1D Elastic Collision in CM frame

AI Thread Summary
The discussion focuses on understanding the derivation of final states in one-dimensional elastic collisions in the center of mass (CM) frame. Participants clarify that the CM frame can be assumed to be inertial due to the absence of external forces, aligning with the first law of motion. The initial and final momenta in the CM frame are defined as zero, leading to the conclusion that the momenta of the two colliding bodies are equal in magnitude but opposite in direction. There is confusion regarding the notation used for momentum, with some participants noting that the definitions provided may not align with standard assumptions. The conversation emphasizes the importance of clear definitions in understanding momentum conservation in different reference frames.
stfz
Messages
35
Reaction score
0
Hi all,
I've been self-studying a first year uni introductory mechanics course, and I'm confused with the derivations involved in calculating the final state of an elastic collision in one dimension, given the initial state.
So basically we have masses of m_1, m_2 with initial velocities v_1i, v_2i respectively that collide elastically.
The method I'm trying to understand is to use transformations to get into the CM frame.
upload_2016-1-11_18-48-45.png

Transforming initial velocities into the CM frame:
upload_2016-1-11_18-49-16.png

Given that momentum is conserved in the inertial coordinate frame (using the impulse approximation, we're assuming that CM frame is inertial... My first question - can someone explain why we can assume that CM frame is inertial? Is this because net force is reckoned equal to force of collision, which is an internal force?

Those assumptions then lead to:
upload_2016-1-11_18-51-9.png

I.e. momentum conservation in the CM frame.
Now the next step I don't get:
upload_2016-1-11_18-51-39.png

I'm assuming that ##p'_i = p'_{1i}+p'_{2i}## and similarly ##p'_f=p'_{1f}+p'_{2f}##.
And ##p'_i=p'_f=0##. But how can we state that ##p'_i = p'_{1i}=-p'_{2i}##, etc. in the above. If my assumption was wrong, what are ##p'_i, p'_f## defined as?

Could someone please help me to understand how this works? I am using this course: https://www.phy.duke.edu/~rgb/Class/intro_physics_1/intro_physics_1.a4.pdf

Thanks in advance! :)
 
Physics news on Phys.org
stfz said:
Hi all,]
Given that momentum is conserved in the inertial coordinate frame (using the impulse approximation, we're assuming that CM frame is inertial... My first question - can someone explain why we can assume that CM frame is inertial? Is this because net force is reckoned equal to force of collision, which is an internal force?
Yes. This is just an application of the first law: the assumption is that there are no external forces on the system.
Now the next step I don't get:
View attachment 94168
I'm assuming that ##p'_i = p'_{1i}+p'_{2i}## and similarly ##p'_f=p'_{1f}+p'_{2f}##.
And ##p'_i=p'_f=0##. But how can we state that ##p'_i = p'_{1i}=-p'_{2i}##, etc. in the above. If my assumption was wrong, what are ##p'_i, p'_f## defined as?

Could someone please help me to understand how this works? I am using this course: https://www.phy.duke.edu/~rgb/Class/intro_physics_1/intro_physics_1.a4.pdf
##p'_i= p'_f=0## applies in the CM frame. ##p_f = p_{1f} + p_{2f} \ne 0 ## and ##p_i= p_{1i} + p_{2i} \ne 0 ## applies in the observer frame. The prime (') should be used to designate the CM frame.

AM
 
Last edited:
stfz said:
Now the next step I don't get:
View attachment 94168
I'm assuming that ##p'_i = p'_{1i}+p'_{2i}## and similarly ##p'_f=p'_{1f}+p'_{2f}##.
And ##p'_i=p'_f=0##. But how can we state that ##p'_i = p'_{1i}=-p'_{2i}##, etc. in the above. If my assumption was wrong, what are ##p'_i, p'_f## defined as?

Could someone please help me to understand how this works? I am using this course: https://www.phy.duke.edu/~rgb/Class/intro_physics_1/intro_physics_1.a4.pdf

Thanks in advance! :)

What he's done is not what you assumed. He's defining ##p'_i## to be ##p'_{1i}##. I can't really see a good reason to introduce another term for this momentum here, but that's what he's done. Just to emphasise that ##p'_{1i} = - p'_{2i}##, perhaps. In any case, it's not the total momentum he's defined, as you assumed.
 
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...
Hello everyone, Consider the problem in which a car is told to travel at 30 km/h for L kilometers and then at 60 km/h for another L kilometers. Next, you are asked to determine the average speed. My question is: although we know that the average speed in this case is the harmonic mean of the two speeds, is it also possible to state that the average speed over this 2L-kilometer stretch can be obtained as a weighted average of the two speeds? Best regards, DaTario
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?
Back
Top