Impulse: Jumping before the elevator crashes

[js732192]

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!

 

[SammyS]

...

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?

 

[js732192]

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?

 

[ideasrule]

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.

 

[rwisz]

Hello! Your calculations and equations are correct. The numbers may seem high because the scenario is quite extreme, with a free fall from a height of 36m. It is important to use the correct units in your calculations, as these can affect the magnitude of the values. Overall, your approach and solution seem reasonable. Keep up the good work!