Elastic collision - should final velocities be asymmetric?

[brotherbobby]

TL;DR

The author of a certain book shows how symmetry arguments can be used to rule out answers we get to in problems. I copy and paste the section where he says that in an elastic collision where a particle travels with a certain speed collides with a stationary particle, the expression for their final velocities should not be symmetric.
I fail to understand why not? On what basis should they $\text{not}$ be symmetric?

The portion from the text : I copy and paste the portion from the text. After discussing a situation where the result should be symmetric, he discusses one where we should not expect the answer to be symmetric. In this case, he means that were we to substitute the values of the masses, that is $m_1\rightarrow m_2$ and $m_2\rightarrow m_1$ in our expression for their final velocities, then the velocities should change, if we have correctly obtained them. If they don't and therefore the velocity expressions are symmetric, we have made a mistake.

Doubt : Why should the expression for the velocities of the particles following the collision be not symmetric? Please note that, according to the author, this is to be answered independently of calculations. There is something about this elastic collision that should make the final velocities of the masses asymmetric.

A hint would be very welcome.

 

[Dale]

Physically, imagine that the initially moving mass is very large and the initially stationary mass is very small. Then after the collision the initially moving mass will have almost the same speed but the initially stationary mass will have a large velocity (roughly twice the initial velocity of the moving mass). If you reverse the masses so that the small mass is initially moving and the large mass is initially stationary then the initially moving mass will have the reverse velocity and the initially stationary mass will remain nearly stationary. The velocities are not symmetric under interchange of the masses.

 

[A.T.]

he says that in an elastic collision where a particle travels with a certain speed collides with a stationary particle, the expression for their final velocities should not be symmetric.
I fail to understand why not? On what basis should they $\text{not}$ be symmetric?

Because he only swaps the masses. If you also swap/mirror the initial velocities its symmetric again.

 

[brotherbobby]

Because he only swaps the masses. If you also swap/mirror the initial velocities its symmetric again.

If you swapped both the masses and the initial velocities, aren't you back to the original problem?

 

[brotherbobby]

Physically, imagine that the initially moving mass is very large and the initially stationary mass is very small. Then after the collision the initially moving mass will have almost the same speed but the initially stationary mass will have a large velocity (roughly twice the initial velocity of the moving mass). If you reverse the masses so that the small mass is initially moving and the large mass is initially stationary then the initially moving mass will have the reverse velocity and the initially stationary mass will remain nearly stationary. The velocities are not symmetric under interchange of the masses.

I chose to take mass $m_1<m_2$, moving with speed $u$ where mass $m_2$ is stationary. After collision, $m_1$ moves with a velocity $v_1$ and $m_2$ with a velocity $v_2$, labelled as $\text{I}$ and $\text{II}$ in the figure to the right.

After doing the algebra, accounting for conservation both in momentum and energy, I find that final velocity $\boxed{v_1 = \frac{m_1-m_2}{m_1+m_2}u}$ and $v_2 = \boxed{\frac{2m_1}{m_1+m_2}u}$.
I present the calculations below for anyone interested, in rough using $\text{Autodesk Sketchbook}^{\circledR}$.

Let us have the expressions for the final velocities again : $\boxed{v_1 = \frac{m_1-m_2}{m_1+m_2}u}$ and $v_2 = \boxed{\frac{2m_1}{m_1+m_2}u}$.

If we swapped $m_1\rightarrow m_2,\quad m_2\rightarrow m_1$, we would have $v'_2 = \dfrac{m_2-m_1}{m_1+m_2}u$ and $v'_1 = \dfrac{2m_2}{m_1+m_2}u$.

Clearly, $v'_2\ne v_2$ and $v'_1\ne v_1$.

This is not an intuitive argument though, which the author was asking for. It does show however that the velocities are not symmetric.

 

[PeroK]

Another way to get those equations is to move to the centre of momentum frame. The only way to balance the energy-momentum equations in the centre of momentum frame is for the masses to rebound from the collision with the same speed and opposite direction. So, we have (I'll use primes for the velocities in this frame):

Before the collision: $m_1$ has velocity $u'_1$ and $m_2$ has velocity $u'_2$.

After the collision: $m_1$ has velocity $v'_1 = -u'_1$ and $m_2$ has velocity $v'_2 = -u'_2$.

Now we have to calculate the velocity of the centre of momentum relative to the original frame, where mass $m_2$ is at rest and $m_1$ moves with velocity $u$. Consider a frame moving with velocity $V$. In that frame, $m_1$ has velocity $u'_1 = u - V$ and $m_2$ has velocity $u'_2 = -V$. And the total momentum is:$$p = m_1(u - V) - m_2V = m_1u - (m_1 + m_2)V$$For $p = 0$ we need:$$V = \frac{m_1u}{m_1+m_2}$$In this frame, therefore, we have:$$v'_1 = - u'_1 = V - u, \ v'_2 = -u'_2 = V$$To convert back from this frame to the original frame, we add $V$ to the velocities in this frame. Hence:
$$v_1 = v'_1 + V = 2V -u = \frac{2m_1u}{m_1+m_2} - u = \frac{m_1-m_2}{m_1+m_2}u$$$$v_2 = 2V = \frac{2m_1}{m_1+m_2}u$$

 

[brotherbobby]

The only way to balance the energy-momentum equations in the centre of momentum frame is for the masses to rebound from the collision with the same speed and opposite direction.

