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What is the magnitude of the girl's change in momentum?

  1. Oct 12, 2016 #1
    1. The problem statement, all variables and given/known data

    A 50-kg ice skater moves across the ice at a constant speed of 2.0 m/s. She is caught by her79-kg partner, and then the pair continues to glide together. He is at rest when he catches her, and immediately afterward they both coast.

    Part A)
    What is their velocity just after the catch? Assume she moves in the positive x-direction before her partner catches her. (solved already)
    Part B)
    A skater follows the girl at a constant speed of 1.0 m/s. What is the magnitude of the girl's change in momentum in the reference frame of the skater?

    2. Relevant equations
    m(u1-v1)

    3. The attempt at a solution
    Part a) solved. answer was 0.78 m/s
    Part B) 50kg(2.0-1.0 m/s)= 50__
    i dont not know if this is the correct answer an i am also unsure the units it should be in
     
  2. jcsd
  3. Oct 12, 2016 #2

    cnh1995

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    This is the initial momentum of the girl w.r.t.the following skater. The problem is asking the "change" in momentum.
     
  4. Oct 12, 2016 #3
    would it be 50(2.78-1)= 86 kg-m/s
     
  5. Oct 12, 2016 #4

    cnh1995

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    I'm not an expert in mechanics but isn't the velocity after being caught by the partner 0.78m/s? You are asked to find the "difference" in momenta.
     
  6. Oct 12, 2016 #5
    someone in my class told me to subtract by 1 since in reference frame of a person moving at velocity 1
     
  7. Oct 12, 2016 #6

    Simon Bridge

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    Why are you subtracting 1m/s from 2.78m/s?

    What is the change in total momentum in part A?
    What is the girl's change in momentum in part A? (what is her initial momentum? What is her final momentum?)
    Now find the girl's change in momentum in the frame of the skater? (what is her initial momentum? what is her final momentum?)
     
  8. Oct 12, 2016 #7
    can you explain more how to do that?
    total momentum for part a would just be (50+79)(0.78)-50(2)=0.62
    girl's change: 50(0.78)-50(2)=-61
     
  9. Oct 12, 2016 #8

    Simon Bridge

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    What is the change in total momentum in part A? What does the law of conservation of momentum say?

    Well done.
    Now picture yourself as the skater ... you will see the partner initially going backwards at 1m/s, do you see how that works?
    When the girl is doing 2m/s across the ice, what speed is she doing wrt the skater?
     
  10. Oct 12, 2016 #9
    Would it be 3 m/s?
     
  11. Oct 12, 2016 #10

    Simon Bridge

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    Why would the partner, stationary on the ice, have speed 1m/s but backwards wrt you the skater?
    If you travel in the same direction as someone faster than you, do they seem to be going faster or slower than they would if you were still?
     
  12. Oct 12, 2016 #11
    If we are both moving in the same direction then they would appear slower then if they were still. Would it then be 1 m/s
     
  13. Oct 12, 2016 #12

    Simon Bridge

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    Well done - you see you already know a lot of physics!
    So the girl's initial speed is 1m/s (wrt the skater).
    What about the final speed?
     
  14. Oct 12, 2016 #13
    Wouldn't her final speed be 0.78-1 so = -0.22 m/s?
     
  15. Oct 12, 2016 #14

    Simon Bridge

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    Well done - what does the negative sign signify?
    What is the change in momentum, then, for the frame of the skater?
     
  16. Oct 12, 2016 #15
    Does the negative sign mean it appears they are moving in opposite directions instead of the same.
    Is the change in momentum 50( -0.22-1)=-61?
     
  17. Oct 12, 2016 #16

    Simon Bridge

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    The negative sign means the girl appears to be going backwards - this makes sense because her final speed is slower than the skater's speed.
    Well done - that is all there way to it.
    All these conservation based problems follow the same pattern - find out the initial and final quantities, then do final minus initial.

    Compare your answer (above) with the change in momentum of the girl you worked out for part A back in post #7. What do you notice?
    In part B (post #1) - what does the problem actually ask you for? (What is the answer?)
    Don't forget the units.

    Aside: back in post #6 I asked you what the change in total momentum was for part A.
    You replied (post #7) that this was 0.62[kgm/s].
    Is this consistent with the law of conservation of momentum?
     
  18. Oct 12, 2016 #17
    So
    so the answer to part B in the op is -61? Is it -62 kg•m/s?
     
  19. Oct 12, 2016 #18
    *-61
     
  20. Oct 12, 2016 #19

    haruspex

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    Yes. There are two ways to know those are the right units.
    All of the input numbers are in SI units, so by the internal consistency of the SI system the result should also be in SI units, and the SI units for momentum are kg m/s. This is a pragmatic rule, but not ideal.
    The more rigorous way is to keep track of units throughout the calculation. You can effectively treat them as variables in the algebra. With this method you can cope with a mix of units.
     
  21. Oct 12, 2016 #20

    Simon Bridge

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    Went to sleep - I was only up because of insomnia, and I was confident you could do it.
    You have the right units, as haruspex says, but you don't quite have the answer part B is asking for.
    These things can trip you up if you miss an important word, which is why it is good to check the wording at the end... here it is:
     
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