{"id":239,"date":"2026-06-15T22:30:34","date_gmt":"2026-06-15T22:30:34","guid":{"rendered":"https:\/\/physicsfundamentalsinfo.com\/blog\/?p=239"},"modified":"2026-06-15T22:30:36","modified_gmt":"2026-06-15T22:30:36","slug":"conservation-of-momentum","status":"publish","type":"post","link":"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/conservation-of-momentum\/","title":{"rendered":"What Is Conservation of Momentum?"},"content":{"rendered":"\n<div class=\"pf-citation\"><div class=\"eyebrow\">Definition<\/div><p>\nConservation of momentum is the physical law stating that the total momentum of an isolated system stays constant: the vector sum of every object&#8217;s momentum (mass \u00d7 velocity) before an interaction equals the total momentum afterward. With no external force, collisions, explosions and recoil only redistribute momentum \u2014 they never create or destroy it.\n<\/p><\/div>\n\n<p>Step off a small boat onto a jetty and the boat darts backwards beneath your foot. Fire a rifle and your shoulder takes the kick. Break a rack of pool balls and a single cue strike scatters fifteen of them across the felt.<\/p>\n\n<p>Every one of these is the same rule quietly balancing its books. Momentum \u2014 the &#8220;quantity of motion&#8221; a moving object carries \u2014 is never simply lost. It only shifts from one object to another, and the totals always match. That single idea is one of the most powerful problem-solving tools in all of physics.<\/p>\n\n<h2>What Is Conservation of Momentum?<\/h2>\n\n<p>Picture two ice skaters drifting toward each other. They grab hands, spin, and push apart. Track every push and shove, and you find the books always balance \u2014 what one skater gains, the other loses.<\/p>\n\n<p><strong>Conservation of momentum<\/strong> is the law that the total momentum of an isolated system never changes. Momentum is the product of an object&#8217;s mass and its velocity, written <em>p = mv<\/em>. Add up the momentum of every object before an event, and you get the exact same total afterward.<\/p>\n\n<p>The key word is <em>isolated<\/em>. As long as no outside force pushes on the system, momentum has nowhere to go. Objects can swap it between themselves through collisions, but the grand total is locked.<\/p>\n\n<p>Momentum is also a <strong>vector<\/strong> \u2014 it has direction. A 2&nbsp;kg cart rolling left does not cancel an identical cart rolling right by accident; their momenta point in opposite directions and sum to zero. Keeping track of sign and direction is half the skill of momentum problems, which is why it pays to be clear about <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/kinematics\/velocity-vs-speed\/\">velocity versus speed<\/a> before you start.<\/p>\n\n<svg role=\"img\" aria-label=\"Diagram showing the total momentum of two carts is equal before and after a one-dimensional collision\" viewBox=\"0 0 720 400\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" style=\"width:100%;height:auto;max-width:680px;display:block;margin:24px auto;\">\n  <defs>\n    <marker id=\"ahg\" markerWidth=\"9\" markerHeight=\"9\" refX=\"7\" refY=\"3\" orient=\"auto\"><path d=\"M0,0 L7,3 L0,6 Z\" fill=\"#C8932A\"><\/path><\/marker>\n  <\/defs>\n  <rect x=\"2\" y=\"2\" width=\"716\" height=\"396\" rx=\"12\" fill=\"#FAF6EE\" stroke=\"#D9CFB8\" stroke-width=\"2\"><\/rect>\n  <text x=\"360\" y=\"36\" text-anchor=\"middle\" font-family=\"Georgia, serif\" font-size=\"20\" fill=\"#0A1628\" font-weight=\"bold\">Total momentum before = total momentum