Help on Sticky Collision Problem for Conservation of Momentum/Energy

[Afro_Akuma]

Due to having an extremely trying week recently, I missed a large portion of the conservation of momentum/energy material. So far, I've been relearning and applying with no problems, but now I've hit a roadblock: "Sticky" collisions. A problem on my assignment requires that I use a sticky collision, and I don't know the formula! Can somebody help me out please?

 

[Nylex]

Do you mean for conservation of momentum when two particles stick together after the collision? If so, they have a common velocity afterwards, so

CLM: m1u1 + m2u2 = (m1 + m2)v.

Other than that, I'm not sure what you mean, so I hope that helps.

 

[dextercioby]

Do you mean for conservation of momentum when two particles stick together after the collision? If so, they have a common velocity afterwards, so

CLM: m1u1 + m2u2 = (m1 + m2)v.

Other than that, I'm not sure what you mean, so I hope that helps.

The relation Nylex gave u must be understood and applied as an equality among vectors,right...??Momentum is a vector.
As for the energy conservation law,that involves the equality between scalars,so it should involve only the moduli of the momentum.It reads
m_{1}\frac{v^2_{1}}{2}+m_{2}\frac{v^2_{2}}{2}=(m_{1}+m_{2})\frac{v^2}{2}

Daniel.

 

[Afro_Akuma]

What I'm looking for is a formula to calculate velocity of two objects at minimum separation. I was told that this was the point at which the objects functioned as a single entity, thereby having the same velocity. Will Nydux's formula work for this?

 

[dextercioby]

What I'm looking for is a formula to calculate velocity of two objects at minimum separation. I was told that this was the point at which the objects functioned as a single entity, thereby having the same velocity. Will Nydux's formula work for this?

Yes it will.Conservation of momentum is essential when discussing kinematics of the problems.

 

[Afro_Akuma]

Using the conservation of energy formula in the next part of the problem gave me an inequality. Was I supposed to use the full formula or simply calculate the kinetic energy separately? 0.0288 W = 0.0192 W ?

 

[dextercioby]

Using the conservation of energy formula in the next part of the problem gave me an inequality. Was I supposed to use the full formula or simply calculate the kinetic energy separately? 0.0288 W = 0.0192 W ?

I'm Sorry,i've given u the wrong formula.That one i gave you applies for the case of elastic collisions.Silly me... The difference u found is nothing but the part from the original KE gets converted into heat.Sorry again.
You should appy only the conservation law for linear momentum,and from the energy law u can find the part of the initial Ke of the system which is lost as heat in the plastic collisions.I haven't worked problems involving plastic collisions in decades.Sorry,again.

 

[Afro_Akuma]

Uh...the problem states that this should be taken as a perfectly elastic, linear collision.

 

[dextercioby]

It's something fishy about it.Obviously the total momentum must conserve (u assume the 2 colliding particle are isolated so,no other forces act on them,or if they do,their result force is null),so from there u find the components of the velocity on the 3 axes.As with the energy,well,apparently it's not conserved,as that would mean that the energy conservation law would not correspond/match with the linear momentum law.So what happens with the lost energy??Definitely is converted into heat.The final resulting particle will have a certain KE (calculated with the velocity found when applying the conservation of momentum and the mass of the resulting particle (which is just the sum of the 2 colliding ones)) and the diffrerence between that and the total amount of KE of the initial colliding particles,must be the quantitiy of heat.

Guys,the ones who haven't forgotten everything about simple physics,please tell me where I'm wrong...

PS.Don't be fooled by the labels "science advisor/homework helper".They don't say we're always right.

 

[Diane_]

I don't think you're wrong, Daniel, but I think there's information we're missing.

Background: a perfectly elastic collision will be one in which kinetic energy is conserved. This is the kind of collision really only seen between uncharged fundamental particles - neutrons, say. In the rest of the world, all collisions are at least slightly inelastic. Think of it as a "second law of thermodynamics" kind of thing.

An inelastic collision, then, is one in which the total kinetic energy after the collision is different - generally less - that it was before.

When someone describes a "sticky" collision, that's almost a textbook case of an inelastic collision. If the two objects do, in fact, stick together even briefly, it will take some energy for them to come apart again, and that will in all probability come out of the kinetic energy. I am therefore not surprised that you found a different kinetic energy after the collision. That's exactly what one would expect.

The only thing that holds true in both elastic and inelastic collisions is conservation of momentum. Linear momentum is absoutely conserved - one of the few quantities in the universe for which that's true. Consequently, that's pretty much the place to start in analyzing any collision.

So - forgive my verbosity on this, but would it be possible for you to post the problem exactly as it's written, including the instructions? It might make it easier for us to figure out what's going on.