h2g2 The Hitchhiker's Guide to the Galaxy: Earth Edition

Hey, guys!
I don't really know, if this is the right place to ask an d if there's andybody who can help me on this.
But taking my physics class and learnign about the conservation of momentum I experianced something that looks to me like a contreadiction between both theoremes. Nobody could give me an answer yet so I turn to the collected genius of the h2g2 society.

In a closed, isolated system one body hits another. Both bodies have the same mass. They stick together and can be considered one body with twice the mass of each single one.

Let's try now to find out the speed of this object!

m1: mass of moving object before collision
V1: its velocity before collision
m2: mass of both objects after collision
V2: its velocity

using the conservation of momentum:

m1 * V1 = m2 * V2
V2 = V1 * m1 / m2
since m1/m2 = ½
V2 = ½ V1

easy, right?
Just to check our result let's use the conservation of energy!

½ * m1 * V1² = ½ * m2 * V2²
m1 * V1² = m2 * V2²
V2 = radical(m1 / m2) * V1

Ooops! Different results!

I really hope, there is someone who can help me. Otherwise I will lose my faith in science *g

You're conservation of momentum equation is flawed, it only has one inital body and velocity, if there are two objects in the closed system, the equation should reflect that, and it doesn't.

"since m1/m2 = ½"
Ummm. no. The numbers there would be properly expessed a subscrips to denote that the first 'm' is a different one than the second 'm'. They aren't really numbers.


I'm under the impression that you are working in one dimentional space due to the lack of incident angles or other vector resolution elements in your equations.

Try this one out and see what you get.

m1 = mass of first particle
m2 = mass of second particle
v1 = velocity of first particle
v2 = velocity of second particle
vf = final velocity of combined particles

conservation of mass in an inelastic collision of two particles using one dimentional physics:

m1*v1 + m2*v2 = (m1 + m2)*vf

conservation of energy in an inelastic collision of two particles using one dimentional physics:

1/2(m1*v1*v1) + 1/2(m2*v2*v2) = 1/2((m1 + m2)*vf*vf)

Try that and see if it helps.

Ahh, I think 'm1/m2 = 1/2' was right, seeing as the first object has half the mass of the second (the second object being, actually, both objects joined togher). It looks like the other object is assumed to be not moving initially. This is getting somewhat confusing... it'd be a lot easier if we use the following:

Object 1: First particle (initially moving)
Object 2: Second particle (initially stationaryObject 3: Both particles joined together after collision, twice the mass of object 1

Therefore:

m3 = m1 + m2

Since we're assuming m2 and m1 to be the same:

m1 x v1 = m3 x v3

m1 x v1 = 2 x m1 x v3

v3 = v1 / 2

Initial energy:

e1 = 1/2 x m1 x v1 x v1

Final energy:

e3 = 1/2 x m3 x v3 x v3

Now, substituting values for m3 and v3:

e3 = 1/2 x m1 x 2 x v1/2 x v1/2

Therefore:

e3 = e1 / 2

And you are absolutely correct, energy has been lost. And that, in fact, is your answer: when two objects collide and stick together, what's known as an 'inelastic collision' has taken place. During an inelastic collision, energy is *always* converted to other forms, usually heat and sound.

In order for an elastic collision, ie one in which no energy is lost, to take place, there is a simple rule:

Speed of approach = Speed of separation

If object two is stationary to start off with, this means that object one has to stop dead during the collision and transfer all its velocity to object two. This is pretty much what happens during a game of snooker/pool... you hit the que ball into another ball, the que ball then stops, and the other ball starts moving.

The thing to remember is that whereas conservation of momentum *always* applies, conservation of energy can be confusing because energy can be converted to other forms.

Hope this helps!

26199

Thanks a lot. Neither me nor my physics teacher was able to find an answer (neither were the people she asked *g). I knew, that there would be energy lost, but I am kind of used to ignore that, because it happens almost allways and it makes things so much more difficult. Usually it doesn't make too much of a difference, but I guess in this case it's important.

Something else:
Do you know a bit about Einstein?

Oh yea, I forgot that conservation of energy only applies to perfectly elastic collisions. However, the two masses do no necissarily have to be the same, and one particle does not have to be stationary. Therefore the first equation I gave will always apply to two particles in an inelastic collison in one dimention while the second only applies to perfectly elastic collisions.

Something else:
What this about a beer?

Thanks a lot for your help. But I don't quite understand the 'something else' - part.

You said something about Einstein, which roughly translates as "one beer" from German.

Well - I AM German. But I can't see that connection. I would just translate it as "one stone". But maybe there's a beer, I don't know of, in Germany called "Stein". Oh! There is a beer "Wahrsteiner". That might be it.

Anyway: I've got another question about physics. It deals with speed of light, relativity and different pastes (?) of time. Do you think you could answer it? Or are you bored anyway

Like I said, "roughly translates as". Maybe it was one beer mug, er something.

I can try to help with the relativity thing, it reallly depends on the question, though.

Hey, thanks!
OK - here it goes.
I understand the following:
There is no such thing as absolute position and therefore no absolute velocity in space. If I am moving in respect to someone else in direction of a light source, the light approaches me still with light speed according to my own mesurements since I am NOT moving in respect to myself and light moves always with the same speed in empty space. The outside spectator though sees the light approaching me with more than light speed (my speed + the approaching light). Therefore time has to be slower for me than for the spectator.

But what happens, if the light source is behind me? Then the same logical reasoning tells me, that MY time moves faster than the spectator's.

And that really throws me off. Do you understand my problem? I'm afraid I'm not very good with logics today. If you don't, I'll write it out with a little more words and explanation.

Thanks anyway.

You think that's confusing, what if someones shining two lights on you from opposite directions?

That's one of the things that's always bugged me about relativity, too.

There's the old "twin paradox" scenario where one twin gets on a ship and travels near the speed of light for awhile, and then returns. Many people say that the one on the ship will be younger, but I disagree because their average relative velocities have been the same ie zero because they start and end in the same place relative to each other, so the will be the same age, and this would seem to be the result of the thing you have brought up, it would seem time would go faster when travalling realtively in one direction, and slower in the opposite.

In other words, sure, I'll go for that.

Hey!
That's interesting. Of course I know this story, but I've never thought about that. It doesn't matter, if one is in the space ship and the other one stays back. Cool!
The only reason for one of the twins to be younger than the other one is, that the one in the space ship would have spent his life almost only surrounded by empty space. And time moves slower close to big masses. (they say, that time doesn't move in a black hole)
How do you come in contact with physics? I am just an interested high school student (yes, they exist).

Hey, now, don't quote me on that. It's just what I think. I have not in any way verified that it is even remotely accurate.

I had a quick net search, and found

http://www.suite101.com/article.cfm/advanced_physics/26998

or, for a more varied (confusing?) interpretation :

http://math.ucr.edu/home/baez/physics/twin_paradox.html

Thank you, the fact that relativistic speeds do not linearly add has cleared it up for me, I hadn't considered that.

So, yes, the stationary twin would be older when the two meet up again.

Thanks from me, too! I only start to understand, but I think, I'm getting part of it.
Highly interesting topic by the way. What about starting a physics question forum?