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Finding momentum; I'm doing something wrong

  1. Aug 7, 2012 #1
    Hello again forum!
    I need a hint or a straight answer as to what I am doing wrong in solving this momentum problem...

    1. The problem statement, all variables and given/known data

    A 5.0 kg ice block is sliding along a smooth floor at 1.0 m/s west when a 0.20 N force directed east acts on it for 4.0 s. What is the magnitude of the block’s final momentum?

    2. Relevant equations

    Please see attached document.

    3. The attempt at a solution

    Please see attached document. I first solved for a change in momentum, then a final velocity and then the regular momentum. The actual answer is supposed to be 4.2.
    What did I overlook this time? :rolleyes:

    Thank you for reading and your seemingly endless patience :)
     

    Attached Files:

  2. jcsd
  3. Aug 7, 2012 #2

    Dick

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    Homework Helper

    You've got the delta p right. But you don't need to know vf at all. Just combine the change in momentum with the initial momentum. And since the force is pointing east and the momentum is pointing west, they will subtract. Won't they?
     
  4. Aug 7, 2012 #3
    Your impulse and your velocity are in opposite direction, so you have to make one or the other negative in order to reflect this.

    I haven't checked to see if this yields the correct answer based on the rest of your work, but it should.


    However, I think that it would be simpler if you use ##\Delta p = p_f - p_i## in order to say that ##p_f = mv_i + \Delta p##. It is in essence the same as what you did, but you changed momentum back to velocity and then back to momentum, which is an extra step.
     
  5. Aug 7, 2012 #4
    Hi, so you're saying I need to....
    Δp = -0.2N x 4.0s
    Δp = -0.8

    [Δp] + [m x v]?

    [-0.8] + [5 x 1]

    = 4.2 :biggrin:

    How can I adjust my Δp to be going in the "opposite" direction? As in, what should be negative, the force or time? LOL
     
  6. Aug 7, 2012 #5
    Does it matter which is negative? Or is it just any one value that has to be negative? And you're right, it works out :smile:
     
  7. Aug 7, 2012 #6

    Dick

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    You only have to worry about momenta, not time or force. It doesn't matter which momentum you adjust to be negative. The question is only asking for the MAGNITUDE of the final momentum. You just have to add them with opposite signs and take the absolute value.
     
  8. Aug 7, 2012 #7
    Well if you call heading east to be positive, vi will be positive, Δp will be negative, and your result will be 4.2
    If you call going west to be positive, vi will be negative, Δp will be positive, and your result will be -4.2 (and negative refers to going east).

    So the answer will have the same meaning, although it could be negative or positive before you interpret it.
     
  9. Aug 7, 2012 #8
    Dick and Villyer, thank you both so much for your time and for caring to explain the mechanism of this problem. Makes good sense :biggrin: I really appreciate your help:smile:
     
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