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Impulse And Momentum

  1. May 31, 2005 #1
    Okay, So... I'm having trouble understanding what I did wrong here. I can't seem to understand how I got this problem wrong.
    Here's the problem statement:

    A 0.160-kg hockey puck is moving on an icy, frictionless, horizontal surface. At t=0 the puck is moving to the right at 2.98 m/s.

    a.) Calculate the magnitude of the velocity of the puck after a force of 24.9 N directed to the right has been applied for 4.80×10−2 s.
    b.) What is the direction of the velocity of the puck after a force of 24.9 N directed to the right has been applied for 4.80×10−2 s.
    c.) If instead, a force of 12.2 N directed to the left is applied from t=0 to t= 4.80×10−2 s, what is the magnitude of the final velocity of the puck?
    d.) What is the direction of the final velocity of the puck in this case?


    So I've got everything right-- except for part c. ...I would imagine that I'd do it the exact same way I did part a. This is what I did for part c:


    [tex]J_{x} = impulse[/tex]
    [tex]J_{x} = p_{2x} - p_{1x}[/tex]
    [tex]J_{x} = F_{x}(t_{2}-t_{2}) = -12.2(.048) = -.5856[/tex]
    [tex]p_{2x} = J_{x} + p_{1x} = -.5856 + (.16)(2.98) = -.1088[/tex]
    [tex]v_{2x} = \displaystyle{\frac{p_{2x}}{m}} = \displaystyle{\frac{-.1088}{.16}} = -.68[/tex]m/s

    Why is this wrong? What did I do wrong?


    Thanks!
     
  2. jcsd
  3. May 31, 2005 #2
    Initial momentum: mv = 0.16*2.98 kgm/s
    Impulse: Ft = -12.2*0.048 kgm/s

    The impulse is the change in momentum.
     
  4. May 31, 2005 #3
    ...Okay, so Ft stands for... impulse?

    If so, I indicated that I already found the value for that one, which would be approximatley -.5856


    I still don't know what I did wrong.
     
  5. Jun 1, 2005 #4
    Nevermind, I'm getting the same answer you did.
     
  6. Jun 1, 2005 #5
    Haha, weird. ...We use the program, 'Mastering Physics' at the University I attend... And the damn thing has been kind of quirky already. Perhaps it could be messing things up, because I keep doing the problem, different ways, and I still get the same answer.

    Who knows.
     
  7. Jun 1, 2005 #6
    I keep rereading it, thinking I missed out on something. This is EXACTLY what I was given:

    A 0.160-kg hockey puck is moving on an icy, frictionless, horizontal surface. At t=0 the puck is moving to the right at 2.98 m/s.
    If instead, a force of 12.2 N directed to the left is applied from t=0 to t= 4.80×10−2 s, what is the magnitude of the final velocity of the puck?

    And this is how I came about my solution... That apppears to be incorrect:


    Can anyone else seem to figure how this is wrong?... There has to be something...
     
  8. Jun 1, 2005 #7
    Hah. I got it.

    what is the magnitude of the final velocity of the puck? As we know, magnitudes are absolute values. So our answer would be 0.68 instead of -0.68.

    Although, usually the program will tell you that you got your negatives and positives mixed up. Strange it didn't do it this time.

    Thanks, by the way!
     
  9. Jun 1, 2005 #8
    Yeah I justn oticed that and was abuot to say. I used masteringphysics for kinematics an E&M, i hated it.
     
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