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Magnatude of adverage force

  1. Oct 14, 2009 #1
    1. The problem statement, all variables and given/known data

    A 4 kg steel ball strikes a wall with a speed
    of 16 m/s at an angle of 50.6◦ with the normal
    to the wall. It bounces off with the same
    speed and angle, as shown in the figure.

    http://img517.imageshack.us/img517/4498/physicspic1.png" [Broken]


    If the ball is in contact with the wall for
    0.277 s, what is the magnitude of the average
    force exerted on the ball by the wall?
    Answer in units of N.



    2. Relevant equations

    Im = mvsin(8)i+mvcos(8)j
    Fm= -mvsin(8)i+mvcos(8)j
    Impulse = Fm - Im
    Force average = Impulse/Time

    8 = THeta

    3. The attempt at a solution

    Im = (4kg)(16m/s)(sin39.4) + (4kg)(16m/s)(cos39.4) = 90.078

    Fm = -(4kg)(16m/s)(sin39.4) + (4kg)(16m/s)(cos39.4) = 8.832

    Impulse = (8.832)-(90.078) = -81.246 Which I also found Impulse = -2(Mass)(Velocity)(Sin(8))

    Then F adv= (-81.246)/ .227s = -357.912 N

    This should be the case, but apparently I am wrong, can some one help me find my mistake, and I am aware I am useing a different angle than what is given, I am using the angle off the wall. So instead of 50.6 degrees, I am using 39.4 degrees. :) Help Plz.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Oct 14, 2009 #2

    kuruman

    User Avatar
    Homework Helper
    Gold Member

    You cannot add the x and y components of the momentum vectors like this. It is meaningless and incorrect. To find the impulse, you need to subtract x-xomponent from x-component and y-component from y-component.
     
  4. Oct 15, 2009 #3
    I have found my error. I plugged in .227 seconds instead of .277 seconds to get the right answer.

    And to kuruman.... I can and it worked. basically, what i did was found the momentum going in minus the momentum going out, the devided by the correct time to get the adverage force.

    so the momentum going in is both a x and y component minus the x( in this case negative) and y.

    So final is

    Then F adv= (-81.246)/ .277s = -293.307 N But since we are looking for the magnatude of the average force we make the answer positive.

    = 293.307 N which is correct.
     
  5. Oct 15, 2009 #4

    kuruman

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    Homework Helper
    Gold Member

    It worked essentially by dumb luck. The correct expression involves the change in momentum in the direction perpendicular to the wall only. That is take

    (4kg)(16m/s)(sin39.4)-(-(4kg)(16m/s)(sin39.4))

    and divide by 0.227 seconds.

    You added and subtracted the other component, (4kg)(16m/s)(cos39.4), so you ended up with the right answer. The fact that you got the right answer in this particular problem does not justify adding x and y components of vectors (as you have done) for all problems that you will have to face in the future. Of course you can do what you want, but don't be surprised if it doesn't work the next time you try something like this.
     
  6. Oct 15, 2009 #5
    Would that be the same as I wrote above in the third part of my equations....

    but what you wrote down is the same thing but without the negative, since we are talking about momentum of average force which should be positive. So I originally had it to start with, but negative because I was using the ball's x component away from the wall as a negative, but simple fix :)

    But thanks for letting me know that I can't add these, I will ask my professor about it and see if he can help explain further.
     
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