Solve for vբ m_1v_1+m_2v_2=m_1v_f+m_2v_f

$m_{1} v_{1} + m_{2} v_{2} = m_{1} v_{f} + m_{2} v_{f}$

Step 1

Rewrite the equation as $m_{1} v_{f} + m_{2} v_{f} = m_{1} v_{1} + m_{2} v_{2}$ .

$m_{1} v_{f} + m_{2} v_{f} = m_{1} v_{1} + m_{2} v_{2}$

Step 2

Factor $v_{f}$ out of $m_{1} v_{f} + m_{2} v_{f}$ .

Factor $v_{f}$ out of $m_{1} v_{f}$ .

$v_{f} m_{1} + m_{2} v_{f} = m_{1} v_{1} + m_{2} v_{2}$

Factor $v_{f}$ out of $m_{2} v_{f}$ .

$v_{f} m_{1} + v_{f} m_{2} = m_{1} v_{1} + m_{2} v_{2}$

Factor $v_{f}$ out of $v_{f} m_{1} + v_{f} m_{2}$ .

$v_{f} (m_{1} + m_{2}) = m_{1} v_{1} + m_{2} v_{2}$

Step 3

Divide each term in $v_{f} (m_{1} + m_{2}) = m_{1} v_{1} + m_{2} v_{2}$ by $m_{1} + m_{2}$ and simplify.

Divide each term in $v_{f} (m_{1} + m_{2}) = m_{1} v_{1} + m_{2} v_{2}$ by $m_{1} + m_{2}$ .

$\frac{v_{f} (m_{1} + m_{2})}{m_{1} + m_{2}} = \frac{m_{1} v_{1}}{m_{1} + m_{2}} + \frac{m_{2} v_{2}}{m_{1} + m_{2}}$

Simplify the left side.

Cancel the common factor of $m_{1} + m_{2}$ .

Cancel the common factor.

$\frac{v_{f} (m_{1} + m_{2})}{m_{1} + m_{2}} = \frac{m_{1} v_{1}}{m_{1} + m_{2}} + \frac{m_{2} v_{2}}{m_{1} + m_{2}}$

Divide $v_{f}$ by $1$ .

$v_{f} = \frac{m_{1} v_{1}}{m_{1} + m_{2}} + \frac{m_{2} v_{2}}{m_{1} + m_{2}}$

Simplify the right side.

Combine the numerators over the common denominator.

$v_{f} = \frac{m_{1} v_{1} + m_{2} v_{2}}{m_{1} + m_{2}}$