Dot Physics

I am so pumped up that MythBusters is back on. Not only do I like the show, but it offers so many blogging opportunities. Their latest show featured car crashing myths. One of the myths from the episode was a redo of the myth where two trucks crash head on simultaneously crushing a smaller car in the middle.

The first test was very similar to the previous time they tested this, but faster. They towed two 18 wheelers to crash together around 50 mph and smash a stationary car. The results were impressive. However, they did not have the result of the car completely contained in the collision.

For the next test, they changed it so that a rocket sled crashed into a stationary car next to a stationary wall. Adam made a claim (or statement) something about if the rocket car was going 100 mph into a stationary car, it would be equivalent to two 50 mph cars colliding. Actually, this is not true. I can’t remember if Adam actually said this or if he really meant it, however it is still and interesting situation.

Let me start with a couple of somewhat similar situations.

Two cars, in space going the same speed, smashing a stationary object

i-26f37b93d5e7f0bc60cdb34ab7a3dd37-case-1a.jpg

Remember that in space, no one can hear you scream. Also, this makes it a little simpler to keep track of all the interactions. I will assume this is far from any massive objects so that gravity can be neglected. Clearly, there is no air resistance. The lines behind the cars are zooming motion they make. Assume they already turned off their rockets. The smiley face in the middle is some target car or object. Finally, I am looking at this in the frame where the target is stationary. Two things must be true:

i-798726de431ebbbb576ed8e10933d9e7-conservation.jpg

The total momentum vector before and after the collision must be the same. This will be a constant value as long as there are no external forces, like this situation. So the total momentum vector is the zero vector. This is because the target object is not moving, and the momentums of the two cars are opposite each other.

For the energy, this will be constant as long as there is no work done on the system. Here the system consists of the two cars and the target. No work done. If these cars are moving at normal car speeds (or even normal rocket speeds) that are much less than the speed of light, then the mass energy will not really change. Before the collision all the energy is essentially kinetic energy. After the collision this energy can be either in kinetic energy of the moving objects or in something I will call structural energy in the deformation of the vehicles.

To look at the different things that can happen, let me simplify by looking at the component of momentum in the direction of motion (since this is 1-d problem). This will give the initial momentum as:

i-8caf5f55a8f79e9e0f014bcf11a936fe-initial-momentum-1.jpg

mc is the mass of the car(s), mt is the mass of the target. The target is at rest, so its component of momentum is zero in that direction. The energy before the collision is:

i-f16d6dd8bc945cefb5e5be7aa387278e-total-energy-1.jpg

I neglected the mass energy of the objects since it doesn’t really change. So, I can have any situation in which the total component of momentum is zero and the total energy is mcv2. One such case would be in the two cars simply “bounce” off the target.

i-c0cd6ca8eff1bda76b38a7378de5d6a2-result-1a.jpg

For this case, it is obvious that the momentum and the energy are the same as before.

Another case is when the two cars smash and stop at the target. In this case, the final momentum would be:

i-5fe45a4f2f3fc194844411b39004dbb4-crash-stop.jpg

Forgive me for not putting units with the velocity – but you see this would indeed give the same initial momentum. What about the energy? Clearly the kinetic energy is zero so:

i-2b40285a1fdb8a83f71079de7bef21cf-estructural.jpg

I am calling structural energy the change in the system of the cars. Clearly it takes some energy to twist all that metal and stuff. This is the key point though. If the mythbusters are trying to smash up this target car, they want to look at how much structural energy (I made that word up) goes into the thing. In this case, it is mcv2 energy goes into changing all the cars (who knows how that energy is divided up). Now for the next case.

