Impulse: Jumping before the elevator crashes

  • Thread starter Thread starter js732192
  • Start date Start date
  • Tags Tags
    Elevator Impulse
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
3 replies · 11K views
js732192
Messages
2
Reaction score
0
Hello!

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!
 
on Phys.org
js732192 said:
...

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!
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?
 
SammyS said:
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?

Would the passenger's speed be 7 + 26.56 = 33.56 m/s?
 
js732192 said:
Would the passenger's speed be 7 + 26.56 = 33.56 m/s?

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.