Perfectly inelastic collision - conceptual - where does the lost KE go?

[geejodi]

Homework Statement

This is a conceptual question for a lab.

In the lab we used two gliders on a nearly frictionless airtrack on a level surface, with photogates set up to measure the velocities of the two gliders before and after the collisions. The second glider was initially stationary, and the first glider was given an initial velocity and made to collide with the second glider.

In one trial, the two gliders had nearly equal mass. Velcro was attached to the ends of the gliders, so when they hit each other, they stuck together afterwards. This was called a perfectly inelastic collision. In lab calculations we proved that in this scenario, momentum of the system was conserved, but the kinetic energy of the system was not conserved.

My question is, where does the lost kinetic energy go?

Homework Equations

p = mv
KE = (1/2)mv^2

The Attempt at a Solution

I have read about inelastic collisions where the kinetic energy is lost to friction or heat, such as in a car crash for example, where the kinetic energy goes partly into crushing the metal of the cars and into heat. However, in this situation there is no friction. I was thinking maybe the kinetic energy lost goes into vibrational energy in the atoms/molecules of the second glider. I have also read something online that talks about kinetic energy being turned into potential, but how can this be in this case?

Thanks!

 

[Dick]

Velcro itself wouldn't work without friction. It's not as simple as the block sliding on a table thing, but it's still friction. You are correct some also goes into vibrational energy and ultimately into heat, but most of it gets dissipated as heat in the velcro. Wonderful stuff.

 

[kwantum0]

you know, i have the same question myself. Mathematically, if you know the masses and the initial and final velocities of two objects before a collision, you will conserve momentum but not energy? To me the idea of heat being lost to friction doesn't make any sense, as we do not include mu, or any other coefficients of friction, nor does the idea of potential energy make any sense to me, as they are still the same height from the floor as before the collision and the system is closed, so no mass has been gained.

m1v + m2v = (m1 + m2)v'

but:
1/2m1v^2 + 1/2m2v^2 > 1/2(m1+m2)v'^2 [?]

i understand mathematically this has to do with the velocity being squared, but i still don't get what happens to the energy...

 

[Mapes]

Hi kwantum0, welcome to PF. The coefficient of friction \mu only appears in one particular model of friction, that of one object sliding against another. But friction can arise through other mechanisms as well. Whenever an object is deformed, useful energy is lost to heat via internal friction. This topic is covered in more advanced physics and materials science classes.

 

[kwantum0]

Thanks Mapes,
that conceptually makes a lot more sense. If energy is absorbed by the mass due to the resistance against deformation (the [+/-]acceleration of the mass to its v') during the collision and then radiated out as heat, then conservation of energy is maintained.

Mathematically then, if you knew the masses and initial velocities, you should be able to know the amount of energy absorbed in the collision, and if you knew the rate of heat radiation of the mass, you could say something about the properties of its composition.

 

[Dick]

You really don't need to know anything about the exact properties of the material. The fact that the collision is perfectly inelastic and momentum is conserved determines the amount of energy that MUST be lost. If the objects don't manage to lose exactly that amount of energy, then the collision won't be perfectly inelastic.