Impulse

### Impulse (Δp) – A central concept in safety engineering

The concept of impulse is one of simplest in all of mechanics. It's just the change in momentum during some process, like a collision:

$$\Delta p = p_{\text{final}} - p_{\text{initial}}$$

Yet when we relate impulse to force (which we'll do below), we get a powerful tool for understanding the forces on humans and animals in accidents like vehicle crashes and falls. It's a way of thinking that led to important safety features like seatbelts and air-bags in cars, stretchy rock climbing ropes, parts of cars that crumple strategically in a collision, and many other devices that have saved many lives. Aahh ... Science!

The basic idea is that when a living thing, like a human, a dog, a cat, livestock ... is involved in a collision, its momentum can change dramatically over a very short time. It can go from a very high value to zero very rapidly. Imagine your body traveling at 60 mi./h when your car slams into a big tree. In a tenth of a second or less, your body will go from a very high momentum to zero momentum. Bodies can only take so much force, and it turns out that the shorter the time it takes to reduce your momentum to zero, the larger the force on you. If we can do things to increase the time it takes to undergo that momentum change, the force will be reduced and you might survive.

The advertisement on the left was run by the U.S. Department of Transportation to encourage people to use seatbelts, one of many devices now in cars meant to reduce the forces that could kill a person in a car collision.

### Impulse and force

The real value of considering Δp is relating it to force. Let's start by remembering that

$$\Delta p = p_{\text{final}} - p_{\text{initial}}$$

Now think about units. The unit of force is the Newton (1N = 1Kg·m·s-2), and if we multiply by time (in seconds), one of those seconds in the denominator divides out, and we get the units of momentum, Kg·m·s-1

So force multiplied by the change in time is just the change in momentum, and we can rearrange to find a new formula for force:

$$F \Delta t = \Delta p \: \: \longrightarrow \: \: F = \frac{\Delta p}{\Delta t}$$

If we break momentum down into velocity and mass, we can further refine our formula:

$$F = \frac{\Delta p}{\Delta t} = \frac{\Delta (mv)}{\Delta t} = \frac{m \Delta v}{\Delta t}$$

Now this is really a remarkable result. It says that for a given momentum change, or equivalently for a given velocity change of some fixed mass – like a person, the force is inversely proportional to the amount of time it takes to undergo that momentum or velocity change.

The longer it takes to change momentum, the less force is applied or experienced. The shorter the time, the more force. Here's a graph of the situation.

#### Impulse and Force

Impulse is the change in momentum during some process, like a collision:

$$\Delta p = p_{\text{final}} - p_{\text{initial}}$$

Force and impulse are related:

$$F = \frac{\Delta p}{\Delta t} = \frac{m \, \Delta v}{\Delta t}$$

Force is inversely proportional to the time it takes to change momentum.

### Why does this matter: Seatbelts and airbags

One of the most dramatic momentum changes that can be undergone by a human body is slamming into the steering wheel (or worse, being ejected from the vehicle through the windshield) during a high-speed crash. Air bags and seatbelts have dramatically reduced the death rate from such crashes.

Consider the crash-test dummy below, pictured in a test of an air bag. Let's say that a person is traveling at 60 mi./h and hits an immovable object.

First we'll convert 60 mi./h to meters/second:

\require{cancel} \begin{align} \left( \frac{60 \cancel{mi}}{\cancel{h}} \right)\left( \frac{1 \cancel{h}}{3600 \, s} \right)&\left( \frac{1609 \, m}{1 \cancel{mi}} \right) \\[5pt] &= \: 26.82 \, \frac{m}{s} \end{align}

Now let's say the average person has a mass of 60 Kg, but we're really only talking about the movement of the torso of a seat-belted driver, so we'll say that the mass is 30 Kg. Then the initial momentum is

\begin{align} p+i &= (30 \, Kg)(26.82 \, m/s) \\[5pt] &= \: 805 \, \frac{Kg \, m}{s} \end{align}

Because the final momentum is zero, the impulse is:

$$\Delta p = 805 \, Kg \, \frac{m}{s}$$

(By definition, this should be negative, but we'll just redefine our coordinate system to make a loss of momentum positive – no harm in that.) Now let's say that our driver's head is 30 cm (0.3 m) from the steering wheel. At 26.8 m/s, the time it would take for the head to contact the wheel is

$$\Delta t = 0.3 \cancel{m} \left( \frac{1 \, s}{26.82 \cancel{m}} \right) = 0.0112 \, s$$

That's 11.2 milliseconds (ms) – pretty fast. With that time, we can calculate the force felt by the head as it comes to a stop against the wheel.

\begin{align} F &= \frac{\Delta p}{\Delta t} = \frac{805 \, \frac{Kg \,m}{s}}{0.0112 s} \\[5pt] &= 72,000 \, N \\[5pt] &= 72 \, KN \end{align}

Now 72 KN is a lot of force. In fact, most human bones break at a force below 10 KN.

