Elastic collision of the rigid balls

In summary, the conversation discusses the concept of covariance matrix before and after an elastic collision. The speaker struggles to find the correct equations due to a brain blockage, but eventually proposes a solution using the notation u for velocities in the center of mass system and vCMS for the center of mass velocity. It is suggested to add a relative velocity VCMS to both v1 and v2 in order to make the total momentum of the initial system equal to zero.
  • #1
skrat
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Homework Statement


I can't believe it how my brains stopped cooperating today.
We have the first ball with ##m_1 ## and ##v_1## and of course the second one with ##m_2## and velocity ##v_2##. Covariance matrix before the collision is ##M=
\begin{bmatrix}
\sigma _1^2 & 0\\
0& \sigma _2^2
\end{bmatrix}## . Calculate the covariance matrix after the elastic collision

Homework Equations


##{\vec x }'=\Phi {\vec x }## if ##\Phi ## is the transformational matrix and if ##{(xyz)}'## indicates the values after the collision.

The Attempt at a Solution


Now my idea was to move to the center of mass system but due to my brain blockage I am not able to find or to work with any given equation. -.-

This is wrong, and I can't find out why:

Lets use notation ##u## for the velocities in the center of mass system and let ##v_{CMS}## be the center of mass velocity. Than I guess ##u_1=v_1-v_{CMS}## and ##u_2=-v_2-v_{CMS}##.

Of course in center of mass system ##m_1u_1-m_2u_2=0## which leaves me with the most stupid thing ever, saying that ##v_{CMS}=\frac{m_1v_1+m_2v_2}{m_1-m_2}##.

Could somebody please help me a bit?
 
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  • #2
If you assume that v1 and v2 are signed values, then the total momentum of the initial system is just:

P = m1*v1 + m2*v2

You can then find a velocity VCMS (the relative velocity of the center of momentum frame) to add to both v1 and v2 such that the momentum becomes zero. That is,

0 = m1*(v1 + VCMS) + m2*(v2 + VCMS)

Solve for VCMS
 
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1. What is an elastic collision?

An elastic collision is a type of collision between two objects in which both the momentum and kinetic energy are conserved. This means that the total momentum and total kinetic energy of the system before and after the collision are equal.

2. How are rigid balls defined in the context of elastic collisions?

Rigid balls are objects that do not deform or change shape during a collision. In an elastic collision, the rigid balls will bounce off each other without any change in their shape or size.

3. What are the equations used to calculate the velocities of the balls in an elastic collision?

The equations used to calculate the velocities of the balls in an elastic collision are the conservation of momentum equation: m1v1i + m2v2i = m1v1f + m2v2f and the conservation of kinetic energy equation: 1/2m1v1i2 + 1/2m2v2i2 = 1/2m1v1f2 + 1/2m2v2f2

4. How does the mass of the balls affect the outcome of an elastic collision?

The mass of the balls does not affect the outcome of an elastic collision. This is because the conservation of momentum and kinetic energy equations take into account the mass of the objects, so the velocities of the balls will adjust accordingly to conserve these quantities.

5. Are there any real-life examples of elastic collisions between rigid balls?

Yes, there are many real-life examples of elastic collisions between rigid balls. Some examples include billiard balls colliding on a pool table, balls bouncing off each other in a game of pinball, and collisions between molecules in a gas. These collisions are important in understanding the behavior of objects and particles in motion.

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