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Astronaut and space shuttle momentum

  1. Jun 17, 2011 #1
    1. The problem statement, all variables and given/known data

    An astronaut is in trouble. He is outside the space shuttle and he is moving away from the space shuttle at speed 4.00 m/s. He has a rocket gas tank with which to change his velocity. The gas is ejected from the tank at relative speed 101.0 m/s; the remaining mass of gas in the tank is 1.6 kg. The mass of the astronaut is 94.0 kg, and that of the rocket tank is 7.0 kg.
    (a) If he exhausts all the gas in the tank, what will be his velocity (relative to the space shuttle)?
    (b) Since the velocity in (a) is still moving away from the space shuttle, the astronaut will be lost unless he can throw the tank away with a high enough speed to recoil toward the shuttle. What minimum final velocity of the tank (relative to the space shuttle) will allow the astronaut to reach the shuttle?

    2. Relevant equations

    3. The attempt at a solution
    The answer to part a is 2.413m/s
    so using convservation of momentum:
    (M_astronaut+M_tank)V_part A=M_astronaut*(4m/s)-(M_tank*Vfinal)


    then I added 2.413 m/s to this velocity because he must overcome the velocity he has traveling away from the shuttle so v=21.311m/s, however my homework program says this is not correct.
  2. jcsd
  3. Jun 17, 2011 #2
    Your problem is that there should be two unknowns on the right hand of the equation. From part a) we know that the astronaut and the tank are now moving together at a speed 2.413. After the astronaut pushes off of the tank we don't know either the velocity of the tank or the velocity of the astronaut. But the final velocity of the astronaut depends on the final velocity of the tank. So, we can solve the equationg to find what minimum speed the tank needs to go so that the astronaut goes in the other direction.
  4. Jun 17, 2011 #3


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    Staff: Mentor

    Let's say that ma is the mass of the astronaut, mt the mass of the tank, vA the velocity away from the shuttle after part A (so vA = 2.413 m/s). Velocities are with respect to the shuttle. After flinging the tank let va be the astronaut's velocity and vt the tank's.

    Your conservation of momentum should then look like:
    [tex] (m_a + m_t) v_A = m_a v_a + m_t v_t [/tex]
    "Breakeven" occurs when the astronaut's relative velocity ends up being zero...
    Last edited: Jun 17, 2011
  5. Jun 17, 2011 #4
    well you can assume that if his v is at least 4 m/s to overcome his initial velocity relative to the ship, so there shouldonly be one unknown and that is the v of the tank
  6. Jun 17, 2011 #5
    No you can't assume that. There is a new set of conditions for b) that only indirectly relate to the initial velocity of the astronaut, tank, and gas in part a). gneill has the right idea...
  7. Jun 17, 2011 #6
    ok so then i have two unknowns what would my second equation be?
  8. Jun 17, 2011 #7
    You don't really need another equation based on the question. Solve this one equation for the velocity of the astronaut and figure out what velocity the tank has to be moving so that the velocity of the astronaut is towards the spaceship. Since we have chosen the direction away from the ship to be the positive direction, we want v_a < 0
  9. Jun 17, 2011 #8
    got it thanks.
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