after<\/text>\n  <text x=\"44\" y=\"84\" font-family=\"Arial, sans-serif\" font-size=\"14\" fill=\"#7A1F2B\" font-weight=\"bold\">BEFORE<\/text>\n  <line x1=\"44\" y1=\"150\" x2=\"676\" y2=\"150\" stroke=\"#D9CFB8\" stroke-width=\"3\"><\/line>\n  <rect x=\"120\" y=\"110\" width=\"96\" height=\"40\" rx=\"6\" fill=\"#0A1628\"><\/rect>\n  <text x=\"168\" y=\"135\" text-anchor=\"middle\" font-family=\"Arial, sans-serif\" font-size=\"14\" fill=\"#FAF6EE\">m\u2081 = 2 kg<\/text>\n  <line x1=\"220\" y1=\"100\" x2=\"292\" y2=\"100\" stroke=\"#C8932A\" stroke-width=\"4\" marker-end=\"url(#ahg)\"><\/line>\n  <text x=\"256\" y=\"92\" text-anchor=\"middle\" font-family=\"Arial, sans-serif\" font-size=\"13\" fill=\"#C8932A\" font-weight=\"bold\">u\u2081 = 3 m\/s<\/text>\n  <rect x=\"430\" y=\"114\" width=\"78\" height=\"36\" rx=\"6\" fill=\"#7A1F2B\"><\/rect>\n  <text x=\"469\" y=\"137\" text-anchor=\"middle\" font-family=\"Arial, sans-serif\" font-size=\"13\" fill=\"#FAF6EE\">m\u2082 = 1 kg<\/text>\n  <text x=\"469\" y=\"100\" text-anchor=\"middle\" font-family=\"Arial, sans-serif\" font-size=\"13\" fill=\"#7A1F2B\" font-weight=\"bold\">u\u2082 = 0<\/text>\n  <text x=\"360\" y=\"188\" text-anchor=\"middle\" font-family=\"Arial, sans-serif\" font-size=\"15\" fill=\"#0A1628\">total p = (2)(3) + (1)(0) = <tspan fill=\"#C8932A\" font-weight=\"bold\">6 kg\u00b7m\/s<\/tspan><\/text>\n  <line x1=\"44\" y1=\"210\" x2=\"676\" y2=\"210\" stroke=\"#D9CFB8\" stroke-width=\"1\" stroke-dasharray=\"5 5\"><\/line>\n  <text x=\"44\" y=\"252\" font-family=\"Arial, sans-serif\" font-size=\"14\" fill=\"#7A1F2B\" font-weight=\"bold\">AFTER<\/text>\n  <line x1=\"44\" y1=\"320\" x2=\"676\" y2=\"320\" stroke=\"#D9CFB8\" stroke-width=\"3\"><\/line>\n  <rect x=\"240\" y=\"280\" width=\"96\" height=\"40\" rx=\"6\" fill=\"#0A1628\"><\/rect>\n  <text x=\"288\" y=\"305\" text-anchor=\"middle\" font-family=\"Arial, sans-serif\" font-size=\"14\" fill=\"#FAF6EE\">m\u2081 = 2 kg<\/text>\n  <line x1=\"340\" y1=\"270\" x2=\"384\" y2=\"270\" stroke=\"#C8932A\" stroke-width=\"4\" marker-end=\"url(#ahg)\"><\/line>\n  <text x=\"362\" y=\"262\" text-anchor=\"middle\" font-family=\"Arial, sans-serif\" font-size=\"13\" fill=\"#C8932A\" font-weight=\"bold\">v\u2081 = 1 m\/s<\/text>\n  <rect x=\"470\" y=\"282\" width=\"78\" height=\"36\" rx=\"6\" fill=\"#7A1F2B\"><\/rect>\n  <text x=\"509\" y=\"305\" text-anchor=\"middle\" font-family=\"Arial, sans-serif\" font-size=\"13\" fill=\"#FAF6EE\">m\u2082 = 1 kg<\/text>\n  <line x1=\"552\" y1=\"270\" x2=\"632\" y2=\"270\" stroke=\"#C8932A\" stroke-width=\"4\" marker-end=\"url(#ahg)\"><\/line>\n  <text x=\"592\" y=\"262\" text-anchor=\"middle\" font-family=\"Arial, sans-serif\" font-size=\"13\" fill=\"#C8932A\" font-weight=\"bold\">v\u2082 = 4 m\/s<\/text>\n  <text x=\"360\" y=\"358\" text-anchor=\"middle\" font-family=\"Arial, sans-serif\" font-size=\"15\" fill=\"#0A1628\">total p = (2)(1) + (1)(4) = <tspan fill=\"#C8932A\" font-weight=\"bold\">6 kg\u00b7m\/s<\/tspan><\/text>\n<\/svg>\n<p style=\"text-align:center;font-size:13px;color:#1F2E47;font-style:italic;margin-top:-6px;\">The carts trade velocity in the collision, yet the total momentum (6&nbsp;kg\u00b7m\/s) is identical before and after.<\/p>\n\n<h2>The Conservation of Momentum Formula<\/h2>\n\n<p>For two objects colliding in a straight line, the law is written as a single, balanced equation. Initial velocities use <em>u<\/em>; final velocities use <em>v<\/em>.