Forces, energy and momentum

Three questions, three tools

Newton's laws relate interactions and acceleration, allowing forces, tensions and time evolution to be found when the model is known. The work–energy theorem relates the sum of work to the change in kinetic energy. Impulse relates a force applied over time to the change in linear momentum. Each method highlights a different structure in the problem.

The first decision is always to define the system. If we study only one cart in a collision, the force from the other cart is external. If we study the pair, collision forces are internal and occur in opposite pairs. The system's total momentum may be conserved even if each cart's momentum changes greatly. We may also need to include Earth to account for gravitational potential energy or momentum transferred to the ground. A conservation statement makes sense only after this boundary is declared.

Linear momentum is a vector quantity

For constant mass in the classical regime, p=mv. Units are kg·m/s, equivalent to N·s. Direction is that of velocity. Two bodies with equal speeds can have opposite momenta; adding magnitudes would erase that information. In one dimension, choose a positive direction and use signed velocities. In two dimensions, add components separately.

The second law can also be written ΣFext=dp/dt. For a constant-mass body, this recovers ma. This form prepares generalizations, but systems with mass entering or leaving require momentum flux to be included; simply differentiating mv and calling any term a propulsion force is not enough. In the following problems, we consider fixed-mass bodies and systems, speeds much smaller than c and approximately inertial frames.

Impulse and contact duration

A force's impulse is J=∫F dt. If force is approximately constant, J=FΔt; using a vector average force, the same expression gives total impulse over the interval. By the impulse theorem, Jext=Δp. On a graph of a force component against time, that component's impulse is the signed area.

For the same momentum change, increasing deceleration time reduces the magnitude of average force. This is why deformable materials can change impact forces. Maximum force is not necessarily equal to average force and depends on pulse shape. A calculated average alone does not tell us about damage or allow interaction details to be ignored. The theorem describes the overall momentum transfer without requiring force to remain constant at every instant.

When total momentum is conserved

When a system's momenta are added, Newtonian internal forces cancel in pairs. The change in total momentum equals total external impulse. If that impulse is zero, momentum is conserved; if it is small compared with internal impulses during a short collision, conservation may be a good approximation over that interval.

An external force need not be literally zero at every instant to have zero total impulse; the criterion is the integrated result. Conservation may also hold for only one component. In a collision on a horizontal table, horizontal external impulse may be negligible while weight and normal force act vertically. Declaring vector conservation requires analyzing all directions; conserving only one component requires stating that limitation. Momentum is not automatically conserved just because a collision occurred.

Kinetic energy and types of collision

In an elastic collision, total kinetic energy is conserved, as well as momentum when external impulse is negligible. In an inelastic collision, some kinetic energy becomes deformation, internal energy, sound or other forms. Total energy remains conserved. In a perfectly inelastic collision, bodies stay together and share the same final velocity.

Momentum and kinetic energy are not interchangeable: the former depends linearly on velocity and retains direction; the latter depends on speed squared and is scalar. In a one-dimensional perfectly inelastic collision, final v=(m₁v₁+m₂v₂)/(m₁+m₂). This equation comes from momentum conservation, not kinetic-energy conservation. Applying it and then imposing conservation of K without another condition creates a physical contradiction. The loss of K must be calculated by comparing states.

Combine stages without mixing assumptions

In a problem involving a collision followed by a climb, separate the stages. During the short contact, conserve momentum if the appropriate external impulse is negligible. Then, during the climb without dissipation, use mechanical-energy conservation to relate speed and height. Do not conserve mechanical energy through an inelastic collision or an isolated body's momentum during a climb under gravity.

The center of mass has a mass-weighted position and velocity Vcm=total p/M for fixed total mass. Internal forces can transform relative motions without changing Vcm if net external force is zero. This viewpoint helps explain recoil and collisions, but does not remove the need to study internal energy. At the end, check momentum before and after, the energy balance and plausible signs. A final velocity between the initial velocities is expected when two bodies stick together without external impulse, since it is a weighted average with positive masses.

Worked examples, without skipping steps

Follow the reasoning first. Then cover the solution and try to reconstruct it in your own words.

Now try it yourself

There is no timer. Write what you know, choose a relationship and justify each step. Hints and solutions are available whenever you need them.

Mistakes worth avoiding

Momentum is conserved in every collision.

It requires zero or negligible total external impulse in the component analyzed.

An inelastic collision violates energy conservation.

Total energy is conserved; some kinetic energy changes form.

Impulse is always maximum force times time.

It is the force integral or average force times time; the maximum does not replace the average.