newtonian mechanics - Why is there a $frac 1 2$ in $frac 1 2 mv^2$?

For elastic collisions of n particles, we know that momentum in the three orthogonal directions are independently conserved:$$ \frac{d}{dt}\sum\limits_i^n m_iv_{ij} =0,\quad j=1,2,3$$

From this, it follows there's also a corresponding scalar quantity conserved:$$\frac{d}{dt}\sum\limits_i^n m_i(v^2_{i1} + v^2_{i2} + v^2_{i3}) = \frac{d}{dt}\sum\limits_i^n m_iv^2_i =0,\quad j=1,2,3$$

So why is there a need to put 1/2 in front of this conserved scalar quantity, kinetic energy?

Answer

The half in the non-relativistic kinetic energy can be traced back to the Work-Energy Theorem$^1$.

Of course, if one is only interested in solving an elastic collision problem for an isolated system of point particles using momentum and kinetic energy conservation, no harm is done by multiplying the energy conservation equation with a factor 2 on both sides of the equation.

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$^1$ Here we assume that the standard formulas for work $W=\int {\bf F}\cdot d{\bf r}$, Newton's 2nd law ${\bf F}=m{\bf a}$, acceleration ${\bf a}=\dot{\bf v}$, velocity ${\bf v}=\dot{\bf r}$, etc, hold without unconventional normalization factors.