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Homework Help: Conservation of momentum or kinetic energy?

  1. Oct 16, 2015 #1
    1. The problem statement, all variables and given/known data

    A 100 kg person in space throws a 1kg sphere away from their body. The sphere starts from rest and ends up travelling 10 m/s away from the person. What is the kinetic energy of the person and the kinetic energy of the sphere? Is kinetic energy conserved?

    2. Relevant equations
    1/2 * M * V^2

    3. The attempt at a solution
    1/2 (1kg) * 10^2= 50Joules for the sphere...and 1/2(100kg)(.1m/s)^2= 5 Joules for the person. No. Kinetic energy is not conserved.

    It is odd to me that kinetic energy is not equal and opposite(?)
    Last edited: Oct 16, 2015
  2. jcsd
  3. Oct 16, 2015 #2


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    Staff: Mentor

    Note that kinetic energy is a scalar value and does not have a direction. Kinetic energies are positive values that sum.

    How did you arrive at the velocities for the bodies? The problem states that the sphere ends up traveling away from the person at 10 m/s, so that makes it a relative velocity (also known as the separation velocity). It's not measured with respect to a "fixed" background. It also makes it rather unlikely that one of the objects can have a velocity of 10 m/s in the initial rest frame if both objects are moving after the throw.

    Whatever energy the system has after the throw should be divided between the two objects, and the total kinetic energy will be their sum.

    Think about that sum before and after the throw...
  4. Oct 16, 2015 #3
    What's happening is two objects starting from rest pushing each other so they travel in opposite direction. Kinetic energy is positive by definition.
    It's not really too weird that kinetic energy isn't conserved. Every action that does work changes energy from one form to another. In this case one could assume it's chemical energy from the human that turn into kinetic energy.
    I believe it should be ##0.5 J## in the second case as well. (That ##v_{human} = 0.1m/s## is correct from conservation of momentum.)

    Edit: gneill has a good point about the wording of the question as well, I didn't notice that it said away from the body when i answered.
  5. Oct 16, 2015 #4
    Thank you both. I see I did miscalculate and it should be .5J for the human. The 10m/s is supposed to be absolute. So the question wasn't properly stated. I calculated the velocity of each object by using conservation of momentum. The sphere is travelling at 10m/s and is 1kg so the human is 100kg should be traveling at .1m/s(?) I just had the assumption that kinetic energy would be equal and opposite in this situation and it just -seems- like magic that more energy is put into the sphere (50J) than the human (0.5J).
  6. Oct 16, 2015 #5
    That is correct. Also if you care about direction it should be ##-0.1m/s## but we only need the absolute value here. As for that thinking kinetic energy should be conserved I understand that it may feel natural but it's not true. I think with things like this it's no good way to really explain it, with time doing more exercises and maybe some physical experiment’s you eventually get used too it.
    I think a quote by Von Neumann fitting here "Young man, in mathematics you don't understand things. You just get used to them."
  7. Oct 16, 2015 #6


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    Staff: Mentor

    I would take the frame of reference to be the one where the two objects are at rest before the throw. Then, knowing that the directions that the objects move after the the throw should be in opposite directions, then one velocity should be positive and the other negative (if you use conservation of momentum to find them). Knowing that, you can easily determine an expression to find the separation velocity from the two values. Use that expression in conjunction with the conservation of momentum expression to solve for the velocities.

    The sphere should be traveling at 10 m/s away from the person from the person's point of view. But in the initial "rest" frame of reference, it will be slightly less, the difference being associated with the person's new velocity.
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