Regarding momentum transfer, couple questions.

In summary: Thanks so much for your help!In summary, Chuck is trying to solve a problem in which two masses collide and their velocities are not automatically equalized. He has to find a way to determine the velocity of the masses at a specific point in time after they collide.
  • #1
Chuckstabler
31
1
Hello,

I've been learning calculus and physics for the last year. Let's say i have 2 point masses mass one (m1) and mass two (m2). Their respective velocity vectors are constrained to one spatial dimension, with the directional component of their velocity vectors being determined by their respective signs (+,-). So i will refer to their respective velocity vectors from here on as simply being a signed magnitude (+2 m/s, -4m/s). Let the velocity vector of point mass one be V1, the velocity vector of point mass 2 be V2. Their respective momentum vectors are therefore Vn*Mn. Now let's assume that the two point masses will collide, and that when they do collide they will stick together.

So here's my question; is there a way that i can determine the rate of momentum transfer with respect to time P' (or the equal and opposite forces) they are experiencing at each moment after they collide until their velocities equalize? This really boils down to whether or not there is a way to determine the equal and opposite forces these two point masses will experience based on their masses and velocity vectors, assuming that from the point of contact onwards they magically stick together. I initially thought I could set up a pair of coupled linear differential equations and find the eigenvalues of the resultant matrix and solve it that way. It soon became apparent that there was no clear way to determine what the contact force should be based on their velocities, but the only way that the system would show the desired behavior (conservation of momentum, asymtotic velocity equilization, collissions with large mass djfferences taking less time) was if the force experienced by m1, p1' = u(v2-v1), where u is some constant, and p2' = -u(v2-v1), and the only constant u I could find that showed the desired behavior (large forces and quick velocity equilization when you have one massive and one very less massive object) was the reduced mass, m1*m2/(m1+m2), so v1'=u/m1(v2-v1) and v2'=u/m2(v1-v2). The end solution appeared to show the correct behavior, with the trajectories through the phase space moving along scalar multiples determined by the initial conditions of one of the eigenvectors towards scalar constants of the second eigenvector (which is a vector pointing at 45 degree or 225 degrees, meaning that the velocities eventually equalize).

Is any of this right?

Thanks for your time :)
 
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  • #2
Hi Chuck:

I am not sure I understand what problem you are trying to solve. The concept of momentum "transfer" seems a bit vague. Are you thinking of two input momenta being transferred to one output momemtum. You seem to understand what the output momentum would be, and the resulting velocity of the combined mass. So I am guessing you are interested in the specific details of motion of the two masses from their time of contact to the time when where there is no further "internal" motion.

You described this as the time when both masses (combined and stuck together) have the same velocity, Another way to say this is the time when their relative velocity becomes zero.

You seem to understand that the problem is not completely specified. You need to specify the process of the mechanism which is reducing their relative velocity to zero. This is really quite open ended. Here is one example.

One of the masses has a spring attached (as part of its mass) with an associated force per unit length parameter. The length of the spring is assumed to be long enough. When the relative velocity reaches zero, assume the spring locks into that fixed position. With this mechanism, the kinetic energy of the two masses' due to their relative velocity is linearly absorbed by the spring at a rate depending on the spring's parameter. This gives you a well defined problem that can be solved.

I hope this is helpful.

Regards,
Buzz
 
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  • #3
Hey, thanks for your reply. I'm pretty bad with trying to convey what's going on in my head so bear with me.

So what i really want to find out is if there is a way to determine the velocity of these two point masses as a function of time starting the moment they collide. It seems to me that they cannot automatically equalize their velocities at the very moment they collide, right? Since they cannot do so, there must be a velocity dependent transfer of momentum starting the moment they collide, or a velocity dependent force acting on both masses in equal and opposite ways right?

I guess another way of thinking of it is this : if you had a stationary IMMOVABLE point mass, and then had a point mass with a relative velocity V, setting time 0 at the moment of impact, what would it's velocity V be if you removed the immovable point mass after time T? Maybe I'm approaching this wrong, but it seems to me to be a reasonable question.

So to reframe the question : At time 0 an impact occurs between a fixed point mass (m1) and a point mass (m2) with a relative velocity of V, with the sign determining its direction (its sign is arbitrary in this case so ill set it to be positive). After time T, the fixed point mass m1 disappears. What would the new velocity of point mass m2 be at time T when m1 disappears? Despite the fact that m2 was stationary during the time interval 0 to T, it's velocity shouldn't be zero. To illustrate, assume the final velocity after T is infact 0, implying that it will become 0 regardless of the time interval. Let the time interval (T - 0 = T, so the interval is just T) approach 0, therefore let T approach 0. So after a non zero infitesimally small unit of time, m2's velocity becomes 0, meaning that some force would have had to have caused a change in velocity of -V, and delta V = force * time / mass2, so -V*m2/T = Force. Taking the limit as T (therefore as the time interval) approaches 0 results a divergent/infinite required velocity. So in this example, an infinite force would be required if the relative velocity becomes 0 at the moment of impact. The alternative is that there is a time dependent transfer of momentum and velocity.
 
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  • #4
If they are really point masses, the duration of the collision will be zero and the velocity will change instantaneously. The velocity as function of time will be a step function and the first derivative of the velocity will be zero every except at a single time when it is undefined. The force on the particles will be zero except at the moment of the collision, when it will be infinite.

This is obviously not a realistic situation, but all that tells us is that ideal point particles (and also perfectly rigid objects with infinite modulus of elasticity) are an idealization not encountered in real life.

If you want to get the velocity as a function of time across the collision you have to know something about the interaction between the two particles. Write the force between them as a function of their separation distance, and as long as that function is continuous you will be be able to find the continuous expression for velocity that you're looking for.

It's actually somewhat surprising how seldom this question comes up. The reason is that in most problems we can get by with a view of the system shortly before the collision and shortly after. We usually care more about where the bouncing ball goes than the precise details of its deformation while it is squashing itself against the floor prior to rebounding.
 
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  • #5
Thanks Nugatory and Buzz. Good to know I'm not just an idiot and was missing something obvious. I thought there was maybe some general relative velocity dependant forcing function that was universal, but evidently that isn't the case. Thanks for being patient, I'm learning all this stuff on my own, never taken a calculus course in my life (I got up to math 11 in high school and only started to find it interesting a year ago, so i started learning calculus.)
 

FAQ: Regarding momentum transfer, couple questions.

1. What is momentum transfer?

Momentum transfer is the movement of momentum from one object to another through contact or a force. It is a fundamental concept in physics that describes the amount of motion an object has.

2. How is momentum transfer calculated?

Momentum transfer is calculated by multiplying the mass of an object by its velocity. This can be represented by the equation p = mv, where p is momentum, m is mass, and v is velocity.

3. What is the difference between momentum and momentum transfer?

Momentum is a property of an object that describes its motion, while momentum transfer is the transfer of momentum from one object to another. In other words, momentum transfer is the change in momentum of an object due to an interaction with another object.

4. What are some examples of momentum transfer?

Some examples of momentum transfer include a game of pool, where the cue ball transfers its momentum to the other balls, and a rocket launching into space, where the burning fuel transfers momentum to the rocket to propel it forward.

5. How does momentum transfer affect collisions?

Momentum transfer is a key factor in collisions. In an elastic collision, where there is no loss of kinetic energy, momentum is transferred from one object to another without any change in magnitude. In an inelastic collision, where some kinetic energy is lost, momentum is still conserved, but the magnitude may change due to the transfer of energy.

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