One car moving at 2v1 crashing into an identical stationary car and target car

i-dedc3faa9e9c72f75ac94dbd524a8d01-case-2pic.jpg

Here I am starting one car with twice the speed of the previous case. So, the initial momentum is:

i-51eda8ccd6a18f383410cee1757757ff-onestationary-momentum.jpg

The initial momentum is NOT zero in this case. Duh, that was obvious. I guess I didn’t even need to say that. Oh well, what about the energy before the collision?

i-a8b625c150068bf979cc8c441ad52018-onestationary-init-energy.jpg

So, by doubling the speed of one of the cars and making the other stationary, the initial energy is twice as much. This happens because the kinetic energy depends on the square of the velocity. Also, the initial momentum is not zero so the final momentum will not be zero. What if all the stuff sticks together during the collision? How much energy would go into structural energy? If everything sticks together, what would be its final speed? According to momentum:

i-44849ee874defd4dbba03fe028f1e530-stuck-together-energy-1.jpg

So, it wouldn’t be stationary. The energy equation for this case would be:

i-9ff09158672f55503af86ef59f872bf0-onestationy-struc-energy.jpg

All the energy can not be transfered into structural energy because the stuff still has to be moving after the collision in order to conserve momentum. Ok, well, what about the mythbusters case? It was different because they didn’t do it in space. Exactly. So let me look at that case and see how it is different.

One moving, the other fixed to the ground and not in space

i-1af8f0b24a92339ff24d7a2657b085e4-collision-on-ground.jpg

If I again take the two cars plus the target as the system, there is a huge difference in this situation: external forces. Not gravity, that doesn’t really do anything since the motion is perpendicular to gravity and the normal force from the ground essentially is opposite of it. However, the force of the ground on the stationary car IS important. This means that for this system, the initial and final momentum are not the same. The momentum principle can be written as:

i-da07edbaf44ee303efab8ac335765b55-momentum-principle-ground.jpg

Where the net force on the system is the force the ground exerts on the stationary car when it gets hit and delta t is the time the collision lasts (short). This is why the final momentum is not the same as the initial. In the frame of the Earth, the final momentum is zero (vector). What about energy?

In this system, the external forces don’t do any work. The force from the ground doesn’t actually move, so there is no work. This means that all the energy of the initial moving car can go into structural energy so that:

i-21e2ddc42bf7e16c8675f294803e7907-estruct-on-ground.jpg

Ok, one more case. Is there anyway to make one object stationary and the other moving and have everything work? What if I did this in space (for simplicity) and had one of the cars stationary? One thing would be to look at the collision in the reference frame of one of the cars. This would be a little different.

Reference frame of one of the moving cars

i-4dcc2ecef6fac81a885d7fead85bc231-frame-b.jpg

If you look at the first case I did, and pretend like you are moving with the right car (which I labeled as “B”), then you would see the other car (A) moving towards you at a speed of 2v1. But you would also see the target moving towards you at a speed v1. This would give a total initial momentum of:

i-de0b3e036e6b883ab75aa7f7886ce733-space-p-collsions.jpg

If everything sticks together, it will be moving at a speed of (in the frame of car B):

i-60bfe4a60a7c96be231b1761b8812c65-car-b-frame-v-2.jpg

So, the final speed of the mangled stuff is the same initial speed in the frame of the target as the two moving cars. This makes sense. If in the frame of the target, everything is stopped at the end and in the frame of the car B, the target is moving towards you at speed v1 then after it should still be v1. So, the two frames agree. What about energy?

i-970e905f729860c069d3928e3895f3ef-carbenergyinitial-1.jpg

From the energy (before and after) equation, I can solve for the structural energy:

i-0a0a25c1d775f56d6d016e2b6b26b013-fixed-equation.jpg

This is the EXACT same structural energy from before. That is good. The two cases should agree on something measurable like that. So, in summary, what is the point of this post? First, making one car stationary and the other moving twice as fast is NOT the same thing. Second, the MythBusters are still awesome. Third, if you move to a different reference frame, stuff should still work.

Update

There was an algebra mistake as pointed out by the commenters (thanks!). I fixed it. Strange this is that I derived that for the last case the E-structural was 2m(v1)^2 and I said this was IDENTICAL as viewed from the other frame (which it isn’t). Odd. When I fixed it, the two are the same. I guess that’s what happens when you blog on vacation.

