# Foundations of Newtonian Mechanics: From the Three Laws to Momentum and Energy Conservation

> The three laws restated as the existence of inertial frames, force versus acceleration, and action-reaction; then one-dimensional motion, momentum and energy conservation.
> https://rikai.mugen-giken.com/en/physics/mechanics/newtonian-mechanics

## 0. Key points

- Newton's first law is not the claim that "a body free of forces moves uniformly in a straight line". It is the claim that **a coordinate system with that property (an inertial frame) exists**. The second law has meaning only in such a frame.
- The equation of motion $m\ddot{\boldsymbol{r}} = \boldsymbol{F}$ is a **second-order** differential equation for the position. That is precisely why prescribing an initial position and an initial velocity determines all subsequent motion uniquely.
- The equations close only once we say what the force actually is (gravity, a spring, friction). The part that specifies the form of the force is not a law but a **constitutive relation**, obtained from experiment.
- Conservation of momentum follows from the third law (action and reaction); conservation of mechanical energy follows from the force being conservative. Both are **consequences** of the equation of motion, not additional axioms.
- In a one-dimensional conservative system, one obtains the motion simply by treating energy conservation as a first-order equation and integrating it (quadrature). Sketching the potential reveals the qualitative behaviour of the motion before any calculation.

## 1. Motivation: why force equals acceleration

Since Aristotle, the relation between force and motion had been read as "force determines velocity". Keep pushing a cart and it moves; let go and it stops. Everyday experience supports this view.

Galileo destroyed it. Watching balls roll on inclined planes, he noticed that they speed up going down, slow down going up, and on a horizontal plane keep moving at the same speed indefinitely — the more so the more friction is reduced. The conclusion is that **no force is needed to maintain uniform rectilinear motion**. The released cart stops not because the force disappeared, but because another force, friction, acts on it.

Restated in the language of the calculus, the reversal reads as follows: what a force fixes is not the velocity $\dot{\boldsymbol{r}}$ but the acceleration $\ddot{\boldsymbol{r}}$; that is, the equation of motion is a **second-order** differential equation for the position. This single fact — the order being two — very nearly settles the character of classical mechanics. The initial value problem for a second-order ordinary differential equation has a unique solution once the initial position $\boldsymbol{r}(t_0)$ and the initial velocity $\dot{\boldsymbol{r}}(t_0)$ are specified. Conversely, knowing the position alone tells us nothing about the future; we must know the velocity at the same time. When Laplace spoke of an intelligence that knows the positions and velocities of all the particles in the universe and can therefore compute the future, the pairing of "position and velocity" was a direct reflection of the equation being of second order.

What Newton did in the *Principia* of 1687 was to write this insight down as a system of axioms and then, combining it with a specific form for the force — universal gravitation — to derive the planetary laws that Kepler had extracted from observation. The claim that celestial motion and terrestrial falling obey the same equation was, for its time, extraordinarily bold. The application to planetary motion is treated in [Planetary Motion and Central Forces](/en/physics/mechanics/central-forces); that the orbits are conic sections is <Ref to="physics/mechanics/central-forces#thm-kepler-first" text="Kepler's first law" />.

In this article we restate the three laws precisely and then verify that **the conservation laws are theorems, not axioms**. Both momentum conservation and energy conservation can be proved from the equation of motion. Once the proofs make visible which hypothesis supports which conservation law, one can also predict what happens when those hypotheses fail — when there is friction, or an external force.

## 2. Preliminaries: point masses, time, trajectories

The simplest object in classical mechanics is a **point mass**: a body whose size may be ignored so that a single point represents it, its state given by a position vector alone. Whether the Earth may be treated as a point mass depends on the scale of the problem (yes for its orbital motion, no for its rotation).

We identify three-dimensional space with the real vector space $\mathbb{R}^3$ and describe the position of the point mass by a function of time $\boldsymbol{r}(t) = (x(t), y(t), z(t))$. We take for granted the vector-space structure treated in [Vector Spaces and Linear Maps](/en/mathematics/linear-algebra/vector-spaces) (<Ref to="mathematics/linear-algebra/vector-spaces#def-vector-space" text="the definition of a vector space" />). When $\boldsymbol{r}$ is twice differentiable, we call

$$
\boldsymbol{v}(t) = \dot{\boldsymbol{r}}(t) = \frac{d\boldsymbol{r}}{dt}, \qquad
\boldsymbol{a}(t) = \ddot{\boldsymbol{r}}(t) = \frac{d^2\boldsymbol{r}}{dt^2}
$$

the **velocity** and the **acceleration** respectively. The dot denotes differentiation with respect to time, a notation going back to Newton. For the definition of the derivative itself (<Ref to="mathematics/calculus/derivatives#def-derivative" />), see [The Derivative and the Basic Rules of Differentiation](/en/mathematics/calculus/derivatives).

We assume that every body carries a **mass** $m > 0$, a positive real number. The mass is intrinsic to the body and is assumed independent of place and of state of motion — within classical mechanics; relativity modifies this assumption.

## 3. The three laws of motion

The three laws are mutually independent assertions. We take them in turn.

<Definition id="def-inertial-frame" title="Inertial frame">
Among the reference systems that assign a pair $(t, \boldsymbol{r})$ of time and spatial coordinates, those in which **every point mass subject to no force from other bodies moves uniformly in a straight line (with acceleration $\boldsymbol{0}$)** are called **inertial frames**.
</Definition>

<Axiom id="ax-first-law" title="First law (law of inertia)">
There exists at least one inertial frame.
</Axiom>

If one states the first law as "a body subject to no force moves uniformly in a straight line", it is nothing more than the second law with $\boldsymbol{F} = \boldsymbol{0}$, and carries no independent content. The content of the first law is instead an existence claim: **there exists a stage (an inertial frame) on which the second law holds**. In a coordinate system fixed to a rotating disc, a point mass subject to no force appears to curve; fictitious forces called centrifugal and Coriolis forces appear. The first law guarantees that not all frames are of this kind.

<Axiom id="ax-second-law" title="Second law (law of motion)">
In an inertial frame, define the **momentum** of a point mass of mass $m$ by $\boldsymbol{p} = m\boldsymbol{v}$. Then, for the force $\boldsymbol{F}$ acting on the point mass,
$$
\frac{d\boldsymbol{p}}{dt} = \boldsymbol{F}
$$
holds. In particular, when $m$ is independent of time this may be written $m\boldsymbol{a} = \boldsymbol{F}$.
</Axiom>

When the mass varies — a rocket expelling fuel, for instance — the form $d\boldsymbol{p}/dt = \boldsymbol{F}$ is the essential one, and using $m\boldsymbol{a} = \boldsymbol{F}$ as it stands gives wrong answers. From here on we take $m$ constant.

