The Timpson Lyceum · University Physics I (2025)

Momentum, Impulse, Collisions, and Rockets – Lecture 8 (University Physics I)

Overview

This lecture introduces momentum as a powerful alternative to energy and force tracking, especially for collisions. While energy methods break down when energy disappears into heat and deformation (a bullet embedding in wood, two cars crumpling together), momentum is conserved through even the most violent internal interactions, making it far easier to track. The lecture builds momentum from Newton's second law, defines impulse as the change in momentum, derives the conservation of momentum from Newton's third law, and applies it to the three types of collisions (inelastic, completely inelastic, and elastic). It closes with center of mass and the physics of rocket propulsion, where continuously expelling mass produces thrust.

Key Concepts

Momentum and impulse

Momentum is a vector defined as mass times velocity: $\vec{p} = m\vec{v}$ . Units are $\text{kg}\cdot\text{m/s}$ , equivalently $\text{N}\cdot\text{s}$ .
Newton's second law (original form): the net force equals the time rate of change of momentum, $\sum \vec{F} = \dfrac{d\vec{p}}{dt}$ . For constant mass this reduces to $\vec{F} = m\vec{a}$ .
Impulse $\vec{J}$ is the change in momentum produced by a force acting over a time interval: $\vec{J} = \int_{t_1}^{t_2}\vec{F}\,dt = \Delta\vec{p} = \vec{p}_2 - \vec{p}_1$ .
Impulse–momentum theorem: for a constant or average force, $\vec{J} = \vec{F}_{\text{avg}}\,\Delta t$ , so $\vec{F}_{\text{avg}} = \dfrac{\vec{J}}{\Delta t}$ .
Momentum vs. kinetic energy: momentum comes from integrating force over time (a vector); kinetic energy comes from integrating force over displacement (a scalar). Two objects pushed by the same force over the same distance gain the same kinetic energy; pushed for the same time, they gain the same momentum. These are different quantities — do not interchange them.

Conservation of momentum

For any system of interacting particles, internal forces cancel in pairs by Newton's third law, so the change in total momentum equals only the sum of external forces: $\dfrac{d\vec{P}}{dt} = \sum \vec{F}_{\text{ext}}$ .
If the net external force is zero, total momentum is constant — it is conserved regardless of how violent the internal interactions are. Lighting a stick of TNT in space does not change the total momentum of its fragments.
In collisions, external forces (friction, gravity vs. normal force) are negligible compared with the impact forces, so momentum is conserved even though kinetic energy may not be.

Types of collisions

Completely inelastic: the objects stick together and move with a common final velocity. Momentum is conserved; kinetic energy is lost to deformation, heat, and sound. $m_A\vec{v}_{A1} + m_B\vec{v}_{B1} = (m_A + m_B)\vec{v}_2$ .
Inelastic: some deformation/partial sticking; momentum conserved, kinetic energy partly lost. This is what real-world collisions usually are.
Elastic (completely elastic): the objects bounce apart with no deformation; both momentum and kinetic energy are conserved (e.g. billiard balls). You get two equations to solve simultaneously.

Elastic-collision results (1-D, target $B$ initially at rest)

Final velocities: $v_{A2} = \dfrac{m_A - m_B}{m_A + m_B}\,v_{A1}$ and $v_{B2} = \dfrac{2m_A}{m_A + m_B}\,v_{A1}$ .
Relative velocity reverses: the speed of approach before equals the speed of separation after, $v_{A1} - v_{B1} = -(v_{A2} - v_{B2})$ .
Limiting cases: a light object hitting a heavy wall ( $m_A \ll m_B$ ) bounces back at nearly the same speed; a heavy object hitting a tiny one ( $m_A \gg m_B$ ) barely changes; equal masses exchange velocities — the incoming object stops and the struck one moves off with the incoming speed.

Center of mass

For point masses, $\vec{r}_{\text{cm}} = \dfrac{\sum_i m_i \vec{r}_i}{\sum_i m_i} = \dfrac{1}{M}\sum_i m_i \vec{r}_i$ , where $M$ is the total mass. It is a mass-weighted average position; compute each component ( $x$ , $y$ , $z$ ) separately.
The total momentum equals the total mass times the velocity of the center of mass: $\vec{P} = M\vec{v}_{\text{cm}}$ .
The center of mass obeys $M\vec{a}_{\text{cm}} = \sum \vec{F}_{\text{ext}}$ — the whole system moves as if all its mass were concentrated there. An exploding projectile's center of mass continues on the original trajectory.

Rocket propulsion

A rocket gains speed by expelling mass backward; faster ejection speed and a greater rate of mass loss both increase the gain (intuitively, throwing balls off a frictionless sled).
Thrust: $F = v_{\text{ex}}\left|\dfrac{dm}{dt}\right|$ , the exhaust (exit) speed times the rate at which mass is expelled.
Rocket velocity equation (constant exhaust speed): $v = v_0 + v_{\text{ex}}\ln\!\dfrac{m_0}{m}$ , where $m_0$ is the initial mass and $m$ the current mass.

Worked Example

Ballistic pendulum — finding a bullet's speed. A bullet of mass $m_B$ is fired into a hanging block of wood of mass $m_W$ . The bullet embeds (completely inelastic collision) and the block-plus-bullet swings up to a maximum height $h$ . Find the bullet's initial speed $v_B$ .

Step 1 — Collision (momentum conserved). The bullet's momentum before equals the combined momentum just after impact, giving the common speed $v_W$ :

\begin{aligned} m_B v_B &= (m_B + m_W)\,v_W \\ v_B &= \frac{m_B + m_W}{m_B}\,v_W \end{aligned}

Step 2 — Swing (energy conserved after impact). Kinetic energy of the combined mass converts to gravitational potential energy at height $h$ :

\begin{aligned} \tfrac{1}{2}(m_B + m_W)\,v_W^{2} &= (m_B + m_W)\,g\,h \\ v_W &= \sqrt{2gh} \end{aligned}

Step 3 — Combine. Substituting $v_W$ back gives the bullet speed in measurable quantities:

\begin{aligned} v_B &= \frac{m_B + m_W}{m_B}\,\sqrt{2gh} \end{aligned}

Note the two phases use different conservation laws: momentum (not energy) for the collision, then energy for the swing — because kinetic energy is lost as the bullet buries itself in the wood.

Summary

Momentum $\vec{p} = m\vec{v}$ is a vector; impulse $\vec{J} = \Delta\vec{p} = \vec{F}_{\text{avg}}\Delta t$ is the change it produces.
Total momentum is conserved when the net external force is zero — internal forces cancel by Newton's third law — which makes collisions easy to analyze even when energy is messy.
All collisions conserve momentum; only elastic collisions also conserve kinetic energy, while inelastic ones lose it to deformation and heat (completely inelastic objects stick together).
For 1-D elastic collisions the relative speed of approach equals the relative speed of separation, and equal masses simply swap velocities.
The center of mass moves as though all mass and all external force acted at that single point, with $\vec{P} = M\vec{v}_{\text{cm}}$ .
A rocket produces thrust $F = v_{\text{ex}}|dm/dt|$ by expelling mass, reaching $v = v_0 + v_{\text{ex}}\ln(m_0/m)$ .