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anonymous

  • 5 years ago

In an isolated system, two roller skaters, each with a mass of 100 kg, collide. Skater 1 was initially at rest, while Skater 2 was moving at 6 m/s. They move off together. What is their speed?

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  1. JamesJ
    • 5 years ago
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    You need to to use conservation of momentum.

  2. JamesJ
    • 5 years ago
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    Let p1 be the momentum of the first skater before the collision on p1' afterwards. Likewise p2 and p2' for skater 2. Then by conservation of momentum \[ p_1 + p_2 = p_1' + p_2' \] Now, use that the definition of momentum to find the joint velocity afterwards.

  3. anonymous
    • 5 years ago
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    so 6m/s?

  4. JamesJ
    • 5 years ago
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    what's the definition of momentum?

  5. anonymous
    • 5 years ago
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    mass x velocity

  6. JamesJ
    • 5 years ago
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    Yes. So p1 = ? and p2 = ?

  7. anonymous
    • 5 years ago
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    0 and 6ms?

  8. JamesJ
    • 5 years ago
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    \[ p_1 = 0 \ kg.m/s \] yes. But \[ p_2 = m_2.v_2 = ... what? \]

  9. anonymous
    • 5 years ago
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    10m/s?

  10. JamesJ
    • 5 years ago
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    what's m2 = ? what's v2 = ?

  11. anonymous
    • 5 years ago
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    3 m/s?

  12. JamesJ
    • 5 years ago
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    \[ v_2 = 6 m/s \] \[ m_2 = 100 kg \] Hence \[ p_2 = m_2.v_2 = (100 kg)(6 m/s) = 6000 kg.m/s \] Therefore the total momentum before the collision is \[ p_1 + p_2 = 0 + 6000 = 6000 \] Now what about afterwards? What do you know about v1' and v2' ?

  13. anonymous
    • 5 years ago
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    is it 6 m/s...that's all i need to know..

  14. JamesJ
    • 5 years ago
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    No. That's wrong. After the collision, they both move off together so their velocities are the same. Call it v: \[ v = v_1' = v_2' \] Hence the total momentum after the collision is \[ p_1' + p_2' = m_1v_1' + m_2v_2' = (m_1 + m_2)v \]

  15. anonymous
    • 5 years ago
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    0!

  16. JamesJ
    • 5 years ago
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    Now, set that equal to the total momentum before the collision and you can solve for v.

  17. anonymous
    • 5 years ago
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    0 m/s!

  18. JamesJ
    • 5 years ago
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    No. That can't possibly be right. If they are both stationary, v = 0, then their total momentum is zero. But we know it must be equal to 6000.

  19. anonymous
    • 5 years ago
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    so 10??

  20. JamesJ
    • 5 years ago
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    Work the algebra here. Set that last expression for the total momentum equal to the total momentum before the collision.

  21. anonymous
    • 5 years ago
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    it has to be one of them, its a multiple choice question.

  22. anonymous
    • 5 years ago
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    can you please just tell me the answer i'll ask my teacher tomorrow

  23. JamesJ
    • 5 years ago
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    You're just guessing the answers. I'm trying to show you how to find it. By Conservation of Momentum, momentum before and after the collision are equal: \[ p_1 + p_2 = p_1' + p_2' \] We know that \[ p_1 + p_2 = 6,000 \ kg.m/s \] We also know that \[ p_1' + p_2' = (m_1 + m_2)v \] where v is their shared velocity after the collision. Hence we have that \[ (m_1 + m_2)v = 6000 \] Now solve for v.

  24. JamesJ
    • 5 years ago
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    **correction: p1 + p2 = 600, because m2.v2 = (100)(6) = 600. Hence you have that \[ (m_1 + m_2)v = 600 \]

  25. anonymous
    • 5 years ago
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    10 and 6

  26. JamesJ
    • 5 years ago
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    Now, what's the value of \[ m_1 + m_2 \]?

  27. anonymous
    • 5 years ago
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    well one is 6...so the other has to be 10!

  28. JamesJ
    • 5 years ago
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    they both weight 100 kg. Hence \[ m_1 + m_2 = ... what? \] and hence what is the solution to \[ (m_1 + m_2)v = 600 \ kg.m/s \]

  29. anonymous
    • 5 years ago
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    if they weigh 100, it must be 6

  30. JamesJ
    • 5 years ago
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    \[ m_1 = m_2 = 100 \ kg \] hence \[ m_1 + m_2 = 200 \ kg \] therefore \[ (200 \ kg) v = 600 \ kg.m/s \] hence \[ v = ...what? \]

  31. anonymous
    • 5 years ago
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    3!!!!!!!

  32. JamesJ
    • 5 years ago
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    v = 3 m/s, yes. Makes intuitive sense as well, as that's half the original velocity.

  33. anonymous
    • 5 years ago
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    god i'm stupid. thank you!

  34. JamesJ
    • 5 years ago
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    Now, do yourself a favor, and write this solution out again on a blank piece of paper. When you can do that without looking at this or anything else, then you understand the problem.

  35. anonymous
    • 5 years ago
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    just did. thank you!

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