Conservation of Linear Momentum vs Conservation of Energy

[kipinator101]

Homework Statement

http://i.imgur.com/aeleeB9.png
Above is a diagram of a simple two-block mechanical system. Block A is initially sliding with some initial velocity and Block B is initially at rest. Block A and block B collide, stick together, and begin moving with some final velocity. Determine the final velocity of Block B.

Homework Equations

In this system, both energy and linear momentum are conserved quanties. It is easy for me to solve the problem using the conservation of linear momentum, but I am unable to solve the problem using conservation of energy. In this, I am asking help.
I asked my professor in class why you must use linear momentum to solve the problem. He said that internal energy was involved and that you must account for that. A question I want an answer to is, how do you determine the initial and final internal energies of the two block system? And are these the only two forms of energy that must be account for (KE and U, provided the blocks are in a vacuum and sliding across a frictionless surface)?

The Attempt at a Solution

Known Solution:

d/dt(P_sys,x)=ƩF_x + Ʃd/dt(m_in)*V_in - Ʃd/dt(m_out)*V_out
→closed system, no external forces
∴ d/dt(P_sys,x)=0
Integrate:
P_final -P_initial = 0
M_block,a*V₀ = (M_block,b+M_block,a) * V_final
→Solve for V_final

 

[BOYLANATOR]

Did the problem say that these blocks were in a vacuum? Energy is always conserved in a closed system but the KINETIC energy of the blocks is only conserved in an elastic collision. In general the objects heat up a little, make a noise and deform.

 

[BOYLANATOR]

Why don't you equate your two expressions for v_f to find out what properties the blocks would have to have in order for both kinetic energy and momentum to be conserved.

 

[kipinator101]

I wrote the problem myself. It can be assumed that the block is in a vacuum and traveling on a friction-less surface. Using conservation of energy, accounting only for Kinetic Energy, yields V_final = sqrt(M_A/(M_A+M_B))*V₀ whereas conservation of linear momentum yields V_f = (M_A/(M_A+M_B)*V₀

 

[BOYLANATOR]

Adding mass at rest to a moving object will decrease its kinetic energy and increase its internal energy. If you equate your two equations you can see that it only works when Mb = 0 i.e. no mass is added.

 

[kipinator101]

What if we assume both masses are a rigid body? (Can we even make that assumption?). If the body is rigid, how can energy be stored as internal energy? Also, linear momentum allows us to determine the final velocity of the two masses at any temperature. With the final velocity, we can calculate the final kinetic energy, and therefore the change in internal energy. Doesn't this imply that the change in internal energy has temperature independence? From what I've seen of Thermodynamics so far, this isn't usually or ever the case.

 

[haruspex]

Did the problem say that these blocks were in a vacuum?

I can't see how that is relevant.

What if we assume both masses are a rigid body? (Can we even make that assumption?). If the body is rigid, how can energy be stored as internal energy?

It depends what you mean by rigid. The normal use of the term only refers to macroscopic behaviour (and is anyway an idealisation - all bodies are at least a little compressible). If you are going to assume the objects cannot hold internal energy then the collision must be 100% elastic and cannot behave as described.

Also, linear momentum allows us to determine the final velocity of the two masses at any temperature. With the final velocity, we can calculate the final kinetic energy, and therefore the change in internal energy. Doesn't this imply that the change in internal energy has temperature independence? From what I've seen of Thermodynamics so far, this isn't usually or ever the case.

It would be true in this case.

 

[kipinator101]

Okay. Let's assume that the collision is 100% elastic. The calculated values of v_final are still significantly different. For linear momentum, we get the ratio of the masses times the initial velocity (Ma/Mb)*V₀, but we get the square root of the ratio of the masses times the initial velocity sqrt(Ma/Mb)*V₀ ... There must be some energy not accounted for.

 

[haruspex]

Okay. Let's assume that the collision is 100% elastic. The calculated values of v_final are still significantly different. For linear momentum, we get the ratio of the masses times the initial velocity (Ma/Mb)*V₀, but we get the square root of the ratio of the masses times the initial velocity sqrt(Ma/Mb)*V₀ ... There must be some energy not accounted for.

Are you still assuming they move together after collision? As I wrote, that is not possible for an elastic collision. If that's not what you mean, please post your complete working.

 

[kipinator101]

No, I'm no longer assuming they stick together. I'm assuming block A collides with block B and then subsequently comes to a complete stop, with block B moving forward with some velocity V_final.

It's pretty straightforward. There are no mass flows or net forces on the system. Therefore the initial linear momentum is equal to the final linear momentum. M_a*V₀ = M_b*V_final
→solve for V_final

With energy, assuming a rigid body for both masses, I can't imagine a change in internal energy. Assuming that kinetic energy is the only form of energy that changes, the initial kinetic energy is equal to the final kinetic energy (again, no mechanical work being done on the system, no heat transfer that I'm aware of, closed system)
1/2*M_a*V₀^2 = 1/2*M_b*V_final^2
→solve for V_final

 

[haruspex]

No, I'm no longer assuming they stick together. I'm assuming block A collides with block B and then subsequently comes to a complete stop, with block B moving forward with some velocity V_final.

But you cannot assume that either. If work is conserved then the subsequent velocities are determined by the conservation laws. There are certain combinations of inputs that will result in A stopping, but that you cannot constrain that to be the case.

 

[kipinator101]

Is not the definition of an elastic collision that all of the [STRIKE]energy[/STRIKE] momentum from Mass A transferred to Mass B?

It depends what you mean by rigid. The normal use of the term only refers to macroscopic behaviour (and is anyway an idealisation - all bodies are at least a little compressible). If you are going to assume the objects cannot hold internal energy then the collision must be 100% elastic and cannot behave as described.

 

[haruspex]

Is not the definition of an elastic collision that all of the energy from Mass A transferred to Mass B?

No! It is that the total KE is conserved. Together with conservation of linear momentum, this leads to a simple relationship between the before and after relative velocities: they're reversed. v_2f - v_1f = v_1i - v_2i.
If the initial state is that one of the masses is at rest and the masses are equal then the impinging mass will come to rest.

 

[kipinator101]

Thank you for clarifying that!

You mentioned that the collision would have to be elastic if the internal energy does not change. However, if the collision is elastic, that means KE is conserved, which would imply that linear momentum conserved isn't in this model.

 

[kipinator101]

Oh! I believe I am wrongly making the assumption that an elastic collision is one where the first body has a final velocity of zero, which is incorrect.