Regarding momentum transfer, couple questions.

AI Thread Summary
The discussion centers on understanding momentum transfer between two colliding point masses, focusing on how their velocities equalize post-collision. The original poster seeks to determine the time-dependent forces acting on the masses as they collide and stick together, questioning the nature of momentum transfer and the forces involved. Responses highlight the need for a defined mechanism to describe how relative velocity decreases to zero, suggesting that without this, the problem remains open-ended. The conversation emphasizes that idealized models, such as point masses, may not accurately reflect real-world interactions, and a continuous force function is necessary for a realistic velocity-time relationship. Overall, the exploration reveals complexities in collision dynamics and the limitations of simplified models in physics.
Chuckstabler
Messages
31
Reaction score
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 :)
 
Physics news on Phys.org
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
 
  • Like
Likes Chuckstabler
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.
 
Last edited:
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.
 
  • Like
Likes Chuckstabler
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.)
 
The rope is tied into the person (the load of 200 pounds) and the rope goes up from the person to a fixed pulley and back down to his hands. He hauls the rope to suspend himself in the air. What is the mechanical advantage of the system? The person will indeed only have to lift half of his body weight (roughly 100 pounds) because he now lessened the load by that same amount. This APPEARS to be a 2:1 because he can hold himself with half the force, but my question is: is that mechanical...
Some physics textbook writer told me that Newton's first law applies only on bodies that feel no interactions at all. He said that if a body is on rest or moves in constant velocity, there is no external force acting on it. But I have heard another form of the law that says the net force acting on a body must be zero. This means there is interactions involved after all. So which one is correct?
Thread 'Beam on an inclined plane'
Hello! I have a question regarding a beam on an inclined plane. I was considering a beam resting on two supports attached to an inclined plane. I was almost sure that the lower support must be more loaded. My imagination about this problem is shown in the picture below. Here is how I wrote the condition of equilibrium forces: $$ \begin{cases} F_{g\parallel}=F_{t1}+F_{t2}, \\ F_{g\perp}=F_{r1}+F_{r2} \end{cases}. $$ On the other hand...
Back
Top