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Physics Final Exam Sample Question

  1. Jun 15, 2012 #1
    1. The problem statement, all variables and given/known data

    A hockey puck of mass 0.200kg is sliding along a smooth, flat section of ice at 18.0m/s
    when it encounters a patch of snow. After 2.50s of sliding through the snow it returns to smooth ice, continuing at speed of 10.0m/s.

    a) What is the change in momentum of the puck?

    I = m(v2 - v1)
    I = 0.200 (10-18)
    I=-1.60kg m/s

    b) What was the work done by the snow on the puck?


    c) what Average frictional force does the snow exert on the puck?
     
  2. jcsd
  3. Jun 15, 2012 #2

    cepheid

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    Welcome to PF,

    What have you done so far on parts b and c of this problem? (Have a look at the PF rules for homework threads: you must show an attempt)

    For b: are you familiar with the work-energy theorem?

    For c: can you see how this can easily be solved once you have the answer to b?
     
  4. Jun 15, 2012 #3
    W = F (d)

    I tried solving for F first by using

    f(t) = m (v)

    F = m(v)/t

    F = 0.200(-8)/2.50

    F = -0.64

    Then I tried to solve for D by multiplying final velocity and time
    which gave me 25m.

    So W = (-0.64)(25)
    = -16J

    But not to sure if that is correct.


    For c) I know the formula is Ff = u (m) (g) but don't know where to start.

    I know m = 0.200kg
    g = 9.80 and not to sure what the U coefficient is
     
  5. Jun 15, 2012 #4
    Solve b by using the work energy theorem, in this case: W = ΔEk, W would be the work done by friction and ΔEk is kinetic energy lost. After finding a value of W, c should be easy, just use W = Fxd.
     
  6. Jun 15, 2012 #5

    ΔEk = W
    ΔEk = 1/2 (m)(v)^2
    1/2(0.200)( ? )

    For V should I use CHANGE of velocity so 10-18 or should I use one of the velocities.

    I think I have it figured out, just that one part is confusing me
     
  7. Jun 15, 2012 #6

    cepheid

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    It's the change in kinetic energy. Ek_after - Ek_before.

    You have enough info to compute each of those Ek's.
     
  8. Jun 15, 2012 #7
    I'm aware I have to use ΔEk formula which is
    1/2 mv^2

    But which velocity do I use to solve for ΔEk.
     
  9. Jun 15, 2012 #8

    cepheid

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    NO. The equation for kinetic energy is (1/2)mv2.

    Delta Ek is the CHANGE in kinetic energy. Therefore it is the difference between the kinetic energy after and the kinetic energy before. Each of those two kinetic energies is computed using the equation (1/2)mv2.
     
  10. Jun 15, 2012 #9
    Update: Now I got

    1/2(0.200)(10)^2 - 1/2(0.200)(18)^2 = Ek

    -22.4 is what I get

    is that the amount of WORK?
     
  11. Jun 15, 2012 #10

    cepheid

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    Okay, so you got it. I haven't checked your numbers, but the answer is yes. That's what the work-energy theorem says. It says that the work done on an object is equal to its change in kinetic energy. So now you know how much work was done on the object by friction to slow it down.
     
  12. Jun 15, 2012 #11
    Sorry about my earlier post, I had the equation written wrong on my formula sheet.
    But okay so now I have -22.4J for work.

    Ff = μ(m)(g)

    Ff = μ(0.200)(9.80)

    Is that on the right track for C?

    I am confused to what μ is
     
  13. Jun 15, 2012 #12


    Work is equal to the change in Kinetic Energy.
    Now, b is trivially easy.

    For c, think about how you calculate friction in the first place and then its easy
     
  14. Jun 16, 2012 #13
    No, you have -22.4J for work, set this equal to F x d, F would be the frictional force that you use for μmg, infact, you don't even need μmg as the question only asks for force. As for d, solve that using kinematics equations, you were given time, intial velocity and final velocity, so use s = (u + v)t/2
     
  15. Jun 16, 2012 #14

    cepheid

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    Instead of doing all that, I think it would be simpler for the OP to use the impulse-momentum theorem. The impulse was computed in part a, and the time interval is given.
     
  16. Jun 16, 2012 #15

    cepheid

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    Also, your equation assumes constant acceleration, which needn't have been true, and was true just by luck (or judicious selection of the numbers) here.
     
  17. Jun 16, 2012 #16
    Thanks everyone, I have it figured out now.
     
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