Is there an elegant or easier way to solve this system of equations?












2














This is a physics problem that involves an inelastic collision and a system of equations that I want to solve. I know the values of $v_1,m_1,v',Delta E$ where $v'$ is the velocity of the two objects after they stick together. I want to solve the equations for either $m_2$(only in terms of $v_1,v',m_1, Delta E$) or for $v_2$ (only in terms of ($v_1,v',m_1, Delta E$) but I am stuck because the terms get very messy.



$$m_1v_1+m_2v_2=(m_1+m_2)v' tag{1}$$



$$frac{1}{2}m_1v_1^2+frac{1}{2}m_2v_2^2=frac{1}{2}m_1(v')^2+frac{1}{2}m_2(v')^2+Delta E tag{2}$$



My attempt so far:



Solving$(1)$ for $v_2$:



$$implies v_2=frac{m_1}{m_2}(v'-v_1)+v' tag{1a}$$



Plug $(1a)$ into $(2)$:



$$implies frac{1}{2}m_1v_1^2+frac{1}{2}m_2(frac{m_1}{m_2}(v'-v_1)+v')^2=frac{1}{2}m_1(v')^2+frac{1}{2}m_2(v')^2+Delta E \ iff frac{1}{2}m_1v_1^2+frac{1}{2}m_2 left[frac{m_1^2}{m_2^2 }(v'-v_1)^2+2frac{m_1}{m_2}v'(v'-v_1)+(v')^2 right]=frac{1}{2}m_1(v')^2+frac{1}{2}m_2(v')^2 +Delta E\ iff frac{1}{2}m_1v_1^2+frac{1}{2} m_2left[ left( frac{m_1}{m_2}right)^2((v')^2-2v'v_1-v_1^2)+2left( frac{m_1}{m_2}right )((v'^2)-v_1v')+(v')^2right] \=frac{1}{2}m_1(v')^2+frac{1}{2}m_2(v')^2+Delta E \ iff ...?$$




Is there an easier way to do this? How do I even continue?







systems-of-equations physics







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    2














    This is a physics problem that involves an inelastic collision and a system of equations that I want to solve. I know the values of $v_1,m_1,v',Delta E$ where $v'$ is the velocity of the two objects after they stick together. I want to solve the equations for either $m_2$(only in terms of $v_1,v',m_1, Delta E$) or for $v_2$ (only in terms of ($v_1,v',m_1, Delta E$) but I am stuck because the terms get very messy.



    $$m_1v_1+m_2v_2=(m_1+m_2)v' tag{1}$$



    $$frac{1}{2}m_1v_1^2+frac{1}{2}m_2v_2^2=frac{1}{2}m_1(v')^2+frac{1}{2}m_2(v')^2+Delta E tag{2}$$



    My attempt so far:



    Solving$(1)$ for $v_2$:



    $$implies v_2=frac{m_1}{m_2}(v'-v_1)+v' tag{1a}$$



    Plug $(1a)$ into $(2)$:



    $$implies frac{1}{2}m_1v_1^2+frac{1}{2}m_2(frac{m_1}{m_2}(v'-v_1)+v')^2=frac{1}{2}m_1(v')^2+frac{1}{2}m_2(v')^2+Delta E \ iff frac{1}{2}m_1v_1^2+frac{1}{2}m_2 left[frac{m_1^2}{m_2^2 }(v'-v_1)^2+2frac{m_1}{m_2}v'(v'-v_1)+(v')^2 right]=frac{1}{2}m_1(v')^2+frac{1}{2}m_2(v')^2 +Delta E\ iff frac{1}{2}m_1v_1^2+frac{1}{2} m_2left[ left( frac{m_1}{m_2}right)^2((v')^2-2v'v_1-v_1^2)+2left( frac{m_1}{m_2}right )((v'^2)-v_1v')+(v')^2right] \=frac{1}{2}m_1(v')^2+frac{1}{2}m_2(v')^2+Delta E \ iff ...?$$




