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https://ophysics.com/e2.html The link above is a collision simulation and I can control the elasticity. I tried to make an equation to solve collisions given % of energy loss using the kinetic and momentum equations thinking that % energy loss is the same as elasticity %, however my numbers to match up. How is % energy loss related to % elasticity?

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The energy prior to and after collision is entirely kinetic energy of the system. If $T'$ is the kinetic energy after collision, and $T$ is the kinetic energy before, then the percent energy loss is just given by

$\frac{T-T'}{T} \times 100 \% $,

where $T'$ will be less than $T$.

If you have blocks $A$ and $B$, then the system kinetic energy prior to collision is given by

$T = \frac{1}{2}m_A {\bf v}_A \cdot {\bf v}_A + \frac{1}{2}m_B {\bf v}_B \cdot {\bf v}_B$,

and after collision as

$T' = \frac{1}{2}m_A {\bf v}'_A \cdot {\bf v}'_A + \frac{1}{2}m_B {\bf v}'_B \cdot {\bf v}'_B$.

The coefficient of restitution $e \in [0,1]$ represents the amount of elasticity in the collision. If $e = 1$, then the collision is elastic. If $e = 0$, the collision is plastic.

For these 1D motions, the coefficient of restitution is

$e = \frac{\text{separation speed}}{\text{approach speed}}$,

where the separation speed is the relative speed between the two particles post-collision and the approach speed is the relative speed between the two particles pre-collision. See how for a plastic collision $e=0$, the separation speed is zero? This means that the particles stick together.

You should be able to combine these equations and perhaps conservation of system linear momentum to relate the percent energy lost to $e$, your measure of the elasticity of the collision.

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