I am sorry - thus far I have only heard of the "center-of-mass" frame. "Center of momentum" frame is new to me. I'd like to convince myself of your statement above so please bear with me. I can see that velocities after rebound may not be equal and opposite if energy was not conserved. However, with conservation of energy, that may be the case. I'd like to be sure of that first before I go over to the rest of your calculation.

 

[PeroK]

I am sorry - thus far I have only heard of the "center-of-mass" frame. "Center of momentum" frame is new to me. I'd like to convince myself of your statement above so please bear with me. I can see that velocities after rebound may not be equal and opposite if energy was not conserved. However, with conservation of energy, that may be the case. I'd like to be sure of that first before I go over to the rest of your calculation.

The centre of momentum frame is another name for centre of mass frame.

If they rebound with their original speeds, then momentum remains zero and energy is conserved. That clearly is a solution to both energy and momentum equations.

If one speed is lower then before, then so must the other speed be lower (to conserve momentum). This gives less kinetic energy. And, likewise, if both speeds are greater, then there is a more kinetic energy. The same speeds before and after is the only possibility in the centre of momentum frame.

 

[brotherbobby]

The centre of momentum frame is another name for centre of mass frame.

If they rebound with their original speeds, then momentum remains zero and energy is conserved. That clearly is a solution to both energy and momentum equations.

If one speed is lower then before, then so must the other speed be lower (to conserve momentum). This gives less kinetic energy. And, likewise, if both speeds are greater, then there is a more kinetic energy. The same speeds before and after is the only possibility in the centre of momentum frame.

Thank you... but surely one must be able to prove it mathematically. That is what am doing. Please bear with me.

 

[PeroK]

Thank you... but surely one must be able to prove it mathematically. That is what am doing. Please bear with me.

A proof doesn't demand formulaic calculations.

Moreover, given that the original question related to understanding "symmetry", it's not a bad idea to understand an elastic collision in the centre of momentum frame by the logic of symmetry. Rather than diving into formulas and calculations.

 

[A.T.]

Thank you... but surely one must be able to prove it mathematically.

In the common center of mass frame the two bodies always have equal but opposite momenta (follows from the definition of center of mass). And they both experience the same but opposite change in momentum during collision (Newton's 3rd).

That is the most symmetric frame of reference for collisions, and swapping the masses swaps the velocities here. But if you use a frame where one mass is initially at rest this is no longer the case.

 

[hutchphd]

In some sense the easiest way to show the asymmetry is to show that there is generally symmetry in the Center of Mass frame and, therefore , not in any other frame.

 

[brotherbobby]

In the common center of mass frame the two bodies always have equal but opposite momenta (follows from the definition of center of mass). And they both experience the same but opposite change in momentum during collision (Newton's 3rd).

That is the most symmetric frame of reference for collisions, and swapping the masses swaps the velocities here. But if you use a frame where one mass is initially at rest this is no longer the case.

Sorry for coming in late.

Are you saying that the elasticity of the collision (K.E. conserved) is not relevant when two identical bodies collide for them to have the opposite velocities (but same speeds) following the collision in their center of mass frame? According to your argument above, all we need is momentum conservation to arrive at the conclusion.

 

[PeroK]

Sorry for coming in late.

Are you saying that the elasticity of the collision (K.E. conserved) is not relevant when two identical bodies collide for them to have the opposite velocities (but same speeds) following the collision in their center of mass frame? According to your argument above, all we need is momentum conservation to arrive at the conclusion.

That must be true. What we mean by symmetry in this case is that the collision looks the same (in some sense) regardless of which mass is which. A car running into a stationary bicycle looks very different from a bicycle running into a stationary car (at the same speed). As far as final velocities compared to initial velocity is concerned. It's not symmetrical.

An example of where you would expect symmetry is in the period of the gravitational orbit of two objects about their common centre of mass. You would expect the formula for the period to be symmetric in the two masses ($m_1$ and $m_2$, say). And, indeed, a symmetry argument could lead to the conclusion that the two masses must have the same orbital period.

 

[nasu]

Sorry for coming in late.

Are you saying that the elasticity of the collision (K.E. conserved) is not relevant when two identical bodies collide for them to have the opposite velocities (but same speeds) following the collision in their center of mass frame? According to your argument above, all we need is momentum conservation to arrive at the conclusion.

Equal in magnitude momenta does not necesarilly means equal speeds. There was no mention of velocities in the post you quote. The momenta are opposite and equal in magnitude (in the COM) at all time, but not the velocities (in general).

 

[PeroK]

Equal in magnitude momenta does not necesarilly means equal speeds. There was no mention of velocities in the post you quote. The momenta are opposite and equal in magnitude (in the COM) at all time, but not the velocities (in general).

The point is that in the COM frame the final velocities are equal and opposite to the initial velocities: $v_1 = -u_1, v_2 = -u_2$. These are not just symmetric in the two masses, but independent of them.

 

[nasu]

The point is that in the COM frame the final velocities are equal and opposite to the initial velocities: $v_1 = -u_1, v_2 = -u_2$. These are not just symmetric in the two masses, but independent of them.

This is so for elastic collisions. The two momenta being equal and opposite in the COM at any time is true for any collision. This is the point I was making, in reply to the new poster's question.

 

[A.T.]

Are you saying that the elasticity of the collision (K.E. conserved) is not relevant when two identical bodies collide for them to have the opposite velocities (but same speeds) following the collision in their center of mass frame? According to your argument above, all we need is momentum conservation to arrive at the conclusion.

If the two colliding masses are equal then in the CoM-frame the velocities are equal but opposite at any given time. Elasticity merely determines if/how the equal magnitude of these velocities changes during collision.