<Axiom id="ax-third-law" title="Third law (action and reaction)">
Writing $\boldsymbol{F}_{ji}$ for the force exerted by point mass $i$ on point mass $j$,
$$
\boldsymbol{F}_{ji} = -\boldsymbol{F}_{ij}
$$
holds (the **weak form**). If moreover $\boldsymbol{F}_{ji}$ is parallel to the line $\boldsymbol{r}_j - \boldsymbol{r}_i$ joining the two point masses, we say that the **strong form** holds.
</Axiom>

The weak form suffices to derive conservation of momentum, but conservation of angular momentum requires the strong form. Gravity and the Coulomb force satisfy the strong form. On the other hand, the magnetic force between moving charges does not satisfy the third law at all; the missing momentum is carried off by the electromagnetic field. It is safest to understand the third law not as a universal truth but as an assumption that depends on the type of force. The viewpoint that recasts conservation laws in terms of a deeper principle — the symmetries of spacetime — is treated in [Symmetries and Conservation Laws (Noether's Theorem)](/physics/mechanics/noethers-theorem) (<Ref to="physics/mechanics/noethers-theorem#thm-noether" text="Noether's theorem" />).

<Remark id="rem-circularity" title="Are force and mass defined circularly?">
Ask what a force is and one is told "mass times acceleration"; ask what mass is and one is told "the reluctance to accelerate under a given force". That is a circle. Mach criticised exactly this point and showed a way out: define the ratio of masses first, using the third law. When two isolated point masses interact, $m_1\boldsymbol{a}_1 = -m_2\boldsymbol{a}_2$ holds, so measuring the ratio of the accelerations fixes the mass ratio $m_2/m_1 = |\boldsymbol{a}_1|/|\boldsymbol{a}_2|$ without any knowledge of the force. Fix a standard kilogram and every mass becomes measurable. Once masses are fixed, the second law defines force, and specific force laws such as "gravity falls off as the $-2$ power of the distance" acquire independent content.
</Remark>

## 4. Constitutive laws for forces

The three laws alone do not determine any motion. Only when we specify what function of position, velocity and time $\boldsymbol{F}$ is does the equation of motion become a differential equation to be solved. Such a specification comes from experiment and is called a **constitutive law**. Here are the standard ones.

**Universal gravitation.** Two point masses of masses $M$ and $m$ separated by a distance $r$ attract each other with a force of magnitude $GMm/r^2$, where $G = 6.674 \times 10^{-11}\ \mathrm{N\,m^2/kg^2}$. Near the surface of the Earth one may take $r \approx R_\oplus$ (the Earth's radius), so the magnitude of the force is $mg$ with $g = GM_\oplus/R_\oplus^2 \approx 9.8\ \mathrm{m/s^2}$ very nearly constant.

**Spring force (Hooke's law).** For an extension $x$ from the natural length, $F = -kx$, with a spring constant $k > 0$. This linearity is not a fundamental law but an approximation. Indeed, if a general potential $U(x)$ has a minimum at $x = x_0$, then Taylor's theorem (<Ref to="mathematics/calculus/mean-value-and-taylor#thm-taylor" />, [The Mean Value Theorem and Taylor's Theorem](/en/mathematics/calculus/mean-value-and-taylor)) gives
$$
U(x) = U(x_0) + \tfrac{1}{2}U''(x_0)(x-x_0)^2 + O\!\left((x-x_0)^3\right)
$$
(the first-order term vanishes because $U'(x_0)=0$). Hence the force is $F = -U'(x) \approx -U''(x_0)(x-x_0)$, and setting $k = U''(x_0)$ recovers Hooke's law. **Near a minimum, every system is approximately a spring.** This is why simple harmonic motion appears in every field.

**Friction.** For sliding friction between solids one commonly uses Coulomb's approximation: a force of magnitude $\mu N$ independent of the speed, directed opposite to the motion, where $N$ is the normal force and $\mu$ the coefficient of kinetic friction. Drag in a fluid is $\boldsymbol{F} = -\gamma\boldsymbol{v}$ (viscous drag) at low speeds, and has magnitude proportional to $v^2$ at high speeds.

**Constraint forces.** The tension in a string or the normal force from a floor is not given in advance; its value is determined by the requirement that a condition be met — that the string not stretch, that the body not penetrate the surface. Eliminating constraint forces from the treatment is one of the chief motivations for [Lagrangian Mechanics](/en/physics/mechanics/lagrangian-mechanics).

<Proposition id="prop-uniqueness" title="Uniqueness for the initial value problem">
Let $m > 0$ and let $\boldsymbol{F} : \mathbb{R} \times \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}^n$, $(t, \boldsymbol{r}, \boldsymbol{v}) \mapsto \boldsymbol{F}(t,\boldsymbol{r},\boldsymbol{v})$, be continuous and moreover Lipschitz continuous in $(\boldsymbol{r},\boldsymbol{v})$ uniformly in $t$: there exists $L > 0$ such that for all $t$ and all $(\boldsymbol{r}_1,\boldsymbol{v}_1), (\boldsymbol{r}_2,\boldsymbol{v}_2)$,
$$
|\boldsymbol{F}(t,\boldsymbol{r}_1,\boldsymbol{v}_1) - \boldsymbol{F}(t,\boldsymbol{r}_2,\boldsymbol{v}_2)| \le L\left(|\boldsymbol{r}_1-\boldsymbol{r}_2| + |\boldsymbol{v}_1-\boldsymbol{v}_2|\right).
$$
Then on an interval $J \ni t_0$ there is at most one $C^2$ solution satisfying
$$
m\ddot{\boldsymbol{r}}(t) = \boldsymbol{F}(t, \boldsymbol{r}(t), \dot{\boldsymbol{r}}(t)), \qquad
\boldsymbol{r}(t_0) = \boldsymbol{r}_0,\quad \dot{\boldsymbol{r}}(t_0) = \boldsymbol{v}_0 .
$$
</Proposition>

<Remark id="rem-determinism" title="When uniqueness fails">
The proof is placed in the Appendix. Drop the Lipschitz condition and uniqueness genuinely breaks down. Norton pointed out that an equation of the form $m\ddot{s} = \sqrt{s}$ (with $s \ge 0$) admits, for the initial condition $s(0)=\dot{s}(0)=0$, the solution $s(t) = t^4/144$ (when $m=1$) besides $s \equiv 0$, and he presented this as a mechanical system by shaping it into a "dome". The right-hand side $\sqrt{s}$ is not Lipschitz at $s=0$, so the hypothesis of <Ref to="prop-uniqueness" /> fails. Determinism in classical mechanics does not come from the laws themselves but from an additional assumption: the smoothness of the force.
</Remark>

## 5. Solving one-dimensional motion

From here we consider motion $x(t)$ along a line. The equation of motion is $m\ddot{x} = F$.

