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] $\displaystyle v_f = \frac{m_1u_im_1u_f}{m_2}$ [2] [2] into [1] $\displaystyle u_f = \frac{m_1m_2u_i  m_1m_2(u_iu_f)}{m_1m_2}$ $\displaystyle (m_1m_2)u_f+m_1m_2(u_iu_f) = m_1m_2u_i$ $\displaystyle (m_1m_2)u_f+m_1m_2u_im_1m_2u_f = m_1m_2u_i$ 0=0, true but.... ....?? How to get in terms of $\displaystyle m_1, m_2, u_i$ 
July 23rd, 2015, 02:35 PM  #2  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,084 Thanks: 848 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
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: 2,843 Thanks: 1484 
Is there an actual momentum application problem you're working on, or is this just a madeup 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 $\displaystyle m_1u_i^2=m_1u_f^2+m_2v_f^2$ $\displaystyle v_f = \sqrt{\frac{m_1u_i^2m_1u_f^2}{m_2}}$ Then substitute into [1] $\displaystyle u_f = \frac{m_1u_i}{m_1}\frac{m_2}{m_1}\cdot\sqrt{\frac{m_1u_i^2m_1u_f^2}{m_2}}$ $\displaystyle u_f = \frac{m_1u_im_2\sqrt{m_1u_i^2m_1u_f^2}}{m_1\sqrt{m_2}}$ $\displaystyle (m_1\sqrt{m_2})u_f = m_1u_im_2\sqrt{m_1u_i^2m_1u_f^2}$ $\displaystyle (m_1\sqrt{m_2})u_f + m_2\sqrt{m_1u_i^2m_1u_f^2}=m_1u_i$ 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  
Math Team Joined: Jul 2011 From: Texas Posts: 2,843 Thanks: 1484  Quote:
start with conservation of momentum ... $\displaystyle m_1u_0=m_1u_f+m_2v_f$ $\displaystyle m_1u_0  m_1u_f = m_2v_f$ $\displaystyle m_1(u_0  u_f) = m_2v_f$ conservation of energy (all terms multiplied by 2 to get rid of the 1/2's) ... $\displaystyle m_1u_0^2=m_1u_f^2+m_2v_f^2$ $\displaystyle m_1u_0^2m_1u_f^2 = m_2v_f^2$ $\displaystyle m_1(u_0^2u_f^2) = m_2v_f^2$ $\displaystyle m_1(u_0u_f)(u_0+u_f) = m_2v_f^2$ take the ratio of the two equations ... $\displaystyle \frac{m_1(u_0u_f)(u_0+u_f) = m_2v_f^2}{m_1(u_0  u_f) = m_2v_f}$ $\displaystyle u_0+u_f = v_f$ substitute $u_0+u_f$ in for $v_f$ in the first momentum equation ... $\displaystyle m_1u_0=m_1u_f+m_2(u_0+u_f)$ $\displaystyle m_1u_0=m_1u_f+m_2u_0+m_2u_f$ $\displaystyle m_1u_0  m_2u_0 = m_1u_f+m_2u_f$ $\displaystyle u_0(m_1m_2) = u_f(m_1+m_2)$ $\displaystyle \frac{u_0(m_1m_2)}{m_1+m_2} = u_f$  

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