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My mom and I were in a car accident. We are ok, but I want to know how fast the car that hit us was going. We were stopped at a light. The car that hit us from behind was a big GMC SUV. Our car was a Cadillac srx and mom looked it up it weighs 4,277 pounds. Then we hit a little car in front of us at 12mph and the air bag went off!

Mom says the car behind us must have going faster than that because she says our car absorbed some kinetic energy. I don't know what that means but I think it was going slower and the hit pushed us to go faster.

How can we figure out what actually happened?

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The asker wants to know the initial velocity of the SUV. –  Photon Apr 24 '14 at 18:39
@Photon well, but if that's all user45198 is asking, this would qualify as a no-effort homework-like question under our policy. I want to think there is something more here. –  David Z Apr 24 '14 at 18:42
Hi thanks for answering my question. It's not my homework I just wondered how fast the car was going that hit us. I wondered if since we were stopped if the car that hit us was going faster or slower than we hit the car in front if us. My mom added the part about kinetic energy because she is trying to teach me about science lol!!! I am ok but got a big bump on my head. –  user45198 Apr 24 '14 at 18:48
@user45198 Sure, we understand this is not a homework assignment, but any sort of question where people just ask us to calculate something for them is what we call "homework-like", and is off topic here. Better questions are those that ask about some conceptual misunderstanding - so, for instance, you calculated the answer yourself but you're confused about why some particular step works. –  David Z Apr 24 '14 at 21:22
I think this is a valid question, given the posters age and experience. He/she want to know if it is possible (and how) to calculate the speed of the car from the evidence. –  ja72 Apr 24 '14 at 23:20

3 Answers 3

To figure this out, you need to know about momentum ($p$). That's a combination of how fast something is moving ($v$, for velocity) and how much it weighs ($m$, for mass). You'll also need to understand algebra, which is just using a letter to mean some number you don't know yet.

$$ p = m\cdot v $$

Momentum is conserved, which means the momentum from both your car and the other car before the crash is the same as the momentum from the two cars after the crash (but which car has how much momentum can change). As a formula, that's this:

$$ m_{you}\cdot v_{you} + m_{them}\cdot v_{them} = p = m_{you}\cdot v'_{you} + m_{them}\cdot v'_{them} $$

We know that your car wasn't moving, so we can get rid of $m_{you} v_{you}$ on the left side. When your cars hit, they probably stuck together (that's called an inelastic collision, and it means your mother was right and the cars absorbed some energy). So they were going the same speed, $v'$. Applying both of those facts to the formula above gets us this:

$$ m_{them} \cdot v_{them} = (m_{you} + m_{them}) \cdot v' $$

This is pretty easy to solve

$$ v_{them} = \frac{m_{you} + m_{them}}{m_{them}} \cdot v' $$

Assuming you hit the next car before you had time to slow down, we know $v'$. We also know your car's mass, so we just need to know the other car's mass. After searching a bit, we'll guess their car weighed about 6500 pounds. Now plug in the numbers:

$$ v_{them} = \frac{4277 lbs + 6500 lbs}{6500 lbs} \cdot 12 mph = 19.896 mph $$

Since we're not quite certain what their car weighed, we'll just round a bit and say they were going at 20 mph.

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Thank goodness someone stepped in to point out that this is a momentum problem, not a KE problem! –  Carl Witthoft Apr 24 '14 at 20:01
@CarlWitthoft: Well, it is a problem of both. There are two unknowns, the velocities of the two objects after collision. So you need two equations to solve them. Conservation of momentum provides one. The other depends on how elastic the collision is. In perfectly elastic one, all energy remains as kinetic and conservation of energy is used. In non-elastic one the objects end up with the same velocity and conservation of energy just tells how much energy was released as sound and heat. Real cases are somewhere between these two. –  Jan Hudec Apr 24 '14 at 20:14
@JanHudec Yes, though car crashes like this are normally pretty far towards the inelastic side. –  Kevin Apr 24 '14 at 20:17
It is likely the driver of the car had their foot on the brake absorbing some momentum before it hit the car in front of them. –  ja72 Apr 24 '14 at 20:20

I'm not sure that there is a good way to figure out exactly what happened unless there is video of the crash, however both statements can be true. There will have been some loss of energy in the collision due to overcoming the kinetic friction of your car (getting it moving from being stopped) as well as the energy absorbed by permanently deforming the car's structure, however car bumpers are also designed to buffer energy.

