[Professor Mega-Mind (OP)]

First , if it's only a little bigger , A .
B would be an average .
This is not my concept though , the velcro is not a true inelastic collision .  There is no "sand" to turn K.E. into thermal energy . 
My whole point is to waste K.E. unidirectionally in order to have the remaining induce unidirectional thrust !
Try to reduce the question volume , I'm having trouble finding time to surf , etc. !
P.

[Kryptid]

Here is something that I don't think you considered: with the ball hitting the sand-filled wall, 45% of the kinetic energy is changed into heat. That means that the ball only has 55% of its original kinetic energy after impact. With the ball hitting the hard wall, only 2% of the kinetic energy is changed into heat, meaning that the ball has 98% of its original kinetic energy after impact. Since the one hitting the hard wall has more kinetic energy than the one hitting the soft wall, the hard wall is actually pushed harder than the soft wall (against your implications).  So the box drifts in the direction of the hard wall, while the ball that hit the hard wall bounces back faster than the one that hit the soft wall. The faster moving ball carries more kinetic energy than the slower moving ball. This means that the total momentum of the balls is in the direction away from the hard wall while the momentum of the box is in the direction of the hard wall. Since the motions are in opposite directions, they cancel out and the momentum is still zero.

[Bored chemist]

, the velcro is not a true inelastic collision .

Yes it is. But feel free to imagine sandbags instead.

Please answer the question

In which scenario does the bigger weight move away faster?
Is V(sticky) bigger or smaller than V(bouncy)?

[alancalverd]

Energy is not a vector. Momentum is.

[Professor Mega-Mind (OP)]

Mr. K. ,
Your picture above is also different from my design . I use only loosely packed sandbags .  When a steel ball hits these , it gives up ALL of it's KE : %55 to the Massive-Wall as kinetic energy  (shove) , %45 to the sandbag as thermal energy  (heat).  The ball stops dead in it's tracks as a result  of this , it is done until it is thrown again .
In regards to the steel ball/wall impact : This too was pictured not like my process .  The ball imparts %2 of it's KE to the steel wall , it converts ~%0 to heat , it then reflects back with %98 of it's KE intact .  The sandbag transfers more KE because of it's much longer contact time , same with the steelball on sandbag impact .
Note - Energy must be radiated or conducted out to make this system work .
P.M.

[Bored chemist]

Mr. K. ,
Your picture above is also different from my design . I use only loosely packed sandbags .  When a steel ball hits these , it gives up ALL of it's KE : %55 to the Massive-Wall as kinetic energy  (shove) , %45 to the sandbag as thermal energy  (heat).  The ball stops dead in it's tracks as a result  of this , it is done until it is thrown again .
In regards to the steel ball/wall impact : This too was pictured not like my process .  The ball imparts %2 of it's KE to the steel wall , it converts ~%0 to heat , it then reflects back with %98 of it's KE intact .  The sandbag transfers more KE because of it's much longer contact time , same with the steelball on sandbag impact .
Note - Energy must be radiated or conducted out to make this system work .
P.M.

OK, since you seem to want to talk about non elastic collisions and elastic ones, please answer my question.

In which scenario does the bigger weight move away faster?
Is V(sticky) bigger or smaller than V(bouncy)?

[Professor Mega-Mind (OP)]

Reply # 80 covered it pretty well .  I'll use pool balls to elucidate .
One pool ball bulls-eyes another .  It stops dead , while the target bounces forward at ~%100 speed .  If you increase the target ball's mass slightly , it will bounce a bit slower , while the shooter-ball bounces backwards slowly .
The "adhesive collision" converts two objects into one instantly .  The "Shooter" object's KE is then distributed proportionately throughout the "new object" .  This results in a significantly lower velocity than the "target" object in the first case .
OK , ya got yer cockamamie answer !  Finished work , gotta long drive now .
P.M.

[Bored chemist]

So, without the dross, you think the ball with the elastic collision moves off slower than the one where they stick together?
And do you think the same would happen with, for example, rail trains (assuming they had frictionless wheels)?
If a small waggon hit a larger one and bounced off (backwards) it would set the bigger one moving.
If they hit + "stuck" together (Imagine, if you like, a big spike on the front of the small wagon sticking into a big bag of sand stuck to the back of the big one.)

