From: Doug Jones (email@example.com)
Date: Thu Feb 21 2002 - 19:25:22 MST
> Suppose I am stepping out of the seat of a high, stopped vehicle (a 3000
> pound mini-van), which is stopped on ice.
> Suppose, further, that I have placed one foot outside the van, on the
> ground, and have begun standing on that leg, when:
> The van I am climbing out of is struck from behind by a 4000 pound car
> traveling at, say 35 miles per hour. The car striking the van comes to
> rest, imparting it's energy to the van, which quickly comes up to a high
> (~35 mph?) speed. Since my torso is still in the doorway, the rear leading
> edge of the door frame strikes me in the back at whatever speed the car is
> traveling, accelerating my 220 pound body like the blow from a giant hammer
> (the impact zone between the door frame and my back being about 1 inch by 8
> inches). Luckily for me, the impact was felt about 2 inches to the right
> of my spine, so I was only badly hurt, but not paralyzed.
> My physics question is: how does one characterize the energy transfer that
> occurred here? I'm especially interested in being able to say how much
> energy was in the blow that I received.
> - Jerome
Let's get sytematic about this- first, we calculate the velocity change
of Jerome's mini-van in the intial collision: this is an inelastic
collision that conserves momentum:
m1*v1 + m2*v2 = (m1+m2)v3
using american provincial units
m1= 3000 lbm
v1= 0 ft/s
m2= 4000 lbm
v2= 35 mph *22/15 ft/s/mph = 51 ft/s
m2v2 = 204000 lbm-ft/s
m1+m2 = 7000 lbm then
v3 = 29.1 ft/s, about 20 mph The mini van slowed the other vehicle
Thus the doorframe hit him at about the same speed as falling from 12.5
feet up; a very serious hit indeed.
The combined 7000 lb vehicles then accelerated him up to essentially the
same speed; the second collison is (204000 lbm-ft/s)/7220 lbm = 28.25
ft/s final velocity (neglecting friction on the ice).
The energy imparted to Jerome's body in the impact was mv^2/2, but we
must use slugs mass to get the units to work properly-
E= 220/32.2 * 28.25^2 /2 = 2727 ft-lb
Guessing at the distance over which the acceleration took place at about
.5 ft (human bodies are squishy), the force was about 5500 lbf and he
pulled about 25 gees. Ouch, indeed. I hope you're recovering well.
-- Doug Jones, Rocket Plumber
This archive was generated by hypermail 2.1.5 : Fri Nov 01 2002 - 13:37:40 MST