Center of Mass Frame for Elastic Collision

Click For Summary
The discussion revolves around solving a physics problem involving an elastic collision between two spheres of different masses and velocities. The initial kinetic energy is calculated, and it is noted that 25% of this energy is lost during the collision. The center of mass velocity (Vcm) is found to be -3.0 m/s, leading to initial velocities for both spheres in the center of mass frame. The user struggles with applying the relevant equations to find the final velocities after the collision, seeking clarification on the momentum conservation and kinetic energy equations. Ultimately, the discussion provides equations and solutions for the final velocities, confirming the calculations made.
heartofaragorn
Messages
33
Reaction score
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!
 
Physics news on Phys.org
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:
To what is pi and pf referring? Is that momentum? I'm afraid I'm still a bit lost.
 
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?
 
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 $.
 
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}]]

Let's see
 
Thank you so much!

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

Similar threads

  • · Replies 8 ·
Replies
8
Views
5K
  • · Replies 17 ·
Replies
17
Views
5K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 9 ·
Replies
9
Views
1K
  • · Replies 22 ·
Replies
22
Views
4K
  • · Replies 10 ·
Replies
10
Views
4K
Replies
3
Views
989
  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K