# Painfull physics

1. Apr 25, 2013

### jess82

just a question the mass of a 1967 mustang travelling at 40 mph hitting a human (me) in 1980 my weight 179 lbs travelling across the front of the mustang at full run, maybe 3 to 5 mph. what would be the actuall force be in pounds that i recieved? and i am familiar with f=ma but i can't establish the mass. plus the motion involved. both objects came together at 90 degree angle. i am fine 3 days later walking and talking, but did do alot of damage to the car. and there was no braking. so a full 40 mph impact. i am lucky just would like to know the forces involved. thanks. jess.

2. Apr 25, 2013

### Danger

I don't think that it can be determined unless you had been wearing extensive instrumentation when it happened. My primary argument in favour of that is the fact that your surface area that made contact is unknown. It's easy enough to calculate how much energy the car could have imparted to you, but that won't be a measure of how much it did impart. Getting clipped off at the knees and then being cammed up the windshield and over the roof is not the same as taking a full torso impact from a bumper.

3. Apr 25, 2013

### physwizard

lol what are you made of iron or something? your story doesn't seem believable.

4. Apr 25, 2013

### Danger

That's sort of what I was getting at with the contact point argument. A '67 Stang doesn't have a lot of frontal area, so I'd expect the forces to be concentrated around thigh-level. At 40mph, that's going to break things. This whole bit about a 90° impact is puzzling, though. The only way that I can think of that happening is if he ran into the side of the car.

5. Apr 25, 2013

### CWatters

You could estimate the acceleration (g force) if you knew the distance over which your body was accelerated. That distance is likely to be quite short unless you were wearing a huge inflatable suit.

Problem is different parts of your body were accelerated over different distances. If your hips were hit they might have accelerated to 40mph over a distance of an inch but your head and sholders probably stayed put and your body folded until it hit the windscreen or the hood. So the distance over which that part accelerated was probably more like a few feet?

How about we assume you went from 0 to 40mph in two feet and do the sums... Excuse me while I switch to metric...

Initial velocity U = 0
Final Velocity V = 40mph = 18m/s
Distance s = 2 feet = 0.6m

Simplified standard equation of motion..

V2 = U2 + 2as
or
a = V2[/SUPB]/2s

= 182/2*0.6
= 270 m/s2

if 1g is 10 m/s2 then you might have experienced 27g during the impact. Parts of you body were probably subject to considerably more. Imagine if you had been bending down to tie your laces at the time and so your whole body had be hit at once or had been accelerated over say half that distance. Half the distance = double the g force.

I believe I read a NASA paper that said 30-50g is survivable but there is a very real risk of serious injury. There are stats available for the survival rates at 30mph and 40mph. You were very lucky. I think only 1% of pedestrians survive a 40mph impact.

6. Apr 25, 2013

### CWatters

You could apply f = ma to that if you want.

7. Apr 25, 2013

### mrspeedybob

3 mph is walking speed, 5 mph is a lazy jog. A flat out run (for me anyway) is 10-12 mph. At least that's what my GPS tells me.

8. Apr 25, 2013

### MostlyHarmless

You would use Momentum. FΔt=mv. Where t is the time you were in contact with the car mass is YOUR mass and v is your veloctiy. Doing this we could get a rough estimate by making some assumptions and simplifications, and we wouldn't care about the motion or mass of the car, just the result of collision. To simplify matters just assume nothing on the car gave and you we're basically just sent flying at some angle from the car at some velocity.

You get hit and and go flying from the car at a velocity v1 at an angle θ. So you're new motion can be described as x and y components. Vy=v1sin Vx=v1cosθ. You're force in y would be given as, $F_{y}={\frac{m(v_{yf}-v_{yi})}{Δt}}$The force in x would be given by, $F_{x}={\frac{m(v_{xf}-v_{xi})}{Δt}}$ I would call the direction of your initial motion y and the car's motion x. So you're initial y velocity was 5mph or 2.2m/s. and you're initial x velocity was 0.

Once you find the x and y components of the force exerted on you by the car you can find the magnitude of that force with this $F={\sqrt{F_x^2+F_y^2}}$ and the direction would be given by $arctan({\frac{F_y}{F_x}})$

So I won't venture to guess how your body rebounded off the car, but you can plug in your estimates into those equations I've layed out and can get a very rough estimate of the kind of force involved. Just be sure to have you're units correct. I'd recommend converting to metric and then you'll get Newtons and just convert that to pounds.

Also, you we're far off when you said F=ma, because that is where the first formula I gave comes from. $a={\frac{Δv}{Δt}}$→$F=m{\frac{Δv}{Δt}}$→$FΔt=mΔv$