Do you think the "sticky" collision would send the combined train off slower than the "bouncy" collision?

[Professor Mega-Mind (OP)]

You're throwing out multiple variables there .  An "adhesive" collision is fundamentally different from a true in-elastic collision , terminology notwithstanding . 
Now , your puzzle :
If a steel ball bulls-eyes a same-weight steel ball , it stops dead as it transmits ~100% of it's kinetic energy to the target ball .  If the target ball is heavier , it absorbs less KE from the shooter-ball , which keeps some KE as it bounces back from the target .  If the target-ball is massive-plate heavy , the shooter-ball will bounce back with ~98% of it's KE , while the target-ball absorbs ~2% . 
 IF you add a "perfect" adhesive to the target ball , then the balls will instantly become one upon contact .  They will then be a larger object with the KE of the 2 objects put together .
 In summary , your final statement is correct .

[Bored chemist]

An "adhesive" collision is fundamentally different from a true in-elastic collision

For a start, no it isn't. For a finish, what do you think the difference is?

[Professor Mega-Mind (OP)]

The first fifteen sentences of reply # 85 explain it nicely .
D.

[Bored chemist]

The first fifteen sentences of reply # 85 explain it nicely .

Well, they don't make sense but never mind

OK, let's do the maths that you are apparently incapable of.
Let's say the small mass is 1KG and the big one is 100KG the small mass  starts of at 1 m/s  the collision is perfectly elastic.

The initial KE is 1/2 Joules (from KE=1/2 M V^2)

First let's look at the non elastic collision case.
We know we can't rely on conservation of energy here. Some energy is lost warming up the sand.
But we have the conservation of momentum.
As before the total  initial momentum is 1, so the final momentum must also be 1
And the combined mass of the (now joined) masses is 101Kg.
So the final velocity is 1/101 m/s

Now let's look at the  inelastic collision, the final KE is also 1/2 J
What we need to do next is work out what the velocities of the two masses are.

We know the initial momentum (1 Kgm/s) and that's conserved, so we know the final momentum.
And the final momentum must be the sum of momenta of the two masses.
1 V_(small)  +100 V_(big) =1 kgm/s
and from the KE we know that
1/2 V_(small)^2  +100/2 (V_(big)^2 =1/2 J
Multiply  both sides by 2 to tidy up
V_(small)^2 + 100V_(big)^2 =1

That's a pair of simultaneous equations.
It's a bit late in the day for me to think about that pair of equations , perhaps someone will stuff them into a solver for us.

In the meantime...
There's a simplifying assumption we can make here.
If the big mass was infinitely big then the ball would bounce off at the same speed as it hit.
Because the big mass is much bigger than the small one (100Kg vs 1Kg) the speed it bounces off must be nearly the same as the initial speed.

So, the initial momentum of the small  mass is 1 Kgm/s
And the final momentum is (nearly)  -1 Kgm/s
(There's a minus sign there because the movement is in the opposite direction.)
So the change in momentum of the small mass is (nearly) -2 Kg m/s.

And if the small mass has lost 2 Kgm/s of momentum and the overall momentum has stayed the same, then the big mass must have gained (nearly)2 Kgm/s
And since it's 100Kg that means the velocity must be (nearly) 2/100 Kgm/s


So, the final velocity of the big mass for the perfectly elastic (bouncy) collision is nearly 0.02 m/s

And the final velocity of the big mass for the perfectly inelastic (sticky) collision is 1/101 i.e. 0.0099 m/s

Now, if you remember,your assertion was this

The "dead" sandbag makes a huge difference also .  It's powerful shove completely outweighs the weak shove of the metal ball . 

And then you contradicted yourself

The "adhesive collision" converts two objects into one instantly .  The "Shooter" object's KE is then distributed proportionately throughout the "new object" .  This results in a significantly lower velocity than the "target" object in the first case .

(The first case was the elastic collision)

So you started off by saying the inelastic (sandbag) case hits harder. (They were the same mass at the same speed)
But you ended up (correctly) saying that the elastic (steel ball) case  sends the target off faster.
(and what I proved above was that the bouncing ball transmits nearly twice as much momentum to the target as a non bouncy one.)


But your "spaceship" example was based on your original idea that the sandbag delivers more impact- and that is false.
Which is why your original idea is wrong.

