What Happens When a Soviet and US Satellite Collide Elastically?

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Homework Help Overview

The discussion revolves around an elastic collision between a Soviet satellite and a US satellite, focusing on the conservation of momentum and energy principles. The specific parameters include the masses and velocities of both satellites prior to the collision.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of conservation laws to analyze the collision, with one participant attempting to derive final velocities using equations for momentum and kinetic energy. Questions arise regarding the treatment of velocities in different directions and the correctness of the calculations.

Discussion Status

Some participants have provided guidance on the equations to use for momentum and energy conservation. There is an ongoing exploration of the implications of the elastic collision, with different interpretations of the results being discussed.

Contextual Notes

Participants are working under the assumption that the collision is perfectly elastic, and there is a noted discrepancy between calculated results and expected answers from an instructor.

hobo
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I have no clue on where to start on this. I have spent time playing with that cart thing and the momentum balls and I have ideas floating around, but nothing concrete.

There is a 248.2kg Soviet satellite traveling due east at a velocity of 1025km/hr and a 149.9kg US satellite traveling due west at a velocity of 1181.8km/hr. The colission is perfectly elastic. What happens afterwards?
 
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Use conservation of momentum (no external forces) and conservation of energy (fully elastic collision)
 
From what you told me, I used

m1v1 + m2v2 = m1v1' + m2v2'
(mv1^2)/2 + (mv2^2)/2 = (mv1'^2)/2 + (mv2'^2)/2

I solved both for V1' then I plugged in the numbers.After that, it came out to

5364792208 - 35930354.61v2' +59675.19v2'^2

I then plugged it into the quadratic equation, getting 328.08333m/s which is v2, and 274.0153711m/s, which is not the answer my teacher gave me
 
Did you take into account that the velocities are in different directions?
 
Now, let's solve this problem in full:
I will assume that you gave me the right information, in particular that the collision was fully ELASTIC.

The equations for two objects, 1 and 2, to be solved are:
[tex]m_{1}v_{1,a}+m_{2}v_{2,a}=m_{1}v_{1,b}+m_{2}v_{2,b}[/tex]

This is conservation of momentum, as you had; b and a refer to initial and final velocities (before and after collision).

Energy conservation:
[tex]\frac{1}{2}m_{1}v_{1,a}^{2}+\frac{1}{2}m_{2}v_{2,a}^{2}=\frac{1}{2}m_{1}v_{1,b}^{2}+\frac{1}{2}m_{2}v_{2,b}^{2}[/tex]

This is also, I believe, what you meant.
Note that the energy equation can be simplified in the following manner:
[tex]m_{1}(v_{1,b}-v_{1,a})(v_{1,b}+v_{1,a})+m_{2}(v_{2,b}-v_{2,a})(v_{2,b}+v_{2,a})=0[/tex]

Rewrite the momentum equation as:
[tex]m_{2}(v_{2,b}-v_{2,a})=-m_{1}(v_{1,b}-v_{1,a})[/tex]

Hence, the energy equation may be written as:
[tex]m_{1}(v_{1,b}-v_{1,a})(v_{1,b}+v_{1,a}-(v_{2,b}+v_{2,a}))=0[/tex]

The root [tex]v_{1,b}=v_{1,a}[/tex] corresponds to no collision, so we have the following system to solve:
[tex]v_{1,b}+v_{1,a}=v_{2,b}+v_{2,a}[/tex]
[tex]m_{1}v_{1,a}+m_{2}v_{2,a}=m_{1}v_{1,b}+m_{2}v_{2,b}[/tex]

Solving for [tex]v_{1,a},v_{2,a}[/tex] yields:
[tex]v_{1,a}=\frac{(m_{1}-m_{2})v_{1,b}+2m_{2}v_{2,b}}{m_{1}+m_{2}}[/tex]
[tex]v_{2,a}=\frac{(m_{2}-m_{1})v_{2,b}+2m_{1}v_{1,b}}{m_{1}+m_{2}}[/tex]

Now, we have simplified the symbolic expressions maximally; it is time to enter in the values!
We set:
[tex]m_{1}=149.9,v_{1,b}=1181.8[/tex]
[tex]m_{2}=248.2,v_{2,b}=-1025[/tex]

The answers should be quite different from the ones you gave
 

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