Dot Physics

Have I not made it clear how much I like the MythBusters? Also, I am totally aware that they are not (nor do they claim to be) scientists. Really, this is what makes their show appealing (maybe?). So here is the problem now. And, it is not just the MythBusters – I see other shows making the same mistake.

If two things are colliding, how do you characterize the collision? No one really gets this right. For this particular episode, the MythBusters were looking at the collision between a bullet and gun. They wanted to see how hard it would be to shoot a gun out of someone’s hand. The question was: what kind of impact would it take to knock the gun out of the hand? No one wants to actually hold a gun while someone shoots it, so they tried to do something comparable. And here is the problem. What has to be the same for it to be a comparable collision. This is what Jamie came up with: a baseball bat. And here is his analysis:

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Let us ignore the error confusing potential and kinetic energy. Clearly, he means kinetic energy. Well, I understand what he is trying to say. He is saying this baseball bat has the potential (when it hits) to create the following collision. But, this is a bad idea. The kinetic energy of an object does not characterize what kind of collision it will have. Suppose I cover the baseball bat with really squishy foam. Further suppose that this doesn’t significantly change the mass (or I shave off some of the wood to keep the mass constant). Also, assume that I could swing it the same speed, then it would have the same kinetic energy. However, this squishy bat collision would be significantly different than a hard bat collision.

So, kinetic energy of an object is not a good method for characterizing a collision. But before I go on. Let me check Jamie’s math – you know, just because. He says the bat has a mass of 800 grams and a speed of 85 mph. Kinetic energy can be calculated as:

i-7be0f278e87ff9e3861bfcc676697b82-2009-12-10_la_te_xi_t_1_2.jpg

I would like to calculate this kinetic energy in Joules. To do that, I need the mass in kg and the speed in m/s. I have a more detailed post on unit conversions, but google calculator does an awesome job also. The mass of the bat is 0.8 kg, and the speed is:

i-e60b50a581d7abd2b9a790dd2aebc731-2009-12-10_speed_conversion.jpg

This gives a kinetic energy for the bat = 578 Joules. Again, I can use google calculator to check this in foot-pounds and I get 426 foot-pounds. Ok, good enough (even though that is a weird unit for energy)

Doing the same for the bullet, I get a kinetic energy of 563 Joules. Close enough.

What about momentum?

When people think about collisions, momentum often comes up. Would this be a good way to characterize a collision? Again, no. The same reason as above. If I have a baseball bat covered with foam, I could make it have the same momentum as the bullet (at a different speed) but it would have a different effect during the collision. How fast would the bat have to be moving to have the same momentum (note that I am pretending like it is a free object and not being held by the bat swinger).

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And now for the bat:

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I know what you are thinking. If an object had the same momentum AND kinetic energy as the bullet, it would have the same collision. Again, this doesn’t work. Think of two objects with the same mass and same speed. These would have the same momentum and same kinetic energy. Now what if one of these is a pillow and one is a brick? Will the interact the same with a gun that they hit? No.

The answer for collisions

I think the best thing to characterize these collisions is momentum of the object and time of impact. For the case of the gun in the hand, really you are asking “how much force would I need to exert on the gun to make it fall out”. This comes straight from the momentum principle. If I look at the forces acting on the thrown object (say it is a bat), then:

i-71ec727eca4b63e631097377b440776b-2009-12-10_la_te_xi_t_1_25.jpg

If I assume the only force acting on the bullet is the target object, then you can see that for the same change in momentums, you can have very different forces (depending on the time). Let me make an extreme example. Suppose I can push a car and run really fast (I can push a car, but I can’t run very fast). And suppose I push this car for 2 minutes while exerting a net force of 300 Newtons (I think I could do that). If the car started from rest, it would have a momentum of: (I am going to drop the vector notation and just deal with the x-direction)

i-5d50adc0f83e10b181050841f624c078-2009-12-10_la_te_xi_t_1_28.jpg

Now suppose you are standing up against a wall and the car crashes into and takes 1 second to stop. Here is a picture:

i-224a6552c3eb054ebf9b40bbf29212aa-2009-12-10_untitled_2.jpg

I guess I should have pushed the car the other way so that my positive x-axis would be to the right. Oh well. Now, what force would be exerted on the person if the car stops in 1 second?

i-2e8475c8ba2b6943dc24ba68e4eb92af-2009-12-10_la_te_xi_t_1_27.jpg

This is the same mistake Jamie makes. He assumes that if you shoot a bullet, the forces from the shooting are the same as when the bullet collides with the gun. So, he made a special gun that shoots from the side to simulate getting hit by a bullet. This is his first version that actually failed to fire (but has the concept).

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Comments

  1. #1 Markk
    December 10, 2009

    I thought the same thing when I watched, but really though, in the bat situation how different are the cases? You are talking about a very short interval for the actual collision in both cases and pretty much inelastic collisions. So even though I agree with you in the sense they maybe should have categorized it as momentum and time of collision, and how elastic the collision was, this I think was close enough to get a feel.
    In the gun shooting sideways, likewise the interval for momentum transfer when the gun went off is still pretty darn short so somewhat similar to an impact.