In this sense, they are kind of like a spring. If the car behind you was much larger, if it was moving slowly, it still would have had a large amount of energy. When it hit your car, it is possible that energy was transferred in to the bumper and then, as the bumper released that energy, it would have pushed your car forward with a portion of that energy. If there was enough energy, it could have easily pushed your car forward faster than the original vehicle was going, so long as it was much larger and so long as your bumper was a big enough spring.

It is important to note that all the energy the car behind you put in to the system had to come out somewhere, some of it was in crushing the car, some of it was heat, some of it was sound and some of it was making your car move. Energy can't be destroyed, so all the energy that went in to the crash had to go somewhere. Friction for example, produces heat.

There is, unfortunately however, no way to tell after the fact how much energy went in to heat vs sound vs deforming your car, nor is it easy to determine how much energy was lost prior to hitting the car in front of you from other factors such as braking, so there isn't any good way to get back to the original speed of the SUV. It could have been slower, it could have been faster, all depends on how much energy was dissipated by sources other than your car taking on momentum.

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you should also note that since it's an inelastic collision (the kinetic energy is not conserved), the total energy is always conserved, so the kinetic energy "lost" in the collision has to convert into some other form: heat, sound, etc. –  Krazer Apr 24 '14 at 18:35
@Krazer: I find it funny that most people tend to ignore the most obvious source of energy dissipation in car crashes: deformation of solids. Good that AJ correctly included it. –  Kyle Kanos Apr 24 '14 at 18:52
@KyleKanos - thanks, my engineering knowledge gets more practice than my physics, though not by much, so I naturally go to the engineering side of it. Last seriously did physics in AP Physics in 2001... :/ –  AJ Henderson Apr 24 '14 at 18:54
Hi thanks for answering my question. It's not my homework I just wondered how fast the car was going that hit us. I wondered if since we were stopped if the car that hit us was going faster or slower than we hit the car in front if us. My mom added the part about kinetic energy because she is trying to teach me about science lol!!! I am ok but got a big bump on my head. – user45198 7 mins ago delete –  user45198 Apr 24 '14 at 18:57
@user45198 - you are welcome. –  AJ Henderson Apr 24 '14 at 19:11

Here are the step to work out what happened, including the missing pieces of information. I am calling your car "Car 1" with mass $m_1=4800\,{\rm lbs}$ and the car that hit your "Car 2" with mass $m_2=6500\,{\rm lbs}$.

  1. Car two is moving with unknown speed $v_2$ and then before they hit you they hit the brakes for a distance of $d$. This is when the police brings out the little wheel and measure the distance of the skid marks. If you knew the impact speed $v_I$ and the brake distance $d$ (in feet) you can work out the traveling speed $v_2 = \sqrt{ v_I^2 + 29.78\; d}$ assuming 1g of deceleration during hard braking.
  2. Here is where the police inspect the damage to the car to estimate the impact speed based on crash tests of cars. Since we do not have this information, I can work out the momentum transfer during the impact instead. With an inelastic contact (all things stick together) the momentum transfer is $J= \frac{v_I}{ \frac{1}{m_1} + \frac{1}{m_2} } = 2761\, v_I$
  3. Your car started moving at this point. Your launch speed was $v_L = \frac{J}{m_1} = 0.572\,v_I$.
  4. I assume your mom was applying the brake at this point for a distance (I estimate) of $b=8\,{\rm ft}$ until it hit the car in front of you. Now we know the impact with the car ahead of you as $v_T = 12\,{\rm mph}$ which allows as to estimate the launch speed as $v_L = \sqrt{ v_T^2 + 29.78\,b} = 19.55 \,{\rm mph}$
  5. So the impact speed is (from step 3) $v_I = 34.0\,{\rm mph}$
  6. And the traveling speed was (from step 2) $v_2 = \sqrt{ 34^2 + 29.78\,d}$ where $d$ again is the reaction braking distance. For a typical crash to a stationary vehicle we can assume this is about $d=50\,{\rm ft}$ making the traveling speed $\boxed{v_1 = 51.4\,{\rm mph}}$.


NOTE: Those magic coefficients I used are because of unit conversions to get the quantities to and from consistent units.

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