<\/p>\n\n<div class=\"pf-formula\">m\u2081u\u2081 + m\u2082u\u2082 = m\u2081v\u2081 + m\u2082v\u2082<\/div>\n\n<p>Every term is built from one simpler quantity \u2014 the momentum of a single object:<\/p>\n\n<div class=\"pf-formula\">p = m\u00b7v<\/div>\n\n<p>Here is what each symbol means, with its SI unit:<\/p>\n\n<ul>\n  <li><strong>p<\/strong> \u2014 momentum of an object, in kilogram-metres per second (kg\u00b7m\/s)<\/li>\n  <li><strong>m\u2081, m\u2082<\/strong> \u2014 the masses of object 1 and object 2, in kilograms (kg)<\/li>\n  <li><strong>u\u2081, u\u2082<\/strong> \u2014 the velocities <em>before<\/em> the interaction, in metres per second (m\/s)<\/li>\n  <li><strong>v\u2081, v\u2082<\/strong> \u2014 the velocities <em>after<\/em> the interaction, in metres per second (m\/s)<\/li>\n<\/ul>\n\n<p>Because velocity is a vector, sign matters. Choose one direction as positive, then anything moving the other way is negative. Get the signs right and the equation does the rest.<\/p>\n\n<p>The same idea scales to any number of objects: the total momentum of the system, <em>\u03a3p<\/em>, is simply the sum of every individual <em>mv<\/em>, and that sum stays fixed.<\/p>\n\n<h2>How Conservation of Momentum Works<\/h2>\n\n<p>Why should the books balance so perfectly? The answer falls straight out of <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/newtons-laws-of-motion\/\">Newton&#8217;s laws of motion<\/a> \u2014 specifically the third law.<\/p>\n\n<p>When two objects collide, they push on each other. Newton&#8217;s third law says those pushes are equal in size and opposite in direction.<\/p>\n\n<ol>\n  <li>During contact, object A pushes B with force <em>F<\/em>. By the third law, B pushes A with force <em>\u2212F<\/em>.<\/li>\n  <li>They touch for exactly the same time, <em>\u0394t<\/em>. So the impulse (force \u00d7 time) on B is <em>F\u00b7\u0394t<\/em>, and the impulse on A is <em>\u2212F\u00b7\u0394t<\/em>.<\/li>\n  <li>Impulse equals change in momentum. So B gains exactly the momentum A loses: <em>\u0394p<\/em><sub>B<\/sub> = \u2212<em>\u0394p<\/em><sub>A<\/sub>.<\/li>\n  <li>Add them: <em>\u0394p<\/em><sub>A<\/sub> + <em>\u0394p<\/em><sub>B<\/sub> = 0. The total momentum does not change.<\/li>\n<\/ol>\n\n<p>That is the whole proof. The internal forces of a collision always come in equal-and-opposite pairs, so they cancel perfectly when you total the system. According to <a href=\"https:\/\/www1.grc.nasa.gov\/beginners-guide-to-aeronautics\/conservation-of-momentum-2\/\" target=\"_blank\" rel=\"noopener\">NASA&#8217;s Glenn Research Center<\/a>, momentum is &#8220;neither created nor destroyed, but only changed through the action of forces.&#8221;<\/p>\n\n<p>Use the interactive lab below to see it for yourself. Set the masses and starting speeds, choose how bouncy the collision is, and watch the total momentum readout hold steady while kinetic energy does not.<\/p>\n\n<div class=\"pf-sim-slot\"><div class=\"pf-sim-slot-header\"><span class=\"icon-dot\"><\/span><span class=\"label\">Momentum &amp; Collisions Lab<\/span><\/div><div class=\"pf-sim-slot-body\"><style>.pf-sim-frame{width:100%;border:none;height:600px}@media(max-width:760px){.pf-sim-frame{height:1000px}}<\/style><iframe src=\"\/labs\/momentum.html\" class=\"pf-sim-frame\" loading=\"lazy\"><\/iframe><\/div><\/div>\n\n<h2>When Is Momentum Actually Conserved?<\/h2>\n\n<p>Here is the catch students often miss: momentum is only conserved for an <strong>isolated system<\/strong> \u2014 one with no net external force acting on it.