### 5.1. Uniformly accelerated motion

When $F$ is a constant $F_0$, the acceleration $a = F_0/m$ is constant too. Integrating $\dot{v} = a$ from $t_0=0$ (by the fundamental theorem of calculus, <Ref to="mathematics/calculus/integration-and-ftc#thm-ftc2" />, [The Fundamental Theorem of Calculus and Definite Integrals](/mathematics/calculus/integration-and-ftc)),

$$
v(t) = v_0 + at, \qquad x(t) = x_0 + v_0 t + \tfrac{1}{2}at^2 .
$$

Substituting $t = (v - v_0)/a$ from the first equation (for $a \ne 0$) into the second,

$$
x - x_0 = v_0\cdot\frac{v-v_0}{a} + \frac{a}{2}\cdot\frac{(v-v_0)^2}{a^2}
= \frac{2v_0(v-v_0) + (v-v_0)^2}{2a} = \frac{v^2 - v_0^2}{2a}
$$

which gives the time-free relation $v^2 - v_0^2 = 2a(x-x_0)$. This is a special case of the energy conservation law proved below (<Ref to="thm-energy-conservation" />). Indeed, multiplying both sides by $m/2$ gives $\frac{1}{2}mv^2 - \frac{1}{2}mv_0^2 = F_0 (x - x_0)$, which is exactly the statement that the change in kinetic energy equals the work done.

<Example id="ex-linear-drag" title="Falling with viscous drag">
Take the downward vertical direction as positive and suppose a body of mass $m$ falls from $v(0)=0$ under gravity $mg$ and viscous drag $-\gamma v$ with $\gamma > 0$. The equation of motion is
$$
m\dot{v} = mg - \gamma v .
$$
Setting $\tau = m/\gamma$ and $v_\infty = mg/\gamma$ gives $\dot{v} = -(v - v_\infty)/\tau$. Putting $w = v - v_\infty$ we get $\dot{w} = -w/\tau$, hence $\frac{d}{dt}\left(w e^{t/\tau}\right) = e^{t/\tau}(\dot{w} + w/\tau) = 0$. Therefore $w(t) = w(0)e^{-t/\tau} = -v_\infty e^{-t/\tau}$, and
$$
v(t) = v_\infty\left(1 - e^{-t/\tau}\right), \qquad
x(t) = v_\infty\left[t - \tau\left(1 - e^{-t/\tau}\right)\right]
$$
(integrating $v$ with $x(0)=0$). As $t \to \infty$ we have $v \to v_\infty$, the **terminal velocity**.

Let us check that we recover free fall in the limit of negligible drag. For small $t/\tau$ we have $e^{-t/\tau} = 1 - \frac{t}{\tau} + \frac{t^2}{2\tau^2} - \cdots$, so
$$
v(t) = v_\infty\left(\frac{t}{\tau} - \frac{t^2}{2\tau^2} + \cdots\right)
= gt\left(1 - \frac{t}{2\tau} + \cdots\right)
$$
(using $v_\infty/\tau = g$). The first term is $gt$, the free-fall velocity. The relative size of the correction is $t/(2\tau)$, so we may say quantitatively that drag can be ignored as long as $t \ll \tau$.
</Example>

### 5.2. Simple harmonic motion

<Theorem id="thm-shm" title="General solution of the harmonic oscillator">
Let $m > 0$, $k > 0$ and set $\omega = \sqrt{k/m}$. If $x : \mathbb{R} \to \mathbb{R}$ is $C^2$ and satisfies
$$
m\ddot{x}(t) = -k\,x(t) \qquad (t \in \mathbb{R}),
$$
then
$$
x(t) = x(0)\cos\omega t + \frac{\dot{x}(0)}{\omega}\sin\omega t
$$
holds for every $t$. Conversely, for any real numbers $A, B$ the function $x(t) = A\cos\omega t + B\sin\omega t$ is a solution of the equation.
</Theorem>

<Proof of="thm-shm">
We first check the converse direction. With $x(t) = A\cos\omega t + B\sin\omega t$ we have
$\dot{x}(t) = -A\omega\sin\omega t + B\omega\cos\omega t$ and
$\ddot{x}(t) = -A\omega^2\cos\omega t - B\omega^2\sin\omega t = -\omega^2 x(t)$,
so $m\ddot{x} = -m\omega^2 x = -kx$ and the equation is satisfied.

Now for uniqueness. Let $x$ be a solution, put $A = x(0)$, $B = \dot{x}(0)/\omega$, and define
$$
u(t) = x(t) - \left(A\cos\omega t + B\sin\omega t\right).
$$
Both $x$ and the bracketed function solve the equation, and the equation is linear in $u$, so $u$ also satisfies $\ddot{u} = -\omega^2 u$. Moreover $u(0) = x(0) - A = 0$ and $\dot{u}(0) = \dot{x}(0) - B\omega = 0$.

Now set
$$
E(t) = \tfrac{1}{2}\dot{u}(t)^2 + \tfrac{1}{2}\omega^2 u(t)^2 .
$$
Since $u$ is $C^2$, $E$ is differentiable, and the product rule gives
$$
\dot{E} = \dot{u}\ddot{u} + \omega^2 u\dot{u} = \dot{u}\left(-\omega^2 u\right) + \omega^2 u \dot{u} = 0 .
$$
Hence $E$ is constant, and $E(0) = \frac{1}{2}\cdot 0 + \frac{1}{2}\omega^2\cdot 0 = 0$, so $E \equiv 0$. As $E$ is a sum of two non-negative terms, both vanish; in particular $\frac{1}{2}\omega^2 u(t)^2 = 0$. Since $\omega > 0$, we get $u(t) = 0$ for every $t$.
</Proof>

This proof anticipates the energy conservation law derived later (<Ref to="thm-energy-conservation" />) and uses it as a tool for uniqueness. In the language of linear algebra, we have shown that the solution space is a two-dimensional vector space with basis $\{\cos\omega t, \sin\omega t\}$.

Rewriting $A\cos\omega t + B\sin\omega t = C\cos(\omega t - \varphi)$ with $C = \sqrt{A^2+B^2}$ and $\tan\varphi = B/A$, the quantity $C$ is the **amplitude**, $\omega$ the **angular frequency** and $\varphi$ the **phase**. The period is $T = 2\pi/\omega = 2\pi\sqrt{m/k}$, and note that it is **independent of the amplitude** (isochronism).

<Example id="ex-spring" title="A spring pendulum in numbers">
A body attached to a spring with $m = 0.50\ \mathrm{kg}$ and $k = 200\ \mathrm{N/m}$ is pulled $0.030\ \mathrm{m}$ from the equilibrium position and released from rest.

The angular frequency is $\omega = \sqrt{k/m} = \sqrt{200/0.50} = \sqrt{400} = 20\ \mathrm{rad/s}$, the period is $T = 2\pi/20 \approx 0.314\ \mathrm{s}$, and the frequency is $1/T \approx 3.18\ \mathrm{Hz}$.

The initial conditions are $x(0) = 0.030\ \mathrm{m}$ and $\dot{x}(0) = 0$, so by <Ref to="thm-shm" /> we have $x(t) = 0.030\cos(20t)\ \mathrm{m}$. Hence the maximum speed is $\omega C = 20 \times 0.030 = 0.60\ \mathrm{m/s}$ and the maximum magnitude of the acceleration is $\omega^2 C = 400 \times 0.030 = 12\ \mathrm{m/s^2}$.

