Understanding relative velocity in 1D elastic collision

[zenterix]

Homework Statement

Consider two particles of masses $m_1$ and $m_2$ interacting via some force.

When we have a 1d collision between these two particles, what is the relationship between the initial and final relative velocities.

Relevant Equations

Consider a lab frame.

It can be shown that the change in kinetic energy is the same in all inertial frames and equals

$$\Delta K=\frac{1}{2}\mu(v_{1,2,f}^2-v_{1,2,i}^2)\tag{1}$$

where $\mu=\frac{m_1m_2}{m_1+m_2}$ and $v_{1,2}=v_1-v_2$.

Consider an elastic collision in one-dimension.

We have a momentum equation

$$m_1v_{1,i}+m_2v_{2,i}=m_1v_{1,f}+m_2v_{2,f}\tag{2}$$

and a kinetic energy equation ($\Delta K=0$)

$$\frac{1}{2}\left (m_1v_{1,i}^2+m_2v_{2,i}^2\right)=\frac{1}{2}\left (m_1v_{1,f}^2+m_2v_{2,f}^2\right)\tag{3}$$

We can rewrite the kinetic energy equation (3) as

$$m_1(v_{1,i}^2-v_{1,f}^2)=m_2(v_{2,f}^2-v_{2,i}^2)\tag{4}$$

$$m_1(v_{1,i}-v_{1,f})(v_{1,i}+v_{i,f})=m_2(v_{2,f}-v_{2,i})(v_{2,f}+v_{2,i})\tag{5}$$

and the momentum equation (2) as

$$m_1(v_{1,i}-v_{1,f})=m_2(v_{2,f}-v_{2,i})\tag{6}$$

We divide (5) by (6) to obtain

$$v_{1,i}+v_{1,f}=v_{2,i}+v_{2,f}\tag{7}$$

Which we can rewrite as

$$v_{1,i}-v_{2,i}=-(v_{1,f}-v_{2,f})\tag{8}$$

$$v_{1,2,i}=-v_{1,2,f}\tag{9}$$

My question is about this equation (9).

If we had started with the change in kinetic energy equation (1) then we'd have

$$\Delta K=\frac{1}{2}\mu(v_{1,2,f}^2-v_{1,2,i}^2)\tag{10}$$

$$=\frac{1}{2}\mu(v_{1,2,f}-v_{1,2,i})(v_{1,2,f}+v_{1,2,i})\tag{11}$$

$$=0$$

Therefore, it seems we have two possibilities.

$$v_{1,2,f}=v_{1,2,i}\tag{12}$$

or

$$v_{1,2,f}=-v_{1,2,i}\tag{13}$$

How come (12) didn't appear when we derived (9)?

 

[PeroK]

Because you divided by $v_{1,2,f} -v_{1,2,i}$, which assumes that is not zero. At that point you excluded the solution that they are equal.

 

[Orodruin]

Division by zero.

 

[zenterix]

Ok, I see the division by zero. The truth is, my doubts are deeper, however.

For some reason, out of all the topics of classical mechanics that I know, collisions seems the most difficult to grasp even though the equations present are the fewest and the easiest out of all the topics.

Here is my understanding.

1) Conservation of momentum arises from the definitions of momentum and force, plus Newton's third law.

2) Kinetic energy is simply an antiderivative of the line integral of force relative to displacement. Change in kinetic energy is by definition a definite line integral of this sort.

3) When we have a system of two particles interacting via a pair of 3rd law conservative forces, then the work done by these two forces can be expressed as the work done on a fictitious mass

$$\mu=\frac{m_1m_2}{m_1+m_2}\tag{1}$$

by one of the original forces, along the trajectory of the relative position vector, $\vec{r}_{21}=\vec{r}_1-\vec{r}_2$.

That is, if we have the following setup

then it can be shown that the work done by the pair of internal conservative forces in moving the system from a state A to a state B is

$$W_c=\int_A^B\vec{F}_{21}\cdot d\vec{r_{21}}\tag{2}$$

We can also show that the force $\vec{F}_{21}$ (or alternatively, but with a change in sign in subsequent expressions, $\vec{F}_{12}$) can be expressed as

$$\vec{F}_{21}=\frac{m_1m_2}{m_1+m_2}\frac{d^2\vec{r}_{21}}{dt^2}\tag{3}$$

Thus, by subbing (3) into (2) we end up with

$$W_c=\frac{\mu}{2}(v_{21,B}^2-v_{21,A}^2)\tag{4}$$

which expresses the work done by these internal conservative forces as the same as the change in kinetic energy of a system with a single particle of mass $\mu$ and velocity $v_{21}$.

