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Homework Help: Variation on momentum q

  1. Aug 9, 2011 #1
    If a 1kg object is moving at 3m/s in a positive direction, and a 12N force is applied in the negative direction, what is the velocity immediately after 2s?

    I'm fairly sure this will be a variation of relevent momentum equations, and/or mixed with kinematics, yet I'm not seeing the correct application. I lean towards answers c or b.

    Obviously mo will be conserved.


    conservation of mo: m1Vi= (m1+m2)vf t=v/a



    a. 12m/s
    b. 6 m/s
    c. 3 m/s
    d. 0 m/s

    Please explain.
     
  2. jcsd
  3. Aug 9, 2011 #2
    I'm afraid you can't use conservation of momentum here - that only applies in the absence of an external force (in the relevant direction), and here the problem says there's a 12 N force. Do you know any other way you might go about solving this?
     
  4. Aug 9, 2011 #3
    What you need is the impulse formula. Whenever time is involved for a force it's almost always Impulse.

    Definition: The Impulse (I) imparted onto a body is the change in it's momentum. Impulse is a vector quantity.

    Derivation: F = ma

    F = m (v-u)/t
    F = mv - mu /t
    Ft = mv -mu



    Unit is obviously Newton seconds.
     
  5. Aug 9, 2011 #4
    Yes, I do believe this is on the right track, so thanks, but

    I believe I'm remiss in something. What is u?

    I'm recalling something like.......Ft=mvf-mvi (change in p)
     
    Last edited: Aug 9, 2011
  6. Aug 10, 2011 #5
    u = initial velocity
     
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