My Math Forum Momentum Equation

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 July 23rd, 2015, 02:25 PM #1 Senior Member     Joined: Dec 2014 From: The Asymptote Posts: 142 Thanks: 6 Math Focus: Certainty Momentum Equation $\displaystyle m_1u_i = m_1u_f + m_2v_f$, ($\displaystyle m_1>m_2$) I want $\displaystyle u_f$ in terms of $\displaystyle m_1, m_2, u_i$ $\displaystyle u_f = \frac{m_1u_i - m_2v_f}{m_1}$ [1] $\displaystyle v_f = \frac{m_1u_i-m_1u_f}{m_2}$ [2] [2] into [1] $\displaystyle u_f = \frac{m_1m_2u_i - m_1m_2(u_i-u_f)}{m_1m_2}$ $\displaystyle (m_1m_2)u_f+m_1m_2(u_i-u_f) = m_1m_2u_i$ $\displaystyle (m_1m_2)u_f+m_1m_2u_i-m_1m_2u_f = m_1m_2u_i$ 0=0, true but.... ....?? How to get in terms of $\displaystyle m_1, m_2, u_i$
July 23rd, 2015, 02:35 PM   #2
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Quote:
 Originally Posted by hyperbola $\displaystyle m_1u_i = m_1u_f + m_2v_f$, ($\displaystyle m_1>m_2$) I want $\displaystyle u_f$ in terms of $\displaystyle m_1, m_2, u_i$ $\displaystyle u_f = \frac{m_1u_i - m_2v_f}{m_1}$ [1] $\displaystyle v_f = \frac{m_1u_i-m_1u_f}{m_2}$ [2] [2] into [1] $\displaystyle u_f = \frac{m_1m_2u_i - m_1m_2(u_i-u_f)}{m_1m_2}$ $\displaystyle (m_1m_2)u_f+m_1m_2(u_i-u_f) = m_1m_2u_i$ $\displaystyle (m_1m_2)u_f+m_1m_2u_i-m_1m_2u_f = m_1m_2u_i$ 0=0, true but.... ....?? How to get in terms of $\displaystyle m_1, m_2, u_i$
Your first equation, along with equations 1 and 2 all say the same thing...they've just been re-arranged. That's why you are getting 0 = 0.

The problem is you need more information. Either you need to know some more variables (m1, m2, etc.) or you need another condition on the variables you have. In many cases you might have energy conservation, for example.

-Dan

 July 23rd, 2015, 03:16 PM #3 Math Team     Joined: Jul 2011 From: Texas Posts: 2,949 Thanks: 1555 Is there an actual momentum application problem you're working on, or is this just a made-up algebra drill?
 July 23rd, 2015, 03:17 PM #4 Senior Member     Joined: Dec 2014 From: The Asymptote Posts: 142 Thanks: 6 Math Focus: Certainty Yes. I am told that the collision is elastic however I am not given any additional information... Perhaps then I make $\displaystyle v_f$ the subject of $\displaystyle m_1u_i^2=m_1u_f^2+m_2v_f^2$ $\displaystyle v_f = \sqrt{\frac{m_1u_i^2-m_1u_f^2}{m_2}}$ Then substitute into [1] $\displaystyle u_f = \frac{m_1u_i}{m_1}-\frac{m_2}{m_1}\cdot\sqrt{\frac{m_1u_i^2-m_1u_f^2}{m_2}}$ $\displaystyle u_f = \frac{m_1u_i-m_2\sqrt{m_1u_i^2-m_1u_f^2}}{m_1\sqrt{m_2}}$ $\displaystyle (m_1\sqrt{m_2})u_f = m_1u_i-m_2\sqrt{m_1u_i^2-m_1u_f^2}$ $\displaystyle (m_1\sqrt{m_2})u_f + m_2\sqrt{m_1u_i^2-m_1u_f^2}=m_1u_i$ My math isn't too sound at this point.... is there, can a square root be expressed as a fraction?? I know exponentially $\displaystyle \sqrt{x}=x^{\frac{1}{2}}$... I need to clean the equation up if I am to express $\displaystyle u_f$ in terms of $\displaystyle u_i, m_1, m_2$
July 23rd, 2015, 03:53 PM   #5
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Quote:
 Yes. I am told that the collision is elastic however I am not given any additional information...
I think that would be an important additional piece of information.

$\displaystyle m_1u_0=m_1u_f+m_2v_f$

$\displaystyle m_1u_0 - m_1u_f = m_2v_f$

$\displaystyle m_1(u_0 - u_f) = m_2v_f$

conservation of energy (all terms multiplied by 2 to get rid of the 1/2's) ...

$\displaystyle m_1u_0^2=m_1u_f^2+m_2v_f^2$

$\displaystyle m_1u_0^2-m_1u_f^2 = m_2v_f^2$

$\displaystyle m_1(u_0^2-u_f^2) = m_2v_f^2$

$\displaystyle m_1(u_0-u_f)(u_0+u_f) = m_2v_f^2$

take the ratio of the two equations ...

$\displaystyle \frac{m_1(u_0-u_f)(u_0+u_f) = m_2v_f^2}{m_1(u_0 - u_f) = m_2v_f}$

$\displaystyle u_0+u_f = v_f$

substitute $u_0+u_f$ in for $v_f$ in the first momentum equation ...

$\displaystyle m_1u_0=m_1u_f+m_2(u_0+u_f)$

$\displaystyle m_1u_0=m_1u_f+m_2u_0+m_2u_f$

$\displaystyle m_1u_0 - m_2u_0 = m_1u_f+m_2u_f$

$\displaystyle u_0(m_1-m_2) = u_f(m_1+m_2)$

$\displaystyle \frac{u_0(m_1-m_2)}{m_1+m_2} = u_f$

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