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Momentum HW problem help

  1. Oct 4, 2008 #1
    1. The problem statement, all variables and given/known data
    Two friends, Al and Jo, have a combined mass of 167kg. At an ice skating rink they stand close together on skates, at rest and facing each other, with a compressed spring between them. The spring is kept from pushing them apart, because they are holding each other. When they release their arms, Al moves off in one direction at a speed of .897 m/s, while Jo moves off in the opposite direction, at a speed of 1.09 m/s. Assuming at friction is negligible, find Al's mass.


    2. Relevant equations

    m1v1 + m2v2=0

    3. The attempt at a solution

    I'm at a loss about how to factor in the combined mass here. You've got v1 and v2 and a combined mass, not individual masses.

    m1(.897) + m2(1.09)=0 (What do you do with these masses now??)
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 4, 2008 #2
    So, m1 + m2 = 167kg, rearrange the equation. m1 = ?
     
  4. Oct 4, 2008 #3
    so m1=167-m2

    (167-m2)(.897) + m2(1.09)=0
    distribute 149.799-.897m2 + 1.09m2=0
    combine like terms 149.799+.193m2=0
    .193m2=-149.799
    m2=-776.16 (which can't be right, because one, it's negative, and two, the masses have to add up to 167, and 776.16 is obviously larger)

    did i make a math error somewhere? because i was sure what you said would work
     
  5. Oct 5, 2008 #4
    You just have a sign problem. If Al moves in one direction at .897m/s, and Jo moves in the 'opposite' direction, the sign of Jo's velocity should be negative (-1.09m/s). Try that and it should work.
     
  6. Oct 5, 2008 #5
    thanks so much!!
     
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