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Imdiate Help needed

  1. Mar 27, 2009 #1
    imdiate Help needed :(

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

    A 2000 kg truck traveling at a speed of 6 m/s makes a 90 turn in a time of 4 s and emerges from this turn with a speed of 4 m/s. What is the magnitude of the average resultant force in KN on the truck during this turn? 2. Relevant equations

    F avr. = Pf - Pi / t
    3. The attempt at a solution

    Pf = m Vf = 2000 * 4 = 8000

    Pi = m Vi = 2000 * 6 = 12000

    Pf - Pi = - 4000 and F avr. in KN = -1

    Is that Right ???
     
    Last edited: Mar 27, 2009
  2. jcsd
  3. Mar 27, 2009 #2
    Re: Easy, But, Take a look :O

    I was wondering if it's correct to use the momentom equation in such a case, where we have only one object ?

    And how can the"90 degree" effect the solution ?
     
  4. Mar 27, 2009 #3

    djeitnstine

    User Avatar
    Gold Member

    Re: imdiate Help needed :(

    Yes that's correct. 90 degrees does not affect anything, it could have been 180 degrees and the resulting average force would have been the same.
     
  5. Mar 27, 2009 #4
    Re: imdiate Help needed :(

    What do you think of what I did ?
     
  6. Mar 27, 2009 #5

    djeitnstine

    User Avatar
    Gold Member

    Re: imdiate Help needed :(

    It is perfectly fine. Another way to think of it is finding the average deceleration (vf-vi/t) then multiplying it by the mass of the truck. However, look closely and it is essentially the same thing you did by conservation of momentum.
     
  7. Mar 28, 2009 #6
    Re: imdiate Help needed :(

    Uha, this gives the same answer, 1000, which is not in the choises,

    Or 1 Kn
    and that

    what cofuses me.
     
  8. Mar 28, 2009 #7
    Re: imdiate Help needed :(

    What are you getting as a result for ((vf-vi)/t)?

    edit: Your answer does seem to be correct, I just mispushed a button when I did it. Odd.

    edit2: Actually, consider the direction of each of the momentums. Think pythagoras.
     
  9. Mar 28, 2009 #8
    Re: imdiate Help needed :(

    Do you mean to find the velocity components ?

    I tried that, but with no difference because the angle is 90 ..

    What do you think?
     
  10. Mar 28, 2009 #9
    Re: imdiate Help needed :(

    Momentum is a vector itself (it has direction), so you don't need to break it down to just velocity. The 90 degrees bit makes it relatively easy to compute the momentum needed. Try drawing a triangle showing the initial and final momentums.
     
  11. Mar 28, 2009 #10
    Re: imdiate Help needed :(

    Yes, and its direction depends on the Vel. direction ??


    Besides, how can I draw the triangle showing the momentom ????
    its a quarter from a circle ?
    Right ?
     
  12. Mar 28, 2009 #11
    Re: imdiate Help needed :(

    Yes, the direction comes from velocity. For simplicity, it may be easiest to assume the initial momentum is pointing north, and after turning, it's going east (although any corodinate system where they're 90 dergees apart will do).

    If you place the two momentum vectors tail to tail at a common origin, what is the size of the vector connecting the two tips?
     
  13. Mar 28, 2009 #12
    Re: imdiate Help needed :(

    Ok, very good, I did that and I got

    a vector R which magnitude is vf - vi

    But when I multlied its magnitude by the mass and divide by 4 which is the time to find F avr.

    I got a very big number ?????
     
    Last edited: Mar 28, 2009
  14. Mar 28, 2009 #13
    Re: imdiate Help needed :(

    How did you conclude that its magnitude was vf-vi?
     
    Last edited: Mar 28, 2009
  15. Mar 28, 2009 #14
    Re: imdiate Help needed :(

    Or,

    I was thinking about that way :
    deltap = pf - Pi
    mVf- mVi
    m(Vf+Vi) assuming that this direction is positive

    then subs.

    2000( 6+4) = 20,000/4 = 5000 N

    What do you think ??
     
  16. Mar 28, 2009 #15
    Re: imdiate Help needed :(

    This is only true if the vectors were in the same direction. In this case, they're perpindicular to one another.

    If you factor out an m, you would have m(vf-vi), not m(vf+vi), although from before, I don't believe you can just subtract them as you're doing.
     
  17. Mar 28, 2009 #16
    Re: imdiate Help needed :(

    Then, just tell me how to the triangles ??

    I mean after I draw the triangle what shall I do next ??
     
  18. Mar 28, 2009 #17
    Re: imdiate Help needed :(

    One of your vectors will point up like "|", and the other sideways like "_", so you have the base and height of a right angled triangle. You want the hypotenuse.
     
  19. Mar 28, 2009 #18
    Re: imdiate Help needed :(

    Then, what does it mean?

    Is it the momentom ???

    Or it's vf- vi
     
  20. Mar 28, 2009 #19
    Re: imdiate Help needed :(

    If the vectors you use are momentum vectors, than the hypotenuse will be momentum as well. (It would be possible to do it with velocities as well, but that's more work.)
     
  21. Mar 28, 2009 #20
    Re: imdiate Help needed :(

    Well, what I did is I multplied this hyp. by the mass and divide by the time

    Right ?
     
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