Newtonian (classical) mechanics is primarily concerned with forces, motion, and the causes of changes in motion. In many contexts—especially macroscopic ones on Earth—it is extremely powerful, in part due to its simplicity. Within this framework, the typical goal is either to find the function that satisfies the second-order differential equation given by Newton’s second law, or to determine the forces required to produce a desired motion. To begin, I will describe Newton’s three laws, which serve as the axioms of Newtonian mechanics.
Newton's First Law
Also known as the Law of Inertia, this law simply states that an object moving at a constant velocity, will remain at the velocity until acted upon by a force.
Newton's Second Law
This law provides the fundamental differential equation of motion. It states that the net force acting on an object is equal to the product of its mass and acceleration: \[F = m a\] In other words, force causes acceleration, and the magnitude of that acceleration depends on the object's mass.
Newton's Third Law
Also known as the law of action and reaction, this law states that for every action, there is an equal and opposite reaction. In other words, forces always occur in pairs: if one object exerts a force on another, the second exerts an equal and opposite force in return. For example, if you push on a wall, the wall pushes back on you with the same magnitude of force.
What Next?
Now that we've set the rules of our playground, we can determine how things move, act on other things, etc. But first I would like to reinterpret Newton's second law as a differential equation. \[F=ma\] As mentioned, the goal is to typically find where an object will be in the future, or maybe the function describing where it will be at any time \(t\); you could even be given the way the object moves and then be asked to find the forces that cause that. In any case, we are concerned with a function \[x(t)\] That gives the position of the object at time \(t\). We know that the derivative of position is velocity. That is, velocity is how fast our position is changing. The rate at which our velocity changes (i.e. the derivative of velocity) is acceleration. Therefore, if we are given that \[F=ma\] We could equivalently write this as \[F=m\ddot{x}(t)\] Where the two dots above the variable \(x\) represent the second time derivative. This represents a second order ODE in time. We can then import methods of solving second order ODEs to this setting. Since \(\sin, \cos\) are eigenfunctions of the second order derivative operator, we expect to see them often.
That is pretty much all that is concerned with Newtonian Mechanics in terms of its principal axioms. However, there are nontrivial considerations for extra things like torque, angular velocity, etc. However, they can all be linked back to these principals, with just specific equations for force.
Also note that we have assumed mass to be constant, which is true for most everyday scenarios. However, in cases like rocket propulsion—where the mass decreases over time as fuel is burned—we must use Newton’s full second law: \[ F = \frac{dp}{dt} \] This expression reduces to the familiar \( F = ma \) form when mass is constant. In the equation above, \( p \) represents the momentum: \[ p = m v, \] where \( v \) is the velocity (i.e., \( \dot{x} \)). If the mass changes with time (that is, if \( m = m(t) \)), then applying the chain rule gives: \[ \frac{d}{dt}(m v) = \dot{m} v + m \dot{v}, \] which can be written as: \[ F = \dot{m} \dot{x} + m \ddot{x}. \] This more general form is used in problems such as rocket motion and other variable-mass systems. For our purposes, however, we will assume the mass remains constant.
Problems
I will go over 5 problems that cover the various different situations you may encounter in Newtonian mechanics. These should be representative of the problems covered in a first year course.
1. Free Fall / Constant Acceleration
A ball of mass \(m\) at rest (zero velocity) is dropped from a height \(h_0\) above the ground. The force due to gravity is a constant: \(-mg\) (this is because the acceleration due to gravity is simply \(-g\) so the force must account for mass). Find the height and velocity of the ball as a function of time.
