Apply Momentum Conservation to Collisions - Grade 11 Physics

SECTION 1

Historical Context & Motivation

For centuries, thinkers have tried to understand what happens when objects crash into one another. Early natural philosophers noticed that something seemed to be preserved during collisions—some quantity that didn't just disappear. The search for that hidden quantity led to one of the most powerful ideas in all of physics: the conservation of momentum. This principle tells us that when two objects collide in an isolated system (one with no outside forces acting on them), the total momentum before the collision equals the total momentum after. Understanding this law allows engineers to design safer cars, physicists to probe subatomic particles, and athletes to perfect their technique in sports like billiards and football.

1644 • Descartes Proposes 'Quantity of Motion'

René Descartes introduced the idea that the total 'quantity of motion' in the universe is constant, laying early groundwork for momentum conservation, though his formula lacked direction.

1668 • Wallis, Wren & Huygens Formalize Collision Laws

The Royal Society challenged three scientists to explain collisions. John Wallis, Christopher Wren, and Christiaan Huygens independently published rules showing that momentum (mass × velocity, with direction) is conserved.

1687 • Newton's Principia Mathematica

Isaac Newton formalized the laws of motion, with the third law (action–reaction) providing the theoretical foundation for why momentum is conserved in all interactions.

1918 • Noether's Theorem Links Symmetry to Conservation

Mathematician Emmy Noether proved that every symmetry of nature corresponds to a conservation law. Momentum conservation arises because the laws of physics are the same everywhere in space—a profound insight still central to modern physics.

The core question driving this topic is deceptively simple: When two objects collide, how do their speeds and directions change, and what rule governs that change? By the end of this lesson, you will be able to answer that question for any collision type using the conservation of momentum.

SECTION 2

Core Principles & Definitions

Before diving into collisions, you need to master a few foundational ideas. Momentum is defined as the product of an object's mass and its velocity. Because velocity is a vector (it has both magnitude and direction), momentum is also a vector. This means direction matters—a ball rolling left has a different momentum than the same ball rolling right at the same speed. The law of conservation of momentum states that the total momentum of all objects in a system stays constant as long as no external net force acts on them. This holds true whether the objects bounce apart, stick together, or shatter into pieces.

Momentum (p)

The product of mass (m) and velocity (v). Measured in kg·m/s. It is a vector quantity, so direction matters.

Isolated System

A group of objects that experience no net external force. Internal forces (like the collision force) are allowed, but outside pushes or friction are negligible.

Elastic Collision

A collision in which both momentum and kinetic energy are conserved. Objects bounce off each other with no energy lost to deformation or heat.

Inelastic Collision

A collision in which momentum is conserved but kinetic energy is not. Some energy converts to sound, heat, or deformation. In a perfectly inelastic collision, the objects stick together.

Impulse

The change in momentum of an object, equal to the net force multiplied by the time interval (F × Δt). Impulse explains why momentum changes for individual objects, even though total system momentum stays the same.

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KEY TAKEAWAY

Think of momentum like money in a closed economy. During a trade (collision), money moves from one person to another, but the total cash in the room stays the same. No one prints new money and no one burns any. If you add up every dollar held by every person before and after the trade, the totals match. Momentum works the same way in an isolated system—it is redistributed, never created or destroyed.

SECTION 3

Visualizing Collisions

The diagram below illustrates the two main types of collisions side by side. On the left, an elastic collision shows two objects bouncing apart after impact, each retaining separate velocities. On the right, a perfectly inelastic collision shows the same two objects sticking together and moving as a single combined mass. In both cases, the total momentum before the collision equals the total momentum after.

Comparison of elastic and perfectly inelastic collisions. The left panel shows objects bouncing apart (both KE and momentum conserved), while the right panel shows objects sticking together (only momentum conserved). The yellow burst represents the moment of impact.

Notice that in the elastic case on the left, the smaller ball (m₂) picks up speed while the larger ball (m₁) slows down or even reverses direction. In the perfectly inelastic case on the right, both masses merge into a single object traveling at a new, shared velocity. The key idea in both diagrams is the same: add up all the momentum arrows before the collision and after the collision, and the totals are identical.

SECTION 4

Mathematical Framework

The mathematics behind momentum conservation is elegant and straightforward. Every collision problem starts from the same master equation and then branches depending on the collision type. Let's build the framework step by step.

