Conservation of Momentum

🔒 Conservation of Momentum — Isolated Systems

Part 1 of 7 — Conservation of Momentum

One of the most powerful principles in all of physics: the total momentum of an isolated system is conserved. This means the total momentum before an interaction equals the total momentum after — no matter how complex the forces between objects are.

$\vec{p}_i = \vec{p}_f$

This single equation lets us solve problems that would be nearly impossible using Newton's laws alone.

What Is an Isolated System?

An isolated system is one where no net external force acts on the system.

External vs. Internal Forces

Examples

Key Insight

By carefully choosing what to include in your "system," you can often make external forces cancel or be negligible.

The Law of Conservation of Momentum

If the net external force on a system is zero, the total momentum of the system remains constant.

$\vec{p}_{\text{total, initial}} = \vec{p}_{\text{total, final}}$

For two objects:

$m_1 \vec{v}_{1i} + m_2 \vec{v}_{2i} = m_1 \vec{v}_{1f} + m_2 \vec{v}_{2f}$

Why It Works — Newton's Third Law

When object A pushes on object B, B pushes back on A with an equal and opposite force:

$\vec{F}_{A \text{ on } B} = -\vec{F}_{B \text{ on } A}$

Since both forces act for the same time $\Delta t$:

$\vec{J}_{\text{on } B} = -\vec{J}_{\text{on } A}$

$\Delta \vec{p}_B = -\Delta \vec{p}_A$

$\Delta \vec{p}_A + \Delta \vec{p}_B = 0$

The total momentum change is zero — momentum is transferred, not created or destroyed.

Simple Example

Two Skaters Push Apart

Two ice skaters face each other and push off. Skater A ($m_A = 60$ kg) and Skater B ($m_B = 80$ kg) are initially at rest.

Before: $p_i = 0 + 0 = 0$

After: $p_f = m_A v_A + m_B v_B = 0$

$60 v_A + 80 v_B = 0$

$v_A = -\frac{80}{60} v_B = -\frac{4}{3} v_B$

If Skater B moves at $+1.5$ m/s:

$v_A = -\frac{4}{3}(1.5) = -2.0 \text{ m/s}$

The lighter skater moves faster in the opposite direction — but the total momentum remains zero.

Concept Check — Conservation of Momentum 🎯

Conservation of Momentum Calculations 🧮

1. A 10 kg ball moving at +6 m/s collides with a 5 kg ball at rest. After the collision, the 10 kg ball moves at +2 m/s. What is the velocity of the 5 kg ball? (in m/s)

2. Two objects ($m_1 = 4$ kg at $+3$ m/s, $m_2 = 2$ kg at $-6$ m/s) collide and stick together. What is their final velocity? (in m/s)

3. A 3 kg object at rest breaks into two pieces. Piece A (1 kg) moves at +12 m/s. What is the velocity of piece B (2 kg)? (in m/s, include sign)

Conservation of Momentum Concepts 🔍

Exit Quiz — Isolated Systems ✅

🎱 Two-Object 1D Momentum Problems

Part 2 of 7 — Conservation of Momentum

The most common AP Physics 1 momentum problems involve two objects interacting in one dimension. In this lesson, we'll master the systematic approach to solving these problems using conservation of momentum.

The General 1D Setup

For two objects in one dimension:

$m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f}$

Problem-Solving Steps

1. Draw a before/after diagram — show velocities with arrows
2. Choose positive direction — typically the direction of the initially moving object
3. Assign signed velocities — positive or negative based on direction
4. Write the momentum equation — substitute known values
5. Solve for the unknown
6. Interpret the sign — positive = positive direction, negative = opposite

Three Scenarios

Example 1: Head-On Collision

A 6 kg ball moving at $+5$ m/s collides head-on with a 4 kg ball moving at $-3$ m/s. After the collision, the 6 kg ball moves at $+1$ m/s. Find the velocity of the 4 kg ball.

