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?

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    $\begingroup$ BTW--It does not follow in general, and kinetic energy is only conserved under special conditions. $\endgroup$ Commented May 5, 2012 at 22:27
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    $\begingroup$ I agree with dmckee. Your "proof" is wrong. The last step $=0$ simply doesn't follow from the previous line, you were sloppy. Kinetic energy is $mv^2/2$ in non-relativistic mechanics because it's the integral of $\int F\cdot ds = \int dp/dt\cdot ds/dt\cdot dt = \int v\cdot dp/dt\cdot dt$ and the integral of $mv$ from $0$ to $v_{max}$ is $mv_{max}^2/2$. Even if your derivation of the conservation law were right, which it's not, it wouldn't follow that the right prefactor is a random number that you find simple. Other conditions on normalization of $E$ are more important than "simplicity". $\endgroup$ Commented May 6, 2012 at 5:48
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    $\begingroup$ @LubošMotl and dmckee yeah my proof is nonsense - I'll leave it in so others learn not to end up embarrassing themselves in the same way. $\endgroup$ Commented May 6, 2012 at 15:53
  • $\begingroup$ Related: physics.stackexchange.com/q/68005 Check the last part of my answer re: the coefficients. $\endgroup$
    – Greg
    Commented Jun 21, 2013 at 3:58
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    $\begingroup$ The 1/2 is purely conventional. We could make kinetic energy be $mv^2$, but then we'd have to change all other forms of energy by the same factor, e.g., by making gravitational PE be $2mgh$. $\endgroup$
    – user4552
    Commented Jun 21, 2013 at 15:43

8 Answers 8


The factor of 1/2 is required from Galilean invariance, so that the energy mixes up with the momentum without a factor. This was understood before relativity, but it is largely conventional before relativity, since you could make the energy mix up with the momentum using some coefficient. Once you have relativity, the 1/2 is no longer optional.

I'll start with relativity. The kinetic energy formula is the extra energy in a moving particle

$$ {m\over \sqrt{1-v^2}} = m + {mv^2\over 2} + \cdots$$

in units where c=1. The one half comes from the expansion of the geometrical square root, and the square-root form in the denominator is uniquely and naturally fixed by requiring that the energy and momentum fit together into a four vector. This is the only natural definition in relativity, and it is justified by geometry.

In Newtonian mechanics, the energy and momentum transform together after a Galilean boost. If you have a closed system with momenta $p_i$ that add up to zero (center of mass frame), the change in the kinetic energy after a boost, which shifts $p_i\rightarrow p_i - m_i v$ is

$$\sum_i {p_i^2\over 2m_i} \rightarrow \sum_i m_i~(v_i - v)^2 = \sum_i {p_i^2\over 2m_i} - \sum_i p_i \cdot v + \sum_i m_i {v^2\over 2} $$

The change has two parts,

$$ (\sum_i p_i) \cdot v $$

This is the mixing of energy and momentum, and this part is the total momentum dot the velocity, and it is zero when you start in the CM frame. The other part is:

$$ (\sum_i m_i ) {v^2\over 2} $$

This part is the total mass times half the square of the velocity, or the total kinetic energy added to the object by boosting it. You see that you get the right answer, the total mass times the center of mass velocity, now that the center of mass is moving. If the object was already moving, it is nontrivial to ensure that you get the right answer--- the new center of mass velocity squared times the total mass squared--- for the new energy.

This requirement, that the energy is consistent under galilean boosts, is what is meant by the mixing of momentum and energy. It is simplest if you take the coefficient 1/2, ultimately because this is the natural nonrelativistic limit of relativity. You would have a factor of 2 in the p mixing term if you didn't take the coefficient 1/2. It is actually a testiment to the insight of the 19th century physicists that they adopted the most natural convention before relativity was discovered.

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    $\begingroup$ Neat, but you really don't need to go into relativity to explain this... Nor could it do much good to someone with zero knowledge of relativity, since you're more likely to confuse than to illuminate. And if the OP is a high school student just getting started with physics, then to invoke relativity is to ask that they take your argument on faith, which leaves them in pretty much the same place they were before your answer---wondering why there is a 1/2 in the kinetic energy expression, just with the minor advantage of knowing that relativity seems to somehow account for it. But interesting! $\endgroup$
    – user273231
    Commented Sep 6, 2020 at 5:27

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.


$^1$ 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.