    Is there an easier way to do this? How do I even continue?







    systems-of-equations physics







    share|cite|improve this question



























      2












      2








      2


      1





      This is a physics problem that involves an inelastic collision and a system of equations that I want to solve. I know the values of $v_1,m_1,v',Delta E$ where $v'$ is the velocity of the two objects after they stick together. I want to solve the equations for either $m_2$(only in terms of $v_1,v',m_1, Delta E$) or for $v_2$ (only in terms of ($v_1,v',m_1, Delta E$) but I am stuck because the terms get very messy.



      $$m_1v_1+m_2v_2=(m_1+m_2)v' tag{1}$$



      $$frac{1}{2}m_1v_1^2+frac{1}{2}m_2v_2^2=frac{1}{2}m_1(v')^2+frac{1}{2}m_2(v')^2+Delta E tag{2}$$



      My attempt so far:



      Solving$(1)$ for $v_2$:



      $$implies v_2=frac{m_1}{m_2}(v'-v_1)+v' tag{1a}$$



      Plug $(1a)$ into $(2)$:



      $$implies frac{1}{2}m_1v_1^2+frac{1}{2}m_2(frac{m_1}{m_2}(v'-v_1)+v')^2=frac{1}{2}m_1(v')^2+frac{1}{2}m_2(v')^2+Delta E \ iff frac{1}{2}m_1v_1^2+frac{1}{2}m_2 left[frac{m_1^2}{m_2^2 }(v'-v_1)^2+2frac{m_1}{m_2}v'(v'-v_1)+(v')^2 right]=frac{1}{2}m_1(v')^2+frac{1}{2}m_2(v')^2 +Delta E\ iff frac{1}{2}m_1v_1^2+frac{1}{2} m_2left[ left( frac{m_1}{m_2}right)^2((v')^2-2v'v_1-v_1^2)+2left( frac{m_1}{m_2}right )((v'^2)-v_1v')+(v')^2right] \=frac{1}{2}m_1(v')^2+frac{1}{2}m_2(v')^2+Delta E \ iff ...?$$




      Is there an easier way to do this? How do I even continue?







      systems-of-equations physics







      share|cite|improve this question















      This is a physics problem that involves an inelastic collision and a system of equations that I want to solve. I know the values of $v_1,m_1,v',Delta E$ where $v'$ is the velocity of the two objects after they stick together. I want to solve the equations for either $m_2$(only in terms of $v_1,v',m_1, Delta E$) or for $v_2$ (only in terms of ($v_1,v',m_1, Delta E$) but I am stuck because the terms get very messy.



      $$m_1v_1+m_2v_2=(m_1+m_2)v' tag{1}$$



      $$frac{1}{2}m_1v_1^2+frac{1}{2}m_2v_2^2=frac{1}{2}m_1(v')^2+frac{1}{2}m_2(v')^2+Delta E tag{2}$$



      My attempt so far:



      Solving$(1)$ for $v_2$:



      $$implies v_2=frac{m_1}{m_2}(v'-v_1)+v' tag{1a}$$



      Plug $(1a)$ into $(2)$:



      $$implies frac{1}{2}m_1v_1^2+frac{1}{2}m_2(frac{m_1}{m_2}(v'-v_1)+v')^2=frac{1}{2}m_1(v')^2+frac{1}{2}m_2(v')^2+Delta E \ iff frac{1}{2}m_1v_1^2+frac{1}{2}m_2 left[frac{m_1^2}{m_2^2 }(v'-v_1)^2+2frac{m_1}{m_2}v'(v'-v_1)+(v')^2 right]=frac{1}{2}m_1(v')^2+frac{1}{2}m_2(v')^2 +Delta E\ iff frac{1}{2}m_1v_1^2+frac{1}{2} m_2left[ left( frac{m_1}{m_2}right)^2((v')^2-2v'v_1-v_1^2)+2left( frac{m_1}{m_2}right )((v'^2)-v_1v')+(v')^2right] \=frac{1}{2}m_1(v')^2+frac{1}{2}m_2(v')^2+Delta E \ iff ...?$$