What if we could increase the time to the steering wheel by, say, a factor of 10. Instead of taking 0.01 s to stop, let's say it took 0.1 s. Now the force would be reduced to:

$$F = 7.2 \, KN = 7200 \, N$$

which is in the range of survivable forces. That's just what an airbag is designed to do: It increases the time it takes to undergo the impulse change.

Airbags work by sensing rapid deceleration of the vehicle, pressurizing the bag explosively when needed, then deflating more slowly, say over several tenths of a second.

Seatbelts have a dual effect. They are designed to keep car occupants inside the car where it's likely to be safer during a crash, and they stretch just a bit, helping to increase the time it takes for the impulse to be absorbed.

### Cars are full of force-reducing measures

The drawing above shows some of the safety features built into most modern vehicles. The driver is surrounded by the passenger "cage," a steel framework designed to maintain a shell of saftey around the passengers.

Inside that cage are the safety restraints – seatbelts, and several airbags, some even on the sides of the car in case of a side impact.

In back and front are "crumple zones," regions of the car that are designed to collapse during a collision, not just transfer energy to the occupants. These zones further lengthen the time it takes to slow down and lose momentum.

In the front of passenger cars, the engine is often engineered to drop below the passenger compartment so that it doesn't intrude into it.

### Practice problems

 1 A car traveling at high speed collides with a small bug which was initially at rest. Did the momentum of the car + bug system change after the crash? Did the momentum of the bug increase, decrease or remain the same after the collision? Did the momentum of the car increase, decrease or remain the same after the collision? Of the car and bug, which experienced a greater change in speed after the collision? Which experienced a greater impulse (Δ p) during the collision? Solution The momentum of the system must remain constant (momentum is conserved.) The momentum of the bug was zero (maybe it was hovering) and after the collision it is moving along with the car, so it increases or decreases, depending on how we set up our coordinate system. The momentum of the car must decrease because some of its momentum had to be transferred to the bug in order to get it to move. After the collision, the speed of the bug is the same as the speed of the car (it's stuck to the windshield). The collision must have reduced the momentum, therefore the speed of the car just a little. The momentum lost by the car is equal to the momentum gained by the bug, so the Δp's are the same, just of different sign. 2 Before collision, a 20 Kg object was moving at 15 m/s. Calculate the impuls that acted on the object if, after the collision, its velocity relative to the original velocity was 10 m/s -10 m/s Solution The impulse is just the momentum change, Δp, so we have \begin{align} \text{(a)} \: \: \Delta p &= (20 \, Kg)(15 - 10)\, m/s \\[5pt] &= 100 \, Kg \, m/s \\[5pt] \text{(b)} \: \: \Delta p &= (20 \, Kg)(15 - (-10)) \, m/s \\[5pt] &= 500 \, Kg \, m/s \end{align} 3 Let's say your mass is 60 Kg and you're traveling at 25 m/s (~56 mi./h) in a car when you suddenly slam into a moose crossing the road, stopping your car in its tracks. You are wearing your seatbelt and it takes your body 0.400 s to come to a stop. What average force did the seatbelt exert on your body? If you had not been wearing your seatbelt, and the steering wheel and windshield stopped your body in about a millisecond (1 × 10-3 s), what average force would those things have exerted on your body? How many times greater is the stopping force of the steering wheel/windshield than the seatbelt? Solution First, we'll calculate your momentum, which is also the total change in momentum (impulse) because the final momentum is zero: $$p = mv = (60 \, Kg)(25 \, m/s) = 1,500 \, Kg \, m/s$$ Now the momentum changes are \begin{align} p_{\text{seatbelt}} &= 1500/0.4 \, s = 3.75 \, KN \\[5pt] p_{\text{wheel,window}} &= 1500/0.001 \, = 1,500 \, KN \end{align} Without the seatbelt, the force is about 400 times greater, and probably not survivable. 4 The mass of a baseball is about 0.145 Kg. A good fastball moves at about 35 m/s. Catching a ball with a stationary glove can hurt. One strategy is to allow the mitt to move backward a little bit during the catch. Calculate the force on the glove if it catches a fastball without moving so that it stops the ball instantaneously (we'll say in 1 ms) while moving backward so that it takes 0.25 s to stop the ball. Solution The momentum of the baseball is $$p = mv = (0.145 \, Kg)(35 \, m/s) = 5.075 Kg \, m/s$$ The force felt with the stationary mitt is \begin{align} \Delta F &= \frac{\Delta p}{\Delta t} \\[5pt] &= \frac{5.075 \, Kg \, m/s}{0.001 \, s} = 5075 \, N \end{align} Now if the catch takes 0.25 s to stop the ball, \begin{align} \Delta F &= \frac{\Delta p}{\Delta t} \\[5pt] &= \frac{5.075 \, Kg \, m/s}{0.25 \, s} = 20.3 \, N \end{align} The force is much less if the mitt is allowed to move backward during the catch.

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