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    $\begingroup$ You could always add a factor of 2 in the Work-Energy Theorem to make the 1/2 disappear. The reason is the Galilean invariance formulas are a little simpler with the 1/2 convention. $\endgroup$
    – Ron Maimon
    Commented May 7, 2012 at 6:40
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    $\begingroup$ Right, it is possible to kick the can further down the road. $\endgroup$
    – Qmechanic
    Commented May 7, 2012 at 9:57
  • $\begingroup$ @Qmechanic can you tell me the physical significance of this term $ \frac{1}{2}$ ? $\endgroup$
    – Ankit
    Commented Oct 9, 2020 at 16:17

The reason is because of the Work-Energy theorem, which states that if a force is applied on an object over a distance, that the integral of the force along the distance, or the work applied on the object, must equal the change in the kinetic energy, as the energy transferred through work must be conserved. So for an object at rest, if a force $F$ is applied in one dimension over a distance $x$, its change in kinetic energy is given by the work done on the object: $$E_k - 0 = \int_{0}^{x} {F \cdot dx}$$ (We have the change in kinetic energy represented by $E_k$ as the final kinetic energy, and $0$ as the initial kinetic energy.) If we manipulate the integral using $F = ma$, it becomes $$E_k = \int_{0}^{x} {ma \cdot dx}$$ We can multiply by $\frac{dt}{dt}$ in the integral, and it becomes $$E_k = \int_{0}^{x} {ma \hspace{2pt} dt \cdot \frac{dx}{dt}}$$ Since we know that $v = \frac{dx}{dt}$, and that $a = \frac{dv}{dt}$, $a \hspace{2pt} dt$ becomes $dv$, and $\frac{dx}{dt}$ becomes $v$. This turns the integral into: $$E_k = \int_{0}^{v} {mv \cdot dv}$$ We now integrate from $0$ to $v$ because we've turned the integral into an integral over position into an integral over velocity. Since integrating a linear function $kx$ with respect to $x$ gives $\frac{1}{2} kx^2$, the expression becomes: $$E_k = \frac{1}{2} mv^2$$

So basically, the Work-Energy theorem as well as manipulating the integral is the reason for the $\frac{1}{2}$ in $\frac{1}{2} mv^2$.

  • $\begingroup$ Perfect. This has got to be the best way to explain the question. And note that the 1/2 comes from integration. It is not an arbitrary choice, as many seem to think! Thanks for posting this. Excellent. $\endgroup$
    – user273231
    Commented Sep 6, 2020 at 5:33
  • $\begingroup$ Yea, it's definitely the integral thanks to v being variable. I noticed E=mc^2 doesn't have a 1/2 factor. v is variable to absorb energy and do work. But c is constant so integrating from 0 to v=c doesn't make sense. Rather, mv • dv becomes mv • v sans integral, and the 1/2 factor goes away. It's kinda neat how the rest mass looks like a swarm of photons trying to escape at v=c and it has exactly the energy you'd expect for constant c, and if you push that mass around with your hands with variable v far slower than c, it has exactly the energy you'd expect for variable v. Upvoted! $\endgroup$ Commented Nov 7, 2023 at 16:09

Kinetic Energy ($\mathrm{KE}$) is equal to work done, $\mathrm{KE} = W$, and work is equal to force ($F$) applied to a body multiplied by the distance ($d$) it travels, or $W = Fd$. Since $F = ma$, the former equation renders $W = mad$.

Providing that acceleration is uniform and the initial velocity is zero, suppose there is a graph with velocity $v$ on the $y$-axis and time ($t$) on the $x$-axis. Furthermore there is a line passing through the origin and has a slope equal to the acceleration of the body. For a given interval of time, the distance traveled by the body is equal to the area under the line and that is $d = vt/2$. If we substitute $d$ into $W = mad$ we get $W = mavt/2$ (notice how the half has appeared). The equation becomes $W = matv/2$ but since $at = v$ we get $$ W = \mathrm{KE} = \frac{1}{2} mv^2. $$

See the below diagram where the equation of the line (i.e. a lineal function) has a slope equal to the acceleration of a body. Recall the y-intercept form of equation for a line or $y = mx + b$, where $m$ represents slope and $b$ represents the y-intercept or initial velocity. Notice also the shape of the area $d$ is that of a right triangle. This means, $$v = at$$

  • $\begingroup$ @Michael Lee: Welcome to Physics Stackexchange! I reformatted your response using Latex-style formatting, as suggested in our help pages in order to increase readability. If you are unfamiliar with latex, you can click the edit button above to see how I did the formatting here. Also, Math Stackexchange has a rather complete guide to all the latex features implemented on this site. $\endgroup$
    – user10851
    Commented Jun 21, 2013 at 4:37
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    $\begingroup$ Βest answer! The 1/2 factor is physically a result of the assumption that the initial velocity of the point mass is zero $υ_{0}=0$ and then slopes to the final value $υ$. The Work done from $υ_{0}=0$ accelerating to $υ$ is equat to the kinetic energy (KE). When no acceleration is assumed but a constant velolity υ the KE=W=m$υ^2$. $\endgroup$
    – Markoul11
    Commented Oct 27, 2023 at 7:48
  • $\begingroup$ Correction to my previous comment, If no accelaration is considered from zero to υ and only a constant velocity value υ from t=0 then there is no change in the kinetic energy of a mass and therefore the previous kinetic energy obtained from zero to u acceleration is conserved therefore the KE of the mass is still at $ \mathrm{KE} = \frac{1}{2} mv^2. $ $\endgroup$
    – Markoul11
    Commented Oct 27, 2023 at 8:55
  • $\begingroup$ Also the same as the translational case presented herein for the 1/2 factor applies also for the rotational kinetic energy of a point mass thus $ K E_{\text {(rotational) }}=\frac{1}{2} L \omega $. physics.stackexchange.com/q/785335/183646 $\endgroup$
    – Markoul11
    Commented Oct 27, 2023 at 10:42

