- #1

js732192

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## Homework Statement

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)?

## Homework Equations

J = F⃗Δt= Δp⃗ = m(v⃗f−v⃗i)

v2=v20+2aΔy

## 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!