The total energy is $\frac{1}{2}kC^2 = \frac{1}{2}\times 200 \times (0.030)^2 = 0.090\ \mathrm{J}$. All of it should become kinetic energy at the instant the body passes through the centre, and indeed $\frac{1}{2}m v_{\max}^2 = \frac{1}{2}\times 0.50 \times (0.60)^2 = 0.090\ \mathrm{J}$ agrees.
</Example>

## 6. Conservation of momentum

We now consider a system of $N$ point masses. We split the force on point mass $i$ (mass $m_i$, position $\boldsymbol{r}_i$) into the **external force** $\boldsymbol{F}_i^{\mathrm{ext}}$ acting from outside the system and the **internal forces** $\boldsymbol{F}_{ij}$ exerted by the point masses $j$ within the system. The second law (<Ref to="ax-second-law" />) reads

$$
m_i \ddot{\boldsymbol{r}}_i = \boldsymbol{F}_i^{\mathrm{ext}} + \sum_{j \ne i} \boldsymbol{F}_{ij} \qquad (i = 1,\dots,N).
$$

<Theorem id="thm-momentum-conservation" title="Conservation of momentum">
In the setting above, suppose the internal forces satisfy the weak form of the third law (<Ref to="ax-third-law" />), $\boldsymbol{F}_{ij} = -\boldsymbol{F}_{ji}$. Define the total momentum of the system by $\boldsymbol{P} = \sum_{i=1}^{N} m_i\dot{\boldsymbol{r}}_i$. Then
$$
\frac{d\boldsymbol{P}}{dt} = \sum_{i=1}^{N}\boldsymbol{F}_i^{\mathrm{ext}}
$$
holds. In particular, if the sum of the external forces vanishes identically, then $\boldsymbol{P}$ is constant in time.
</Theorem>

<Proof of="thm-momentum-conservation">
Sum the equations of motion over $i$:
$$
\frac{d\boldsymbol{P}}{dt} = \sum_{i} m_i\ddot{\boldsymbol{r}}_i
= \sum_i \boldsymbol{F}_i^{\mathrm{ext}} + \sum_{i}\sum_{j \ne i}\boldsymbol{F}_{ij} .
$$
The double sum runs over all ordered pairs $(i,j)$ with $i \ne j$. Grouping it by unordered pairs $\{i,j\}$, each unordered pair contributes exactly two terms, $\boldsymbol{F}_{ij}$ and $\boldsymbol{F}_{ji}$. By the weak form of the third law their sum is $\boldsymbol{F}_{ij} + \boldsymbol{F}_{ji} = \boldsymbol{0}$. Hence the entire double sum vanishes, which gives the first identity.

If the sum of the external forces is $\boldsymbol{0}$, then $d\boldsymbol{P}/dt = \boldsymbol{0}$, so each component is a constant function and $\boldsymbol{P}$ is constant.
</Proof>

<Corollary id="cor-center-of-mass" title="Motion of the centre of mass">
Let $M = \sum_i m_i$ be the total mass and $\boldsymbol{R} = \frac{1}{M}\sum_i m_i \boldsymbol{r}_i$ the centre of mass. Under the hypotheses of <Ref to="thm-momentum-conservation" />,
$$
M\ddot{\boldsymbol{R}} = \sum_{i}\boldsymbol{F}_i^{\mathrm{ext}} .
$$
That is, the centre of mass moves exactly as a single point mass carrying the whole mass and subject to the sum of the external forces.
</Corollary>

<Proof of="cor-center-of-mass">
Differentiating $M\boldsymbol{R} = \sum_i m_i\boldsymbol{r}_i$ once with respect to $t$ gives $M\dot{\boldsymbol{R}} = \sum_i m_i\dot{\boldsymbol{r}}_i = \boldsymbol{P}$ (both $m_i$ and $M$ are constants, so linearity of differentiation applies directly). Differentiating once more gives $M\ddot{\boldsymbol{R}} = d\boldsymbol{P}/dt$, and applying <Ref to="thm-momentum-conservation" /> yields the conclusion.
</Proof>

Thanks to this corollary one can say at once that when a firework bursts, the centre of mass of all the fragments continues along the original parabola. However complicated the internal forces of the explosion, they have no effect whatsoever on the motion of the centre of mass.

<Example id="ex-collision" title="Perfectly inelastic collision">
On a smooth horizontal surface a cart of mass $m_1 = 2.0\ \mathrm{kg}$ travelling at $v_1 = 3.0\ \mathrm{m/s}$ collides with a stationary cart of mass $m_2 = 4.0\ \mathrm{kg}$ and the two stick together. No external force acts horizontally, so by <Ref to="thm-momentum-conservation" /> the momentum is conserved across the collision:
$$
m_1 v_1 + m_2\cdot 0 = (m_1+m_2)v' \implies v' = \frac{2.0\times 3.0}{6.0} = 1.0\ \mathrm{m/s}.
$$
What about the kinetic energy? Before the collision it is $\frac{1}{2}\times 2.0\times 3.0^2 = 9.0\ \mathrm{J}$ and afterwards $\frac{1}{2}\times 6.0 \times 1.0^2 = 3.0\ \mathrm{J}$, so $6.0\ \mathrm{J}$ has been lost. The lost part went into deformation and heat.

**Momentum is conserved but kinetic energy is not.** This asymmetry matters. Momentum conservation follows from the third law alone and is therefore indifferent to the details of the internal forces, whereas conservation of mechanical energy requires, as the next section shows, the additional condition that the force be conservative.
</Example>

## 7. Work and energy

<Definition id="def-work" title="Work and kinetic energy">
When a point mass moves under a force $\boldsymbol{F}$ along a $C^1$ path $\boldsymbol{r}:[t_1,t_2] \to \mathbb{R}^3$, we define the **work** done by the force during this interval to be
$$
W = \int_{t_1}^{t_2} \boldsymbol{F}(t)\cdot\dot{\boldsymbol{r}}(t)\,dt ,
$$
and the **kinetic energy** of the point mass to be $K = \frac{1}{2}m|\boldsymbol{v}|^2$.
</Definition>

<Theorem id="thm-work-energy" title="Work-energy theorem">
Let a point mass of constant mass $m$ satisfy the equation of motion $m\ddot{\boldsymbol{r}} = \boldsymbol{F}$ in an inertial frame, with $\boldsymbol{r}$ of class $C^2$ on $[t_1,t_2]$ and $\boldsymbol{F}$ continuous. Then
$$
K(t_2) - K(t_1) = \int_{t_1}^{t_2}\boldsymbol{F}\cdot\dot{\boldsymbol{r}}\,dt .
$$
That is, the change in kinetic energy equals the work done by the force during that interval.
</Theorem>

<Proof of="thm-work-energy">
Differentiate $K(t) = \frac{1}{2}m\,\boldsymbol{v}(t)\cdot\boldsymbol{v}(t)$. The inner product is a sum of products of components, so applying the product rule componentwise gives
$$
\frac{dK}{dt} = \frac{1}{2}m\left(\dot{\boldsymbol{v}}\cdot\boldsymbol{v} + \boldsymbol{v}\cdot\dot{\boldsymbol{v}}\right) = m\,\boldsymbol{a}\cdot\boldsymbol{v}
$$
(using symmetry of the inner product). Substituting the equation of motion $m\boldsymbol{a} = \boldsymbol{F}$ (<Ref to="ax-second-law" />) gives $\dfrac{dK}{dt} = \boldsymbol{F}\cdot\boldsymbol{v}$.