Comments

  1. #1 Anonymous Coward
    April 10, 2009

    I’m confused by your “point of this post” in which you say that the one car stationary and the other moving at 2v is NOT the same thing (as two cars moving at speed v in opposite directions).

    I agree that this is true when the stationary car is buttressed by an immovable wall, but you seem to be implying that it’s also true in the absence of a wall. Is that what you mean to imply?

  2. #2 Lurker #753
    April 11, 2009

    AC:
    Yes, that is what he means.

    Energy scales with square of velocity. 2x speed = 4x energy, 2x mass = 2x energy

    1 car @ ~70 would roughly match energy 2 @ 50. (sqrt(2)=~1.4

  3. #3 PhysicsDude
    April 11, 2009

    I think you are confused about what your equations say.

    If you simplify the problem by assuming m_t is tiny (set m_t = 0), you’ll see that the second scenario (two cars, one-stationary, one moving at 100mph) has collision energy m_c v_1^2, the same as the first scenario (two cars colliding at 50mph). This makes perfect sense. Both these scenarios are identical from the cars’ relative viewpoints. The only reason these two scenarios are slightly different is because the “target” has different relative movement in each.

    The only scenario thats get you energy “2 m_c v_1^2″ is the hitting a stationary wall scenario. This scenario results in twice as much collision energy as any of the others.

  4. #4 PhysicsDude
    April 11, 2009

    In particular, in the second scenario, plugging in “m_t = 0″ gives you a factor of (1/2) which makes E_structural = m_c (v_1)^2

  5. #5 Uncle Al
    April 11, 2009

    The Mythbusters are engineer craftsmen not scientists. Point a finger at their technical consultants and editors. Yeah, Adam blew it.

    Had they put a hollow low-mass frangible aeroshell in front of the rocket-accelerated flat plate the Cd would have been much smaller. Air resistance varies as v^2 and has a trans-sonic discontinuity. Why waste energy pushing air aside when it could be accelerating the impactor?

    Steel melts around 1550 C, aluminum melts around 660 C. Clean aluminum surfaces (before oxidation) are soft and sticky. Intense deformation and heating therefrom of an aluminum car body or engine block might lead to “welding”.

  6. #6 PhysicsDude
    April 11, 2009

    Under heading “Reference frame of one of the moving cars”, you make an algebra mistake at the very last step. The result is m_c v_1^2, not 2 m_c v_1^2.

  7. #7 PhysicsDude
    April 11, 2009

    In fact, Adam’s claim IS true. You should really double check your work.

  8. #8 Lurker #753
    April 11, 2009

    PhysicsDude:
    Um, if Adam’s claim is true, there should exist a reference frame from which the events look identical, and I can’t find it. Help?
    Obviously, the effect of a 100mph car won’t be the same in reality, since the model does not discuss where the structural energy ends up. The car is not massless, and the driving truck will suffer disproportionate damage as it provides all the energy to accelerate it into the other truck. (Historical example in how coinage was struck: the obverse die is mounted in a stand face up; a blank is laid on it; the reverse die is inverted over it, and then the stack is struck from above. The result is that reverse dies were replaced more often (roughly 2x, IIRC) than obverse.)

  9. #9 marc
    April 11, 2009

    It doesn’t seem to make sense that a 100/0mph pair aren’t the same as a 50/50mph, after all, as you point out, the only difference between the two situations is effectively whether the observer is moving. I think that the problem you have is that KE is not invariant between frames of reference.

  10. #10 Lurker #753
    April 12, 2009

    marc: no, I’m happy with subtracting off KE depending on reference frame, but there are real differences: in the first case, the r.v. of the target w.r.t. to the right hand car is 50, and in the second, it’s 0; likewise the r.v. of the left-hand car w.r.t the target is 100, not 50.

