# Impulse: Jumping before the elevator crashes

• js732192
In summary: You need to use the relative velocity formula, but you should find that v1 is the same as it was for the first part of the problem.Okay, so using the relative velocity formula, I got that the passenger's speed just before impact is 19.56 m/s.c) Using J = Δp⃗ = m(v⃗f−v⃗i), I got that J = -1300 kgm/s.d) Using J = F⃗Δt, I got that F = 260000 N.In summary, after a cable snaps and the safety system fails, an elevator cab free falls from a height of 36m and collides with the bottom of the
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!

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.

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!

## What is "Impulse: Jumping before the elevator crashes"?

"Impulse: Jumping before the elevator crashes" is a thought experiment that explores the concept of impulse and its effect on a person's survival in a potentially life-threatening situation.

## What is impulse?

Impulse is the product of an object's mass and its velocity. It is a measure of the object's momentum and its tendency to continue moving in a particular direction.

## How does jumping affect the outcome of the elevator crash?

Jumping before the elevator crashes can potentially increase a person's chances of survival. By jumping, the person increases their impulse, which can potentially reduce the impact of the crash.

## Is jumping always the best course of action in an elevator crash?

No, jumping may not always be the best course of action in an elevator crash. It depends on the specific circumstances and variables involved, such as the height of the fall and the person's physical abilities.

## How does this thought experiment relate to real-life situations?

This thought experiment may relate to real-life situations by highlighting the importance of impulse and quick decision making in emergency situations. It also emphasizes the role of physics in understanding and potentially altering the outcome of such scenarios.

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