Since $\boldsymbol{r}$ is $C^2$ and $\boldsymbol{F}$ is continuous, the right-hand side is continuous on $[t_1,t_2]$ and $K$ is $C^1$. By the fundamental theorem of calculus,
$$
K(t_2)-K(t_1) = \int_{t_1}^{t_2}\frac{dK}{dt}\,dt = \int_{t_1}^{t_2}\boldsymbol{F}\cdot\dot{\boldsymbol{r}}\,dt .
$$
</Proof>

<Definition id="def-conservative" title="Conservative forces and potentials">
A force field $\boldsymbol{F}: D \to \mathbb{R}^3$ defined on a region $D \subset \mathbb{R}^3$ is called **conservative** if there is a $C^1$ function $U : D \to \mathbb{R}$ with
$$
\boldsymbol{F}(\boldsymbol{r}) = -\nabla U(\boldsymbol{r}) \qquad (\boldsymbol{r} \in D).
$$
This $U$ is called the **potential energy**. It is unique up to an additive constant (when $D$ is connected).
</Definition>

A conservative force is determined by position alone and involves neither velocity nor time. Friction depends on the direction of motion — it is a function of velocity — and so falls outside this definition. For multivariable differentiation and the gradient $\nabla$, see <Ref to="mathematics/calculus/multivariable-differentiation#def-differentiable" text="the definition of total differentiability" /> ([Differentiation of Functions of Several Variables](/mathematics/calculus/multivariable-differentiation)).

<Proposition id="prop-1d-conservative" title="Every continuous force in one dimension is conservative">
Let $I \subset \mathbb{R}$ be an interval, let $F : I \to \mathbb{R}$ be continuous, fix $x_0 \in I$ and set
$$
U(x) = -\int_{x_0}^{x} F(\xi)\,d\xi .
$$
Then $U$ is $C^1$ with $U'(x) = -F(x)$; that is, $F$ is conservative.
</Proposition>

<Proof of="prop-1d-conservative">
Since $F$ is continuous on $I$, the fundamental theorem of calculus says that $x \mapsto \int_{x_0}^{x}F(\xi)d\xi$ is differentiable with derivative equal to $F(x)$, and this derivative is continuous because $F$ is. Multiplying by $-1$ gives $U'(x) = -F(x)$ with $U'$ continuous, i.e. $U$ is $C^1$.
</Proof>

In one dimension, every force that is a continuous function of position is conservative. Energy conservation can fail only when the force depends on velocity (friction, drag) or depends explicitly on time (the system is being shaken from outside). In three dimensions the situation changes: a function of position alone need not be conservative — one needs $\nabla \times \boldsymbol{F} = \boldsymbol{0}$.

<Theorem id="thm-energy-conservation" title="Conservation of mechanical energy">
Let $m > 0$ be constant, let $U$ be a $C^1$ function on a region $D$, and suppose a point mass traces a $C^2$ trajectory inside $D$ satisfying the equation of motion
$$
m\ddot{\boldsymbol{r}}(t) = -\nabla U(\boldsymbol{r}(t)).
$$
Then the **mechanical energy**
$$
E(t) = \tfrac{1}{2}m|\dot{\boldsymbol{r}}(t)|^2 + U(\boldsymbol{r}(t))
$$
is constant in time.
</Theorem>

<Proof of="thm-energy-conservation">
As shown in the proof of <Ref to="thm-work-energy" />, $\dfrac{dK}{dt} = \boldsymbol{F}\cdot\dot{\boldsymbol{r}}$. On the other hand, the chain rule (<Ref to="mathematics/calculus/multivariable-differentiation#thm-chain-rule" />) gives
$$
\frac{d}{dt}U(\boldsymbol{r}(t)) = \nabla U(\boldsymbol{r}(t))\cdot\dot{\boldsymbol{r}}(t).
$$
Using the hypothesis $\boldsymbol{F} = -\nabla U$,
$$
\frac{dE}{dt} = \frac{dK}{dt} + \frac{d}{dt}U(\boldsymbol{r}(t))
= \left(-\nabla U\right)\cdot\dot{\boldsymbol{r}} + \nabla U\cdot\dot{\boldsymbol{r}} = 0 .
$$
Since $E$ is differentiable on an interval with identically vanishing derivative, the mean value theorem shows it is a constant function.
</Proof>

<Corollary id="cor-quadrature" title="Quadrature for one-dimensional conservative systems">
Consider one-dimensional motion governed by $m\ddot{x} = -U'(x)$ with $U$ of class $C^1$, and set $E = \frac{1}{2}m\dot{x}^2 + U(x)$. On an interval where $E - U(x) > 0$ and $\dot{x} > 0$, we have
$$
\dot{x} = \sqrt{\frac{2\left(E - U(x)\right)}{m}}, \qquad
t - t_1 = \int_{x(t_1)}^{x(t)}\sqrt{\frac{m}{2\left(E-U(\xi)\right)}}\,d\xi .
$$
That is, the motion is determined by a single integration (a quadrature).
</Corollary>

<Proof of="cor-quadrature">
By <Ref to="thm-energy-conservation" /> we have $\frac{1}{2}m\dot{x}^2 = E - U(x)$. Dividing both sides by $m/2$ and taking the square root, choosing the positive sign because $\dot{x} > 0$ by hypothesis, gives the first identity. Next, separate variables in the first identity as $\dfrac{dx}{\sqrt{2(E-U(x))/m}} = dt$ and integrate from $t_1$ to $t$. The left-hand side becomes a substitution integral in $x$, and since $E - U > 0$ the integrand is continuous, so the second identity follows.
</Proof>

### 7.1. Reading the motion off the potential diagram

Because $E - U(x) = \frac{1}{2}m\dot{x}^2 \ge 0$, the point mass can only enter the region where $U(x) \le E$. At points where equality holds the velocity vanishes and the motion turns around; such points are called **turning points**. Simply drawing the horizontal line $U = E$ on the graph of the potential tells us whether the motion is bounded or unbounded and where it turns around.