      Is there an easier way to do this? How do I even continue?







      systems-of-equations physics




      systems-of-equations physics






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      share|cite|improve this question













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      edited Dec 8 at 18:51

























      asked Dec 8 at 18:04









      Nullspace

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          begin{align}
          m_1v_1+m_2v_2&=(m_1+m_2)v'
          tag{1}label{1}
          ,\
          frac{1}{2}m_1v_1^2+frac{1}{2}m_2v_2^2
          &=
          frac{1}{2}m_1(v')^2+frac{1}{2}m_2(v')^2+Delta E
          tag{2}label{2}
          .
          end{align}



          begin{align}
          eqref{1}:quad
          m_2(v_2-v')&=
          m_1(v'-v_1)
          tag{3}label{3}
          ,\
          eqref{2}:quad
          m_2(v_2^2-v'^2)&=
          m_1(v'^2-v_1^2)+2Delta E
          tag{4}label{4}
          .
          end{align}



          eqref{3}$times (v_2+v')-$eqref{4}$quadRightarrow$



          begin{align}
          0&=
          m_1(v'-v_1)(v_2+v')-(m_1(v'^2-v_1^2)+2Delta E)
          tag{5}label{5}
          ,\
          end{align}



          begin{align}
          v_2 &= v_1+frac{2Delta E}{m_1(v'-v_1)}
          tag{6}label{6}
          .
          end{align}






          share|cite|improve this answer





















          • That's awesome! I would have never multiplied the third equation by $(v_2+v')$ that's such a cool way to do it. Thank you very much! (How do I get better at this?)
            – Nullspace
            Dec 8 at 19:58













          Your Answer





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          1 Answer
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          1 Answer
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          active

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          active

          oldest

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          1














          begin{align}
          m_1v_1+m_2v_2&=(m_1+m_2)v'
          tag{1}label{1}
          ,\
          frac{1}{2}m_1v_1^2+frac{1}{2}m_2v_2^2
          &=
          frac{1}{2}m_1(v')^2+frac{1}{2}m_2(v')^2+Delta E
          tag{2}label{2}
          .
          end{align}



          begin{align}
          eqref{1}:quad
          m_2(v_2-v')&=
          m_1(v'-v_1)
          tag{3}label{3}
          ,\
          eqref{2}:quad
          m_2(v_2^2-v'^2)&=
          m_1(v'^2-v_1^2)+2Delta E
          tag{4}label{4}
          .
          end{align}



          eqref{3}$times (v_2+v')-$eqref{4}$quadRightarrow$



          begin{align}
          0&=
          m_1(v'-v_1)(v_2+v')-(m_1(v'^2-v_1^2)+2Delta E)
          tag{5}label{5}
          ,\
          end{align}



          begin{align}
          v_2 &= v_1+frac{2Delta E}{m_1(v'-v_1)}
          tag{6}label{6}
          .
          end{align}






          share|cite|improve this answer





















          • That's awesome! I would have never multiplied the third equation by $(v_2+v')$ that's such a cool way to do it. Thank you very much! (How do I get better at this?)
            – Nullspace
            Dec 8 at 19:58


















          1














          begin{align}
          m_1v_1+m_2v_2&=(m_1+m_2)v'
          tag{1}label{1}
          ,\
          frac{1}{2}m_1v_1^2+frac{1}{2}m_2v_2^2
          &=
          frac{1}{2}m_1(v')^2+frac{1}{2}m_2(v')^2+Delta E
          tag{2}label{2}
          .
          end{align}



          begin{align}
          eqref{1}:quad
          m_2(v_2-v')&=
          m_1(v'-v_1)
          tag{3}label{3}
          ,\
          eqref{2}:quad
          m_2(v_2^2-v'^2)&=
          m_1(v'^2-v_1^2)+2Delta E
          tag{4}label{4}
          .
          end{align}



