Solving for Unknowns in Conservation of Energy and Momentum Equations

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In summary, the conversation discusses the process of solving for the ratio of E'/E, starting from the equations E = v^2=E' = v'2 + AV^2 and p=v=p' = v' + AV = v. The conversation also mentions using conservation of energy and momentum to simplify the equations and solve for V^2. The final solution involves rearranging terms and plugging in V^2 to find the ratio of v' and v' in relation to E'/E.
  • #1
middlephysics
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Homework Statement



Could someone help me get from these two equations to E'/E=[(A-1)/(A+1)]2

http://postimg.org/image/m5zi9ha19/

Homework Equations



E = v^2=E' = v'2 + AV^2
p=v=p' = v' + AV = v


from conservation of energy, momentum

E'/E= (v'2 + AV^2)/(v^2)

The Attempt at a Solution



Three or four pages in my binder with no progress.
 
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  • #2
$$mv = mv' + AmV$$ can be written as $$mv - mv'= AmV$$ and this gives $$(mv - mv')^2= (Am)^2 V^2$$
You can use this in the equation for the conservation of energy and get rid of V, so just v' and v' are left as unknown, and their ratio is related to E'/E.
 
  • #3
mfb said:
$$mv = mv' + AmV$$ can be written as $$mv - mv'= AmV$$ and this gives $$(mv - mv')^2= (Am)^2 V^2$$
You can use this in the equation for the conservation of energy and get rid of V, so just v' and v' are left as unknown, and their ratio is related to E'/E.

Ok from what you wrote I've solved for $$V^2=(mv-mv')^2/(Am)^2$$.

Now I then plugged that V^2 into my $$E'/E= (v'2 + AV^2)/(v^2)$$

I am left with something like

$$(v'(A-1)+v^2-2vv')/(Av^2)$$

so far so good?
 
  • #4
mfb said:
$$mv = mv' + AmV$$ can be written as $$mv - mv'= AmV$$ and this gives $$(mv - mv')^2= (Am)^2 V^2$$
You can use this in the equation for the conservation of energy and get rid of V, so just v' and v' are left as unknown, and their ratio is related to E'/E.

I've figured it out, thank you
 

Related to Solving for Unknowns in Conservation of Energy and Momentum Equations

1. What is the meaning of "two equations, a few unknowns"?

"Two equations, a few unknowns" refers to a mathematical system that contains two equations with a limited number of unknown variables. The goal is to solve for the unknown variables by finding their values that satisfy both equations.

2. Why are two equations necessary in this type of problem?

Two equations are necessary because they provide enough information to solve for the unknown variables. With only one equation, there would be an infinite number of solutions that could satisfy the equation, making it impossible to find the specific values of the unknown variables.

3. How do you solve a system of equations with a few unknowns?

The most common method is to use substitution or elimination. Substitution involves solving one equation for one variable and substituting that into the other equation. Elimination involves manipulating the equations to eliminate one variable and solve for the other. Both methods ultimately lead to finding the values of the unknown variables.

4. Can a system of equations with a few unknowns have more than one solution?

Yes, it is possible for a system of equations with a few unknowns to have more than one solution. This occurs when the equations are dependent on each other, meaning one equation can be derived from the other. In this case, there are infinite solutions that satisfy both equations.

5. How can I check if my solution to a system of equations with a few unknowns is correct?

You can check your solution by plugging the values of the unknown variables into both equations and seeing if they satisfy both equations. If the values work for both equations, then your solution is correct. You can also graph the equations and see if the intersection point corresponds to your solution.

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