In addition to the other answers, bear in mind that $ E = \frac 1 2 m v^2 $ is a quadratic form.

Quadratic forms arise whenever you integrate an $f(x) = k\cdot x$ function.


$$ E_{kin} = \int m v\; \mathrm d v = \tfrac 1 2 m v ^ 2 \\ A_\circ = \int \tau r \; \mathrm d r = \tfrac 1 2 \tau r^2 \quad \tau = 2\pi \\ E_{spring} = \int k x \;\mathrm d x = \tfrac 1 2 k x^2 \\ v = \int a t \;\mathrm d t = \tfrac 1 2 a t^2$$

In Special Relativity velocity's additive group is not the Reals, so you get a different result of that integration.

  • $\begingroup$ $m v^2$ is also a quadratic form in your definition with $m\rightarrow m/2$... $\endgroup$
    – rubenvb
    Commented Jul 19, 2013 at 11:59
  • $\begingroup$ @rubenvb I am not sure I follow, honestly. $ m $ in $ \int m v \:\mathrm d v $ is assumed to be constant? $\endgroup$ Commented Jul 19, 2013 at 16:06
  • $\begingroup$ Excellent. This is the sort of insight that helps us all. Such a shame that we can go through years of physics and calculus and still fail to grasp these basic general principles... Thanks for sharing. $\endgroup$
    – user273231
    Commented Sep 6, 2020 at 5:36

The same reason why distance is half the acceleration I have my own ways of answering the question since i don't know calculus. Anyway

$w = mda$, assuming you know distance and acceleration(for now)

$d = \frac{1}{2}at^2$

$\frac{2d}{a}= t^2$

$\sqrt{\frac{2d}{a}}= t$

multiplying the time t by acceleration a gives us the velocity

$v = \sqrt{\frac{2d}{a}}a$

$v = \frac{\sqrt{{2d}}}{\sqrt{a}} a$

$v = \sqrt{2d}\sqrt{a}$

$v = \sqrt{2da}$ and da equals work over mass so

$v = \sqrt{\frac{2w}{m}}$

$v^2 =\frac{2w}{m} $

$mv^2 =2w$

$w = \frac{1}{2}mv^2$


Kinetic energy isolates the energy associated directly with the vector motion of a specific object, and excludes the offsetting momentum to other objects which must necessarily exist, in the opposite direction, in order to satisfy Newton's third law.

The sum of all of the kinetic energy created across all objects equals the amount of energy so applied, but since any single object (used very loosely) can only "own" the one-half of the energy associated with its total greater momentum-neutral motion, that is therefore also all that it can transmit by passing on its own momentum in a collision.

Other forms of energy capture the total energy required to create motion in a momentum neutral fashion. Notably, gravitational potential energy includes both the motion associated with an object falling down to earth, and the much smaller, but still momentum-equal, motion of the earth rising up to meet the object. Thus the existence of the one-half convention. Related: Gravity and KE

  • $\begingroup$ Why does the total energy of the momentum-neutral motion need to be distributed equally among both the objects? $\endgroup$
    – udiboy1209
    Commented Jul 19, 2013 at 13:18
  • $\begingroup$ It's not, it's distributed by mass. $\endgroup$ Commented Jul 19, 2013 at 15:05

In classical physics: $$\begin{align*} W&=\int_{\vec s_1}^{\vec s_2}\vec F\cdot d\vec s\\ &=\int_{t_1}^{t_2}m\frac{d\vec v}{dt}\cdot\vec v dt\\ &=m\int_{t_1}^{t_2}\frac12\frac{d}{dt}\left(\vec v\cdot\vec v\right)dt\\ &=\frac12m\int_{t_1}^{t_2}\frac{d}{dt}\left(||\vec v||^2\right)dt\\ &=\frac12m||\vec v(t_2)||^2-\frac12m||\vec v(t_1)||^2\\ \end{align*}$$ $$\frac{d}{dt}\left(\vec v\cdot\vec v\right)=\frac{d\vec v}{dt}\cdot\vec v+\vec v\cdot\frac{d\vec v}{dt}=2\frac{d\vec v}{dt}\cdot\vec v$$


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