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Impulse: Jumping before the elevator crashes

  1. Jun 30, 2011 #1
    Hello!
    1. The problem statement, all variables and given/known data
    After the cable snaps and the safety system fails, an elevator cab free falls from a height of 36m. During the collision at the bottom of the elevator shaft, a 90 kg passenger is stopped in 0.005 s. Assume that neither the passenger nor the cab rebounds. What are the magnitudes of the:

    a) impulse and b) average force on the passenger during the collision?

    If the passenger were to jump upward with a speed of 7.0 m/s relative to the cab floor just before the cab his the bottom of the shaft, what are the magnitudes of the:

    c) impulse and d) average force (assuming the same stopping time)?

    2. Relevant equations

    J = F⃗Δt= Δp⃗ = m(v⃗f−v⃗i)
    v2=v20+2aΔy


    3. The attempt at a solution

    a) I used this formula v2=v20+2aΔy with y=-36m and v2=0 to solve for v1. I got that v1=26.56 m/s

    I then used J = Δp⃗ = m(v⃗f−v⃗i) using vf= 0 and v1=26.56 to get that J=-2390.7 kgm/s

    b) I used J = F⃗Δt with J=-2390.7 and t=0.005s to get that F= -478136.38 N (480000 N)

    c) I used J = Δp⃗ = m(v⃗f−v⃗i) again, but I was confused as to whether or not vi= 7 m/s (or if it was more complex than that). But I used vi=7 m/s to get that J = -630 kgm/s

    d) J = F⃗Δt with the new J to find that F=126000 N

    Am I on the right track? The numbers just seem so high...But I guess the dude's falling from quite a height.

    Thanks!
     
  2. jcsd
  3. Jun 30, 2011 #2

    SammyS

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    Hi js732192. Welcome to PF !

    a & b look good.

    For c & d:

    The passenger jumps upward at a speed of 7m/s, relative to the cab floor, just before the cab hits the bottom ...

    If the cab's speed is 26.56 m/s (downward) just before it hits the bottom, what is the passengers speed just before impact?
     
  4. Jun 30, 2011 #3
    Would the passenger's speed be 7 + 26.56 = 33.56 m/s?
     
  5. Jul 1, 2011 #4

    ideasrule

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    No. By jumping upwards, he's trying to move away from the ground, so it doesn't make sense that he's approaching the ground faster.
     
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