggests that we needn’t only use the unit $\frac{kg*m}{s}$ to express momentum since the change in momentum is equivalent to $F \Delta t$ , expressed in the unit $N-s$ .

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.

Impulse experienced by ball. The graph on the right shows that the force on the baseball is not constant.

#### 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.

a. In an earlier example, we showed that the change in momentum $\Delta p$ was:

$\Delta p = p_f-p_i = -6.808 - (+6.808) = -13.62 \rightarrow -13.6 \frac{kg*m}{s}$ or a magnitude of $13.6 \frac{kg*m}{s}$

(Note-the negative sign indicates that the direction of the force on the ball is opposite to the ball’s incoming direction.)

Since $\Delta p = F \Delta t$  ,the impulse the ball experiences is -13.6 N-s

b. The magnitude of the average force is $\bar F = \frac{\Delta p}{\Delta t} = \frac{13.6 \ N-s}{0.70 \times 10^{-3}}=19428=1.9 \times 10^4 \ N$

#### 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: $m=0.145 - kg$ and $v_i=46.95 \frac{m}{s}$ , in addition to the two displacements- 1.00-cm and 10.00-cm- to solve the problem.

##### 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,

$v_{avg} &= \frac{(0+46.95)}{2} = 23.475 \rightarrow 23.48 \frac{m}{s}\\x &= v_{avg} \Delta t \rightarrow \Delta t = \frac{x}{v_{avg}} = \frac{0.010 \ m}{23.475 \frac{m}{s}} = 4.26 \times 10^{-4} \rightarrow 4.3 \times 10^{-4} \ s$

Recall that the change in momentum was calculated already as $-6.81 \frac{kg*m}{s}$ . We now use $F = \frac{\Delta p}{\Delta t} = \frac{-6.81 \frac{kg*m}{s}}{4.26 \times 10^{-4} \ s}=-15,986 \rightarrow -1.60 \times 10^4 \ N$

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 $\rightarrow \Delta t = \frac{0.10 \ m}{23.475 \frac{m}{s}}=4.26 \times 10^{-3} \rightarrow 4.3 \times 10^{-3} \ s$

And as before $F=\frac{\Delta p}{\Delta t} = \frac{-6.81 \frac{kg*m}{s}}{4.26 \times 10^{-3} \ s} = -1,599 \rightarrow -1.60 \times 10^3 \ N$ .  Even though the force in Case 2 is ten times smaller, you wouldn’t want to do this without a really good catcher’s mitt!

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 $v_1$  with which you hit both the mattress and the concrete is the same, since you fall through the same displacement. Let’s define velocity $v_1$ as the initial velocity of impact and the final velocity as zero, since the mattress and concrete have brought you to rest. In each case, the change in your velocity is the same as you’re brought to rest, and your mass is the same, as well. Therefore, the change in your momentum (and therefore impulse) is the same in both cases. The only difference is, since the mattress “gives” more than the concrete, it will take more time for the mattress to bring you to rest than the concrete.  As we saw above, if the change in momentum is constant, force is decreased the longer it takes to come to rest.