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-   -   Impulse, momentum and efficiency to find height formula (http://www.physicsforums.com/showthread.php?t=558902)

Neutrinoftw Dec10-11 03:52 PM

Impulse, momentum and efficiency to find height formula
 
http://a6.sphotos.ak.fbcdn.net/hphot...15117382_n.jpg
1. The problem statement, all variables and given/known data

A ramp system is set up. Due to friction the ramp has an efficiency of 82%. Two cars of equal mass are allowed to collide and car 1 starts from rest at height h1. The two stick together and coast to a height of h2. Derive a formula relating h1 to h2.

2. Relevant equations

I=Δp, p=m*v, eff=(output/input)*100%, GPE=mgh,

3. The attempt at a solution

Unfortunately, I cannot figure out almost anything based on this question. Even my friend's father, who is an engineer, could not understand. Just looking for some good help if any one is available.

Spinnor Dec10-11 05:37 PM

Re: Impulse, momentum and efficiency to find height formula
 
First things first, what is the speed car 1 just before impact? It's kinetic energy is not mgh_1 but some fraction of that.

gneill Dec10-11 06:08 PM

Re: Impulse, momentum and efficiency to find height formula
 
Quote:

Quote by Spinnor (Post 3661818)
First things first, what is the speed car 1 just before impact? It's kinetic energy is not mgh_1 but some fraction of that.

First things first, but not necessarily in that order*

What is the definition of efficiency to apply? Final system energy over initial system energy? Final velocity over ideal velocity? Something else?


* Doctor Who

BetoG93 Dec10-11 06:15 PM

Re: Impulse, momentum and efficiency to find height formula
 
I would calculate kinetic energy (it is not the same as the potential energy, since the efficiency is less than 100%). Then, I would find speed, and later calculate the speed after the inelastic collision. Then, I would use conservation of energy.

BetoG93 Dec10-11 06:19 PM

Re: Impulse, momentum and efficiency to find height formula
 
Gneill, I think that efficiency here refers to energy; in other words, only 82% initially available energy is useful, and the other 18% is converted to other forms of energy.

gneill Dec10-11 08:00 PM

Re: Impulse, momentum and efficiency to find height formula
 
Quote:

Quote by BetoG93 (Post 3661884)
Gneill, I think that efficiency here refers to energy; in other words, only 82% initially available energy is useful, and the other 18% is converted to other forms of energy.

Okay, suppose it is so. Can you suggest how the efficiency might have been determined? Was it measured by running a single cart down one side and up the other? Is it independent of total path length? (if you change the initial height h1, does it affect the result?) Does it depend upon the slopes of either side? Normal force (hence friction) depends upon angle, so can "efficiency" be independent of slopes?

Perhaps the OP has been given a particular working definition of efficiency that can be applied here?

Neutrinoftw Dec12-11 01:39 AM

Re: Impulse, momentum and efficiency to find height formula
 
For this question I have been told to use Input/Output for efficiency, if that helps at all.

PeterO Dec12-11 03:45 AM

Re: Impulse, momentum and efficiency to find height formula
 
Quote:

Quote by gneill (Post 3662017)
Okay, suppose it is so. Can you suggest how the efficiency might have been determined? Was it measured by running a single cart down one side and up the other? Is it independent of total path length? (if you change the initial height h1, does it affect the result?) Does it depend upon the slopes of either side? Normal force (hence friction) depends upon angle, so can "efficiency" be independent of slopes?

Perhaps the OP has been given a particular working definition of efficiency that can be applied here?

In real life, the amount of energy lost during a situation like this is very hard to quantise - though perhaps easier to measure.

In MANY cases, the situation is idealised by assuming there is no friction.

In this case it looks like a more representative, but equally random, assumption has been made and so when moving up and/or down this slope, instead of 100% of the PE converting to KE on the way down the hill, only 82% of the PE ends up as KE.
I would assume that after the collision, only 82% of the KE after collision becomes the PE when they stop.


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