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Rifle with one barrel and cartridge in middle

  1. Jun 26, 2013 #1
    I have always wondered about equal and opposite. If I had a rifle with a cartridge in the middle of a barrel, with two projectiles, equal mass, facing in opposite directions, would the two projectiles have the same velocity as a single projectile fired from a conventional rifle? All things being equal.
     
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  3. Jun 26, 2013 #2

    sophiecentaur

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    I think the problem needs to be specified more tightly, for a definitive answer but I am assuming the same amount of energy is supplied in both cases and that the energy conversion from the powder is 100% efficient.
    Firstly, Momentum will be conserved in both cases. The momentum for the bullet will be equal and opposite to the momentum of the (Massive / Infinte) gun. This means that the gun would not move, so it would get none of the energy from the propellant.
    For the two-bullet case, the momentum of the bullets will be equal and opposite. They will share the energy, so their speeds will be 1/√2 of the speed of the single bullet.
    A similar (and easier / more ideal to analyse) scenario would be to compare what happens with a compressed (massless, of course) coil spring and a mass on each end and the same spring, squashed against a rock, with just one mass. The same thing applies as with the gun.

    There is another scenario, more commonly discussed and that it the difference between two vehicles colliding and one vehicle colliding with a massive wall. All the same arguments apply here too and you have to be careful to define the situation very exactly if you want the 'right' answer.
     
  4. Jun 26, 2013 #3

    Nugatory

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    No. There are several ways to look at this problem, starting with conservation of energy: in both cases, the energy generated by the burning charge is the same, but in one case it all goes into one projectile, and in the other case we're dividing the same amount of energy between two projectiles. (I'm leaving out the effect of the recoil, which in the first case transfers a small amount of energy to the shoulder of the shooter, but that doesn't affect the calculation significantly).

    Another way of thinking about it: in the two-bullet case, the forces on each bullet are equal and opposite as we'd expect, but in both cases they're generated by the pressure of the combustion gases. In the two bullet case, that pressure falls off more rapidly because the volume occupied by those gases is increasingly more rapidly as the bullets move apart in the barrel. Less pressure, less force, less acceleration, lower muzzle velocity.

    The above is hand-waving, but it's pretty good hand-waving. To really do this problem right without hand-waving and including the recoil in the first case, you can play with the formulas for kinetic energy ##E_k=\frac{mv^2}{2}## (the total energy of everything moving must be the same for both cases, because all that energy comes from the same amount of propellant) and momentum ##p=mv## whose total for everything moving must be zero. You can't get the two cases to come up with the ##v## without breaking conservation of energy or momentum, or both.
     
  5. Jun 26, 2013 #4
    Thank you, I always thought the recoil had the same amount of energy as the projectile, hence my question. btw, Mythbusters did the two car crash, very enlightening.
     
  6. Jun 26, 2013 #5

    sophiecentaur

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    Yes. That's the intuitive thing but, of course, the share of energy depends on relative masses.
    I read a great New Scientist article, yonks ago, on designing the optimum baseball bat. It involved a compromise between having as massive a bat as possible and yet keeping it light enough for the player to be able to get it going fast enough. Just another example of good ol' basic mechanics.
     
  7. Jun 26, 2013 #6

    Nugatory

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    Same amount of momentum, but not energy.
     
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