Why is energy conserved in an elastic collision? I proved it is not?

[flyingpig]

Homework Statement

Let's say I hang a ball mass m from the ceiling and my hand holds another ball mass M in my hand and it is d meter above the hanging ball.

I throw the ball up with velocity of v_{1} and as the ball I throw up hits the ball hanging on the ceiling, the ball collides elastically. What is the velocity of the ball mass m?

So this is how I did it

Mv_{1} + m(0) = M(-v_{1}') + mv_{2}'

Mv_{1} + Mv_{1}' = mv_{2}'

\frac{M( v_{1} + v_{1}')}{m} = v_{2}'

\frac{1}{2}Mv_{1}^2 = \frac{1}{2}Mv_{1}'^2 + \frac{1}{2}mv_{2}'^2

Mv_{1}^2 - Mv_{1}'^2 = mv_{2}'^2

M(v_{1}^2 - v_{1}'^2) = mv_{2}'^2

\frac{M}{m}(v_{1}^2 - v_{1}'^2) = v_{2}'^2

\sqrt{\frac{M}{m}(v_{1}^2 - v_{1}'^2)} = v_{2}'

Clearly v_{2}' are not equal, so what did I do wrong?

 

[gneill]

Without knowing what v1' is, you cannot say that the two are not equal for the particular instance.

In fact, if you take the results of an actual test case:

m1 = 4 kg
v1 = 10 m/s

m2 = 3 kg
v2 = 0 m/s

then a "standard" analysis will find that:

v1' = 1.4286 m/s
v2' = 11.4286 m/s

Plug these into your expressions (keeping in mind that you used a "-v1'" in the derivation of the first one, so we need to reverse the sign there):

(m1/m2)*(v1 - v1') = 11.4286 m/s
sqrt((m1/m2)*(v1^2 - v1'^2)) = 11.4286 m/s

So they both give the same value for the result.

 

[Gear300]

You could have used the same momentum equation if the collision was inelastic. The elastic case is the case for which the kinetic energy is conserved as well as the momentum. It might be best in general not to associate elastic collisions with non-stick collisions by assumption; non-stick collisions can also be inelastic if the energy can dissipate elsewhere.

 

[flyingpig]

Without knowing what v1' is, you cannot say that the two are not equal for the particular instance.

In fact, if you take the results of an actual test case:

m1 = 4 kg
v1 = 10 m/s

m2 = 3 kg
v2 = 0 m/s

then a "standard" analysis will find that:

v1' = 1.4286 m/s
v2' = 11.4286 m/s

Plug these into your expressions (keeping in mind that you used a "-v1'" in the derivation of the first one, so we need to reverse the sign there):

(m1/m2)*(v1 - v1') = 11.4286 m/s
sqrt((m1/m2)*(v1^2 - v1'^2)) = 11.4286 m/s

So they both give the same value for the result.

Uhh what did you do to find v1'?

 

[flyingpig]

Please tel me how you got those weird numbers!

 

[gneill]

I solved the equations using conservation of momentum, with help from a frame-changing trick. You should be able to find how to solve 1D collision problem with a little bit of web surfing.

 

[flyingpig]

I solved the equations using conservation of momentum, with help from a frame-changing trick. You should be able to find how to solve 1D collision problem with a little bit of web surfing.

I don't know what you are talking about, you got two unknowns, you don't know v1' or v2'

 

[gneill]

Newton's laws give you more to go on than just the simple expression of conservation of momentum. Like the fact that the center of mass moves inertially, and in the center of momentum frame the collision results in both objects simply bouncing (reversing direction).

These facts allow you to solve for the two unknowns.

 

[flyingpig]

Newton's laws give you more to go on than just the simple expression of conservation of momentum. Like the fact that the center of mass moves inertially, and in the center of momentum frame the collision results in both objects simply bouncing (reversing direction).

These facts allow you to solve for the two unknowns.

Could you just show it to me?

 

[gneill]

Suppose you have two masses colliding, m1 and m2. Their initial velocities are v1 and v2. After colliding, we wish to know their velocities v1’ and v2’.

The first thing to do is change the observer’s frame of reference to that of the center of momentum frame. That is, we find a velocity U to add to all of our velocities such that

m1*(v1 + U) = -m2*(v2 + U)

Solving for U :

U = -(m1*v1 + m2*v2)/(m1 + m2)

Now our velocities for m1 and m2 become:

u1 = v1 + U
u2 = v2 + U

After the collision in the COM frame, the velocities are simply negated:

u1’ = -u1
u2’ = -u2

Transform back to the original frame of reference:

v1’ = u1’ - U
v2’ = u2’ – U

In a more succinct form:

v1’ = -(v1 + 2U)
v2’ = -(v2 + 2U)

 

[flyingpig]

Suppose you have two masses colliding, m1 and m2. Their initial velocities are v1 and v2. After colliding, we wish to know their velocities v1’ and v2’.

The first thing to do is change the observer’s frame of reference to that of the center of momentum frame. That is, we find a velocity U to add to all of our velocities such that

m1*(v1 + U) = -m2*(v2 + U)

Wait, but my collision is elastic...

 

[gneill]

Suppose you have two masses colliding, m1 and m2. Their initial velocities are v1 and v2. After colliding, we wish to know their velocities v1’ and v2’.

The first thing to do is change the observer’s frame of reference to that of the center of momentum frame. That is, we find a velocity U to add to all of our velocities such that

m1*(v1 + U) = -m2*(v2 + U)

Wait, but my collision is elastic...

Why do you think that's a problem?

 

[flyingpig]

Suppose you have two masses colliding, m1 and m2. Their initial velocities are v1 and v2. After colliding, we wish to know their velocities v1’ and v2’.

The first thing to do is change the observer’s frame of reference to that of the center of momentum frame. That is, we find a velocity U to add to all of our velocities such that

m1*(v1 + U) = -m2*(v2 + U)




Why do you think that's a problem?

Otherwise energy not conserved. I am not very good with frame of references...

Can you tell me what my algebra is then? What is the meaning?