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Homework Help: Rest Mass of Compound Object after Collision

  1. Dec 11, 2015 #1
    1. The problem statement, all variables and given/known data
    A meteorite is approaching earth at very high speed. In order to avoid an impact on earth, the world space agency has launched two space missions: one mission sends the astronaut Albert to the meteorite approaching earth. The other mission, lead by the space commander Trebla, sets off to another meteorite of similar mass to deviate it from its original course and make it collide with the first one so that they end up in an orbit around earth rather than crashing into it.

    The first meteorite is elongated with roughly cylindrical shape. Its length is 150 m and its rest mass is ~0.50 Mt (1 * 10^9 kg). Its velocity relative to earth (at the time when the collision is expected) is 0.60c in the direction along its long axis. The second meteorite is roughly spherical with a diameter of 50 m. Its rest mass is 0.41 Mt and due to the alteration of its course it will have a velocity of 0.65 c relative to earth and flying in the exact opposite direction than the first meteorite.

    Albert’s job is to apply specifically developed foam to the expected collision area. It has a very low density and its mass is negligible compared to the meteorite’s mass, but it will absorb the shock so that we end up with a perfectly inelastic collision. This means: after the collision we have one compound object and no material escapes.

    c) What is the rest mass and velocity of this compound meteorite?

    2. Relevant equations

    [itex]E = \gamma mc^2 [/itex]

    [itex]E^2 = p^2 c^2 + m^2 c^4[/itex]

    Conservation of Energy

    Conservation of Momentum

    3. The attempt at a solution

    I've tried using conservation of energy from the Earth's frame because that's the relation of the velocities we are given. Thus [itex] E_1 + E_2 = E_f [/itex] , where [itex]E_f[/itex] is the energy of the compound object after the collision. However, because the momentum is nonzero after the collision (by conservation of momentum) we know that the the compound object is still moving after the collision in the Earth's frame. By that logic the final energy must be expressed as [itex]E_f = \gamma Mc^2 [/itex], where there is still a gamma present because the compound object is moving relative to the Earth. My classmates have disagreed and said that you can solve it through [itex] E_1 + E_2 = Mc^2 [/itex] where M is the compound mass. I'm confused because wouldn't that require the compound object to be in the Earth's frame if you are using the velocities of the meteors relative to the Earth's frame? I thought you would have to use the CM frame, at least.
  2. jcsd
  3. Dec 11, 2015 #2
    With an inelastic collision, there will be no conservation of energy. In fact, that energy-absorbing foam is perhaps the most magical thing about this entire unlikely scenario.

    So you are only looking to conserve momentum.

    p = γm0v
  4. Dec 11, 2015 #3
    Please correct me if I'm wrong, but I thought total energy is always conserved, but kinetic energy is lost during the collision. Hence why I use total relativistic energy, [itex]E = \gamma mc^2 [/itex].
  5. Dec 11, 2015 #4
    Kinetic energy is conserved in an elastic collision - where the objects bounces off each other. But that magic foam is advertised to absorb all of that enormous extra kinetic energy yielding the collision inelastic.
  6. Dec 11, 2015 #5
    It was my understanding that by inelastic collision, all we know is that the bodies stick together and there is a maximum kinetic energy loss. I didn't think that implied all kinetic energy was lost, otherwise the compound object would have zero momentum after the collision, which we know cannot be so because the momentum before the collision is non-zero.
  7. Dec 11, 2015 #6
    You don't loose all of the kinetic energy, but you do loose some. So you cannot use equations that presume that kinetic energy is conserved.
  8. Dec 11, 2015 #7
    Why does [itex]E = \gamma mc^2 [/itex] presume that kinetic energy is conserved? The total energy is conserved because the kinetic energy is converted to mass during the collision, by my understanding.
  9. Dec 11, 2015 #8


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    It's also interesting to calculate the temperature of this "compound meteorite". If a meteorite implies a solid mass, this is quite a misnomer.
  10. Dec 11, 2015 #9
    Could you elaborate? I don't really understand.
  11. Dec 11, 2015 #10


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    Yes, good. Total energy E is conserved (even in an inelastic collision).
    Your classmates are wrong, conceptually. However, I believe that in this particular problem the numbers work out such that ##\gamma## of the compound object is very close to 1 so that your classmates will end up with an answer that essentially agrees with the correct answer.