          eqref{3}$times (v_2+v')-$eqref{4}$quadRightarrow$



          begin{align}
          0&=
          m_1(v'-v_1)(v_2+v')-(m_1(v'^2-v_1^2)+2Delta E)
          tag{5}label{5}
          ,\
          end{align}



          begin{align}
          v_2 &= v_1+frac{2Delta E}{m_1(v'-v_1)}
          tag{6}label{6}
          .
          end{align}






          share|cite|improve this answer





















          • That's awesome! I would have never multiplied the third equation by $(v_2+v')$ that's such a cool way to do it. Thank you very much! (How do I get better at this?)
            – Nullspace
            Dec 8 at 19:58
















          1












          1








          1






          begin{align}
          m_1v_1+m_2v_2&=(m_1+m_2)v'
          tag{1}label{1}
          ,\
          frac{1}{2}m_1v_1^2+frac{1}{2}m_2v_2^2
          &=
          frac{1}{2}m_1(v')^2+frac{1}{2}m_2(v')^2+Delta E
          tag{2}label{2}
          .
          end{align}



          begin{align}
          eqref{1}:quad
          m_2(v_2-v')&=
          m_1(v'-v_1)
          tag{3}label{3}
          ,\
          eqref{2}:quad
          m_2(v_2^2-v'^2)&=
          m_1(v'^2-v_1^2)+2Delta E
          tag{4}label{4}
          .
          end{align}



          eqref{3}$times (v_2+v')-$eqref{4}$quadRightarrow$



          begin{align}
          0&=
          m_1(v'-v_1)(v_2+v')-(m_1(v'^2-v_1^2)+2Delta E)
          tag{5}label{5}
          ,\
          end{align}



          begin{align}
          v_2 &= v_1+frac{2Delta E}{m_1(v'-v_1)}
          tag{6}label{6}
          .
          end{align}






          share|cite|improve this answer












          begin{align}
          m_1v_1+m_2v_2&=(m_1+m_2)v'
          tag{1}label{1}
          ,\
          frac{1}{2}m_1v_1^2+frac{1}{2}m_2v_2^2
          &=
          frac{1}{2}m_1(v')^2+frac{1}{2}m_2(v')^2+Delta E
          tag{2}label{2}
          .
          end{align}



          begin{align}
          eqref{1}:quad
          m_2(v_2-v')&=
          m_1(v'-v_1)
          tag{3}label{3}
          ,\
          eqref{2}:quad
          m_2(v_2^2-v'^2)&=
          m_1(v'^2-v_1^2)+2Delta E
          tag{4}label{4}
          .
          end{align}



          eqref{3}$times (v_2+v')-$eqref{4}$quadRightarrow$



          begin{align}
          0&=
          m_1(v'-v_1)(v_2+v')-(m_1(v'^2-v_1^2)+2Delta E)
          tag{5}label{5}
          ,\
          end{align}



          begin{align}
          v_2 &= v_1+frac{2Delta E}{m_1(v'-v_1)}
          tag{6}label{6}
          .
          end{align}







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 8 at 19:38









          g.kov

          6,0971718




          6,0971718












          • That's awesome! I would have never multiplied the third equation by $(v_2+v')$ that's such a cool way to do it. Thank you very much! (How do I get better at this?)
            – Nullspace
            Dec 8 at 19:58




















          • That's awesome! I would have never multiplied the third equation by $(v_2+v')$ that's such a cool way to do it. Thank you very much! (How do I get better at this?)
            – Nullspace
            Dec 8 at 19:58


















          That's awesome! I would have never multiplied the third equation by $(v_2+v')$ that's such a cool way to do it. Thank you very much! (How do I get better at this?)
          – Nullspace
          Dec 8 at 19:58






          That's awesome! I would have never multiplied the third equation by $(v_2+v')$ that's such a cool way to do it. Thank you very much! (How do I get better at this?)
          – Nullspace
          Dec 8 at 19:58




















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