# Question #83c7e

Oct 14, 2015

I found that depends on the ratio of the two masses only.

#### Explanation:

I tried considerng Conservation of Momentum $p$ (along $x$):
Before: $0$
After: ${m}_{1} {v}_{1} + {m}_{2} {v}_{2}$
and together:
$0 = {m}_{1} {v}_{1} + {m}_{2} {v}_{2}$

so: ${m}_{1} {v}_{1} = - {m}_{2} {v}_{2}$

squaring both and rearranging:

${v}_{1}^{2} / {v}_{2}^{2} = {m}_{2}^{2} / {m}_{1}^{2}$

Kinetic Energy is $K = \frac{1}{2} m {v}^{2}$
So:
${K}_{1} / {K}_{2} = \frac{\cancel{\frac{1}{2}} {m}_{1} {v}_{1}^{2}}{\cancel{\frac{1}{2}} {m}_{2} {v}_{2}^{2}}$,
substituting: ${v}_{1}^{2} / {v}_{2}^{2} = {m}_{2}^{2} / {m}_{1}^{2}$
${K}_{1} / {K}_{2} = \frac{\cancel{{m}_{1}}}{\cancel{{m}_{2}}} \cdot {m}_{2}^{\cancel{2}} / {m}_{1}^{\cancel{2}} = {m}_{2} / {m}_{1}$