Before: $p_i = (6)(+5) + (4)(-3) = 30 - 12 = +18$ kg·m/s

After: $p_f = (6)(+1) + (4)(v_{2f}) = 6 + 4v_{2f}$

Conservation: $18 = 6 + 4v_{2f}$

$v_{2f} = \frac{18 - 6}{4} = +3 \text{ m/s}$

The 4 kg ball reverses and moves at $+3$ m/s (in the positive direction).

Example 2: Perfectly Inelastic Collision

A 2 kg block moving at $+8$ m/s collides with a 6 kg block at rest. They stick together. Find the final velocity.

$m_1 v_{1i} + m_2 v_{2i} = (m_1 + m_2) v_f$

$(2)(8) + (6)(0) = (2 + 6)v_f$

$16 = 8v_f$

$v_f = +2 \text{ m/s}$

Energy Check

• $KE_i = \frac{1}{2}(2)(8)^2 = 64$ J
• $KE_f = \frac{1}{2}(8)(2)^2 = 16$ J
• Lost: $64 - 16 = 48$ J (converted to heat, sound, deformation)

In a perfectly inelastic collision, kinetic energy is always lost (converted to other forms). This is the maximum KE loss for a given set of initial conditions.

Concept Check — 1D Momentum Problems 🎯

Two-Object Problem Practice 🧮

1. A 4 kg cart at +5 m/s collides with a 6 kg cart at −2 m/s. They stick together. What is the final velocity? (in m/s, to 1 decimal)

2. A 0.010 kg bullet at +400 m/s embeds in a 2.0 kg block at rest. What is the final speed of the block+bullet? (in m/s, to 3 significant figures)

3. A 8 kg ball at +3 m/s hits a 2 kg ball at +1 m/s. The 8 kg ball slows to +2 m/s. What is the final velocity of the 2 kg ball? (in m/s)

Collision Types 🔍

Exit Quiz — 1D Problems ✅

💣 Recoil Problems (Explosion Type)

Part 3 of 7 — Conservation of Momentum

In recoil (or "explosion") problems, objects that start together push apart. Internal forces propel them in opposite directions, but the total momentum of the system is conserved.

Examples: a gun firing a bullet, fireworks exploding, skaters pushing off each other, a nucleus undergoing radioactive decay.

The Recoil Principle

When the system starts at rest, the total initial momentum is zero:

$p_i = 0$

By conservation of momentum:

$p_f = m_1 v_1 + m_2 v_2 = 0$

$m_1 v_1 = -m_2 v_2$

What This Means

• The two objects always move in opposite directions
• The lighter object moves faster
• The ratio of speeds is inverse to the ratio of masses:

$\frac{|v_1|}{|v_2|} = \frac{m_2}{m_1}$

Energy Source

In explosions/recoil, where does the kinetic energy come from?

• Chemical energy (gunpowder, fuel)
• Elastic potential energy (spring, muscle contraction)
• Nuclear energy (radioactive decay)

Internal energy is converted to kinetic energy, but momentum remains zero.

Classic Examples

Gun Recoil

A 5 kg rifle fires a 0.010 kg bullet at 800 m/s. Find the recoil velocity of the rifle.

$0 = (5)(v_r) + (0.010)(800)$ $0 = 5v_r + 8$ $v_r = -1.6 \text{ m/s}$

The rifle recoils at 1.6 m/s — much slower than the bullet because it is much heavier.

Two Skaters

Skater A (50 kg) and Skater B (75 kg) push off from rest.

$0 = (50)(v_A) + (75)(v_B)$ $v_A = -\frac{75}{50} v_B = -1.5 v_B$

If Skater B moves at $+2$ m/s: $v_A = -3$ m/s.

Energy Analysis

$KE_{\text{total}} = \frac{1}{2}(50)(3)^2 + \frac{1}{2}(75)(2)^2 = 225 + 150 = 375 \text{ J}$

This 375 J came from the chemical energy in the skaters' muscles.

