1D Elastic Collision in CM frame

In summary, the conversation is about using transformations to get into the center of mass (CM) frame in an elastic collision in one dimension. The assumption is that the CM frame is inertial due to the absence of external forces on the system. The next step involves defining the initial and final momenta in the CM frame, where the initial momentum is equal to the final momentum, which is equal to zero. The confusion arises from the author's introduction of a new term for the initial momentum and its relationship to the momenta of the individual masses.
  • #1
stfz
35
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! :)
 
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  • #2
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:
  • #3
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.
 

FAQ: 1D Elastic Collision in CM frame

1. What is a 1D Elastic Collision in CM frame?

A 1D Elastic Collision in CM frame refers to a type of collision between two objects where both objects are in motion along a straight line and the total kinetic energy of the system is conserved. In this frame of reference, the center of mass (CM) is at rest, making calculations easier.

2. How is the center of mass calculated in a 1D Elastic Collision?

The center of mass in a 1D Elastic Collision can be calculated by taking the weighted average of the positions of the two objects. This can be represented by the formula: CM = (m1x1 + m2x2) / (m1 + m2), where m1 and m2 are the masses of the two objects and x1 and x2 are their respective positions.

3. What is the conservation of momentum in a 1D Elastic Collision?

The conservation of momentum in a 1D Elastic Collision states that the total momentum of the system before and after the collision must remain constant. This means that the sum of the individual momentums of the two objects before the collision must be equal to the sum of their momentums after the collision.

4. How is the coefficient of restitution calculated in a 1D Elastic Collision?

The coefficient of restitution, denoted by the symbol e, is a measure of the elasticity of a collision. It can be calculated using the formula: e = (v2f - v1f) / (v1i - v2i), where v1i and v2i are the initial velocities of the two objects and v1f and v2f are their final velocities.

5. What is the difference between a 1D Elastic Collision and a 1D Inelastic Collision?

In a 1D Elastic Collision, the total kinetic energy of the system is conserved, while in a 1D Inelastic Collision, some of the kinetic energy is lost. This means that in an inelastic collision, the objects stick together after colliding, while in an elastic collision, they bounce off each other. Additionally, the coefficient of restitution is equal to 1 in an elastic collision, while it is less than 1 in an inelastic collision.

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