3) In a 1d elastic collision between two particles, if we simply consider conservation of momentum then we have an equation such as

$$m_1v_{1,i}+m_2v_{2,i}=m_1v_{1,f}+m_2v_{2,f}\tag{5}$$

which has infinite solutions for the two unknowns $v_{1,f}$ and $v_{2,f}$.

Why is it that most of these solutions don't make sense?

Because (5) by itself doesn't say there is a collision.

When we impose $\Delta K=0$ what we are saying is that the work done by the internal conservative forces between an initial state and a final state is zero.

$$W_c=\frac{\mu}{2}(v_{2,1,f}^2-v_{2,1,i}^2)=0\tag{6}$$

This can only happen if the magnitude of the relative velocity is the same in both states because as we've shown above, work in the system is essentially work on $\mu$ along $\vec{r}_{21}$. That is, work in the system can be represented as the change in kinetic energy of a particle with mass $\mu$

In one dimension, this means that either

1) $\mu$'s velocity didn't change or

2) It's magnitude didn't change but its sign changed.

In terms of the original particles,

1) There is no change in the relative velocity of the particles. The individual particle velocities can be anything as long as the relative velocity is the same in both states.

2) We have any of the pairs of velocities from 1) but with the signs flipped.

Notice that if we use only the kinetic energy equation but not the momentum equation then we also allow infinite solutions: each solution has the same magnitude of relative velocity but the individual velocities can take on infinite pairs of values.

It is only when we add the momentum equation that we narrow the solutions to only two.

 

[zenterix]

Now, honestly, I am not sure why I used all this reasoning about the particles interacting via a pair of conservative forces. I guess it is because this is the sequence of reasoning presented in the book I am reading. After all, if there are forces then there is acceleration. So I am confused.

I mean, is it the case that these collision scenarios consider only an infinitesimal interval of time from right before the collision to after the collision?

 

[andrewkirk]

3) In a 1d elastic collision between two particles, if we simply consider conservation of momentum then we have an equation such as

$$m_1v_{1,i}+m_2v_{2,i}=m_1v_{1,f}+m_2v_{2,f}\tag{5}$$

which has infinite solutions for the two unknowns $v_{1,f}$ and $v_{2,f}$.

Why is it that most of these solutions don't make sense?

Because (5) by itself doesn't say there is a collision.

Say rather that (5) doesn't say there is an interaction, rather than a collision.
We can make physical sense out of any solution to (5) with the following interaction.
Mass 1 uses a spring-loaded device to fire a small particle towards Mass 2, which bounces off 2 and returns to 1, reloading the spring device as it returns (assume the device has some sort of very fine, ratcheted, locking mechanism to implement this). We can achieve any increase in relative velocity by varying the speed at which 1 fires the particle. Conversely, we can achieve any decrease in relative velocity by making 2 fire a particle at 1 instead.
So all the solutions to (5) have a physical realisation.
But all but one of those solutions do not also conserve KE. Physically, this is realised by the particle returning to the spring at a lower relative speed than that at which it was fired, so it doesn't put as much PE into the spring as was there at the start. Where has the missing PE gone? It has converted into KE (assuming the particle collided elastically with the other mass), increasing the KE of the system.
The only solution that also preserves KE is the one involving an elastic collision between the two masses themselves.

 

[zenterix]

How does the center of mass frame work?

We know the velocity of the center of mass

$$v_{cm}=\frac{m_1v_1+m_2v_2}{m_1+m_2}$$

and we know the velocities of $m_1$ and $m_2$ in the CM frame because

$$v_1'=v_1-v_{cm}=\frac{m_2}{m_1+m_2}(v_1-v_2)$$

$$v_2'=v_2-v_{cm}=\frac{m_1}{m_1+m_2}(v_2-v_1)$$

However, I don't quite comprehend the following

The momentum is zero by definition of this frame.

How do we know that "the only way for both momentum and kinetic energy to be the same before and after the collision is either the objects have same velocity (a miss) or to reverse the direction of velocities"?

Where does $v_{1,f}'=-v_{1,i}'$ come from?

The change in kinetic energy in any inertial frame is

$$\Delta K_{cm}=\frac{1}{2}\mu (v_{12,f}^2-v_{12,i}^2)=0$$

Thus

$$v_{12,f}^2=v_{12,i}^2$$

$$v_{12,f}=\pm v_{12,i}$$

So, for the case of $v_{12,f}=-v_{12,i}$ we have

$$v_{1,f}-v_{2,f}=-(v_{1,i}-v_{2,i})$$

 

[zenterix]

Here is an answer to the above

In the CM frame, we have

$$m_1v_{1,i}'+m_2v_{2,i}'=m_1v_{1,f}'+m_2v_{2,f}'=0\tag{1}$$

From (1), we have

$$m_1v_{1,i}'=-m_2v_{2,i}'\tag{2}$$

$$m_1v_{1,f}'=-m_2v_{2,f}'\tag{3}$$

We impose the condition $\Delta K=0$.