<Figure caption="Potential curve and energy level. The gap between them is the kinetic energy; where they meet is a turning point.">
<svg viewBox="0 0 500 300" width="100%" role="img" aria-label="Diagram of a potential curve, an energy level and the two turning points">
  <line x1="45" y1="30" x2="45" y2="275" stroke="currentColor" stroke-width="1.5" />
  <line x1="45" y1="160" x2="480" y2="160" stroke="currentColor" stroke-width="1.5" />
  <text x="52" y="26" fill="currentColor" font-size="13">U</text>
  <text x="470" y="152" fill="currentColor" font-size="13">x</text>
  <polyline
    points="60,39 74,120 94,186 113,221 132,236 152,240 180,235 209,224 248,208 296,192 344,180 402,171 460,166"
    fill="none" stroke="var(--sl-color-accent)" stroke-width="2.5" stroke-linejoin="round" />
  <line x1="70" y1="200" x2="430" y2="200" stroke="currentColor" stroke-width="1.5" stroke-dasharray="6 4" />
  <text x="36" y="205" fill="currentColor" font-size="13" text-anchor="end">E</text>
  <circle cx="100" cy="200" r="4" fill="currentColor" />
  <circle cx="270" cy="200" r="4" fill="currentColor" />
  <text x="96" y="222" fill="currentColor" font-size="13" text-anchor="middle">x₁</text>
  <text x="274" y="222" fill="currentColor" font-size="13" text-anchor="middle">x₂</text>
  <line x1="185" y1="200" x2="185" y2="236" stroke="currentColor" stroke-width="1.5" />
  <text x="192" y="222" fill="currentColor" font-size="12">K</text>
  <text x="330" y="146" fill="currentColor" font-size="12">U → 0 (level at which binding ends)</text>
</svg>
</Figure>

At the energy $E$ shown in the figure, the point mass oscillates between $x_1$ and $x_2$ (bounded motion). Raise $E$ until the dashed line passes above the horizontal asymptote $U \to 0$ and the right-hand turning point disappears: the point mass escapes to infinity (unbounded motion). The energy at this boundary is the condition for the binding to break; for a celestial body it gives the escape velocity.

<Example id="ex-escape" title="Escape velocity from the Earth">
The potential of a body of mass $m$ at distance $r$ from the centre of the Earth (mass $M$, radius $R$) is $U(r) = -GMm/r$, with the zero taken at infinity. If it is launched vertically upward from the surface with speed $v$, then, neglecting air resistance and the Earth's rotation, <Ref to="thm-energy-conservation" /> tells us that
$$
E = \frac{1}{2}mv^2 - \frac{GMm}{R}
$$
is conserved. Reaching infinity (where $U \to 0$ as $r \to \infty$, with $\frac{1}{2}m\dot{r}^2 \ge 0$) requires and is guaranteed by $E \ge 0$, that is
$$
v \ge \sqrt{\frac{2GM}{R}} .
$$
Putting in numbers: $GM = 3.986\times 10^{14}\ \mathrm{m^3/s^2}$ and $R = 6.371\times 10^{6}\ \mathrm{m}$, so
$$
\frac{2GM}{R} = \frac{2 \times 3.986\times 10^{14}}{6.371\times 10^{6}} = 1.251\times 10^{8}\ \mathrm{m^2/s^2},
\qquad v \ge 1.119\times 10^{4}\ \mathrm{m/s} \approx 11.2\ \mathrm{km/s}.
$$
Note that the mass $m$ drops out. A ball and a rocket need the same speed.
</Example>

<Example id="ex-friction" title="With friction: energy is not conserved">
A body of mass $m = 2.0\ \mathrm{kg}$ slides along a horizontal surface with initial speed $v_0 = 6.0\ \mathrm{m/s}$. With a coefficient of kinetic friction $\mu = 0.30$ and $g = 9.8\ \mathrm{m/s^2}$, the normal force is $N = mg$ and the friction force has magnitude $\mu mg$, directed opposite to the motion. The equation of motion is $m\dot{v} = -\mu mg$, so the acceleration is the constant $-\mu g = -2.94\ \mathrm{m/s^2}$.

Substituting $v = 0$ into the relation $v^2 - v_0^2 = 2a(x-x_0)$ for uniformly accelerated motion, the stopping distance is
$$
d = \frac{v_0^2}{2\mu g} = \frac{36}{2 \times 0.30 \times 9.8} = \frac{36}{5.88} \approx 6.1\ \mathrm{m}.
$$
Let us balance the energy books. The initial kinetic energy is $\frac{1}{2}\times 2.0 \times 6.0^2 = 36\ \mathrm{J}$, and the work done by friction is $-\mu mg\,d = -0.30\times 2.0\times 9.8\times 6.1 \approx -36\ \mathrm{J}$; the two match, as <Ref to="thm-work-energy" /> requires. But friction depends on the direction of the velocity, so it is not a conservative force, and there is no potential storing those $36\ \mathrm{J}$. The mechanical energy turns into heat and leaves the framework of mechanics.
</Example>

<Figure caption="How the conservation laws follow from the three laws">
<Mermaid code={`flowchart TD
  A["Second law: m a = F"] --> B["Equation of motion for one point mass"]
  C["Third law: action and reaction (weak form)"] --> D["Internal forces cancel"]
  B --> D
  D --> E["Momentum conservation (external forces sum to 0)"]
  B --> F["Work-energy theorem"]
  G["Conservative force: F = -grad U"] --> H["Mechanical energy conservation"]
  F --> H
  H --> I["Solvable by quadrature in one dimension"]`} />
</Figure>

The diagram shows that the conservation laws are consequences of the equation of motion. There is, however, a route in the opposite direction, deriving the conservation laws from a more fundamental principle: homogeneity of space yields conservation of momentum and homogeneity of time yields conservation of energy. That is Noether's theorem ([Symmetries and Conservation Laws (Noether's Theorem)](/physics/mechanics/noethers-theorem)). Its formulation requires [Lagrangian Mechanics](/en/physics/mechanics/lagrangian-mechanics).

## 8. Exercises

<Exercise id="exr-vertical-throw" difficulty="Easy">
A ball is thrown vertically upward from the ground with initial speed $v_0 = 20\ \mathrm{m/s}$. Neglecting air resistance and taking $g = 9.8\ \mathrm{m/s^2}$, find the height $h$ of the highest point and the time $t_1$ needed to reach it. Then obtain the same $h$ from energy conservation.
<Solution>
Take upward as positive. The only force is gravity, so $m\ddot{y} = -mg$ and the acceleration is the constant $a = -g$. From the formulas for uniformly accelerated motion, $v(t) = v_0 - gt$. At the highest point $v = 0$, so
$$
t_1 = \frac{v_0}{g} = \frac{20}{9.8} \approx 2.04\ \mathrm{s}.
$$
Substituting into $y(t) = v_0 t - \frac{1}{2}gt^2$ gives the height
$$
h = \frac{v_0^2}{g} - \frac{1}{2}\cdot\frac{v_0^2}{g} = \frac{v_0^2}{2g} = \frac{400}{19.6} \approx 20.4\ \mathrm{m}.
$$

Energy gives the same result. Gravity $-mg$ is a function of position alone, so by <Ref to="prop-1d-conservative" /> it is conservative, with $U(y) = mgy$ (indeed $U'(y) = mg = -(-mg)$). By <Ref to="thm-energy-conservation" />,
$$
\tfrac{1}{2}mv_0^2 + 0 = \tfrac{1}{2}m\cdot 0^2 + mgh \implies h = \frac{v_0^2}{2g},
$$
which is $20.4\ \mathrm{m}$, independent of the mass.
</Solution>
</Exercise>

