# Separate a variable

## 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.

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

• SammyS
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.

$$m_1v_1=m_1v_1'cos(a)+\frac{m_1}{2v_2'}cos(B)$$
$$0=m_1v_1'sin(a)-\frac{m_1v_2'sin(B)}{2}$$
$$m_1v_1^2=m_1v_1'^2+\frac{m_1v_2'^2}{4}$$

• SammyS
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?

m1v1^2=m1v1'^2+(m1v2'^2)/4
Are you sure about that 4? It looks inconsistent.