Multi-Piece Explosions

When an object breaks into more than two pieces, we apply conservation of momentum in each direction.

Example: Three-Piece Explosion

A 6 kg object at rest explodes into three pieces:

• Piece 1 (2 kg): $v_x = +3$ m/s, $v_y = 0$
• Piece 2 (1 kg): $v_x = 0$, $v_y = +6$ m/s
• Piece 3 (3 kg): Find $v_x$ and $v_y$

x-direction: $0 = (2)(3) + (1)(0) + (3)(v_{3x})$ $v_{3x} = -2 \text{ m/s}$

y-direction: $0 = (2)(0) + (1)(6) + (3)(v_{3y})$ $v_{3y} = -2 \text{ m/s}$

Piece 3 moves at $(-2, -2)$ m/s, with speed $|v_3| = \sqrt{4 + 4} = 2\sqrt{2} \approx 2.83$ m/s.

Concept Check — Recoil Problems 🎯

Recoil Calculations 🧮

1. A 3.0 kg rifle fires a 0.020 kg bullet at 600 m/s. What is the recoil speed of the rifle? (in m/s)

2. An 80 kg astronaut in space throws a 2 kg tool at 10 m/s. What is the astronaut's recoil speed? (in m/s)

3. A 10 kg object at rest explodes into two pieces. Piece A (4 kg) moves at +15 m/s. What is the speed of piece B? (in m/s)

Round all answers to 3 significant figures.

Recoil Concepts 🔍

Exit Quiz — Recoil ✅

🧭 2D Momentum Conservation

Part 4 of 7 — Conservation of Momentum

Momentum is a vector, and conservation applies to each component independently. When objects collide or interact in two dimensions, we apply conservation of momentum separately in the $x$ and $y$ directions.

2D Conservation Equations

In two dimensions, the single vector equation:

$\vec{p}_i = \vec{p}_f$

becomes two scalar equations:

$\textbf{x-direction: } m_1 v_{1ix} + m_2 v_{2ix} = m_1 v_{1fx} + m_2 v_{2fx}$

$\textbf{y-direction: } m_1 v_{1iy} + m_2 v_{2iy} = m_1 v_{1fy} + m_2 v_{2fy}$

Velocity Components

If an object moves at speed $v$ at angle $\theta$ from the $x$-axis:

$v_x = v\cos\theta, \quad v_y = v\sin\theta$

Problem-Solving Strategy

1. Set up a coordinate system ($x$ and $y$ axes)
2. Break all velocities into components
3. Apply conservation of momentum in $x$ and $y$ separately
4. Solve the two equations (may need both simultaneously)
5. Find magnitude and direction of the final velocity if needed

Example: 2D Collision

A 2 kg ball moving at 5 m/s in the $+x$ direction collides with a 3 kg ball at rest. After the collision, the 2 kg ball moves at 3 m/s at 30° above the $x$-axis. Find the velocity of the 3 kg ball.

x-components: $p_{ix} = (2)(5) + (3)(0) = 10 \text{ kg}\cdot\text{m/s}$ $p_{fx} = (2)(3\cos 30°) + (3)(v_{2x})$ $10 = (2)(2.598) + 3v_{2x}$ $10 = 5.196 + 3v_{2x}$ $v_{2x} = 1.60 \text{ m/s}$

y-components: $p_{iy} = 0$ $p_{fy} = (2)(3\sin 30°) + (3)(v_{2y})$ $0 = (2)(1.5) + 3v_{2y}$ $v_{2y} = -1.0 \text{ m/s}$

Speed: $v_2 = \sqrt{1.60^2 + 1.0^2} = \sqrt{3.56} = 1.89$ m/s

Direction: $\theta = \arctan(-1.0/1.60) = -32°$ (below $x$-axis)

2D Perfectly Inelastic Collision

When objects stick together in 2D:

$m_1 \vec{v}_1 + m_2 \vec{v}_2 = (m_1 + m_2)\vec{v}_f$

Example

Ball A (2 kg) moves at 4 m/s in the $+x$ direction. Ball B (3 kg) moves at 3 m/s in the $+y$ direction. They collide and stick.

x: $(2)(4) + (3)(0) = 5v_{fx}$ → $v_{fx} = 1.6$ m/s

y: $(2)(0) + (3)(3) = 5v_{fy}$ → $v_{fy} = 1.8$ m/s

Speed: $v_f = \sqrt{1.6^2 + 1.8^2} = \sqrt{5.80} = 2.41$ m/s

Direction: $\theta = \arctan(1.8/1.6) = 48.4°$ above the $x$-axis

Concept Check — 2D Momentum 🎯

2D Momentum Calculations 🧮

1. A 5 kg ball at 6 m/s ($+x$) collides with a 5 kg ball at rest. They stick together. What is the $x$-component of the final velocity? (in m/s)

2. For the same collision, what is the final speed? (in m/s)

3. A 3 kg object at 4 m/s ($+x$) and a 1 kg object at 8 m/s ($+y$) collide and stick. What is the final speed? (in m/s, to 3 significant figures)

2D Momentum Concepts 🔍

Exit Quiz — 2D Conservation ✅

⚠️ When Is Momentum NOT Conserved?

Part 5 of 7 — Conservation of Momentum

Momentum conservation is powerful, but it doesn't apply to every situation. Understanding when and why momentum is not conserved is just as important as knowing how to use it.

The answer comes down to one thing: external forces.

The Role of External Forces

The complete version of Newton's Second Law for a system:

$\vec{F}_{\text{net, external}} = \frac{d\vec{p}_{\text{total}}}{dt}$

• If $\vec{F}_{\text{net, ext}} = 0$: momentum is conserved ✅
• If $\vec{F}_{\text{net, ext}} \neq 0$: momentum is NOT conserved ❌

Common External Forces

When Can We Still Use Conservation?

Even with external forces, momentum conservation can be useful in these situations:

1. During Very Short Collisions

If the collision time $\Delta t$ is very small (milliseconds), even large external forces produce negligible impulse:

$J_{\text{ext}} = F_{\text{ext}} \Delta t \approx 0$

So momentum is approximately conserved during the instant of collision.

2. In One Direction Only

If an external force acts only vertically (like gravity), horizontal momentum is still conserved:

$p_x \text{ is conserved} \quad \text{(even if } p_y \text{ is not)}$

3. By Choosing a Larger System

If friction from the floor acts on sliding blocks, include the Earth in your system — then gravity and normal forces become internal! (Impractical, but theoretically valid.)

Example

A ball on a table collides with another ball. Friction acts on both, but during the brief collision instant ($\Delta t \approx 0.001$ s), the friction impulse is negligible. Momentum is conserved during the collision, but not after (friction gradually reduces momentum).

Example: Gravity as External Force

A 2 kg ball is thrown horizontally at 10 m/s off a cliff. After 3 seconds:

Horizontal momentum: $p_x = (2)(10) = 20$ kg·m/s — CONSERVED (no horizontal external force, ignoring air resistance)

Vertical momentum: $p_y = (2)(0 + 10 \times 3) = 60$ kg·m/s downward — NOT conserved (gravity is an external force adding downward impulse)

The impulse from gravity: $J_y = mg\Delta t = (2)(10)(3) = 60$ kg·m/s — exactly the change in vertical momentum!

Key Takeaway

Always check: Is there a net external force? If yes, momentum is NOT conserved in that direction. But it may still be conserved in the perpendicular direction.

