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Momentum Problem on a particle

  1. Feb 29, 2008 #1
    1. The problem statement, all variables and given/known data
    Force F_x =(10N){sin({(2pi(t))/4.0s) (where t in (m/s) is exerted on a 420 g particle during the interval 0 less than or equal to T is less than or equal to 2 seconds.

    If the particle starts at rest what is its speed at t=2 seconds?
    2. Relevant equations

    Jx= area under the Fx(t) cureve between Ti and Tf

    Pfi-Pix=Jx

    3. The attempt at a solution
    This problem looks fairly easy but I am having trouble setting it up. I took the integral of the force over the time interval and got .54817 Ns which should be the impulse. I then used this to find Pfx by the equation above. I then used Vfx=(Pfx/m) to find the final velocity which came out to be 2.19 m/s but its wrong, any insight as to what I did wrong?
     
  2. jcsd
  3. Feb 29, 2008 #2
    integral of F(t)= impulse= change of momentum= Final momentum, since intial velocity is zero.
    Final momentum devide by the mass, .420kg, would equal the velocity.

    I do not have a calculator right now but I am sure this is it :)

    oh by the way, when integrating force, you might want to include absolute value around F(t)
     
    Last edited: Feb 29, 2008
  4. Feb 29, 2008 #3
    It keeps on saying my answer is wrong. What am I missing?
     
  5. Feb 29, 2008 #4
    Use this:
    [tex]F=m\frac{dv}{dt}[/tex]

    You know integration, right ?
     
  6. Feb 29, 2008 #5
    Yea I think Im doing it right, after I integrate it I get 12.73 then you say times that by the mass, so 12.73 x .42kg= 5.35. That should be right, could someone double check.
     
  7. Feb 29, 2008 #6

    Dick

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    I integrated Fdt from t=0 to t=2 and got (40/pi)*N*sec. Why are you getting .54817Ns? Can you show details?
     
  8. Feb 29, 2008 #7
    I got this on integrating:
    [tex]0.042v=-\frac{2}{\pi}cos(\frac{\pi t}{2})+\frac{2}{\pi}[/tex]

    for t=2 it gives v=30.33 m/s.
     
  9. Feb 29, 2008 #8

    Dick

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    Ok, impulse=40/pi*Ns=12.73*Ns=m*v. That's good. Don't you want to divide the impulse by the mass to get velocity?
     
    Last edited: Feb 29, 2008
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