<\/p>\n\n<p>Internal forces (the objects pushing each other) always cancel. External forces do not. Gravity, friction, a wall, or your hand reaching in can all add or remove momentum from the system.<\/p>\n\n<p>So why does the law still work for a real pool break or car crash? Because collisions are <em>fast<\/em>. The forces between the objects are enormous compared with friction, and they act over such a tiny time that the external impulse is negligible. Momentum is conserved to an excellent approximation in that split second.<\/p>\n\n<p>In practice, friction and air resistance are external forces, so an isolated system is an idealisation. That is exactly why a rolling ball eventually stops \u2014 the ground and air are quietly draining its momentum. Read more about that drag in our guide to <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/what-is-friction\/\">what friction is<\/a>.<\/p>\n\n<h2>Elastic vs Inelastic Collisions<\/h2>\n\n<p>Momentum is conserved in <em>every<\/em> collision. <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/kinetic-energy-formula\/\">Kinetic energy<\/a> is not \u2014 and that distinction is what separates the two main collision types.<\/p>\n\n<h3>Elastic collisions<\/h3>\n\n<p>In a perfectly elastic collision, both momentum <em>and<\/em> kinetic energy are conserved. Objects bounce cleanly apart with no energy lost to heat or sound. Two billiard balls or gas molecules come very close to this ideal.<\/p>\n\n<h3>Inelastic collisions<\/h3>\n\n<p>In an inelastic collision, momentum is conserved but some kinetic energy is converted into heat, sound, or permanent deformation. When the objects stick together and move as one, it is <strong>perfectly inelastic<\/strong> \u2014 the maximum possible energy loss.<\/p>\n\n<div class=\"pf-table-scroll\" style=\"display:block;width:100%;max-width:100%;overflow-x:auto;-webkit-overflow-scrolling:touch;margin:1.5em 0;\">\n<table style=\"width:100%;border-collapse:collapse;word-break:break-word;\">\n  <thead>\n    <tr style=\"background:#0A1628;color:#FAF6EE;\">\n      <th style=\"padding:10px;border:1px solid #D9CFB8;text-align:left;\">Property<\/th>\n      <th style=\"padding:10px;border:1px solid #D9CFB8;text-align:left;\">Elastic<\/th>\n      <th style=\"padding:10px;border:1px solid #D9CFB8;text-align:left;\">Inelastic<\/th>\n      <th style=\"padding:10px;border:1px solid #D9CFB8;text-align:left;\">Perfectly inelastic<\/th>\n    <\/tr>\n  <\/thead>\n  <tbody>\n    <tr>\n      <td style=\"padding:10px;border:1px solid #D9CFB8;\"><strong>Momentum conserved?<\/strong><\/td>\n      <td style=\"padding:10px;border:1px solid #D9CFB8;\">Yes<\/td>\n      <td style=\"padding:10px;border:1px solid #D9CFB8;\">Yes<\/td>\n      <td style=\"padding:10px;border:1px solid #D9CFB8;\">Yes<\/td>\n    <\/tr>\n    <tr style=\"background:#F5F2EA;\">\n      <td style=\"padding:10px;border:1px solid #D9CFB8;\"><strong>Kinetic energy conserved?<\/strong><\/td>\n      <td style=\"padding:10px;border:1px solid #D9CFB8;\">Yes<\/td>\n      <td style=\"padding:10px;border:1px solid #D9CFB8;\">No (some lost)<\/td>\n      <td style=\"padding:10px;border:1px solid #D9CFB8;\">No (maximum lost)<\/td>\n    <\/tr>\n    <tr>\n      <td style=\"padding:10px;border:1px solid #D9CFB8;\"><strong>After the collision<\/strong><\/td>\n      <td style=\"padding:10px;border:1px solid #D9CFB8;\">Bounce apart<\/td>\n      <td style=\"padding:10px;border:1px solid #D9CFB8;\">Move separately, slower<\/td>\n      <td style=\"padding:10px;border:1px solid #D9CFB8;\">Stick and move together<\/td>\n    <\/tr>\n    <tr style=\"background:#F5F2EA;\">\n      <td style=\"padding:10px;border:1px solid #D9CFB8;\"><strong>Coefficient of restitution e<\/strong><\/td>\n      <td style=\"padding:10px;border:1px solid #D9CFB8;\">e = 1<\/td>\n      <td style=\"padding:10px;border:1px solid #D9CFB8;\">0 &lt; e &lt; 1<\/td>\n      <td style=\"padding:10px;border:1px solid #D9CFB8;\">e = 0<\/td>\n    <\/tr>\n    <tr>\n      <td style=\"padding:10px;border:1px solid #D9CFB8;\"><strong>Everyday example<\/strong><\/td>\n      <td style=\"padding:10px;border:1px solid #D9CFB8;\">Billiard balls, Newton&#8217;s cradle<\/td>\n      <td style=\"padding:10px;border:1px solid #D9CFB8;\">Most real-world crashes<\/td>\n      <td style=\"padding:10px;border:1px solid #D9CFB8;\">Clay ball hitting a wall; coupling train cars<\/td>\n    <\/tr>\n  <\/tbody>\n<\/table>\n<\/div>\n\n<h2>Real-World Examples of Conservation of Momentum<\/h2>\n\n<p>This is not just a textbook rule \u2014 it shapes everything from spaceflight to sport. Here are five places it shows up.<\/p>\n\n<h3>1. Rocket propulsion<\/h3>\n\n<p>A rocket throws hot exhaust gas downward at high speed. To keep total momentum constant, the rocket gains equal momentum upward. There is nothing to &#8220;push against&#8221; \u2014 the gas <em>is<\/em> the push. This is why rockets work in the vacuum of space.<\/p>\n\n<figure style=\"margin:32px auto;max-width:640px;text-align:center;\">\n  <img decoding=\"async\" src=\"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-content\/uploads\/2026\/06\/01_58_51a_remotesite-2-frame-8-scaled.avif\"\n       alt=\"Rocket launch demonstrating conservation of momentum as exhaust is expelled downward\"\n       loading=\"lazy\"\n       style=\"width:100%;height:auto;border-radius:4px;\" \/>\n  <figcaption style=\"font-size:13px;color:#1F2E47;font-style:italic;margin-top:8px;\">A rocket gains upward momentum equal to the downward momentum of its exhaust.<\/figcaption>\n<\/figure>\n\n<h3>2. Recoil of a gun<\/h3>\n\n<p>Before firing, the rifle-and-bullet system is at rest, so its total momentum is zero. The bullet leaves with forward momentum, so the rifle must recoil backward with an equal and opposite amount. The bullet is light and fast; the rifle is heavy and slow.<\/p>\n\n<svg role=\"img\" aria-label=\"Recoil diagram: a light object moves fast one way while a heavy object moves slowly the other way, so the total momentum stays zero\" viewBox=\"0 0 720 300\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" style=\"width:100%;height:auto;max-width:680px;display:block;margin:24px auto;\">\n  <defs>\n    <marker id=\"ahw\" markerWidth=\"9\" markerHeight=\"9\" refX=\"7\" refY=\"3\" orient=\"auto\"><path d=\"M0,0 L7,3 L0,6 Z\" fill=\"#C8932A\"><\/path><\/marker>\n  <\/defs>\n  <rect x=\"2\" y=\"2\" width=\"716\" height=\"296\" rx=\"12\" fill=\"#FAF6EE\" stroke=\"#D9CFB8\" stroke-width=\"2\"><\/rect>\n  <text x=\"360\" y=\"38\" text-anchor=\"middle\" font-family=\"Georgia, serif\" font-size=\"19\" fill=\"#0A1628\" font-weight=\"bold\">Recoil: momentum still sums to zero<\/text>\n  <circle cx=\"360\" cy=\"150\" r=\"9\" fill=\"none\" stroke=\"#1F2E47\" stroke-width=\"2\" stroke-dasharray=\"3 3\"><\/circle>\n  <text x=\"360\" y=\"190\" text-anchor=\"middle\" font-family=\"Arial, sans-serif\" font-size=\"13\" fill=\"#1F2E47\">start: at rest, p = 0<\/text>\n  <rect x=\"150\" y=\"128\" width=\"96\" height=\"48\" rx=\"6\" fill=\"#7A1F2B\"><\/rect>\n  <text x=\"198\" y=\"157\" text-anchor=\"middle\" font-family=\"Arial, sans-serif\" font-size=\"13\" fill=\"#FAF6EE\">heavy<\/text>\n  <line x1=\"150\" y1=\"200\" x2=\"78\" y2=\"200\" stroke=\"#C8932A\" stroke-width=\"5\" marker-end=\"url(#ahw)\"><\/line>\n  <text x=\"150\" y=\"224\" text-anchor=\"middle\" font-family=\"Arial, sans-serif\" font-size=\"13\" fill=\"#C8932A\" font-weight=\"bold\">\u2212p (slow)<\/text>\n  <rect x=\"520\" y=\"138\" width=\"52\" height=\"34\" rx=\"5\" fill=\"#0A1628\"><\/rect>\n  <text x=\"546\" y=\"160\" text-anchor=\"middle\" font-family=\"Arial, sans-serif\" font-size=\"12\" fill=\"#FAF6EE\">light<\/text>\n  <line x1=\"572\" y1=\"200\" x2=\"648\" y2=\"200\" stroke=\"#C8932A\" stroke-width=\"5\" marker-end=\"url(#ahw)\"><\/line>\n  <text x=\"572\" y=\"224\" text-anchor=\"middle\" font-family=\"Arial, sans-serif\" font-size=\"13\" fill=\"#C8932A\" font-weight=\"bold\">+p (fast)<\/text>\n  <text x=\"360\" y=\"270\" text-anchor=\"middle\" font-family=\"Arial, sans-serif\" font-size=\"16\" fill=\"#0A1628\" font-weight=\"bold\">(+p) + (\u2212p) = 0<\/text>\n<\/svg>\n<p style=\"text-align:center;font-size:13px;color:#1F2E47;font-style:italic;margin-top:-6px;\">In recoil and explosions, the pieces fly apart with momenta that still add up to the original total.<\/p>\n\n<h3>3. Newton&#8217;s cradle<\/h3>\n\n<p>Lift one ball and release it, and a single ball swings out the far end at the same speed. Lift two, and two swing out. The desk toy is a near-elastic chain that passes momentum (and energy) straight through the line of stationary balls.<\/p>\n\n<h3>4. Billiards and the break<\/h3>\n\n<p>When the cue ball strikes the pack, its momentum spreads among every ball it touches. Add up all the scattered momenta as vectors and you recover the cue ball&#8217;s original momentum almost exactly \u2014 a vivid demonstration on a felt table.<\/p>\n\n<h3>5. Ice skaters pushing apart<\/h3>\n\n<p>Two stationary skaters push off each other and glide backward in opposite directions. They started with zero total momentum, so they must end with zero \u2014 the lighter skater simply moves away faster than the heavier one.<\/p>\n\n<h2>Common Misconceptions About Conservation of Momentum<\/h2>\n\n<h3>&#8220;Momentum and kinetic energy are the same thing.&#8221;<\/h3>\n\n<p>They are not. Momentum (<em>mv<\/em>) is a vector and is conserved in every collision. Kinetic energy (\u00bd<em>mv<\/em>\u00b2) is a scalar and is only conserved in elastic collisions. A common exam slip is to assume energy is always conserved \u2014 it usually is not.<\/p>\n\n<h3>&#8220;Momentum is always conserved, no matter what.&#8221;<\/h3>\n\n<p>Only for an isolated system. If an external force \u2014 friction, gravity, a wall \u2014 acts on the system, its total momentum changes. The law applies to the system as a whole, not to a single object being pushed.<\/p>\n\n<h3>&#8220;The heavier object always &#8216;wins&#8217; a collision.&#8221;<\/h3>\n\n<p>Mass alone does not decide the outcome \u2014 momentum does, and that depends on mass <em>and<\/em> velocity. A light, fast object can carry more momentum than a heavy, slow one.<\/p>\n\n<h3>&#8220;In an explosion, momentum is created from nothing.&#8221;<\/h3>\n\n<p>No. If the object started at rest, the fragments fly off with momenta that cancel to zero. The chemical energy creates kinetic energy, but the total momentum is unchanged.<\/p>\n\n<h2>How Conservation of Momentum Connects to Newton&#8217;s Laws, Impulse and Energy<\/h2>\n\n<p>Momentum sits at the centre of a web of ideas you have likely already met. Seeing the links makes each one easier.