    The interesting part for me was how the bullet shattered and would have caused so much shrapnel damage. Pretty nasty in any event!

  2. #2 Dale Basler
    December 10, 2009

    How could we measure the force of bullet on gun? How about a ballistics pendulum?

    Hang the gun from a string and shoot it. Measure the time of collision with high speed camera and calculate the speed after collision based on how high it swings.

    Still, I think this would be too high since the bullet pushes the gun and hand. So, now what? Add the some mass to the gun to compensate for the hand? But bigger hands might have better grips– aagh!

    So we ballpark the new force and then go back to the double handle gun (because, let’s face it, it was cool) but this time can we tune in the powder in the bullet so if fires with less force (is that even possible?- I don’t know enough about ammo).

  3. #3 Rhett Allain
    December 10, 2009

    @Dale,

    Even if you measured the time of the collision between the bullet and the block of the pendulum (which would be a very short time), that would tell you the force the bullet exerts on the block. It would not tell you the force the bullet would exert hitting a gun.

  4. #4 Adam_Y
    December 10, 2009

    There is something missing from your analysis because the last time they categorized a collision using your method they got the answer wrong. I want to say its because in order to measure the force imparted by the object you have to measure its momentum after the collision.

  5. #5 Terry Harding
    December 10, 2009

    A problem I had was that the pieces of bullet kept going in roughly the same direction as the original bullet. This must have carried a significant chunk of the total energy/momentum so that it was not actually delivered to the target gun.

  6. #6 Dale Basler
    December 10, 2009

    No, not a block. Use the GUN as the pendulum. Shoot a bullet at a gun hanging from a string(s).

  7. #7 Rhett Allain
    December 10, 2009

    @Dale,

    OH! I get it. However, that still wouldn’t tell you about the forces involved in the collision. It would just tell you about the momentum after the collision. If you have a long collision bullet (squishy) and a hard one that were otherwise identical, the pendulum gun would behave the same way.

  8. #8 Dale Basler
    December 10, 2009

    Then we divide the momentum by time of the collision which we find from the high speed video.

    Of course, this would give you average force during the collision not the peak force.

  9. #9 Jaime
    December 11, 2009

    I’m afraid they get it very right by looking at kinetic energy…

    To shoot a gun out of someone’s hand you basically have to exert a force which is higher than the hand’s grasp, over a distance that is larger than the fingers’ range of motion. Call them F_h and x_f. If you are going to get that done shooting an object at the gun, the single most important thing to consider is the kinetic energy of the object shot, which has to exceed F_h.x_f.

    Think of the bullet as a mass M, with a spring with constant k in front. Make k infinite, and you have a perfectly rigid body, make k small, and you are shooting a marshmallow… If you wanted to take into account dissipation of energy during the collision, you could add a damper in parallel with the spring, but I’ll forgo that.

    With this simple model, very rigid objects will exert a very high force initially, but the corresponding reaction will drain their kinetic energy correspondingly fast. And unless they have enough energy, they will be stopped before freeing the gun from the hand.

    On the other hand, very elastic bodies will initially not generate enough force to move the gun. But the correspondingly small reaction force will do little to stop the elastic bullet, which will continue to deform against the contact point, increasing the force exerted. Eventually, if the energy stored is sufficiently high, the force exerted will overcome F_h. And if there’s enough energy in store, the force will remain above F_h while travelling x_f until the gun is released from the hand.

    Comparable situations arise when ionizing an atom, and the standard measure is an energy of ionization. Or when molecules evaporate from a liquid into a vapor, and again you have an energy, the latent heat. Resilience, the ability of a material to withstand impacts, which is a measure of (lack of) brittleness is also measured in energy units, the area below the strain-stress curve for the material.

    And more down to Earth, if you get hit with a padded bat, it may not break your skull or your ribs, because the padding will reduce the peak impact force. But it will send you flying just as if it was a wooden bat. Get shot in the chest with a rubber ball, and you’ll fall on you back, just as if it were a real bullet, only you’ll live to tell it.

    It’d be interesting to model the thing more rigurously, taking into account loss of energy due to plastic deformation of the projectile. That is, adding the damper I commented earlier. But my guess is that it will not make things change too significantly.

  10. #10 D. C. Sessions
    December 11, 2009

    Always assuming that “shooting the gun from his hand” is primarily a phenomenon in gross physics.

    From all the analysis I read ages ago (hey, growing up in the 50s with westerns!) the main reason that people drop weapons which are struck by bullets [1] is that the bullet->gun impact is pretty elastic and very short. The result is high-frequency acoustic propagation in the gun (much higher than when firing) and the main lossy element is the person holding the gun.

    In other words, it hurts like hell. Reflexive release of the hand and the gun falls.

    [1] It has happened — primarily to riflemen in combat.