    What else is conserved besides E? You should be able to find the rest mass of the compound object without using the relation [itex]E_f = \gamma Mc^2 [/itex].
  12. Dec 11, 2015 #11
    As I said, that light weight foam that's absorbing all the extra energy without dispersing is really miraculous.
    He's noting that the energy released from this inelastic collision would vaporize the meteorite. Actually, it would be enough to vaporize whatever cities is orbited over while cooling down.
    There are two energy conversions here. The one you mention is the mass equivalent of the kinetic energy. The other is the conversion of kinetic energy to heat - or some other stored energy. If the meteors were strong enough to survive the collision and then simply bounced off each other, then the total kinetic energy (some of it appearing as mass) would be conserved. That would be an elastic collision. In that case, you kinetic energy equation would be useful.

    But we have this magic foam that absorbs the energy and keeps everything stuck together. So most of the kinetic energy will be converted to heat - or whatever else that foam does. At least in the first instant on this collision, it is perfectly inelastic, so no heat or other energy is lost and therefor, no equivalent mass is lost. So the relativistic rest mass of the new meteor will be the sum of the relativistic rest masses of each old meteor. But it should be noted, that that includes a lot of mass-equivalent energy stored in the new meteor.

    The momentum also will not change. So the momentum of the new meteor will be the sum of the momentums of the old meteors.
    So with meteors a and b combining to c -
    For momentum:
    [itex] \rho_a = \lambda_a m0_a v_a = 1.25 \times 0.5 Mt 0.6c = 0.375 Mt c = 1.12425 \times 10^{17} Kg m/s[/itex]
    [itex] \rho_b = \lambda_b m0_b v_b = ... (your turn)[/itex]
    [itex] \rho_c = \rho_a + \rho_b = 0.375 Mt c + ...[/itex]
    For mass:
    [itex] m_a = \lambda_a m0_a = 1.25 \times 0.5 Mt = 0.625 Mt[/itex]
    [itex] m_b = ...[/itex]
    [itex] m_c = m_a + m_b = ...[/itex]
    For the new meteor
    [itex] \rho_c = \lambda_c m0_c v_c[/itex]
    so [itex] v_c = \rho_c / (\lambda_c m0_c)[/itex]

    I'll give you one final clue with that last equation: Since the meteor is said to be left orbiting the Earth, lambda must be pretty close to 1.
    so taking lambda=1: [itex] v_c = \rho_c / m0_c[/itex] (in units of c)
    Last edited: Dec 11, 2015
  13. Dec 11, 2015 #12
    I just did these computations. The final velocity of the new meteor will exceed the Earth's escape velocity - so it will not end up in orbit.
    Bad thing for the problem - good thing for planet Earth.
  14. Dec 11, 2015 #13

    I think I have solved it.

    Solving for initial mass and momentum, we can make use of [itex]E_2=p^2 c^2+m^2 c^4[/itex] because this formula has a special property- it relates momentum, energy, and rest mass, which is what I'm trying to solve for. We know energy and momentum are conserved so we can solve for [itex]m[/itex], which evaluates to ~1.165 Mt. And you're right, after the collision the velocity is relatively small, so the method others used approximates the correct answer. But it would definitely break down at higher velocities.

    To solve for velocity, I used conservation of energy (relativistic) instead of conservation of momentum because there is only one [itex]v[/itex] to solve for. This evaluated to
    ~6.26 × 10^6 m/s

    Would you say this procedure is correct?
    Last edited: Dec 11, 2015
  15. Dec 11, 2015 #14


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    Looks good. Another approach to get the final speed: What does the ratio p/E represent for a particle?
  16. Dec 11, 2015 #15


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    As you said earlier, total energy must be conserved. Most of the incoming kinetic energy is converted into heat. As .Scott said, the heat liberated is more than enough to completely vaporize both meteorites. I calculated that the temperature would be on the order of 200 MeV, (~10^12 K) which is enough to convert the whole mass into a quark-gluon plasma.
  17. Dec 11, 2015 #16


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    Yes, but don't forget that "specifically developed foam". If it can completely absorb the impact to make it perfectly inelastic, then it should have no problem holding everything together so you get "one compound object and no material escapes". :wink:.
  18. Dec 11, 2015 #17


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    Right! Really good stuff that foam.
  19. Dec 12, 2015 #18
    [itex]p = \gamma m v [/itex]
    [itex]E = \gamma m c^2 [/itex]

    [itex]\frac{p}{E} = \frac{ \gamma m v}{ \gamma m c^2}[/itex]
    [itex]\frac{p}{E} = \frac{v}{c^2}[/itex]

    Amazing stuff.
  20. Dec 12, 2015 #19


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    Good work.
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