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Conservation of Linear momentum - confusing question?

  1. Dec 5, 2014 #1
    1. The problem statement, all variables and given/known data
    pic.png

    2. Relevant equations
    Momentum = mass x velocity

    3. The attempt at a solution
    Answer is A. when mass decreases, speed should increase for momentum to be constant. and when mass increases, velocity decreases. i could understand the graph up the before t2 with this. but at t2, why does speed of smith increases and that of jones decrease. shouldn't it be like in answer E???
     
  2. jcsd
  3. Dec 5, 2014 #2
    You're thinking of momentum conservation in an imprecise way. It's not just "when mass decreases, speed should increase for momentum to be constant. and when mass increases, velocity decreases."
    The correct way of thinking about this is just that the total momentum in the system is conserved, considering that momentum is a vector. Here, instead of talking with vector notation and all that I'll just be saying things like "momentum to the right", but I'm hoping you'll get what you mean- it's very intuitive to think of momentum in this way.

    If we think of just the "Smith-ball" system, then Smith's speed increases after he throws the ball. This is because initially the momentum of the system is some quantity of momentum to the right, and the ball (which is thrown backwards) will acquire momentum to the left, so Smith's momentum to the right must increase to compensate, and conserve the total "momentum to the right" of the system. Since Smith's momentum increases, and his mass even diminishes, his speed must increase. Later, when considering the Smith-ball system again: the ball was thrown by Jones and it acquired momentum to the right; its speed must be greater than Smith's speed if it will reach him. When Smith catches the ball, the ball's speed will decrease in order for the ball to stick with Smith, so Smith's speed will increase accordingly to compensate for the ball's loss of speed, and thus maintain the total momentum in the system constant.
    It's 3 in the morning so I'm hoping I didn't spout out any nonsense above.. Hopefully this was helpful.
     
  4. Dec 6, 2014 #3

    haruspex

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    Well, why do you think it should be the other way? Jones throws a ball forward. Why would that make Jones go faster?
    No, his mass does not decrease, nor does the mass of the Smith + ball system decrease. If you want to define a component of the system as being Smith plus whatever he holds at any instant, the mass of that will decrease but so might its momentum. E.g. if Smith simply drops the ball he will not go any faster.
    I note it was 3am for you.
     
  5. Dec 9, 2014 #4
    thanks, but it's becoming confusing again. could you explain clearly what happens at each step represented in the graph? How momentum is conserved? momentum of what increases or decreases? and how does speed increase and decrease as illustrated in the graph?

    Thanks in advance. plz could try to give a clear explanation
     
  6. Dec 9, 2014 #5

    haruspex

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    Smith is in front of Jones, skating at speed s (but I'll call it v here to avoid confusion with S for Smith), and holding a ball. Smith +ball have a certain momentum: (mS+mb)v. If Smith were simply to drop the ball, each would continue at the same speed v, Smith with momentum mSv, the ball with momentum mbv. But Smith throws the ball towards Jones. The ball now has a lower speed measured in the original direction (probably negative, but that doesn't matter). By conservation of momentum, the total momentum of Smith+ball does not change (mS+mb)v = mSvS+mbvb. Since mbvb < mbv, it follows that mSvS > mSv. This is a consequence of the equality of action and reaction. Just as Smith pushed on the ball to throw it, the ball pushed on Smith with an equal and opposite force, making Smith go faster.
    You can follow the same detailed steps for Jones catching and returning the ball. Each time a ball is thrown one way, the thrower accelerates in the opposite direction.
     
  7. Dec 9, 2014 #6
    smith is in front - is this an assumption?
    also, smith is skating backwards, towards Jones?
    what happens when one of them catches the ball - in terms of the force on them?
     
  8. Dec 9, 2014 #7

    haruspex

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    They start off skating at the same speed, in the same direction. Smith is in front, facing Jones, and therefore is skating backwards, away from Jones.
    If you catch a ball thrown to you, you have to push on the ball to slow it down. Therefore the ball also pushes on you. If you are skating forward at the time and the ball comes from in front of you, it will tend to slow you down.
     
  9. Dec 9, 2014 #8

    PeroK

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    Here's a suggestion. Think about the problem with s = 0. The skaters are facing each other, but not moving.

    1) Smith throws the ball to Jones. What happens to Smith?

    2) Jones catches the ball. What happens to Jones?

    I'd imagine the ball as something heavy. A bowling ball or a medicine ball!
     
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