# Loss of mechanical energy in a plastic collision

1. Dec 4, 2008

### fluidistic

Hi all,
1. The problem statement, all variables and given/known data
I'm stupefied. In continuation to my last thread (https://www.physicsforums.com/showthread.php?t=277124), I wanted to see if any perfectly plastic collision would make the total mechanical energy of the system go down to a factor of $$\frac{1}{\sqrt 6}$$. I got a strange result and I know it cannot be possible so there's at least one error. If you could find it out... I'd be grateful.

Say you have initially 2 particles in motion. The first one has a mass $$m$$ and a speed of $$v_1$$. Second one has a mass $$M$$ and a speed of $$v_2$$.
They collide and after the collision they remain attached. They form a new particle of mass $$M+m$$ and speed $$v_3$$. (I assume speed instead of velocity to simplify but I might be wrong by doing this however).
The linear momentum is conserved. Putting away vectors, I have that $$P_i=P_f \Leftrightarrow mv_1+Mv_2=(M+m)v_3 \Leftrightarrow v_3=\frac{mv_1+Mv_2}{M+m}$$.
As the mechanical energy is not conserved I still want to find out the energy before and after the collision in order to compare them.
I have that $$E_i=\frac{mv_1^2+Mv_2^2}{2}$$ while $$E_f=\frac{(m+M)v_3^2}{2}=\frac{(mv_1+Mv_2)^2}{2(m+M)}$$.
I want to know how much times the initial energy is greater than the final. So $$E_i=\alpha E_f \Leftrightarrow \alpha= \frac{mv_1^2+Mv_2^2}{2} \cdot \frac{2(m+M)}{(mv_1+Mv_2)^2}$$ by expanding I finally get that $$\alpha=\frac{m^2v_1^2+Mmv_2^2+Mmv_1^2+M^2v_2^2}{m^2v_1^2+2mv_1Mv_2+M^2v_2^2}$$. Note that the numerator and the denominator are almost equal, they only differ by the term "$$Mm(v_2^2+v_1^2)$$" in the numerator and "$$2mv_1Mv_2$$" at the denominator. But if you set $$v_1$$ and $$v_2$$ to be $$\frac {1m}{s} }$$, there are equal and as a consequence the mechanical energy is conserved...which is obviously wrong. Where did I go wrong?
Thank you very much.

Last edited: Dec 5, 2008
2. Dec 4, 2008

### rl.bhat

V1 and V2 cannot be equal. In that case, they won't collide with each other.
And it is better to wright Ef = alpha*Ei. In that case percentage change in the kinetic energy = Mm( V1 - V2)^2/(M+m)(MV1^2 +mV2^2)
The factor mentioned in the problem leads to 59% loss of energy. For given masses of M and m, depending on V1 and V2 it is possible.

3. Dec 5, 2008

### fluidistic

Hi rl.bhat and thanks for the input.
I don't follow you, how did you reach Mm( V1 - V2)^2/(M+m)(MV1^2 +mV2^2) ?
How do you know it's worth a 59% of loss of mechanical energy? By the way this result confirms my anterior one : $$59$$%$$=\left( 1-\frac{1}{\sqrt 6} \right)100$$. If I'm not wrong this mean that for any perfectly plastic collision between 2 rigid bodies, the same amount of mechanical energy will convert into heat and as you pointed out it's around 59%. How surprising! But I don't understand HOW does it convert. There's no friction as far as I know nor any force that could produce a negative work. Or maybe there must be friction when they collide in order to get them attached...

4. Dec 5, 2008

### physics girl phd

There can be internal friction in the materials (atoms/molecules rubbing against each other). Think about the worst case (dropping a bag of rice). Then if the materials are really distorted, you've broken some internal bonds between atoms/molecules... which requires energy...

5. Dec 5, 2008

### fluidistic

I agree. However if I consider only rigid bodies then the bag of rices is excluded.
I can't print an idea in my head. In case of 2 disks (see the reference of my last thread in post 1), they collide at exactly 1 point so there cannot be some dynamic friction nor even a normal force. They simply stick together and the system loses about 60% of its kinetic energy (or 40%? I'm not sure rl.bhat is right with 59% since he didn't explain how he reached this result). Obviously the energy transform into heat, but HOW is that possible? That's why I can't get it... There is no internal force when they collide in the case of the 2 disks, nor dynamic friction. I don't see any way to convert the energy. I'm really at a loss to understand this.

6. Dec 5, 2008

### physics girl phd

If the two object collide and stick, isn't there going to be an impulse (change in momentum) and with an impulse applied over some time interval, a force? And when you push any two objects with some temperature together, isn't there going to be microscopic effects at the surfaces causing them to blend together -- in the case of two different rigid metals, an alloy? (In this case, the metal as a perhaps less floppy bag of rice isn't necessarily a bad analogy)

If you're talking about microscopic particles, like atoms of subatomic particles... if they fuse isn't there energy stored in some form of the binding forces?

7. Dec 5, 2008

### fluidistic

Thank you for your answer. Okay, I see I cannot approach the problem in a simple way. I mean by that by considering rigid bodies as "perfect rigid bodies" which in fact don't exist. And ok, I must consider internal forces in order to understand how this work.
I feel confused about what I'm learning. I don't know if it's me or if it's because of non possible situations (2 disks stick together instantly after being in contact. It cannot be instantly in reality I think. Because if it's instantly then any force acting on the system doesn't have time to change the momentum or to do a negative work so that some mechanical energy is transformed).

8. Dec 5, 2008

### rl.bhat

I got the expression by using (Ei-Ef)/Ei*100
I got 59% from the factor given in the problem. Depending upon the velocities of the objects that much loss of KE is possible.