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# Coefficient of Restitution? watch

1. What is this? I don't have an M2 book and couldn't find anything on mathsnet about it.
2. The coefficient of restitution or Newtons Law of Restitution/Newtons Experimental Law, explains how the speed of the particles after the collision depends on what the particles are made of (E.G. how much they squish when they collide). If the particles are very squishy (scuse the terminology) then they will both be moving fairly slowly after the impact, because work is done when they squash together.

Speed of seperation of particles divided by the speed of approach of the particles.

The equation is:

e = (V2 - V1)/(U1 - U2)

Where e will be between 0 and 1.
If e is 1 then the collision is perfectly elastic, and if e is 0 then the particles join together.

Another use for e would be to find out the final speed of a particle, if you know u and you know e.

v = eu
3. Cool, thanks. I thought it was the same as "e", but wasn't sure.

One more question. Am I allowed to solve COM quesitons by using vectors? There were two rectangles and I had to show the COM lied on the line dividing them, so I found each one's COM as a position vector from a stated point, took the areas of both and used these to do weighted averages to find their overall COM as a position vectoy. Then showed this was on the line dividinig them. Is this allowed?
4. (Original post by mik1a)
Cool, thanks. I thought it was the same as "e", but wasn't sure.

One more question. Am I allowed to solve COM quesitons by using vectors? There were two rectangles and I had to show the COM lied on the line dividing them, so I found each one's COM as a position vector from a stated point, took the areas of both and used these to do weighted averages to find their overall COM as a position vectoy. Then showed this was on the line dividinig them. Is this allowed?
Sounds fine to me.
5. Yeah thats fine. This is how i would do it though, try using this table for your calculations, i find it much easier.

I would put in the mass or volume times density in the mass cells, then work out a relative mass.

E.G Object 1 = 4Pi X Density and Object 2 = 2Pi X Density so Total Mass = 6Pi X Density

Therefore Relative Masses would be 2,1,3 respectively.

Then put in the COM distances from a fixed point into the x,y,z rows for each of the objects, and finally use the standard com equation with the x,y,z values and relative masses to get the COM(x), COM(y), and COM(z) values.
Attached Images

6. I don't know any M2 equations... thats sort of the problem, I don't have my M2 book so my methods are totally different from those in the markscheme! But I'm guessing the COM equations is basically weighted averages again, based on the masses of each piece.

What about centre of gravity? How does this differ from the centre of mass?

Because if, say, you are inside a star and half the radius distance from the centre, you can't just find the star's centre of mass and use Newton's gravitational equations to find the force, because you will be pulled out from the star by all the molecules beyond you from the centre of the star. Or am I wrong?
7. (Original post by mik1a)
What about centre of gravity? How does this differ from the centre of mass?
?
Same thing
8. (Original post by mik1a)
I don't know any M2 equations... thats sort of the problem, I don't have my M2 book so my methods are totally different from those in the markscheme! But I'm guessing the COM equations is basically weighted averages again, based on the masses of each piece.

What about centre of gravity? How does this differ from the centre of mass?

Because if, say, you are inside a star and half the radius distance from the centre, you can't just find the star's centre of mass and use Newton's gravitational equations to find the force, because you will be pulled out from the star by all the molecules beyond you from the centre of the star. Or am I wrong?
COM Equation:
Take E to be sigma sum signs, m to be mass of individual items, and x to be distance (replace with y and z for other parts).

x = Emx/Em

With the last thing you are wrong there, because you have a resulatant force acting upon you, which is still towards the centre of gravity of the star, although the force of gravity is less, because there is now less mass beneath you.
9. (Original post by Valen)
COM Equation:
Take E to be sigma sum signs, m to be mass of individual items, and x to be distance (replace with y and z for other parts).

x = Emx/Em

With the last thing you are wrong there, because you have a resulatant force acting upon you, which is still towards the centre of gravity of the star, although the force of gravity is less, because there is now less mass beneath you.
Is that not what I said.. perhaps I phrased it wrong, because that's what I meant - instead of pulling you in, the parts 'above' you will be pulling you outwards.
10. You still get pulled in because the resultant is still acting towards the centre, you would (if you survived, and the convection currents etc didnt have any effect) end up at the centre of gravity, and then you would be weightless! Also dead, but weightless nevertheless.
11. Ok, one more question.

A golf ball is hit into the air at a speed 49 m s^-1 at an angle 30* from the horizontal. The hole is 170m from the tee.

Find the magnitude and direction of the velocity of the ball when it is directly above the hole.

