Separate Variable Homework: Solutions & Equations

AI Thread Summary
The discussion revolves around a physics homework problem involving momentum and energy equations, specifically focusing on variables related to collisions. The original poster presents several equations but lacks clarity on the specific question being asked. Respondents emphasize the need for context and suggest using LaTeX for better readability of the equations. There is also a concern about having four unknowns while only providing three equations, indicating a potential inconsistency in the problem setup. Overall, the thread highlights the importance of clear problem statements and proper formatting in mathematical discussions.
zade70
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Homework Statement


m1v1=m1v1'cosa+m1/2v2'cosB
0=m1v1'sina-(m1v2'sinB/2)
m1v1^2=m1v1'^2+(m1v2'^2)/4

Homework Equations


The solution in my book is v2'=2v1sqrt(3)

The Attempt at a Solution


I thought to separate v1' at the firts and put it at the second, but I don't know how to change sin and cos then.
 
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Hi Zade:

This problem statement does not state a problem. What is the question whose answer is sought?
Also, the equations you show would be easier to read with a few spaces added to separate parts.

Regards,
Buzz
 
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Three things: First, got to have some context. We can't help you if we don't know what the question is. Second, learn latex it's not hard at all, your equations are impossible to interpret with certainty. Third, are these your equations? If not, you may quote this message and see how it was typed. Make changes as necessary and repost with context.$$m_1v_1=m_1v_1'cos(a)+\frac{m_1}{2v_2'}cos(B)$$
$$0=m_1v_1'sin(a)-\frac{m_1v_2'sin(B)}{2}$$
$$m_1v_1^2=m_1v_1'^2+\frac{m_1v_2'^2}{4}$$
 
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MostlyHarmless said:
Three things: First, got to have some context. We can't help you if we don't know what the question is. Second, learn latex it's not hard at all, your equations are impossible to interpret with certainty. Third, are these your equations? If not, you may quote this message and see how it was typed. Make changes as necessary and repost with context.
$$m_1v_1=m_1v_1'cos(a)+\frac{m_1}{2v_2'}cos(B)$$ $$0=m_1v_1'sin(a)-\frac{m_1v_2'sin(B)}{2}$$ $$m_1v_1^2=m_1v_1'^2+\frac{m_1v_2'^2}{4}$$
In the first equation, I think OP means (actually it's what he writes using standard Order of Operations)
##\displaystyle m_1v_1=m_1v_1'\cos(a)+\frac{m_1}{2}v_2'\cos(B) ##​
... but, yes, this looks much better with LaTeX.

Using subscript / superscript , etc. from the "blue line" it's easy to make math somewhat readable.
m1v1 = m1v1'⋅\cos(α) + (m1/2)⋅v2'⋅cos(β)​

(3dr eqn.):
m1v12 = m1v1'2 + (m1v2'2)/4​
 
You seem to have four unknowns—##v_1', v_2', \alpha, \text{and }\beta##—but only three equations. Have you given us all of the information?
 
zade70 said:
m1v1^2=m1v1'^2+(m1v2'^2)/4
Are you sure about that 4? It looks inconsistent.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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