DEFINITION OF MOMENTUM

p = m × v

Where p is momentum (kg·m/s), m is mass (kg), and v is velocity (m/s). Momentum is a vector, so assign positive and negative signs based on direction.

CONSERVATION OF MOMENTUM (GENERAL)

m₁v₁ᵢ + m₂v₂ᵢ = m₁v₁f + m₂v₂f

The subscript i means 'initial' (before collision) and f means 'final' (after collision). The left side is total momentum before; the right side is total momentum after. They must be equal.

PERFECTLY INELASTIC COLLISION

m₁v₁ᵢ + m₂v₂ᵢ = (m₁ + m₂)vf

When the objects stick together, they share a single final velocity vf. This simplifies the general equation because v₁f = v₂f = vf. This type of collision loses the maximum amount of kinetic energy.

ELASTIC COLLISION — ADDITIONAL CONDITION

½m₁v₁ᵢ² + ½m₂v₂ᵢ² = ½m₁v₁f² + ½m₂v₂f²

In an elastic collision, kinetic energy (KE) is also conserved. This gives you a second equation to work with alongside the momentum equation, allowing you to solve for both unknown final velocities.

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Sign Convention Tip

Always choose a positive direction at the start of a problem—usually to the right. Objects moving in that direction get positive velocities; objects moving opposite get negative velocities. Forgetting this step is the number-one mistake students make when applying momentum conservation.

SECTION 5

Collision Types in Detail

Collisions exist on a spectrum. At one end, perfectly elastic collisions preserve every joule of kinetic energy—think of ideal billiard balls clicking together. At the other end, perfectly inelastic collisions convert the maximum possible kinetic energy into other forms like heat and sound—picture two lumps of clay slamming together and sticking. Most real-world collisions fall somewhere in between, and we call them partially inelastic. A car crash, for example, is partially inelastic: the cars bounce apart somewhat, but a lot of energy goes into crumpling metal and generating noise.

The three collision types arranged on a spectrum of kinetic energy loss. All three types conserve momentum. The bottom bar illustrates the range from zero energy loss (elastic) to maximum energy loss (perfectly inelastic).

Property	Elastic	Partially Inelastic	Perfectly Inelastic
Momentum conserved?	Yes ✓	Yes ✓	Yes ✓
KE conserved?	Yes ✓	No ✗ (partial loss)	No ✗ (max loss)
Do objects stick?	No — they bounce apart	No — they separate with deformation	Yes — they move as one mass
Unknowns to solve	Two final velocities (need 2 equations)	Two final velocities (need extra info)	One final velocity (1 equation suffices)

Summary of collision type characteristics

SECTION 6

Worked Example

Let's apply momentum conservation to a concrete problem. We'll solve a perfectly inelastic collision first because it's the most straightforward type—only one unknown final velocity.

Perfectly Inelastic Collision — Football Tackle

1
Step 1 — Read the Problem

A 90 kg football player running east at 5.0 m/s tackles a 70 kg player running west at 3.0 m/s. After the tackle, the two players move together. What is their final velocity?
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Step 2 — Choose a Sign Convention

Let east be the positive direction. Therefore: v₁ᵢ = +5.0 m/s (player 1, moving east) and v₂ᵢ = −3.0 m/s (player 2, moving west, so negative).
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Step 3 — Write the Conservation Equation

Since they stick together, use the perfectly inelastic formula: m₁v₁ᵢ + m₂v₂ᵢ = (m₁ + m₂)vf.
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Step 4 — Substitute Known Values

(90 kg)(+5.0 m/s) + (70 kg)(−3.0 m/s) = (90 kg + 70 kg) × vf
5
Step 5 — Calculate Each Side

Left side: 450 kg·m/s + (−210 kg·m/s) = 240 kg·m/s. Right side: 160 kg × vf.

240 kg·m/s = 160 kg × vf
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Step 6 — Solve for vf

Divide both sides by 160 kg: vf = 240 ÷ 160 = +1.5 m/s.

vf = +1.5 m/s (east)
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Step 7 — Interpret the Result

The positive sign tells us the combined players slide eastward after the tackle. The 90 kg player 'wins' directionally because he had more momentum going in. The final speed (1.5 m/s) is much lower than either original speed because momentum was shared across the combined mass.

