A Cadillac Escalade is stopped at a red light at a four-way intersection. Light changes to green, operator of the Escalade accelerates at a normal rate and is broadsided by a Ford Mustang traveling at 20mph. Mustang impacts the Escalade on the drivers side door area. Mustang also impacts without braking.
Is it plausible to say that the Mustang pushes the Escalade at least 25 feet and wedges the Escalade between itself and a utility pole?
What was the condition of the pavement? Loose gravel, raining, snowing???
Yeah, I could see that happening at 20mph. Even with the smaller Mustang and the bigger Escalade.
pavement is asphalt, not wet. Sunny day
“Tonight. On a very expensive episode of Mythbusters…”
It’s hard for me to believe it. The Escalade’s pretty heavy, around 5,700 pounds curb weight, but the Mustang’s no featherweight at 3,750 pounds for a six cylinder. Even if the Mustang transferred all of its momentum to the Escalade it would end up going around 13 miles per hour, but it couldn’t transfer all the energy, because A, it kept moving with the Escalade and pinned it against a pole, and B, with a 20 mph collision a lot of that energy is going to go into crumple zones. I doubt that half the energy got to the larger vehicle, but let’s be generous and say all of it stayed with the Escalade and it was moving 13 miles per hour.
Sliding sideways at 13 mph is pretty much the same as going forwards at 13 mph with your brakes locked. These guys have a sitethat calculated the braking distance at 15 mph to be 11 feet. That neglects the perception time, which isn’t relevent here because we aren’t waiting for the driver to step on the brakes.
So, if the Mustang stopped instantly, which it didn’t, and if nothing was absorbed in crumple zones, which it couldn’t, there still wouldn’t be enough energy at 20 mph to do this.
I doubt it.
While I realize that anecdotes don’t equal cites, I was moving at about five MPH into an intersection when I was hit in the front driver-side quarterpanel by a minivan going about 50 MPH. My Volvo S70 was moved barely five feet to the right.
The speed is really not 20MPH but 20 plus the speed of the Escalade vectored an an angle to the pole that is on the corner. If nobody is braking and the wheels are turning, the Escalade is not truly being slid sideways. Rather it’s moving with a combination of sliding and rolling at an angle. I think it’s plausible.
I agree with Al Bundy, especially when you allow for this. In many model years, the Mustang’s front end slanted down like a wedge. As it broadsided the higher Escalade, it would have lifted some of its weight off the near-side tires of the Escalade, thus reducing its resistance to going sideways.
Bill Door, you’re mixing up energy and momentum in that argument. Some (in fact, a very significant amount) of the energy will be lost to heat, in the crumple zones and elsewhere. But momentum is also conserved, and unlike energy, momentum can’t “hide” in non-mechanical forms like heat. The impact can be very straightforwardly and accurately modeled as a maximally-inelastic momentum-conserving collision.
I don’t know what “a normal rate of acceleration” for the Escalade is, nor how far or how long it was accelerating, so I’ll just assume that it was initially at rest, for the sake of simplicity (this shouldn’t be too bad, anyway, since it seems like the perpendicular component is most significant here). Before the collision, then, all of the momentum was in the Mustang, which had an momentum of P = mv = 3750*20 pound miles per hour, or 75000 pmph. Immediately after the collision, the two cars are effectively a single body with a mass of 9450 pounds. For a body with 9450 pounds of mass and a momentum of 75000 pmph, its speed must be 75000/9450, or 7.94 mph.
OK, now we have the speed of the ugly 8-wheeled car right after the collision. What kind of friction coefficient would we need for this pile of wreckage to slide 25 feet? We do this using conservation of energy. The object has a kinetic energy of 1/2 * mv^2. This energy is dissipated over a distance x, via a frictional force of mumg (where mu is the coefficient of friction). If we assume that this was the maximum extent of the sliding, then all of the energy is dissipated this way, so the work (force times distance) is equal to the total initial energy: mumgx = 1/2 * m*v^2, or canceling out the mass and solving for mu, mu = v^2/(2gx).
At this point, I’m going to take the liberty of converting everything to metric, since I don’t know the value of g in miles per square hour off the top of my head. v = 3.55 m/s, x = 7.62 m, and g = 9.8 m/s^2. This means that, for the mangled cars to slide as far as stated, the coefficient of friction must have been at most 0.084 . This might be plausible if the Escalade were lifted completely off the ground and the whole mess was basically rolling on the Mustang’s wheels, but wheels sliding on pavement generally have a mu somewhere in the vicinity of 1, and the Escalade’s wheels would be sliding sideways, not rolling, so if any significant amount of the Escalade’s weight was still resting on its wheels (as I would expect), you wouldn’t be able to slide that far.