Equations for Calculating Momentum in Pole Vaulting Collisions

  • Thread starter Thread starter StephenDoty
  • Start date Start date
  • Tags Tags
    Physics Sports
Click For Summary

Homework Help Overview

The discussion revolves around the physics of momentum in the context of pole vaulting, specifically examining the nature of collisions and the application of momentum equations when the pole vaulter releases the pole.

Discussion Character

  • Conceptual clarification, Assumption checking, Exploratory

Approaches and Questions Raised

  • The original poster questions whether the collision is elastic or inelastic and seeks to determine the appropriate momentum equation to use for calculating the pole vaulter's momentum upon release. Other participants suggest considering energy conservation instead of momentum due to the nature of the pole's interaction with the vaulter.

Discussion Status

The discussion is exploring different interpretations of the mechanics involved in pole vaulting. Some participants have provided guidance on using conservation of energy, while others are still considering the momentum equations. There is no explicit consensus on the approach to take.

Contextual Notes

Participants are navigating assumptions about the nature of the collision and the relevant physics principles, including potential energy stored in the pole and the dynamics of the pole vaulter's motion.

StephenDoty
Messages
261
Reaction score
0
Is the collison of the olympic sport polevolting elastic or inelastic? The pole collides with the Earth and stays at that point until the polevolter let's go. So wouldn't you use the momentum equations to see how much momentum the polevolter has once he let's go of the pole?

Which one of the following equations would I use to find the momentum of the polevolter once he has let go of his pole? mv1+mv2=mv1+mv2 or mv1+mv2=(m+m)v

Thank you.

Stephen
 
Physics news on Phys.org
StephenDoty said:
So wouldn't you use the momentum equations to see how much momentum the polevolter has once he let's go of the pole?

Hi Stephen! :smile:

It's not a collison … it's exactly like swinging on a rope (but upside down :wink: ).

So you'd just use conservation of energy (including some "spring" energy for the bending of the pole). :smile:
 
so you would have 1/2mv1^2 + 1/2kx^2 +mgh0= 1/2mv2^2 + 1/2kx^2 +mgh1?
 
Yup! :biggrin:
 

Similar threads

Replies
10
Views
6K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 12 ·
Replies
12
Views
5K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 16 ·
Replies
16
Views
4K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
10
Views
3K
  • · Replies 7 ·
Replies
7
Views
7K