Topic 4.3 Conservation of Linear Momentum: state that the total momentum of an isolated system is conserved, and apply it to recoil, explosions and interactions in one and two dimensions.

What this topic is asking

The College Board (Topic 4.3) wants you to state that the total momentum of an isolated system is conserved, to explain why internal forces cannot change it, and to apply conservation of momentum to recoil, explosions and interactions in one and two dimensions. Conservation of momentum is one of the most useful tools in mechanics, because it lets you relate before-and-after states without any knowledge of the (often complicated) forces during the interaction.

The condition for conservation

The justification is Newton's second law for the system, $\vec{F}_{ext} = \dfrac{d\vec{p}_{total}}{dt}$. If the net external force is zero, the total momentum does not change. The forces inside the system, however violent, come in third-law pairs and cancel when summed over the whole system, so they cannot alter the total. This is why two colliding objects conserve their combined momentum even though each feels a large force: those forces are internal to the two-object system.

Internal forces cannot change total momentum

This is the conceptual heart of the topic. When you choose the colliding (or exploding, or recoiling) objects together as the system, the interaction forces become internal and cancel in pairs, leaving the total momentum fixed. An astronaut who throws a tool, a cannon that fires a shell, a firework that bursts, all start with some total momentum (often zero) and end with the same total, the pieces sharing it out so the vector sum is unchanged. The skill is to draw the system boundary so the forces you do not want to track become internal.

Recoil and explosions

For an interaction starting from rest, the total momentum is zero before and therefore zero after. The fragments must carry equal and opposite momenta. A cannon firing a shell recoils so that $m_{cannon}\vec{v}_{cannon} = -m_{shell}\vec{v}_{shell}$; because the cannon is much heavier, it recoils slowly. Notice that although the momenta are equal in magnitude, the kinetic energies are not: the light, fast shell carries far more kinetic energy than the heavy, slow cannon, since $K = p^2/2m$ favors the smaller mass. This asymmetry is a favorite exam point.

Two-dimensional interactions

When motion is two-dimensional, momentum conservation is a vector equation, so it holds separately on each axis:

Resolve every velocity into components, write a conservation equation for $x$ and another for $y$, and solve the two together. This handles glancing collisions, fragments scattering in a plane, and any interaction where the objects do not all move along one line. Keeping the components separate is the key to avoiding errors.

Try this

Q1. A $0.040$ kg bullet is fired at $500$ m/s from a $3.0$ kg rifle initially at rest. Calculate the rifle's recoil speed. [2 points]

• Cue. $0 = (0.040)(500) + (3.0)v \Rightarrow v = -20/3.0 = -6.7$ m/s, so $6.7$ m/s backward.

Q2. Explain why momentum can be conserved in a collision even though kinetic energy is lost. [2 points]

• Cue. Momentum conservation needs only zero net external force; the lost kinetic energy becomes heat and deformation, which does not affect the (vector) momentum balance.