  11. #11 Alfredo Louro
    April 12, 2009

    You and Adam are both wrong. There’s an algebra error in your last case. In fact, the result is the same as in the first case. Adam is also wrong in claiming that the collision against a wall is equivalent to two cars travelling in opposite directions at half the speed, as you have demonstrated. The two cases that Adam is comparing are not related by a coordinate tranformation.

  12. #12 Rhett
    April 12, 2009

    I totally believe that I could have made an algebra mistake – but for the life of me, I can’t see it. Maybe it is one of those cases where you just see the same mistake over and over. Anyway, I think the main points I am trying to make are:

    1.Doubling the speed is not double the kinetic energy.
    2. Things should still work out in a different reference frame (the E-structural should be the same because the things get bent the same amount).
    3. In the case of the stationary car on earth, there are external forces on the system.

  13. #13 Alfredo Louro
    April 12, 2009

    Your very last equation: (1/2)(4m_c – 2m_c) = m_c, not 2m_c.

  14. #14 Rhett
    April 12, 2009

    Alfredo and others,

    thanks for pointing out my mistake. I fixed it.

  15. #15 Anonymous Coward
    April 14, 2009

    Perhaps you should also change your answer to “what is the point of the post?” from the incorrect
    “making one car stationary and the other moving twice as fast is NOT the same thing.”
    to something like
    “making one car stationary and the other moving twice as fast IS the same thing in the absence of a wall, but NOT the same thing in the presence of a wall”
    ?

  16. #16 Robert Carnegie
    April 16, 2009

    If you are observing from car A2 which you are driving alongside car A1 at the same velocity when the collision occurs, you should be watching the road.

  17. #17 PhysicsDude
    April 19, 2009

    Energy levels are not frame invariant. You cannot compare the total energy from one frame of reference to another. You can only compare the energy differences.

    For example:
    Suppose there are two 1 kg objects traveling in space at a constant 1 meter/sec away from each other. From the frame of reference of their center, the total energy is:
    (1/2)(1 kg)(1 m/s)^2 + (1/2)(1 kg)(1 m/s)^2 = 1 joule

    From the frame of reference of one of the objects, the total energy is:
    (1/2)(1 kg)(2 m/s)^2 = 2 joule

    Same exact system, but both observers see a different absolute energy level. The only thing observers in all frame of references can agree upon is the change of energy of the system.

  18. #18 PhysicsDude
    April 19, 2009

    (EOC = Energy of Collision)

    1)
    Two cars with velocity “v” colliding head on:
    EOC = (1/2)mv^2 + (1/2)mv^2 = mv^2
    (they stick together, and come to rest)

    2)
    One car with velocity “2*v” hitting a stationary car:
    EOC = (1/2)m(2v)^2 – (1/2)(2m)(v)^2 = mv^2
    (they stick together, and continue moving at velocity v)

    3)
    One car with velocity “2*v” hitting a wall:
    EOC = (1/2)m(2v)^2 = 2mv^2
    (everything comes to a complete stop)

    Cases #1 and #2 have identical collision energies.
    (in fact, #2 is just the same collision as #1 viewed from the frame of reference of either car)

    Case #3 has twice as much energy, because the resulting pile of mass comes to a complete stop from velocity “2*v”.

  19. #19 Carol Braun
    May 9, 2010

    Below your fourth gray box, you write: “Forgive me for not putting units with the velocity.”

    I don’t see why you’d need forgiveness! All the velocities you plug in are equal to zero. When v = 0, units are irrelevant. In fact, it seems strange to write v = 0 mph, v = 0 m/s, etc. It’s like being asked to simplify the expression 5*x – (2+3)*x and leaving your answer as 0*x. Just do the multiplication by zero: Fully simplify your answer! Make the units (x) go away!

  20. #20 Rhett Allain
    May 9, 2010

    @Carol,

    Even if the velocity is zero, I think it should still have units. Mostly I say this to be a good example for students.

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