Center of Mass Frame for Elastic Collision

In summary: You've been so patient and helpful. I appreciate it!In summary, the problem involves two spheres of equal radius and masses sliding towards each other on a frictionless surface with initial velocities of 15.0 m/s and -9.00 m/s. During the collision, 25% of the total initial kinetic energy is lost. Using the given equations, the final velocities in the center of mass reference frame can be calculated to be -3.0 m/s and -12.0 m/s for sphere 1 and 2, respectively. However, there is a change in kinetic energy which adds complexity to the problem, but by setting up and solving two equations, the final velocities can be determined to be 3.94 m
  • #1
heartofaragorn
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0

Homework Statement


Two spheres of equal radius and masses 5.00 kg and 15.0 kg are sliding towards each other along the same straight line across a frictionless, horizontal surface. Vi1 = 15.0 m/s and Vi2= -9.00 m/s, respectively. During the collision, 25% of the total initial kinetic energy is lost...use the results above to find the final velocities of the two objects in the center of mass reference frame.


Homework Equations


Kf = 1/2 m1 v1f ^2 + 1/2 m2 v2f ^2
Kf = 1/2 M Vcm ^2 + 1/2 m1 u1f ^2 + 1/2 m2 u2f^2
m1 u1f + m2 u2f = 0


The Attempt at a Solution


This was a multi-step problem in which my professor gave us the above equations. I already solved for Vcm, which is -3.0 m/s. This would give sphere 1 a ui of -12.0 m/s and sphere 2 a ui of 12.0 m/s. I calculated Ki to be 1170 Joules, and calculated the value of Kf to be 75% of the value, or 877.5 Joules. However, I am unsure as to how to use the above equations in order to solve for uf and vf, since there is a change in kinetic energy. I tried several things, like plugging in random numbers, solving for one of the variables and plugging it back in, and have so far been stumped. Please help!
 
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  • #2
you have these things:

Kf >> v1_f and v2_f
pf >> v1_f and v2_f

pf = pi
Kf = 0.75Ki
so two unknowns and two equations!

Hopefully this may help you
 
Last edited:
  • #3
To what is pi and pf referring? Is that momentum? I'm afraid I'm still a bit lost.
 
  • #4
YEs

$$0.5\,\left( 5\,{vI}^{2}+15\,{vF}^{2}\right) =877.5$$
$$5\,vI+15\,vF=-60$$
$$[[vI=-\frac{3\,\sqrt{105}+6}{2},vF=\frac{\sqrt{3}\,\sqrt{5}\,\sqrt{7}-6}{2}],[vI=\frac{3\,\sqrt{105}-6}{2},vF=-\frac{\sqrt{3}\,\sqrt{5}\,\sqrt{7}+6}{2}]]$$

oops, here are two equations and their solution using maxima
0.5*(5*vI^2+15*vF^2)=877.5;
5*vI+15*vF=-60
[[vI=-(3*sqrt(105)+6)/2,vF=(sqrt(3)*sqrt(5)*sqrt(7)-6)/2],[vI=(3*sqrt(105)-6)/2,vF=-(sqrt(3)*sqrt(5)*sqrt(7)+6)/2]]

Is that right answer?
 
  • #5
rootX said:
$$0.5\,\left( 5\,{vI}^{2}+15\,{vF}^{2}\right) =877.5$$
$$5\,vI+15\,vF=-60$$
$$[[vI=-\frac{3\,\sqrt{105}+6}{2},vF=\frac{\sqrt{3}\,\sqrt{5}\,\sqrt{7}-6}{2}],[vI=\frac{3\,\sqrt{105}-6}{2},vF=-\frac{\sqrt{3}\,\sqrt{5}\,\sqrt{7}+6}{2}]]$$

For LaTex on the forum, use [ tex ]...[ /tex ] tags, without the spaces inside the square brackets, instead of $.
 
  • #6
[tex]0.5\,\left( 5\,{vI}^{2}+15\,{vF}^{2}\right) =877.5[/tex]
[tex]5\,vI+15\,vF=-60[/tex]
[tex][[vI=-\frac{3\,\sqrt{105}+6}{2},vF=\frac{\sqrt{3}\,\sqrt {5}\,\sqrt{7}-6}{2}],[vI=\frac{3\,\sqrt{105}-6}{2},vF=-\frac{\sqrt{3}\,\sqrt{5}\,\sqrt{7}+6}{2}]][/tex]

Let's see
 
  • #7
Thank you so much!

(My maxima makes it for me :D, but puts $$)
 
  • #8
Wow, thank you for all of the help!
 

1. What is the Center of Mass Frame for Elastic Collision?

The Center of Mass Frame for Elastic Collision is a reference frame used to analyze the motion of particles during an elastic collision. In this frame, the total momentum of the system is equal to zero, making it easier to analyze the collision and calculate the velocities of the particles before and after the collision.

2. How is the Center of Mass Frame for Elastic Collision calculated?

The Center of Mass Frame for Elastic Collision is calculated by finding the center of mass of the system, which is the average position of all the particles involved in the collision. This can be done by taking the weighted average of the positions of each particle, where the weights are equal to the particles' masses.

3. Why is the Center of Mass Frame for Elastic Collision useful?

The Center of Mass Frame for Elastic Collision is useful because it simplifies the analysis of the collision by reducing the number of variables involved. By eliminating the total momentum of the system, the velocities of the particles before and after the collision can be easily determined and compared.

4. Can the Center of Mass Frame for Elastic Collision be used for inelastic collisions?

No, the Center of Mass Frame for Elastic Collision is only applicable for elastic collisions, where there is no loss of kinetic energy. In inelastic collisions, some kinetic energy is converted to other forms of energy, making the analysis more complex.

5. How does the Center of Mass Frame for Elastic Collision relate to the conservation of momentum?

The Center of Mass Frame for Elastic Collision is based on the principle of conservation of momentum, which states that the total momentum of a closed system remains constant. In this frame, the total momentum is always equal to zero, which means that the momentum before and after the collision must also be equal, in accordance with the conservation of momentum.

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