Conservation of momentum - trampoline

AI Thread Summary
The discussion centers on the conservation of momentum and energy in a trampoline scenario where Joe drops from a height and bounces back up. The force exerted by the trampoline is modeled as F = kx, where k is the spring constant. While the theoretical model suggests Joe bounces back to the original height due to energy conservation, real-world factors like energy loss to heat and sound are acknowledged as limitations. Momentum is not conserved when considering Joe alone due to external forces like gravity and the normal force from the trampoline, but it is conserved when the Earth is included in the system. The conversation highlights the complexities of understanding momentum and energy conservation in practical applications.
Pottesur
Messages
2
Reaction score
0
English isn't my main language, so I apologize in advance if something is unclear.

We are leaving air resistance out of this problem!

1. Homework Statement

We are going to describe the force F from the trampoline on the Joe as F = kx, k is a spring constant. This is a model.

1. Joe drops from a window at 3 meters above the trampoline, hits the trampoline with stiff legs, and bounces back up. How high up does Joe go now?

2. Based on your own experience with trampolines: is this model a good model for vertical trampoline jumping? Are there any sides to the model which are particularly weak?

3. Is momentum (size and direction) conserved when Joe hits the trampoline?

Homework Equations

The Attempt at a Solution


1. Conservation of energy. Joe will have potential energy as he drops because of gravity (mgh), which will transfer to kinetic energy as he falls towards the trampoline. When he hits the trampoline, the kinetic energy will be zero and the energy will be stored as elastic potential energy in the springs. He bounces back up to three meters when the elastic potential energy is released to kinetic energy.

2. Wouldn't some of the energy in "the real world" be lost to heat and/or sound? And the spring constant is probably different from trampoline to trampoline. Not quite sure what the right answer here is.

3. This is where I've really been struggling, particularly with what I am supposed to view as my system. If I look at Joe as the system, momentum is not conserved because there are external forces acting on him, right? When he is in mid-air gravity is acting on him, and when he hits the trampoline there is a normal force acting on him. Are both these forces considered external forces? But if I look at velocity, and conservation of energy, his velocity will be the same just before and after the jump. But because they differ in direction, is this also a reason momentum isn't conserved? Or is it a consequence of external forces acting on Joe? What if I consider the person and the Earth as a system, is momentum then conserved?

I am finding momentum quite hard to understand. Hopefully someone here can help me out :)
 
Physics news on Phys.org
Pottesur said:
English isn't my main language, so I apologize in advance if something is unclear.

We are leaving air resistance out of this problem!

1. Homework Statement

We are going to describe the force F from the trampoline on the Joe as F = kx, k is a spring constant. This is a model.

1. Joe drops from a window at 3 meters above the trampoline, hits the trampoline with stiff legs, and bounces back up. How high up does Joe go now?

2. Based on your own experience with trampolines: is this model a good model for vertical trampoline jumping? Are there any sides to the model which are particularly weak?

3. Is momentum (size and direction) conserved when Joe hits the trampoline?

Homework Equations

The Attempt at a Solution


1. Conservation of energy. Joe will have potential energy as he drops because of gravity (mgh), which will transfer to kinetic energy as he falls towards the trampoline. When he hits the trampoline, the kinetic energy will be zero and the energy will be stored as elastic potential energy in the springs. He bounces back up to three meters when the elastic potential energy is released to kinetic energy.

2. Wouldn't some of the energy in "the real world" be lost to heat and/or sound? And the spring constant is probably different from trampoline to trampoline. Not quite sure what the right answer here is.

3. This is where I've really been struggling, particularly with what I am supposed to view as my system. If I look at Joe as the system, momentum is not conserved because there are external forces acting on him, right? When he is in mid-air gravity is acting on him, and when he hits the trampoline there is a normal force acting on him. Are both these forces considered external forces? But if I look at velocity, and conservation of energy, his velocity will be the same just before and after the jump. But because they differ in direction, is this also a reason momentum isn't conserved? Or is it a consequence of external forces acting on Joe? What if I consider the person and the Earth as a system, is momentum then conserved?

I am finding momentum quite hard to understand. Hopefully someone here can help me out :)

You have analysed the issue fairly well. You noticed that momentum is a vector, so has a direction. That's good. Therefore, Joe's momentum is reversed by a trampoline jump.

Yes, if you include the Earth, then momentum of the Joe-Earth system is conserved.
 
  • Like
Likes Pottesur
Hello Pottessur, ##\qquad## :welcome: ##\qquad## !

your english is excellent.
1. agree
2. agree. And different spring constants ##k## can easily be accommodated
3. agree. If you want momentum conservation, all the forces have to be within the system

[edit] slow typist. PeroK was a lot faster :smile: .
 
  • Like
Likes Pottesur
Pottesur said:
momentum quite hard to understand
If it helps: the time derivative of momentum is force $$\vec {\bf F} = {d\vec{{\bf p}}\over dt}$$
 
  • Like
Likes Pottesur
Ahh, thank you guys so much for your feedback! I really appreciate it. I now have a bit more confidence in myself, haha :biggrin:
 
Pottesur said:
When he hits the trampoline, the kinetic energy will be zero
Well, not straightaway. It is gradually converted to elastic PE.
Pottesur said:
the spring constant is probably different from trampoline to trampoline.
Yes, but that is not a flaw in the model. The model encompasses that possibility.
Pottesur said:
Wouldn't some of the energy in "the real world" be lost to heat and/or sound?
Yes, heat mostly, but how would you have to amend the F=kx model?
There's also air resistance.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top