# How is the elastic collision equation derived?

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.

${m}_{1}$ - Mass of object 1; $\setminus q \quad$ ${m}_{2}$ - Mass of object 2;
${v}_{1 i}$ - velocity of object 1 before collision;
${v}_{2 i}$ - velocity of object 2 before collision;
${v}_{1 f}$ - velocity of object 1 after collision;
${v}_{2 f}$ - velocity of object 2 after collision;

Momentum Conservation:
${m}_{1} {v}_{1 i} + {m}_{2} {v}_{2 i} = {m}_{1} {v}_{1 f} + {m}_{2} {v}_{2 f}$

Rearrange this by bring all therms with ${m}_{1}$ on one side and terms with ${m}_{2}$ on the other side,
${m}_{1} \left({v}_{1 i} - {v}_{1 f}\right) = {m}_{2} \left({v}_{2 f} - {v}_{2 i}\right)$ ............... ( 1 )

$\setminus \frac{{m}_{1} \left({v}_{1 i} - {v}_{1 f}\right)}{{m}_{2} \left({v}_{2 f} - {v}_{2 i}\right)} = 1$ ........... ( 2 )

Kinetic Energy Conservation:
$\frac{1}{2} {m}_{1} {v}_{1 i}^{2} + \frac{1}{2} {m}_{2} {v}_{2 i}^{2} = \frac{1}{2} {m}_{1} {v}_{1 f}^{2} + \frac{1}{2} {m}_{2} {v}_{2 f}^{2}$

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',

${m}_{1} \left({v}_{1 i}^{2} - {v}_{1 f}^{2}\right) = {m}_{2} \left({v}_{2 f}^{2} - {v}_{2 i}^{2}\right)$
${m}_{1} \left({v}_{1 i} - {v}_{1 f}\right) \left({v}_{1 i} + {v}_{1 f}\right) = {m}_{2} \left({v}_{2 f} - {v}_{2 i}\right) \left({v}_{2 f} + {v}_{2 i}\right)$
$\setminus \frac{{m}_{1} \left({v}_{1 i} - {v}_{1 f}\right)}{{m}_{2} \left({v}_{2 f} - {v}_{2 i}\right)} . \left({v}_{1 i} + {v}_{1 f}\right) = \left({v}_{2 f} + {v}_{2 i}\right)$,

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

${v}_{2 f} = {v}_{1 i} + {v}_{1 f} - {v}_{2 i}$ .................... ( 3b )

Substitute Equation ( 3b ) in Equation ( 1 ) to eliminate ${v}_{2 f}$

${m}_{1} \left({v}_{1 i} - {v}_{1 f}\right) = {m}_{2} \left(\left({v}_{1 i} + {v}_{1 f} - {v}_{2 i}\right) - {v}_{2 i}\right)$

Rearrange this and solve for ${v}_{1 f}$:

${v}_{1 f} = \left(\setminus \frac{{m}_{1} - {m}_{2}}{{m}_{1} + {m}_{2}}\right) {v}_{1 i} + \left(\setminus \frac{2 {m}_{2}}{{m}_{1} + {m}_{2}}\right) {v}_{2 i}$ ......... ( 4 )

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

${v}_{2 f} = \left(\setminus \frac{2 {m}_{1}}{{m}_{1} + {m}_{2}}\right) {v}_{1 i} + \left(\setminus \frac{{m}_{2} - {m}_{1}}{{m}_{1} + {m}_{2}}\right) {v}_{2 i}$ ......... ( 5 )