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Finding velocities of two people on skates

  1. Jun 27, 2010 #1
    1. The problem statement, all variables and given/known data
    A 66.0 kg person throws a 0.0430 kg snowball forward with a ground speed of 31.0 m/s. A second person, with a mass of 56.0 kg, catches the snowball. Both people are on skates. The first person is initially moving forward with a speed of 2.50 m/s, and the second person is initially at rest. What are the velocities of the two people after the snowball is exchanged? Disregard the friction between the skates and the ice.
    a. thrower
    b. catcher

    3. The attempt at a solution
    I do not even know where to begin
  2. jcsd
  3. Jun 27, 2010 #2


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    As with all of the three problems that you posted consecutively, begin by conserving linear momentum.
  4. Jun 27, 2010 #3
    mv + mv = mv + mv
    (66.0 kg)(31.0 m/s) + (56.0 kg)(0 m/s) = (66.0 kg)v1f + (56.0 kg)v2f

    2046 m/s = 66v1f + 56v2f
    (1924 m/s = -(v1f - v2f) )66

    2046 m/s = 66v1f + 56v2f
    126984 m/s = -66v1f + 66v2f

    129030 m/s = 122v2f
    v2f = 1057.6 m/s

    1924 m/s = - v1f + 1057.6 m/s
    v1f= -866.4 m/s

    these numbers seem very high
  5. Jun 28, 2010 #4


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    What are you saying here? What is moving at 31.0 m/s, the 66-kg person or the snowball? There are three distinct times here.

    1. Time "before": Both persons are at rest, one is holding the snowball.
    2. Time "in-between": The snowball is flying through the air, the person who threw it is moving, the other person is not.
    3. Time "after": The snowball is caught and both persons are moving.

    You need to write two momentum conservation equations, one linking "before" and "in-betwen" and one linking "in-between" and "after". Your system is two persons and one snowball.
  6. Jun 28, 2010 #5


    Staff: Mentor

    Per the given information, the first person is moving at 2.5 m/s.
    The thrower is still moving.
  7. Jun 28, 2010 #6


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    Sorry, I missed that part, bu tit does not change the strategy for answering the question.
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