<Exercise id="exr-amplitude" difficulty="Standard">
For motion governed by $m\ddot{x} = -kx$ with initial conditions $x(0) = x_0$ and $\dot{x}(0) = v_0$, express the amplitude $C$ (the maximum value attained by $x(t)$) in terms of $x_0$, $v_0$ and $\omega = \sqrt{k/m}$. Then verify that the result is consistent with energy conservation.
<Solution>
By <Ref to="thm-shm" />, $x(t) = x_0\cos\omega t + \dfrac{v_0}{\omega}\sin\omega t$. Put $A = x_0$, $B = v_0/\omega$, $C = \sqrt{A^2+B^2}$, and choose $\varphi$ with $\cos\varphi = A/C$ and $\sin\varphi = B/C$ (such a $\varphi$ exists because for $C > 0$ the pair $(A/C, B/C)$ lies on the unit circle). The addition formula then gives
$$
x(t) = C\left(\cos\varphi\cos\omega t + \sin\varphi\sin\omega t\right) = C\cos(\omega t - \varphi).
$$
The range of $\cos$ is $[-1,1]$ and there is a $t$ with $\omega t - \varphi = 0$, so the maximum is exactly $C$. Hence
$$
C = \sqrt{x_0^2 + \frac{v_0^2}{\omega^2}} .
$$

Now the consistency with energy. By <Ref to="thm-energy-conservation" />, $E = \frac{1}{2}mv_0^2 + \frac{1}{2}kx_0^2$ is constant. On the other hand, at the instants when $x = \pm C$ we have $\dot{x} = 0$, so $E = \frac{1}{2}kC^2$. Equating the two,
$$
\tfrac{1}{2}kC^2 = \tfrac{1}{2}mv_0^2 + \tfrac{1}{2}kx_0^2
\implies C^2 = x_0^2 + \frac{m}{k}v_0^2 = x_0^2 + \frac{v_0^2}{\omega^2},
$$
in agreement with the previous result.
</Solution>
</Exercise>

<Exercise id="exr-inelastic" difficulty="Standard">
Two point masses of masses $m_1, m_2$ move along a smooth line with velocities $v_1, v_2$, collide and stick together. No external force acts. Find the velocity $v'$ after the collision and show that the kinetic energy lost is
$$
\Delta K = -\frac{1}{2}\mu\left(v_1 - v_2\right)^2, \qquad \mu = \frac{m_1m_2}{m_1+m_2}
$$
(the quantity $\mu$ is called the reduced mass).
<Solution>
The sum of the external forces is $0$, so by <Ref to="thm-momentum-conservation" /> the momentum is conserved:
$$
m_1v_1 + m_2v_2 = (m_1+m_2)v' \implies v' = \frac{m_1v_1+m_2v_2}{m_1+m_2}.
$$
Put $M = m_1+m_2$. The change in kinetic energy is
$$
\Delta K = \tfrac{1}{2}Mv'^2 - \tfrac{1}{2}m_1v_1^2 - \tfrac{1}{2}m_2v_2^2
= \frac{(m_1v_1+m_2v_2)^2}{2M} - \frac{m_1v_1^2 + m_2v_2^2}{2}.
$$
Bringing the right-hand side over a common denominator and computing the numerator,
$$
\begin{aligned}
(m_1v_1+m_2v_2)^2 - M\left(m_1v_1^2+m_2v_2^2\right)
&= m_1^2v_1^2 + 2m_1m_2v_1v_2 + m_2^2v_2^2 \\
&\quad - \left(m_1^2v_1^2 + m_1m_2v_2^2 + m_1m_2v_1^2 + m_2^2v_2^2\right) \\
&= 2m_1m_2v_1v_2 - m_1m_2v_1^2 - m_1m_2v_2^2 \\
&= -m_1m_2\left(v_1-v_2\right)^2 .
\end{aligned}
$$
Therefore
$$
\Delta K = \frac{-m_1m_2(v_1-v_2)^2}{2M} = -\frac{1}{2}\mu(v_1-v_2)^2 \le 0 .
$$
The loss is determined by the relative velocity $v_1 - v_2$ alone, and there is no loss only when the relative velocity is $0$ (the two travelling side by side at the same velocity). For the numbers in <Ref to="ex-collision" />, $\mu = 8.0/6.0 \approx 1.33\ \mathrm{kg}$ and $\Delta K = -\frac{1}{2}\times 1.33 \times 3.0^2 = -6.0\ \mathrm{J}$, agreeing with the value computed directly.
</Solution>
</Exercise>

<Exercise id="exr-quadrature-period" difficulty="Hard">
Using the quadrature formula of <Ref to="cor-quadrature" />, compute the period of a point mass of mass $m$ moving with energy $E > 0$ in the potential $U(x) = \frac{1}{2}kx^2$, and verify that $T = 2\pi\sqrt{m/k}$, i.e. that the period does not depend on $E$.
<Solution>
Solving $U(x) = E$ gives the turning points $x = \pm A$ with $A = \sqrt{2E/k}$. The motion is a back-and-forth between $-A$ and $A$, and by symmetry the period is four times the time needed to move from $0$ to $A$. Indeed, $U$ is even, so the equation is invariant under $x \mapsto -x$, and it is also invariant under time reversal $t \mapsto -t$; hence the four stretches $0 \to A$, $A \to 0$, $0 \to -A$, $-A \to 0$ all take the same time.

By <Ref to="cor-quadrature" />, while $x$ moves from $0$ to $A$ (where $\dot{x} > 0$ and $E - U > 0$ for $x < A$),
$$
\frac{T}{4} = \int_{0}^{A}\sqrt{\frac{m}{2\left(E - \frac{1}{2}k\xi^2\right)}}\,d\xi
= \int_{0}^{A}\sqrt{\frac{m}{k}}\cdot\frac{d\xi}{\sqrt{\frac{2E}{k} - \xi^2}}
= \sqrt{\frac{m}{k}}\int_{0}^{A}\frac{d\xi}{\sqrt{A^2-\xi^2}} ,
$$
where we used $2(E - \frac{1}{2}k\xi^2) = k\left(\frac{2E}{k} - \xi^2\right) = k(A^2-\xi^2)$. Substituting $\xi = A\sin\theta$ gives $d\xi = A\cos\theta\,d\theta$ and $\sqrt{A^2-\xi^2} = A\cos\theta$ (since $\cos\theta \ge 0$ for $0 \le \theta \le \pi/2$), so
$$
\int_{0}^{A}\frac{d\xi}{\sqrt{A^2-\xi^2}} = \int_{0}^{\pi/2}d\theta = \frac{\pi}{2}
$$
(the integrand diverges as $\xi \to A$, but the integral after substitution is finite: it converges as an improper integral). Therefore
$$
\frac{T}{4} = \frac{\pi}{2}\sqrt{\frac{m}{k}}, \qquad T = 2\pi\sqrt{\frac{m}{k}} .
$$
The amplitude $A$ cancelled, so the period is independent of the energy. This is the isochronism of the harmonic oscillator. If $U$ is proportional to something other than $x^2$ — say $x^4$ — then $A$ survives and the period does depend on the amplitude.
</Solution>
</Exercise>

