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Momentum help please

  1. Nov 3, 2007 #1
    1. The problem statement, all variables and given/known data

    A 2200 kg truck traveling north at 31 km/h turns east and accelerates to 50 km/h.

    (a) What is the magnitude of the change in the linear momentum of the truck?

    (b) What is the direction of the change in the linear momentum of the truck (in degrees)?

    2. Relevant equations

    p=mv

    3. The attempt at a solution

    I found the momentum for each speed and then subtracted the smaller one from the larger one, but it was incorrect. What am I doing wrong?

    And for part (b) I have no clue.
     
    Last edited: Nov 3, 2007
  2. jcsd
  3. Nov 3, 2007 #2

    hage567

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    Homework Helper

    Did you convert your units properly? Can you show your calculation?
     
  4. Nov 3, 2007 #3
    i need help with this too.
     
  5. Nov 3, 2007 #4
    i have been trying to work it out but it is wrong.
     
  6. Nov 3, 2007 #5
    Please Help!
     
  7. Nov 3, 2007 #6
    physicsbhelp do you go to St. Andrews school by any chance?
     
  8. Nov 3, 2007 #7
    um i have no clue what school is st.adnrews?
     
  9. Nov 3, 2007 #8
    do you by any chance know
    how to complete this problem?
     
  10. Nov 3, 2007 #9
    no i still don't have it. I tried everything.
     
  11. Nov 3, 2007 #10
    alright then.
     
  12. Nov 3, 2007 #11
    Maybe make the velocities into vector components. If you draw a picture with the vectors you will get a triangle. Subtract the components of the vectors, final minus the initial components, which will give you change in velocity and mass remains constant, so no change in mass. To calculate the magnitude of the change in momentum, do the square root of the components of the resultant change in momentum, or you can factor out the mass and mutiply that by square root of change in velocity components, which is the hypotenuse of the triangle.
     
    Last edited: Nov 3, 2007
  13. Nov 4, 2007 #12
    but i do not know how to find out the componenets of the vectors?
     
  14. Nov 4, 2007 #13
    [tex]\Delta[/tex]P=m[tex]\Delta[/tex]v

    [tex]\Delta[/tex]v=v2-v1

    draw a triangle, one leg going up 31km/h, and at the tip draw the other leg going east 50km/h. Draw a line from the origin to the point where it is at 50km/h. There your have a right triangle. you hypotenuse is the magnitude of the change in velocity. Look at the equation, change in momentum is mass times change in velocity. There is your magnitude. How do you find the angle between the first leg and and the hypotenuse leg where they have a common point at the origin. remember it is a right triangle.
     
    Last edited: Nov 4, 2007
  15. Nov 5, 2007 #14
    okay so i got part a) by doing the pythagorean theorem but i do not get part c
    i think it has something to do with the inverse of the tangent but im not sure
    CAN SOMEONE PLEASE HELP!!!!!!!!!!!!!
     
  16. Nov 5, 2007 #15
    okay so i got part a) by doing the pythagorean theorem but i dont know how to get part b)!!!!!!!!!!!!!!!! i know it has something to do with the inverse of the tangent but i dont really know!
    PLEASE HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
     
  17. Nov 5, 2007 #16
    how did u get part a? I haven't and this is due tomorrow...
     
  18. Nov 5, 2007 #17
    tangent inverse of the slope.
     
  19. Nov 5, 2007 #18
    thank you. but is the slope the momentum of the north divided by the momentum of the east??????
     
  20. Nov 5, 2007 #19
    is the answer around like 40 degrees?
     
  21. Nov 5, 2007 #20
    and antineutron you are a big help by the way!
     
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