<\/p>\n\n<h3>Newton&#8217;s second law, restated<\/h3>\n\n<p>Newton originally wrote his second law in terms of momentum: force equals the rate of change of momentum. Our guide to <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/newtons-second-law\/\">Newton&#8217;s second law<\/a> shows how <em>F = ma<\/em> is just the constant-mass version of <em>F = \u0394p\/\u0394t<\/em>.<\/p>\n\n<h3>Impulse \u2014 the momentum &#8220;kick&#8221;<\/h3>\n\n<p>Impulse is force applied over time, and it equals the change in momentum. This is why airbags and crumple zones save lives: they stretch out <em>\u0394t<\/em>, which lowers the force needed to change your momentum.<\/p>\n\n<div class=\"pf-formula\">J = F\u00b7\u0394t = \u0394p = mv \u2212 mu<\/div>\n\n<ul>\n  <li><strong>J<\/strong> \u2014 impulse, in newton-seconds (N\u00b7s), equal to kg\u00b7m\/s<\/li>\n  <li><strong>F<\/strong> \u2014 average force, in newtons (N)<\/li>\n  <li><strong>\u0394t<\/strong> \u2014 contact time, in seconds (s)<\/li>\n  <li><strong>\u0394p<\/strong> \u2014 change in momentum, in kg\u00b7m\/s<\/li>\n<\/ul>\n\n<h3>Energy \u2014 the other great conservation law<\/h3>\n\n<p>Momentum and energy are independent bookkeepers. A collision can conserve momentum while losing kinetic energy to heat and sound. To see how energy moves and transforms, visit our explainer on <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/what-is-energy-in-physics\/\">what energy is in physics<\/a>.<\/p>\n\n<p>At the deepest level, conservation of momentum reflects a symmetry of nature: the laws of physics are the same everywhere in space. That link between symmetry and conservation, formalised by Noether&#8217;s theorem, is touched on in this overview of <a href=\"https:\/\/en.wikipedia.org\/wiki\/Momentum\" target=\"_blank\" rel=\"noopener\">momentum<\/a>.<\/p>\n\n<h2>Worked Problems<\/h2>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 1<\/div><div class=\"pf-problem-question\">A 2.0 kg trolley moving at 3.0 m\/s strikes a stationary 1.0 kg trolley. They stick together. Find their common velocity after impact.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<strong>Solution:<\/strong><br>\nStep 1: Conservation of momentum (perfectly inelastic): m\u2081u\u2081 + m\u2082u\u2082 = (m\u2081 + m\u2082)v<br>\nStep 2: Substitute: (2.0)(3.0) + (1.0)(0) = (2.0 + 1.0)v<br>\nStep 3: Solve: 6.0 = 3.0v \u2192 v = 2.0 m\/s<br>\n<strong>Answer: 2.0 m\/s in the original direction<\/strong>\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 2<\/div><div class=\"pf-problem-question\">A 4.0 kg rifle fires a 0.020 kg bullet at 400 m\/s. Find the recoil velocity of the rifle.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<strong>Solution:<\/strong><br>\nStep 1: Total momentum before firing is zero: 0 = m_bullet\u00b7v_bullet + m_rifle\u00b7v_rifle<br>\nStep 2: Substitute: 0 = (0.020)(400) + (4.0)(v_rifle) \u2192 0 = 8.0 + 4.0\u00b7v_rifle<br>\nStep 3: Solve: v_rifle = \u22128.0 \/ 4.0 = \u22122.0 m\/s<br>\n<strong>Answer: 2.0 m\/s backward (the minus sign shows the opposite direction)<\/strong>\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 3<\/div><div class=\"pf-problem-question\">A 3.0 kg ball moving at 4.0 m\/s hits a stationary 2.0 kg ball. After the collision the 3.0 kg ball moves at 0.80 m\/s in the same direction. Find the velocity of the 2.0 kg ball.