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Cart Momentum Question

  1. Oct 12, 2008 #1
    1. The problem statement, all variables and given/known data
    A 478.0 g cart is released from rest 1.16 m from the bottom of a frictionless, 31.0° ramp. The cart rolls down the ramp and undergoes a collision with a rubber block at the bottom. After the cart bounces, how far does it roll back up the ramp if the maximum force applied is 302.0 N for 22.9 ms?


    2. Relevant equations

    pf=pi+impulse
    Vf2=Vi2+2ad
    p=m*v

    3. The attempt at a solution
    First I found my initial impulse:
    Vf2=Vi2+2ad
    Vf2=0+2(9.8cos31)(1.16)
    Vf=4.41m/s
    but since its moving in the negative direction, it is -4.41m/s

    p = m*v
    p = (0.478)(-4.41)
    =-2.11

    The I calculate my impulse (the area under the graph):
    0.5(302)(0.0229)
    =3.46 kg m/s

    Then I calculated my final impulse:
    pf=pi+impulse
    pf=-2.11+3.46
    =1.352

    I then found the distance it travels up after it gains that momentum from hitting the rubber block:
    p=m*Vi
    1.352=0.478*Vi
    Vi=2.83

    Vf2=Vi2+2ad
    0=2.832+2(-9.81cos31)d
    d=0.477m

    However, this answer does not match my answer key...

    Thanks in advanced
     
    Last edited: Oct 12, 2008
  2. jcsd
  3. Oct 12, 2008 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Welcome to PF!

    Hi MightyMan11! Welcome to PF! :smile:

    Isn't it sin31º?
     
  4. Oct 13, 2008 #3
    Re: Welcome to PF!

    AHHH...
    Looks like it.

    Thanks for the help and the welcome :smile:
     
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