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Relative velocities - Velocity of impact

  1. Apr 2, 2013 #1
    1. The problem statement, all variables and given/known data
    https://dl.dropbox.com/u/51681259/physics_diag.png [Broken]

    Object C & P are moving at the same constant velocity to the right, in the configuration of the image. The friction between object C and the ground is 0. Suddenly the friction increases to μ=0.9 between the ground and object C. This makes object C decelerate to a stop over 4m. At what velocity does object P impact the right internal side of object C. There is no friction between C & P.

    d = 1m (distance between p right side and c internal right side)
    μ = 0.9
    s = 4m (stopping distance)

    2. Relevant equations
    I can fine object c's initial velocity by combining a couple of formulas:
    F_net=ma
    F_friction=μmg
    v^2=u^2+2as

    I find the initial velocity of C & P as being 8.4m/s.

    Then I can find the acceleration of C, relative to the ground, after the increase in fiction which turns out to be:
    -8.82m/s/s

    And that means that relative to C, P accelerates forward at 8.82m/s/s. <IS THAT CORRECT??>

    Then I take p's relative acceleration and the initial velocity and put them through v^2=u^2+2as
    and find:
    v^2=8.4^2+(2*8.82*4)

    v=9.39
    So my answer is that P impacts C at 9.39 m/s. Is this even close to correct?
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Apr 2, 2013 #2

    mfb

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    Staff: Mentor

    This looks fine.

    Both C and P are never moving faster than 8.4m/s, and they are moving in the same direction. There is no way the impact can occur with more than 8.4m/s.

    The calculation has two problems:
    - the initial relative velocity between P and C is zero, not 8.4m/s
    - the distance is just 1m and not 4m.
     
  4. Apr 2, 2013 #3
    Sorry! That was so stupid!!

    v^2=0^2+(2*8.82*1)
    v^2 = 17.64
    v=4.2 m/s

    Is that correct?
     
  5. Apr 3, 2013 #4

    mfb

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    2016 Award

    Staff: Mentor

    That looks good.
     
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