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Basic Momentum Question

  1. Mar 6, 2016 #1
    1. The problem statement, all variables and given/known data
    A 0.30kg puck A, moving at 5m/s [W], undergoes a collision with a 0.40 puck B which is initially at rest. Pick A moves off at 4.2m/s [W 30deg N]. Find the final velocity of Puck B.

    2. Relevant equations
    pi=pf
    mvi=mvf

    3. The attempt at a solution

    Let [N] and [W] be positive

    Puck A
    vix = 5m/s
    m1= 0.30kg
    vf=4.2m/s
    vfx : 4.2cos30=3.6m/s
    vfy : 4.2sin30=2.1 m/s

    Puck B
    vi=0m/s
    m2=0.40kg

    Solve for v2fx

    m1v1ix + m2v2ix = m1v1fx + m2v2fx

    (0.30)(5)+(0.40)(0)=(0.30)(3.6)+(0.40)v2fx
    2.43= v2fx

    Solve for v2fy

    m1v1iy + m2v2iy = m1v1fy + m2v2fy

    (0.30)(0)+(0.40)(0)=(0.30)(2.1)+(0.40)v2fy
    -1.58 = v2fy

    Find the resultant v2f

    |v2f| = sqrt((2.43)2+(1.58)2)
    |v2f| = 2.9 m/s

    tantheta = (1.58/2.43)
    theta = tan-1(1.58/2.43)
    theta = 33deg

    Final answer: velocity of Puck B is 2.9m/s [33deg SW].

    Can someone tell me if I'm doing this correctly?:
     
    Last edited: Mar 6, 2016
  2. jcsd
  3. Mar 6, 2016 #2

    haruspex

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    That step looks wrong.

    By the way, you should keep some more decimal digits during the calculation or you may find the final result is rather inaccurate.
     
  4. Mar 6, 2016 #3
    Corrected it, I got v2fx=3.47 m/s and theta now = 25 degrees.

    Does that seem correct now?
     
  5. Mar 6, 2016 #4

    haruspex

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    Still wrong. Please post all your working.
     
  6. Mar 6, 2016 #5
    Ok one second
     
  7. Mar 6, 2016 #6
    Ok one second
     
  8. Mar 6, 2016 #7
    (0.30)(5)+(0.40)(0)=(0.30)(3.6)+(0.40)v2fx
    1.5+0=1.08+0.40v2fx
    0.42=0.40v2fx
    0.42/0.40=v2fx

    1.05= v2fx

    | v2 | = sqrt((1.05)2+(1.58)2)
    = 1.9m/s

    tantheta = (1.58/1.05)
    theta = tan-1(1.58/1.05)
    theta = 56 degrees

    Answer: velocity of puck B is 1.05m/s [56deg SW]

    How about now? Is my process wrong, or is it the math- this is what I get for doing my work at 6am lol.[/QUOTE]
     
  9. Mar 6, 2016 #8

    haruspex

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    That all looks ok now. "56 deg SW" is not a clear way of stating it. Better is "56 deg S of W".

    Edit: You should get into the habit of doing very rough mental estimates to check your answers.
    In the equation you were handling wrongly, you could see the puck A lost 1.4m/s of its x velocity, and it weighs only 3/4 of puck B. So puck B's x velocity must be 3/4 of 1.4m/s.
     
  10. Mar 6, 2016 #9
    Ok thanks so much for you help!
     
  11. Mar 6, 2016 #10
    Is that just a shorter way of performing the calculation?
     
  12. Mar 6, 2016 #11

    haruspex

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    whether it is shorter depends on exactly how you solved the equation. If you multiplied out all the terms first then yes it is. But more generally the numbers might have been a lot more awkward. The point is to do a mental check in terms of the approximate ratios and differences. E.g. if the masses had been 0.31 and 0.39 I would still have called it 3/4.
     
  13. Mar 6, 2016 #12
    Ok thanks for your help!
     
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