Concept Check — External Forces 🎯

External Force Analysis 🧮

1. A 5 kg ball falls for 4 seconds. How much momentum does gravity add? (in kg·m/s, use $g = 10$ $m/s^{2}$)

2. Two 3 kg blocks collide on a surface with friction $\mu_k = 0.2$. If the collision lasts 0.005 s, what is the impulse from friction on the system during the collision? (in N·s, use $g = 10$ $m/s^{2}$)

3. For the same system in problem 2, if the blocks slide together for 2 s after colliding, what impulse does friction deliver during this time? (in N·s)

Round all answers to 3 significant figures.

Conservation Conditions 🔍

Exit Quiz — External Forces ✅

🔧 Problem-Solving Workshop

Part 6 of 7 — Conservation of Momentum

Let's practice solving a variety of momentum conservation problems — from basic 1D collisions to recoil, 2D, and problems with external forces. This workshop focuses on building your problem-solving confidence for the AP exam.

Problem-Solving Checklist

Before writing any equation, always ask:

1. ✅ Is the system isolated? (No net external force → momentum conserved)
2. ✅ What is the system? (Choose objects wisely to eliminate external forces)
3. ✅ Is this 1D or 2D? (How many equations do I need?)
4. ✅ Do objects stick together? (Perfectly inelastic → one final velocity)
5. ✅ Is energy also needed? (Elastic collision → both $p$ and $KE$ conserved)
6. ✅ What are my knowns and unknowns? (List them before solving)

The Master Equation (1D)

$m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f}$

For perfectly inelastic: replace right side with $(m_1 + m_2)v_f$

Problem 1: Ballistic Pendulum 🎯

A 0.010 kg bullet embeds in a 2.0 kg wooden block hanging from a string. The block+bullet swings upward to a height of 0.20 m. What was the bullet's speed? (Use $g = 10$ $m/s^{2}$)

Hint: Use two steps — conservation of momentum during collision, then conservation of energy during swing.

Problem 2: Head-On Collision 🚗

Car A (1500 kg) travels east at 20 m/s. Car B (2000 kg) travels west at 15 m/s. They collide and lock bumpers.

1. What is the total initial momentum of the system? (in kg·m/s, take east as positive)

2. What is the final velocity of the wreckage? (in m/s, to 3 significant figures, include sign)

3. In which direction does the wreckage move? (type "east" or "west")

Problem 3: Recoil in Space 🚀

An astronaut (80 kg) floating at rest in space fires a thruster that ejects 0.50 kg of gas at 2000 m/s relative to the astronaut.

Problem 4: Mixed Practice 📝

1. A 6 kg block at +4 m/s collides with a 2 kg block at −8 m/s. They stick together. What is the final velocity? (in m/s)

2. A 0.15 kg ball at +20 m/s collides with a 0.15 kg ball at −10 m/s. After collision, the first ball moves at +5 m/s. What is the velocity of the second ball? (in m/s, include sign)

3. A 50 kg cannon at rest fires a 5 kg ball at 100 m/s. What is the cannon's recoil speed? (in m/s)

Problem Type Identification 🔍

Exit Quiz — Problem Solving ✅

🎓 Synthesis & AP Review

Part 7 of 7 — Conservation of Momentum

Let's synthesize everything about conservation of momentum: isolated systems, 1D and 2D problems, recoil, and external forces. This review prepares you for AP-level questions on this critical topic.

Key Concepts Summary

When Momentum Is / Isn't Conserved

AP Review — Conceptual 🎯

AP Review — Quantitative 📝

AP Calculation Practice 🧮

1. A 0.050 kg bullet at 400 m/s embeds in a 4.95 kg block at rest. What is the block+bullet velocity? (in m/s)

2. The block+bullet slides along a surface with $\mu_k = 0.40$. How far does it slide before stopping? (in m, use $g = 10$ $m/s^{2}$)

3. A 60 kg person standing on a 15 kg skateboard at rest throws a 5 kg ball at 12 m/s horizontally. What is the recoil speed of the person+skateboard? (in m/s)

Round all answers to 3 significant figures.

Comprehensive Review 🔍

Final Exit Quiz — Conservation of Momentum ✅