Solution
So we know the force acting on our ball. Which is \(-mg\), then we know that \[-mg = m\ddot{x}(t)\] We then see that \[\ddot{x}(t) = -g\]
Here \(x(t)\) represents the height of the ball at time \(t\). The goal is to then find this function. Qualitatively, \(g\) is constant, therefore velocity is linear in time. In other words, we can directly integrate this twice. \[\int \ddot{x}(t) dt = \int -g\ dt = -g \int dt = -gt + C\] We know due to the Fundamental Theorem of Calculus that \[\int\ddot{x}(t) = \dot{x}(t)\] Therefore, we have the velocity as a function of time, \[\dot{x}(t) = -gt + C\] To make things clearer, we just let \(v(t) = \dot{x}(t)\), so \[v(t) = -gt + C\] At time \(t=0\), we said that the ball was at rest, i.e. \(v(0) = 0\). Therefore, \[v(0) = -g\cdot 0 + C = 0 \implies C = 0\] Now we have our final equation of velocity, \[v(t) = -gt\] Integrating this function, we find the height of the ball as a function of time. We can use the velocity equation with or without the constant, since the conditions will give us the same result. However, I will replace \(C\) with \(v_0\) where that represents our initial velocity, since we showed above that they this constant will equal the initial velocity. \[\int \dot{x}(t) \ dt = \int -gt + v_0 \ dt\] \[\implies x(t) = -\frac{1}{2}gt^2 + v_0 t + C\] Note that \(C\) is a constant and not related to the \(C\) used earlier. Finally, using our initial condition that the height of the ball at time \(t=0\) is \(h_0\), we have \[ x(0) = -\frac{1}{2}g\cdot(0)^2 + v_0 \cdot 0 + C = h_0\] So we see that \(C=h_0\). Therefore, our final equation of motion is \[x(t) = -\frac{1}{2}gt^2 + v_0t + h_0\] In our case \(v_0 = 0\), so this reduces to \[x(t) = -\frac{1}{2}gt^2 + h_0\]
Another question might be to determine how long this equation is physically valid for. For instance, if we let \(t\to \infty\) eventually, the value will become negative indicating the ball is below the ground, not physically viable. This is straight forward, we would just want to find when \(x(t) \geq 0\). \[-\frac{1}{2}gt^2 + h_0 \geq 0 \implies h_0 \geq \frac{1}{2}gt^2\] \[\implies 2h_0 \geq gt^2 \implies t \leq \sqrt{\frac{2h_0}{g}}\] So this equation is valid for \(0 \leq t \leq \sqrt{\frac{2h_0}{g}}\)
Another question might be to find the time at which the ball hits the ground (or reaches any arbitrary point for that matter). This is also straightforward, just set the equation for height equal to \(0\) (or whichever point we want to solve for) and solve for \(t\). \[x(t) = 0 \implies -\frac{1}{2}gt^2+h_0 = 0\] \[\implies gt^2 = 2h_0 \implies t = \sqrt{\frac{2h_0}{g}}\] We could have also deduced this from solving for the valid times.
2. Inclined Plane with Friction
A block of mass \(m\) slides down a plane inclined at angle \(\theta\) with coefficient of kinetic friction \(\mu_k\). The net force along the incline is \[ F_{\text{net}} = m g \sin\theta - \mu_k m g \cos\theta, \] giving the acceleration \[ a = g (\sin\theta - \mu_k \cos\theta). \] You can solve for the velocity as a function of time or the distance traveled along the incline.
3. Mass–Spring System (Simple Harmonic Motion)
A mass \(m\) is attached to a spring with spring constant \(k\). The restoring force is \[ F = - k x, \] leading to the differential equation \[ m \ddot{x} + k x = 0 \quad \implies \quad \ddot{x} + \frac{k}{m} x = 0. \] The general solution is \[ x(t) = A \cos(\omega t) + B \sin(\omega t), \quad \omega = \sqrt{\frac{k}{m}}, \] where \(A\) and \(B\) depend on initial conditions. This models simple oscillatory motion, and can be visualized with an animation of the spring.
4. Atwood Machine
Two masses \(m_1\) and \(m_2\) are connected by a light string over a frictionless pulley. Using Newton's second law for each mass: \[ m_1 a = m_1 g - T, \quad m_2 a = T - m_2 g, \] where \(T\) is the tension and \(a\) is the acceleration. Solving these simultaneously gives \[ a = \frac{(m_1 - m_2) g}{m_1 + m_2}, \quad T = \frac{2 m_1 m_2 g}{m_1 + m_2}. \] This demonstrates forces in systems of connected bodies.
5. Rocket Motion (Variable Mass)
A rocket of initial mass \(m_0\) ejects fuel at a constant velocity \(u\) relative to the rocket. Using the full form of Newton's second law: \[ F = \frac{d}{dt}(m v), \] and assuming no external forces, the velocity of the rocket as a function of its decreasing mass is \[ v = u \ln \frac{m_0}{m}. \] This illustrates variable-mass systems and the importance of momentum conservation beyond the \( F = ma \) approximation.