[Professor Mega-Mind (OP)]

HEAVIER , not faster , heavier .
Also , I don't see the momentum of the electrons , atoms , and molecules up there .  Think of heat as dispersed , uni-directional momentum .
P.

[Professor Mega-Mind (OP)]

One important issue ; math error .
Paragraph "So the important. ...." , and following paragraph .  The momentum transferred to the massive plate is ~2% of 1kg.m/s. , not -2kg.m/sec. .  I believe that you are transposing photon absorption /reflection dynamics upon bulk matter collision dynamics .  Not the same thing , at all .  Try your formulas with that 2% figure in place , then congratulate me on my breakthrough !
P.M.

[Kryptid]

Think of heat as dispersed , uni-directional momentum .

You were calling it omni-directional earlier...

Any  waste heat produced is omni-directional

[Professor Mega-Mind (OP)]

Woops , habit .  Kind of obvious I meant in all directions  ( omni ) .
D.

[Kryptid]

Woops , habit .  Kind of obvious I meant in all directions  ( omni ) .
D.

If you did mean omni-directional, then why are you bothering to complain about the lack of molecular momentum calculations? If you propose that the heat is omni-directional, then it will have no effect on the net momentum and won't change Bored Chemist's calculations.

[Bored chemist]

The momentum transferred to the massive plate is ~2% of 1kg.m/s. , not -2kg.m/sec. .

No
You keep making the same mistake.
The momentum transfer is 2 Kgm/s

I believe that you are transposing photon absorption /reflection dynamics upon bulk matter collision dynamics . 

Well spotted.
The reflection of photons (with virtually no loss of everyg- perfectly elastic) gives rise to twice the momentum transfer as absorption of photons (with total loss of energy and thus totally inelastic).

What you don't seem to understand is that this is also the outcome for macroscopic collisions of other things.

Not the same thing , at all . 

It's exactly the same thing. Why would it be different? Momentum is momentum.

Try your formulas with that 2% figure in place , then congratulate me on my breakthrough !

So, what you are saying is that your "breakthrough" depends on getting the maths wrong.

[Professor Mega-Mind (OP)]

I shouldn't need to say it buut ; photons are NOT bulk matter .  The differences are many-fold , especially in matters of reflection .  I will recap : Glass on steel 90/10 .  10% of kinetic energy is transferred  to the Massive Plate , 90% of KE is retained by glass ball after rebound  . Now, Beanbag on Steel : 55/45/0.
55% of kinetic energy is transferred  to Massive Plate , 45% becomes molecular momentum during "bonding" , .002% is retained as rebound KE . 
 The imbalance in striking force is made manifest here .  Proper manipulation can then use this to create a unidirectional thrust engine , or ensemble .
OK , momenta A is not momenta B!
......P.M.

[yor_on]

a different type of shield is necessary ... Constructed mainly of high-tech , impact-absorbing materials

I recently saw some samples of shielding that had been subjected to simulated micrometeorite impacts that you expect to encounter in Earth orbit.

The idea of having two thin layers of protection seems much more effective than a single thick layer.

The idea of the first layer is to partially melt/vaporize the micrometeorite (and slow it a bit). What hits the second layer is a partially melted spray of particles, spread out over a larger area, so it is much less likely to penetrate the next layer.

The composition of the first layer is not so important - in fact, if the first layer is particularly tough, it may provide very tough debris that can puncture the second layer. So it is probably best if the first layer is something like aluminium that is light and has a fairly low melting point, so it splatters rather than remain intact.

This photo shows a 5mm thick layer of aluminium, hit by the ball bearing. The impact created a crater through most of the depth (and probably cracked and weakened the remainder of the depth).
 [ Invalid Attachment ]
This photo shows the effect of two layers of 0.5mm each, and the splatter pattern on the second layer.
 [ Invalid Attachment ]
From an exhibition at Scienceworks, Melbourne Australia.

Actually Evan, the Israelis came up with one better if I remember correct, I think we use it for our tanks in Sweden too. You put 'directed explosives' on the outside of the material, If something hits that it will explode creating a counterforce outwards. Now, I wouldn't recommend hanging dynamite staves (or nitroglycerin) outside the Spacecrafts hull, but there should be some possibilities, well, maybe :)