Collision : When a gnome hits a giant. (Or any non-gnome)

[Tipx]

First of all : There is no gnome or giants in my question/post, it's just that the title would have been too long :
- I'm having a hard time understanding what's happening when an object collides with another object of a different mass.

Here are my premises, followed by the points I don't understand :

In frictionless, gravityless environment, there are a few pool ball size object :
1- Light : It's mass near zero.
2- Medium 1 : Has a mass of Ma.
3- Medium 2 : Has a mass of Ma.
4- Big : Has a mass of 1.5 Ma.
5- Duper : Has a mass of a googol Ma.

If the collisions are totally elastic, I would expect that if Medium 1 hits Medium 2, which didn't move, head on (the normal of the collision being parallel to the relative movement of the 2 objects) Medium 1 stops and Medium 2 "bounces" at the same speed Medium 1 was moving. If the collision was not head on, I would expect Medium 1 to transmit the part of his motion that was parallel to the collision's normal, and keep 100% of the component perpendicular to that collision.

If a Medium collides with Light, I'd expect Light to bounce near the speed of light (I had to phrase it that way, it was too obvious.)
If an object (1, 2, 3 or 4) hits Duper, I'd expect the said object to bounce back at the same speed it had before, with Duper still sitting there, unaffected.

If a Medium hits Big, in a head on collision, or not, I have no clue how it would result.

Can anyone point me in the right direction please? I read stuff here and there about collisions, but I can't find a place where the whole concept, using vectors. I'm all confused about if I must use the concepts of inertia and such. If there are some explanation for inelastic collisions, I wouldn't mind either.

Thanks

 

[AlephZero]

The neat way to answer all of these is to measure things relative to the center of mass of the two particles.

The forces between the particles in the collision are equal and opposite, so the resultant force on the two particles is zero. Therefore, the motion of the center of mass is not changed by the collision.

If the collision is perfectly elastic, the velocities relative to the CM are just reversed in direction by the collision.

To work through your "equal mass" case this way: Suppose the speeds of the particles before collision are v and 0.
The velocity of the CM is (mv + m.0)/2m = v/2
To find the relative velocities before collision, subtract the velocitiy of the CM. They are v - v/2 = +v/2 and 0 - v/2 = -v/2
The relative velocities after collision are reversed, that is -v/2 and +v/2.
To find the absolute velocties after collision, add the CM velocity, and you get -v/2 + v/2 = 0 and +v/2 + v/2 = v, whish is what you said the answer was.

Your result for the "heavy and light" case was wrong. The light particle after the collision will be twice the speed of the heavy particle, not "infinitely fast". Work through it yourself to check that.

BTW, don't you mean there are gno nomes in this question

 

[Tipx]

Thanks for your reply AlephZero. Let me me see if I get this straight :

If the collision is perfectly elastic, the velocities relative to the CM are just reversed in direction by the collision.

Work through it yourself to check that.

Here I go :
m1 = 2kg
v1 = 0m/s
m2 = 5kg
v2 = 3m/s

2 would be moving at (3 - 15/7 = 6/7)m/s relatively to the CM, so after the impact, it would move at (-6/7)m/s relatively to CM, so it would, in absolute, have a velocity of (9/7)m/s, while, using the same parameters, 1 would end up moving at (30/7)m/s?

If I understand correctly, and if I "mix" this with something else I read about oblique collisions, to go with a vectorial approach, I would :
1- Decompose the initial velocity relatively to the normal of the collision into a parallel component, and a perpendicular component.
2- Relativise the parallel component to the CM's velocity parallel component.
3- "Reverse the direction".
4- "Absolutise" the Parallel component by adding back the CM's velocity parallel component.
5- Add back the initial perpendicular component
6- Smile

Sounds correct?

 

[AlephZero]

2 would be moving at (3 - 15/7 = 6/7)m/s relatively to the CM, so after the impact, it would move at (-6/7)m/s relatively to CM, so it would, in absolute, have a velocity of (9/7)m/s, while, using the same parameters, 1 would end up moving at (30/7)m/s?

Yes, I thnk that's right.

If I understand correctly, and if I "mix" this with something else I read about oblique collisions, to go with a vectorial approach, I would :
1- Decompose the initial velocity relatively to the normal of the collision into a parallel component, and a perpendicular component.
2- Relativise the parallel component to the CM's velocity parallel component.
3- "Reverse the direction".
4- "Absolutise" the Parallel component by adding back the CM's velocity parallel component.
5- Add back the initial perpendicular component
6- Smile

Sounds correct?

If you ignore friction, then yes, the veolcities perpendicular to the motion of the CM are unchanged.

But if you do include friction, and the bodies are not points so they can also rotate about their own centers of mass, everything gets a lot more complicated!

 

[Tipx]

But if you do include friction, and the bodies are not points so they can also rotate about their own centers of mass, everything gets a lot more complicated!

Oh yes indeed!
I already though about the moment of inertia, but I'm not sure I want to get into that, so I'm supposing the world is frictionless, rotationless!

Thanks a lot for your help AlephZero, it's MUCH appreciated!