💡

Check Your Work

Always verify: total momentum before = (90)(5.0) + (70)(−3.0) = 450 − 210 = 240 kg·m/s. Total momentum after = (160)(1.5) = 240 kg·m/s. ✓ They match, so the answer is consistent.

SECTION 7

Strengths & Limitations of Momentum Conservation

The conservation of momentum is an incredibly versatile tool, but like any physics principle, it has conditions that must be met and situations where it's more useful than others. Understanding both its power and its boundaries will help you decide when to reach for this equation and when you might need additional tools.

Strengths	Limitations
Works for all collision types — elastic, inelastic, and perfectly inelastic.	Only applies to isolated systems. If significant external forces (like friction or gravity along the collision direction) act during the collision, momentum is not conserved.
You don't need to know the details of the collision forces themselves — just before-and-after information.	In elastic collisions, you get two unknowns but only one momentum equation, so you must also use kinetic energy conservation.
Applies in one, two, or three dimensions. For 2D, use separate x and y momentum equations.	Tells you nothing about energy loss by itself. You must separately calculate KE before and after to determine energy changes.
Scales from subatomic particles to galaxies — it's truly universal.	Requires accurate measurements of mass and velocity, which can be difficult in real lab settings with friction.

Strengths and limitations of applying momentum conservation

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KEY TAKEAWAY

Momentum conservation is like a perfectly balanced seesaw: it tells you the total weight on each side must match, but it doesn't tell you how the riders are arranged. To get the full picture of a collision, you often need a second principle—like conservation of kinetic energy—to pin down all the unknowns. Think of momentum conservation as one powerful lens; sometimes you need a second lens (energy conservation) to bring the entire image into focus.

SECTION 8

Connection to Two-Dimensional & Advanced Collisions

So far, we've focused on one-dimensional collisions where everything moves along a single line. In reality, many collisions happen in two dimensions—think of a cue ball striking another billiard ball at an angle, or two cars colliding at an intersection. The good news is that momentum conservation still applies; you just apply it separately to the x-direction and the y-direction. This is a direct extension of the 1D approach and a topic you'll explore more deeply in AP Physics.

Feature	1D Collisions (This Lesson)	2D Collisions (Advanced)
Directions considered	One axis (e.g., left–right)	Two axes (x and y)
Number of equations	1 momentum equation (+ 1 energy equation if elastic)	2 momentum equations (x and y) + 1 energy equation if elastic
Math tools needed	Algebra with signed numbers	Trigonometry (sin, cos) to resolve vectors into components
Typical example	Head-on car collision on a highway	Two cars colliding at a 90° intersection
Momentum conserved?	Yes, along the single axis	Yes, independently along each axis

Comparing 1D and 2D collision analysis

The fundamental principle never changes: total momentum is conserved in every direction for an isolated system. If you master the 1D framework now, stepping up to 2D later is just a matter of applying the same equation twice—once for each axis—and using trigonometry to break vectors into components. In college-level physics and engineering, these ideas extend even further to rotational collisions, where angular momentum is conserved alongside linear momentum.

SECTION 9

Practice Problems

Work through these five problems in order. They start with conceptual reasoning and build up to more complex scenarios. For calculation problems, show your work step by step—set up the equation, substitute values, and solve.

Question 1: CONCEPTUAL

A 2 kg ball moving at 3 m/s to the right hits a 2 kg ball that is stationary. After the collision, the first ball stops completely. Without doing any math, predict what happens to the second ball and explain your reasoning using momentum conservation.

Show answer

The second ball must move to the right at 3 m/s. Before the collision, only the first ball has momentum: p = 2 kg × 3 m/s = 6 kg·m/s to the right. After the collision, the first ball is at rest (0 momentum), so the entire 6 kg·m/s must belong to the second ball. Since the second ball also has a mass of 2 kg, its velocity must be 6 ÷ 2 = 3 m/s to the right. The momentum was completely transferred from one ball to the other—this is what happens in a head-on elastic collision between equal masses.

Question 2: BASIC CALCULATION

A 0.50 kg air-hockey puck traveling at 4.0 m/s to the right collides with a 0.30 kg puck at rest. If the two pucks stick together (perfectly inelastic collision), what is their final velocity?

