Can Two Particles Have Multiple Collisions in a Single Interaction?

[sdfsfasdfasf]

Homework Statement

If two particles collide (with coefficient of restitution between 0 and 1), is it possible for the first particle to have a speed less than the second particle (and be not have its direction of motion changed)?

Relevant Equations

Conservation of linear momentum
separation / approach = e

I am stuck with this problem.
Intuition tells me the answer is no, but I am struggling to prove it.
If we consider two particles travelling in the same direction, the 2nd particle will gain velocity (impulse is in same direction to velocity), and the first particle will lose velocity (if it finishes with v>0 then its direction of motion is unchanged, yet its going slower than it was), if v<0 then it clearly goes in the other direction. Is it possible for the 1st particle to slow down so little that it ends up going faster than the 2nd one?

In either case, if 1st particle left with less velocity than 2nd (but didn't change direction of motion ) then there would be an immediate second collision (this is what is making me apprehensive).


Help appreciated, thanks for reading.

 

[BvU]

Homework Statement: If two particles collide (with coefficient of restitution between 0 and 1), is it possible for the first particle to have a speed less than the second particle (and be not have its direction of motion changed)?
Relevant Equations: Conservation of linear momentum
separation / approach = e

I am stuck with this problem.
Intuition tells me the answer is no, but I am struggling to prove it.
If we consider two particles travelling in the same direction, the 2nd particle will gain velocity (impulse is in same direction to velocity), and the first particle will lose velocity (if it finishes with v>0 then its direction of motion is unchanged, yet its going slower than it was), if v<0 then it clearly goes in the other direction. Is it possible for the 1st particle to slow down so little that it ends up going faster than the 2nd one?

In either case, if 1st particle left with less velocity than 2nd (but didn't change direction of motion ) then there would be an immediate second collision (this is what is making me apprehensive).


Help appreciated, thanks for reading.

Could you also read what you posted ?
It seems to me that you mix up 1 and 2 now and then. Make an annotated sketch and post it.


And please render the complete problem statement verbatim

$\$

 

[sdfsfasdfasf]

Could you also read what you posted ?
It seems to me that you mix up 1 and 2 now and then. Make an annotated sketch and post it.


And please render the complete problem statement verbatim

$\$

I wrote 0<v1<v2 as I know that in general v1<v2 is possible for all the trivial cases where v1<0. Hope this clears stuff up. Clearly u1>u2 etc.

 

[PeroK]

That scenario looks like the default option to me! Have you ever watched a game of snooker?

 

[sdfsfasdfasf]

I am so stupid. Let me correct the image.

 

[Mister T]

A bowling ball makes a head-on collision with a marble.

 

[sdfsfasdfasf]

I see. The issue I have now is that, when I am calculating the coefficient of restitution, I would usually write v2 - v1 for the separation velocity this works if v1 is positive or negative, but only providing that v2 is greater than v1, if its the other way around then clearly I am missing a negative sign.

 

[PeroK]

I see. The issue I have now is that, when I am calculating the coefficient of restitution, I would usually write v2 - v1 for the separation velocity this works if v1 is positive or negative, but only providing that v2 is greater than v1, if its the other way around then clearly I am missing a negative sign.

I don't know what you are trying to do. Note that velocity is dependent on the reference frame. We can analyse any collision in the centre of momentum frame, where the objects rebound with equal and opposite momenta. Then, we simply switch to a frame where both objects are moving in the same direction after the collision. That's your required end-state solution. The initial state that produced it can be calculated by transforming the initial velocities to that reference frame.

 

[sdfsfasdfasf]

I will try explain it better. If after the collision v1 = 1 and v2 = 4 then clearly the separation speed is 3, and in the case of v1 = -1 and v2=4, the seperation speed is 4 - (-1) = 5. In both cases its v2 -v1, as we know that v2 > v1 so this works. If the opposite is true, i,e that v1 >v2, say v1 = 3 and v2 = 1 then the separation speed is 3-1=2 not 1-3= -2. When I am given a problem with unknown velocities, and I write down e = v2 -v1/ approach speed, then clearly I will get the wrong answer if v1>v2 (I assumed v2 - v1 >0, iff v2 > v1).

 

[sdfsfasdfasf]

Here A and B collide with velocities 2u and u, resulting in velocities v and w, then B and C go on to collide. What I mean by "wrong sign" is that if we assumed (wrongly) that x > y then we would write e = (x-y) / w and that 4mw = 4mx + 2my, resulting in the wrong solution for x and y (in terms of w, e etc). The correct system of equations is y -x / w = e and 4mw = 4mx + 2my. How are we meant to know that y>x (its intuitive but there must be a genuine proof somwhere).

 

[PeroK]

How are we meant to know that y>x (its intuitive but there must be a genuine proof somwhere).

$y > x$ (strange symbols to use for velocities!) is a constraint of the system. The constraint is that the first object collides with the second one and cannot pass through it. $x > y$ is mathematically possible. There is the case where the first object misses the second one and there is no collision. That is a (mathematical) solution in all collision problems.

 

[sdfsfasdfasf]

Thank you so much PeroK. I had a feeling it was a "physics" issue rather than a maths one.

 

[sdfsfasdfasf]

$y > x$ (strange symbols to use for velocities!) is a constraint of the system. The constraint is that the first object collides with the second one and cannot pass through it. $x > y$ is mathematically possible. There is the case where the first object misses the second one and there is no collision. That is a (mathematical) solution in all collision problems.

Is there a way to prove this fact or is it just a "use your eyes" type thing.

 

[Steve4Physics]

... The constraint is that the first object collides with the second one and cannot pass through it. $x > y$ is mathematically possible.

Is this always a necessary constraint? E.g a bullet passing through an object?

 

[PeroK]

Is there a way to prove this fact or is it just a "use your eyes" type thing.

It's a physical constraint. Like the assumption that the collision is one-dimensional. There are lots of other solutions where the objects move in two or three dimensions after the collision. You've chosen to set $v_y = v_z = 0$ after the collision. That's called a physical constraint.

 

[sdfsfasdfasf]

I see, what would happen if x < y ?

 

[PeroK]

I see, what would happen if x < y ?

I guess you mean $y > x$. You get mathematical solutions that are not physically acceptable in this case. Or, solutions to an alternative problem with different constraints.

For example, in a collision between two galaxies, one galaxy can pass through another.

Or, a bullet/projectile passing through a block.

 

[sdfsfasdfasf]

I guess you mean $y > x$. You get mathematical solutions that are not physically acceptable in this case. Or, solutions to an alternative problem with different constraints.

For example, in a collision between two galaxies, one galaxy can pass through another.

Or, a bullet/projectile passing through a block.

Interesting! Similar to "negative time" in suvat!