## References

- I. Newton, *Philosophiæ Naturalis Principia Mathematica*, 1687 — the opening "Axioms, or Laws of Motion". A Japanese translation is Saruhito Nakano (trans.), *Principia: Shizen Tetsugaku no Sūgakuteki Genri*, Kodansha (in Japanese).
- Kiyoshi Harashima, *Rikigaku*, Shokabo (in Japanese) — Chapters 1–3. A standard Japanese textbook, careful about the meaning of the three laws and about one-dimensional solution methods.
- L. D. Landau and E. M. Lifshitz, *Rikigaku (Mechanics, 3rd revised ed.)*, Tokyo Tosho (in Japanese) — Chapters 1 and 2. Starts from the principle of least action and derives the conservation laws from symmetries.
- D. Kleppner and R. Kolenkow, *An Introduction to Mechanics*, 2nd ed., Cambridge University Press, 2014 — Chapters 2–5 (Newton's laws, momentum, energy).
- V. I. Arnold, *Mathematical Methods of Classical Mechanics*, 2nd ed., Springer, 1989 — Chapter 1. Gives a mathematical formulation of Galilean transformations and of the principle of determinacy.
- J. D. Norton, "The Dome: An Unexpectedly Simple Failure of Determinism", *Philosophy of Science* 75 (2008) — an example in which determinism in classical mechanics fails once the Lipschitz condition is violated.

## Appendix: Proof of uniqueness

**Strategy.** We prove <Ref to="prop-uniqueness" />. We rewrite the second-order equation as a first-order system and apply Grönwall's inequality to the difference of two solutions.

Put $\boldsymbol{y} = (\boldsymbol{r}, \boldsymbol{v}) \in \mathbb{R}^{2n}$ and define
$$
\boldsymbol{G}(t, \boldsymbol{y}) = \left(\boldsymbol{v},\ \frac{1}{m}\boldsymbol{F}(t,\boldsymbol{r},\boldsymbol{v})\right).
$$
The original equation is then equivalent to $\dot{\boldsymbol{y}} = \boldsymbol{G}(t,\boldsymbol{y})$ with $\boldsymbol{y}(t_0) = (\boldsymbol{r}_0,\boldsymbol{v}_0)$. Indeed, that $\boldsymbol{y} = (\boldsymbol{r},\boldsymbol{v})$ satisfies this system means $\dot{\boldsymbol{r}} = \boldsymbol{v}$ and $\dot{\boldsymbol{v}} = \boldsymbol{F}/m$, which is the same as $m\ddot{\boldsymbol{r}} = \boldsymbol{F}$.

**Lipschitz continuity of $\boldsymbol{G}$.** For $\boldsymbol{y}_i = (\boldsymbol{r}_i,\boldsymbol{v}_i)$ with $i = 1,2$, estimating the norm by the sum of the components,
$$
\left|\boldsymbol{G}(t,\boldsymbol{y}_1) - \boldsymbol{G}(t,\boldsymbol{y}_2)\right|
\le |\boldsymbol{v}_1-\boldsymbol{v}_2| + \frac{1}{m}\left|\boldsymbol{F}(t,\boldsymbol{r}_1,\boldsymbol{v}_1)-\boldsymbol{F}(t,\boldsymbol{r}_2,\boldsymbol{v}_2)\right|
\le \left(1 + \frac{L}{m}\right)\left(|\boldsymbol{r}_1-\boldsymbol{r}_2| + |\boldsymbol{v}_1-\boldsymbol{v}_2|\right)
$$
(using the assumed Lipschitz condition). Writing the coefficient on the right as $\Lambda = 1 + L/m$ and measuring $|\boldsymbol{y}_1 - \boldsymbol{y}_2|$ by the componentwise sum norm $|\boldsymbol{r}_1-\boldsymbol{r}_2| + |\boldsymbol{v}_1-\boldsymbol{v}_2|$, we obtain $|\boldsymbol{G}(t,\boldsymbol{y}_1)-\boldsymbol{G}(t,\boldsymbol{y}_2)| \le \Lambda|\boldsymbol{y}_1-\boldsymbol{y}_2|$.

**Passing to the integral form.** Let $\boldsymbol{y}_1, \boldsymbol{y}_2$ be solutions on $J$ with the same initial value. Both are $C^1$, so the fundamental theorem of calculus gives, for $t \ge t_0$,
$$
\boldsymbol{y}_i(t) = \boldsymbol{y}_i(t_0) + \int_{t_0}^{t}\boldsymbol{G}(s,\boldsymbol{y}_i(s))\,ds .
$$
Since the initial values agree, taking the difference gives
$$
\varphi(t) := \left|\boldsymbol{y}_1(t)-\boldsymbol{y}_2(t)\right|
\le \int_{t_0}^{t}\left|\boldsymbol{G}(s,\boldsymbol{y}_1(s))-\boldsymbol{G}(s,\boldsymbol{y}_2(s))\right|ds
\le \Lambda\int_{t_0}^{t}\varphi(s)\,ds
$$
(using the triangle inequality for integrals and then the Lipschitz estimate).

**Grönwall's inequality.** Set $\psi(t) = \int_{t_0}^{t}\varphi(s)\,ds$. Since $\varphi$ is continuous, $\psi$ is $C^1$ with $\psi' = \varphi$, and the inequality above reads $\psi'(t) \le \Lambda\psi(t)$ with $\psi(t_0)=0$. Hence
$$
\frac{d}{dt}\left(e^{-\Lambda t}\psi(t)\right) = e^{-\Lambda t}\left(\psi'(t) - \Lambda\psi(t)\right) \le 0 ,
$$
so $e^{-\Lambda t}\psi(t)$ is non-increasing for $t \ge t_0$, whence $e^{-\Lambda t}\psi(t) \le e^{-\Lambda t_0}\psi(t_0) = 0$, i.e. $\psi(t) \le 0$. On the other hand $\varphi \ge 0$ gives $\psi \ge 0$, so $\psi \equiv 0$; then again $\varphi \le \Lambda\psi = 0$ together with $\varphi \ge 0$ yields $\varphi \equiv 0$. Therefore $\boldsymbol{y}_1 = \boldsymbol{y}_2$ for $t \ge t_0$.

**Backwards in time.** For $t \le t_0$, reverse time by setting $\tilde{\boldsymbol{y}}(t) = \boldsymbol{y}(2t_0 - t)$. Then $\tilde{\boldsymbol{y}}$ satisfies $\dot{\tilde{\boldsymbol{y}}}(t) = -\boldsymbol{G}(2t_0-t, \tilde{\boldsymbol{y}}(t))$, whose right-hand side is Lipschitz with the same constant $\Lambda$. Applying the same argument on the side $t \ge t_0$ and returning to the original variable gives agreement for $t \le t_0$. This establishes uniqueness on all of $J$.