The acceleration on the ball is:

a = -9.8j m s^-2

v = INT a dt = -9.8tj + 49 cos 30 i + 24.5j
v = (24.5√3)i + (24.5 - 9.8t)j

When the ball is directly above the hole,

s = 170
v = 24.5√3
t = 170/24.5√3 ~ 4 s

So:

v = (24.5√3)i + (24.5 - 9.8*4)j
v = (24.5√3)i + (-14.7)j
|v| = √[(24.5√3)²+(-14.7)²] = √[(1800.75)+(216.09)] = 44.9 m s^-1

The markscheme does it a completely different way, I don't understand it at all.
Damn this self teaching... I really need the textbook. Is there any other way apart from integration that you can use to find this?

edit -bah never mind I guess newtons equations will do it as well! (boring...)
12. LOL - use the following equations, as your acceleration is constant. (Oh and remember to only use on the j part, as the acceleration on the i part is 0)

v = u + at

s = ut + 1/2 at^2

v^2 - u^2 = 2as
13. (Original post by mik1a)
... if, say, you are inside a star and half the radius distance from the centre, you can't just find the star's centre of mass and use Newton's gravitational equations to find the force, because you will be pulled out from the star by all the molecules beyond you from the centre of the star. Or am I wrong?
I'm sure I read somewhere that if you are actually inside a star, or any other heavenly body, no mattter where you are, then the resultant of the gravitational forces acting on you is such that you become weightless!
So, if you're halfway to the centre of the earth, say. And you do a summation of all the gravitational forces acting on you, then they all cancel out.
Anyone else heard of this??
14. (Original post by Fermat)
I'm sure I read somewhere that if you are actually inside a star, or any other heavenly body, no mattter where you are, then the resultant of the gravitational forces acting on you is such that you become weightless!
So, if you're halfway to the centre of the earth, say. And you do a summation of all the gravitational forces acting on you, then they all cancel out.
Anyone else heard of this??
There is no gravity at the centre of the earth - the earth's mass pulls you in every direction at once, giving a net force of zero. If you go halfway there's less gravity than at the surface but still some.

If the Earth were hollow there would be no gravity anywhere inside.
15. Well say you're in a box for simplicity - halfway inside (a quarter of the length from the edge), the gravitational force from that whole half of the box cancels out. You only feel the pull of the mass on the other side. So I don't think that would be true. In the centre of a star though, (say the star was at 300K and made from 80% nitrogen 20% oxygen for argument's sake) you'd be weightless. If you started halfway in, you'd 'drop' to the middle because you are more dense than the gas forming the star.

16. OK, I still think I'm right so far. I did a quick workout on it. It's a bit rough and very approx - but what I've done does seem to give some credibility to my suggestion

Suppose some body, of mass m is inside the earth somewhere. At a point P, say. Now project a double-cone from the point P, with the slant length of one cone = a, and the slant length of the second cone = b.
The conical angle for both cones is ø.

Approximate this first with a 2-dimensional layout, as shown in the Fig.
We have a mass m at the apices of two "sectors", of side lengths a and b as shown.
Let the mass of the large "sector" be A1 - its area, and the mass of the smaller "sector" be A2 - its area.
The COM of the large mass will be at (approx) 2/3 of the distance from the apex of the "sector" to the circumferential edge. Ditto for the smaller mass.

A1 = ½r²ø = ½b²ø
A2 = ½r²ø = ½a²ø

Gravitatoinal force is given by

F = km1.m2/d²

Gravitational force exerted by larger "sector" on the mass m is F1, where

F1 = kA1.m/[(2/3)b]²
F1 = (9km/4)(A1/b²)

Gravitational force exerted by smaller "sector" on the mass m is F2, where

F2 = kA2.m/[(2/3)a]²
F2 = (9km/4)(A2/a²)

F1/F2 = (A1/A2)(a²/b²)

but

A1/A2 = ½b²ø/ ½a²ø
A1/A2 = b²/a²

Substituting for A1/A2 in the expressoin for F1/F2,

F1/F2 = (b²/a²)(a²/b²)
F1/F2 = 1
======

i.e. the gravitational forces cancel out !!
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17. i did further maths A-Level. I thought i was good at maths. Ive just been reading some of this forum and this thread particularly. I've now realised im a very small fish in a pond full of bloody geniuses - and the odd idiot.

wow. respect to you all. i wish maths excited me as much as it does you all. it used to, then it got annoying.
18. Oh dear, it wasn't true after all
I just did a search and found the info below. There is some cancelling out, but not all. Some gravity force still applies.
Scroll to the bottom of this.

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