Show answer

Using the perfectly inelastic formula: m₁v₁ᵢ + m₂v₂ᵢ = (m₁ + m₂)vf. Substituting: (0.50)(4.0) + (0.30)(0) = (0.50 + 0.30)vf → 2.0 = 0.80 × vf → vf = 2.5 m/s to the right. The combined puck moves in the same direction as the original moving puck, but slower, because the extra mass means the same momentum is now spread across a larger combined mass.

Question 3: INTERMEDIATE

A 1500 kg car traveling east at 20 m/s collides head-on with a 1000 kg car traveling west at 15 m/s. After the collision, the 1500 kg car moves east at 5.0 m/s. Find the velocity of the 1000 kg car after the collision. Is this collision elastic or inelastic?

Show answer

Choose east as positive. Before: p_total = (1500)(+20) + (1000)(−15) = 30,000 − 15,000 = 15,000 kg·m/s. After: 15,000 = (1500)(+5.0) + (1000)(v₂f) → 15,000 = 7,500 + 1000v₂f → v₂f = +7.5 m/s (east). To check elasticity, compare KE before and after. KE_before = ½(1500)(20²) + ½(1000)(15²) = 300,000 + 112,500 = 412,500 J. KE_after = ½(1500)(5²) + ½(1000)(7.5²) = 18,750 + 28,125 = 46,875 J. Since 46,875 ≠ 412,500, this collision is inelastic—most of the kinetic energy was lost to deformation and heat.

Question 4: APPLIED

A 10 g bullet traveling at 400 m/s embeds itself in a 2.0 kg wooden block resting on a frictionless table. What is the velocity of the block-bullet system immediately after impact? Then calculate what percentage of the bullet's original kinetic energy was lost.

Show answer

Convert the bullet mass: 10 g = 0.010 kg. This is a perfectly inelastic collision (bullet embeds in block). Applying conservation of momentum: (0.010)(400) + (2.0)(0) = (0.010 + 2.0)vf → 4.0 = 2.01 × vf → vf ≈ 1.99 m/s. For energy loss: KE_before = ½(0.010)(400²) = 800 J. KE_after = ½(2.01)(1.99²) ≈ 3.98 J. Energy lost = 800 − 3.98 = 796 J, which is about 99.5% of the original kinetic energy. This dramatic energy loss is typical when a small, fast object embeds in a much larger mass—almost all the energy converts to heat, sound, and deformation of the wood.

Question 5: CRITICAL THINKING

Two identical 50 kg ice skaters stand facing each other at rest on a frictionless rink. Skater A pushes Skater B, and afterward Skater B moves at 2.0 m/s to the right. Describe Skater A's motion and explain why this scenario, even though it looks like an 'explosion,' still obeys conservation of momentum. How is the kinetic energy created if no external work is done on the system?

Show answer

Before the push, total momentum is zero (both are at rest). After the push, momentum must still be zero. If Skater B moves at +2.0 m/s, then Skater A must move at −2.0 m/s (to the left): (50)(−2.0) + (50)(+2.0) = 0 ✓. This 'explosion' (reverse collision) conserves momentum because the push force between the skaters is an internal force—Newton's third law ensures both feel equal and opposite forces for the same duration, producing equal and opposite impulses. The kinetic energy came from internal energy—the chemical energy in the skaters' muscles. This is analogous to how a firecracker at rest can send fragments flying in opposite directions, with the chemical energy of gunpowder becoming kinetic energy, while total momentum remains zero throughout.

SUMMARY

Lesson Summary

Summary

Momentum is the product of mass and velocity (p = mv), and it is a vector quantity that depends on direction. The law of conservation of momentum states that the total momentum of an isolated system remains constant before and after a collision. This principle applies universally to elastic collisions (where kinetic energy is also conserved), inelastic collisions (where some kinetic energy is lost), and perfectly inelastic collisions (where objects stick together and maximum kinetic energy is lost).

To solve collision problems, always start by defining a positive direction, then set up the equation m₁v₁ᵢ + m₂v₂ᵢ = m₁v₁f + m₂v₂f. For perfectly inelastic collisions, simplify the right side to (m₁ + m₂)vf. For elastic collisions, pair the momentum equation with the kinetic energy conservation equation to find both final velocities. Always verify your answer by checking that total momentum before equals total momentum after. These ideas extend naturally to two-dimensional collisions and serve as a foundation for advanced topics like angular momentum and particle physics.