Comparing CollisionsDate: 05/20/2002 at 23:56:10 From: Steve Ginkel Subject: kenetic energy If there are two cars with the same weight going head on at each other at exactly 35 mph, will the crash be equivalent to a 70 mph crash? That is, would it be like one car hitting a brick wall at 70 mph? Date: 05/21/2002 at 09:42:28 From: Doctor Rick Subject: Re: kenetic energy Hi, Steve. The relative velocity between the cars is the same as the relative velocity between the car and the brick wall. Thus the crash happens just as fast, and that is significant. But to decide which crash does more damage, we need to consider how much energy a car absorbs in the collision; that's the best measure we can get of the amount of damage sustained. In the head-on collision, the total momentum is zero. If all the kinetic energy is absorbed by the cars, they will end up stationary. The total momentum after the collision must be the same as before the collision, namely zero, and ending up with both cars stationary is consistent with this. They might also bounce off each other, retaining some kinetic energy and reducing the energy absorbed, but I'm considering the worst case. The amount of kinetic energy absorbed in this case is (1/2)m(35mi/hr)^2 + (1/2)m(35 mi/hr)^2 = 1225m where m is the mass of each car. That's in strange units (pound-miles squared per hour squared), but since all I want to do is compare two numbers, I don't need to convert this to more conventional units. We can suppose, in the head-on collision, that the cars will absorb equal amounts of energy, 612.5m units each. This is not necessarily the case, unless the crash is perfectly symmetrical, but it's a reasonable assumption for simplicity's sake. Now we want to consider the car hitting the wall. To get there, though, let's first think about a similar case: the car going 70 mph smashes into a stationary car of mass M. Only the one car is moving at the start, so the initial kinetic energy is (1/2)m(70 mi/hr)^2 = 2450m The momentum going into the collision is m*70 mi/hr + M*0 mi/hr Let's say the cars are welded together by the impact, so that they move together afterward, with velocity v. The final momentum is (m+M)*v The final momentum must equal the initial momentum, so we can solve for v: v = 70 mi/hr * m/(m+M) The final kinetic energy, then, is (1/2)(m+M)(70*m/(m+M))^2 = (1/2)70^2 * m^2/(m+M) If the two cars are the same mass m (as in the head-on collision), the final kinetic energy is (1/4)70^2 m = 1225m. Thus the total amount of energy that must be absorbed in the impact is 2450m - 1225m = 1225m That's the same as in the head-on collision. Thus we can say that a head-on collision is just as bad as a car going twice as fast hitting a stationary *car*. Again it's reasonable to suppose that each car absorbs half the energy, or 612.5m units. On the other hand, consider the case in which the second car's mass M is much larger than m. Then the denominator (m+M) is not much different from M alone, and the final kinetic energy is (1/2)70^2 * m^2/M The larger the mass M, the smaller the final kinetic energy, and the more energy must be absorbed in the collision. If we consider the wall to be a "car" of essentially infinite mass (because it's attached to the earth), the final kinetic energy is zero, and the full 2450m units of energy must be absorbed in the collision. Thus a head-on collision is only half as bad as hitting a *wall* at twice the speed. I have made assumptions here: for one thing, that the cars stick together rather than bouncing away, in which case there would be some kinetic energy at the end and not all the energy would go into damaging the cars. Also, I assumed that the car and the wall absorb the same amount of energy -- that is, the collision hurts the wall as much as the car -- because that's more or less what will happen when two similar cars collide. When it comes to the car hitting the wall, it isn't so easy to say how the total energy absorbed in the collision is divided -- which would sustain more damage, a car or a wall. If the wall "crumples" more easily than the car, thus absorbing much of the energy, then the car could suffer less damage than my total-energy calculations indicate. I could imagine a "wall" (not brick or concrete, but like a big air bag) absorbing *all* the energy, so the car could drive away from the accident! On the other hand, if the wall is untouched and the car absorbs all the energy, then the wall acts like a mirror: the car hitting the wall is just like a car hitting its reflection -- a car traveling toward it at the same speed. The car absorbs the entire 2450m units of energy, 4 times the energy absorbed by each car in the 35 mph head-on. Those are the extreme possibilities: the energy absorbed by the car when it hits the wall at 70 mph is somewhere between 0 and 4 times the energy it would absorb in a 35 mph head-on collision with an identical car. That's about all I can say without actually doing the experiment! - Doctor Rick, The Math Forum http://mathforum.org/dr.math/ |
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