How is the elastic collision equation derived?

1 Answer
Feb 11, 2016

Answered for one-dimensional case ...

Explanation:

In all collisional interactions momentum remain conserved. Collisions are called elastic collisions if, in addition to momentum conservation, kinetic energy remain conserved too. To derive the elastic collision equations we make use of the Momentum Conservation condition and Kinetic Energy Conservation condition.

m_1m1 - Mass of object 1; \qquad m_2 - Mass of object 2;
v_{1i} - velocity of object 1 before collision;
v_{2i} - velocity of object 2 before collision;
v_{1f} - velocity of object 1 after collision;
v_{2f} - velocity of object 2 after collision;

Momentum Conservation:
m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}

Rearrange this by bring all therms with m_1 on one side and terms with m_2 on the other side,
m_1(v_{1i}-v_{1f})=m_2(v_{2f}-v_{2i}) ............... ( 1 )

\frac{m_1(v_{1i}-v_{1f})}{m_2(v_{2f}-v_{2i})} = 1 ........... ( 2 )

Kinetic Energy Conservation:
1/2 m_1v_{1i}^2+1/2m_2v_{2i}^2=1/2 m_1 v_{1f}^2+1/2m_2v_{2f}^2

Rearrange this by bring all therms with m_1 on one side and terms with m_2 on the other side and cancel the common factor of '1/2',

m_1(v_{1i}^2-v_{1f}^2)=m_2(v_{2f}^2-v_{2i}^2)
m_1(v_{1i}-v_{1f})(v_{1i}+v_{1f}) = m_2(v_{2f}-v_{2i})(v_{2f}+v_{2i})
\frac{m_1(v_{1i}-v_{1f})}{m_2(v_{2f}-v_{2i})}.(v_{1i}+v_{1f}) = (v_{2f}+v_{2i}),

Recognise that the first term on the LHS is just '1' [ Equation ( 2 ) ]
v_{1i}+v_{1f} = v_{2i}+v_{2f} ..................... ( 3a )

v_{2f} = v_{1i}+v_{1f}-v_{2i} .................... ( 3b )

Substitute Equation ( 3b ) in Equation ( 1 ) to eliminate v_{2f}

m_1(v_{1i}-v_{1f}) = m_2((v_{1i}+v_{1f}-v_{2i})-v_{2i})

Rearrange this and solve for v_{1f}:

v_{1f} = (\frac{m_1-m_2}{m_1+m_2})v_{1i}+(\frac{2m_2}{m_1+m_2})v_{2i} ......... ( 4 )

Substitute Equation ( 4 ) in Equation ( 3b ) and rearrange the terms to get v_{2f} as

v_{2f} = (\frac{2m_1}{m_1+m_2}) v_{1i} + (\frac{m_2-m_1}{m_1+m_2}) v_{2i} ......... ( 5 )