However, we don't use the expression $\Delta K=\frac{1}{2}\mu(v_{12,f}^2-v_{12,i}^2)$ (for reasons I don't yet understand), but we instead use the usual

$$\frac{1}{2}\left (m_1v_{1,i}'^2+m_2v_{2,i}'^2\right)=\frac{1}{2}\left (m_1v_{1,f}'^2+m_2v_{2,f}'^2\right)\tag{4}$$

$$m_1v_{1,i}'^2+m_2v_{2,i}'^2=m_1v_{1,f}'^2+m_2v_{2,f}'^2\tag{5}$$

$$\frac{(m_1v_{1,i}')^2}{m_1}+\frac{(m_2v_{2,i}')^2}{m_2}=\frac{(m_1v_{1,f}')^2}{m_1}+\frac{(m_1v_{2,f}')^2}{m_2}\tag{6}$$

Using (2) and (3) we get

$$(m_1v_{1,i}')^2=(m_1v_{1,f}')^2\tag{7}$$

and so

$$v_{1,i}'=\pm v_{1,f}'\tag{8}$$

Now, if $v_{1,i}'=v_{1,f}$ then the momentum equation (1) gives us $v_{2,i}=v_{2,f}$.

Thus, in this case, we have that the relative velocity does not change. There is no collision.

If $v_{1,i}=-v_{1,f}$ then bu adding (2) and (3) we obtain

$$-m_2v_{2,f}'-m_2v_{2,i}'=0$$

$$-m_2(v_{2,f}'+v_{2,i})=0$$

$$v_{2,f}=-v_{2,i}$$

So in this case the initial velocities change sign.

 

[PeroK]

We impose the condition $\Delta K=0$.

However, we don't use the expression $\Delta K=\frac{1}{2}(v_{12,f}^2-v_{12,i}^2)$ (for reasons I don't yet understand),

KE has a mass term, and in general $m_1 \ne m_2$.

 

[zenterix]

KE has a mass term, and in general $m_1 \ne m_2$.

Indeed it should be $\Delta K=\frac{1}{2}\mu (v_{12,f}^2-v_{12,i}^2)$.

A typo on my part. Doesn't seem to change my question at this point which is: why can't we use this expression to obtain the solutions for $v_{1,f}$ and $v_{2,f}$ in the CM frame?

 

[PeroK]

Indeed it should be $\Delta K=\frac{1}{2}\mu (v_{12,f}^2-v_{12,i}^2)$.

A typo on my part. Doesn't seem to change my question at this point which is: why can't we use this expression to obtain the solutions for $v_{1,f}$ and $v_{2,f}$ in the CM frame?

$\mu$ is not a mass term, it is a dimensionless ratio of masses.

Sorry, that's wrong!

 

[haruspex]

is it the case that these collision scenarios consider only an infinitesimal interval of time from right before the collision to after the collision?

It's not about the time taken, it is about other forms the energy may take. During an elastic collision, the forces arise because energy is shifted from KE to EPE and back. Only when that form of internal energy is back to its prior value can you apply the conservation law to KE.

$\mu$ is not a mass term, it is a dimensionless ratio of masses.

No, it is a mass term, a product of masses divided by a sum.

 

[haruspex]

How do we know that "the only way for both momentum and kinetic energy to be the same before and after the collision is either the objects have same velocity (a miss) or to reverse the direction of velocities"?

Clearly both of those options satisfy the conservation laws.
Draw two graphs of the one velocity against the other. A straight line represents conservation of momentum, while an ellipse represents conservation of energy. They can only intersect in two places.

 

[PeroK]

Also, the algebra implies there are only two solutions.

 

[Orodruin]

Clearly both of those options satisfy the conservation laws.
Draw two graphs of the one velocity against the other. A straight line represents conservation of momentum, while an ellipse represents conservation of energy. They can only intersect in two places.

Why have I never thought about it like that?!? This is brilliant!

That the velocities are reversed in the CoM frame also becomes obvious as the ellipse is centered on and the line is through the origin!

This is why PF is great! Every once in a while you stumble across some insight that you somehow overlooked throughout your entire physics carreer!

 

[robphy]

Possibly helpful:

https://www.desmos.com/calculator/zcmmqkayop

used in my post at
https://physics.stackexchange.com/q...ions-and-calculating-velocities/602860#602860