Separate Variable Homework: Solutions & Equations

In summary, the equations show that if v1 is at the origin, then v2 is 2v1sqrt(3). If v1 is not at the origin, then v2 is v1sqrt(3).
  • #1
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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  • #2
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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  • #3
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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  • #4
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​
 
  • #5
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?
 
  • #6
zade70 said:
m1v1^2=m1v1'^2+(m1v2'^2)/4
Are you sure about that 4? It looks inconsistent.
 

What is separate variable homework?

Separate variable homework is a type of math problem that involves solving a single equation with two variables by isolating each variable on one side of the equation. It is commonly used in algebra and calculus courses.

Why is separate variable homework important?

Separate variable homework is important because it helps students develop critical thinking and problem-solving skills. It also lays the foundation for more complex mathematical concepts and equations.

What are the steps for solving a separate variable homework problem?

The steps for solving a separate variable homework problem are as follows:1. Identify the two variables in the equation.2. Use algebraic manipulation to isolate one variable on one side of the equation.3. Repeat step 2 for the second variable.4. Check your solution by substituting the values back into the original equation.5. Simplify your solution if necessary.

What are some common mistakes when solving separate variable homework?

Some common mistakes when solving separate variable homework include:- Not identifying the two variables correctly.- Incorrectly isolating one or both variables.- Making errors while simplifying the solution.- Forgetting to check the solution by substituting the values back into the original equation.

How can I improve my skills in solving separate variable homework?

To improve your skills in solving separate variable homework, you can:- Practice regularly with a variety of problems.- Review the steps for solving these types of problems.- Seek help from a teacher or tutor if you encounter difficulties.- Check your work and learn from your mistakes.

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