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Conservation law of momentum... Tricky question

  1. Nov 24, 2015 #1
    1. The problem statement, all variables and given/known data
    A ball rolls at a velocity of 3.5m/s toward a 5.0kg ball at rest. They collide and stick and move off at 2.5 m/s. What was the mass of the moving ball?



    2. Relevant equations
    Pa + Pb = 0
    M(a)V(a) = - M(b)V(b)



    3. The attempt at a solution
    So.. The ball at rest had zero momentum... Therefore how are we supposed to find the momentum in total if we are only given 3 numbers to work w/ ... We can't determine the moving ball momentum.. Soo
     
  2. jcsd
  3. Nov 24, 2015 #2

    haruspex

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    You are given four velocities and one mass. There's only one unknown, and you have an equation.
     
  4. Nov 24, 2015 #3
    But .... If the second ball at rest has a momentum of 0kg°m/s .... Then M(a)3.5m/s (a)= 5.0kg(b)0m/s(b)

    So saying (?)*(3.5m/s) + (5.0kg)*( 0 m/s) = 2.5 m/s..
     
  5. Nov 25, 2015 #4
    Check the relevant equation ##P_a + P_b = 0##. Are you sure that's correct? Shouldn't it be something else:
    $$P_i = P_f \implies P_{a_i} + P_{b_i} = P_{a_f} + P_{b_f}$$
    The ##i## denotes "initial" and ##f## denotes "final".
     
  6. Nov 25, 2015 #5

    haruspex

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    The term on the right has the wrong dimension. You've left out a factor.
     
  7. Nov 25, 2015 #6
    Yes 2.5m/s is only the velocity and the momentum has the units of kg*m/s
     
  8. Nov 25, 2015 #7

    haruspex

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    So correct it. What factor did you leave out?
     
  9. Nov 25, 2015 #8
    Mass... So it would look like (3.5m/s)(?) + (0m/s)(5kg) = (5+?kg) + ( 2.5m/s)
     
  10. Nov 25, 2015 #9

    haruspex

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    Kg+m/s?!
     
  11. Nov 25, 2015 #10
  12. Nov 25, 2015 #11

    haruspex

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    Ok, so write out the equation correctly and solve it.
     
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