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<strong>Solution:<\/strong><br>\nStep 1: m\u2081u\u2081 + m\u2082u\u2082 = m\u2081v\u2081 + m\u2082v\u2082<br>\nStep 2: Substitute: (3.0)(4.0) + (2.0)(0) = (3.0)(0.80) + (2.0)v\u2082<br>\nStep 3: Solve: 12 = 2.4 + 2.0v\u2082 \u2192 2.0v\u2082 = 9.6 \u2192 v\u2082 = 4.8 m\/s<br>\n<strong>Answer: 4.8 m\/s in the original direction<\/strong>\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 4<\/div><div class=\"pf-problem-question\">A 1.0 kg cart moving right at 5.0 m\/s collides head-on with a 2.0 kg cart moving left at 2.0 m\/s. They couple together. Find the velocity of the pair.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<strong>Solution:<\/strong><br>\nStep 1: Take right as positive. Total momentum = m\u2081u\u2081 + m\u2082u\u2082<br>\nStep 2: Substitute: (1.0)(+5.0) + (2.0)(\u22122.0) = 5.0 \u2212 4.0 = 1.0 kg\u00b7m\/s<br>\nStep 3: Divide by total mass: v = 1.0 \/ (1.0 + 2.0) = 0.33 m\/s<br>\n<strong>Answer: 0.33 m\/s to the right<\/strong>\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 5<\/div><div class=\"pf-problem-question\">In an elastic collision, a 0.50 kg ball moving at 6.0 m\/s strikes an identical stationary 0.50 kg ball head-on. Find both final velocities and verify energy is conserved.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<strong>Solution:<\/strong><br>\nStep 1: For an elastic head-on collision between equal masses, the velocities simply exchange.<br>\nStep 2: So v\u2081 = 0 and v\u2082 = 6.0 m\/s.<br>\nStep 3: Check momentum: before = (0.50)(6.0) = 3.0; after = (0.50)(0) + (0.50)(6.0) = 3.0 \u2713<br>\nStep 4: Check KE: before = \u00bd(0.50)(6.0)\u00b2 = 9.0 J; after = \u00bd(0.50)(6.0)\u00b2 = 9.0 J \u2713<br>\n<strong>Answer: the first ball stops; the second moves off at 6.0 m\/s<\/strong>\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 6<\/div><div class=\"pf-problem-question\">A 0.15 kg ball hits a wall at 20 m\/s and rebounds at 15 m\/s. If the contact lasts 0.020 s, find the average force the wall exerts on the ball.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<strong>Solution:<\/strong><br>\nStep 1: Impulse-momentum theorem: F\u00b7\u0394t = \u0394p = m(v \u2212 u). Take the incoming direction as positive, so rebound is negative.<br>\nStep 2: \u0394p = (0.15)(\u221215 \u2212 20) = (0.15)(\u221235) = \u22125.25 kg\u00b7m\/s<br>\nStep 3: F = \u0394p \/ \u0394t = \u22125.25 \/ 0.020 = \u2212262.5 N<br>\n<strong>Answer: about 263 N, directed away from the wall (the minus sign shows direction)<\/strong>\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 7<\/div><div class=\"pf-problem-question\">A 3.0 kg object at rest explodes into two pieces. A 1.0 kg piece flies east at 8.0 m\/s. Find the velocity of the remaining 2.0 kg piece.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<strong>Solution:<\/strong><br>\nStep 1: Total momentum before the explosion is zero: 0 = m\u2081v\u2081 + m\u2082v\u2082<br>\nStep 2: Substitute (east positive): 0 = (1.0)(+8.0) + (2.0)v\u2082<br>\nStep 3: Solve: v\u2082 = \u22128.0 \/ 2.0 = \u22124.0 m\/s<br>\n<strong>Answer: 4.0 m\/s west<\/strong>\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 8<\/div><div class=\"pf-problem-question\">A 2.0 kg ball moving at 3.0 m\/s collides elastically and head-on with a stationary 1.0 kg ball. Find both final velocities using the elastic-collision equations.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<strong>Solution:<\/strong><br>\nStep 1: For an elastic head-on collision with target at rest: v\u2081 = [(m\u2081\u2212m\u2082)\/(m\u2081+m\u2082)]u\u2081 and v\u2082 = [2m\u2081\/(m\u2081+m\u2082)]u\u2081<br>\nStep 2: v\u2081 = [(2.0\u22121.0)\/(3.0)](3.0) = (1\/3)(3.0) = 1.0 m\/s<br>\nStep 3: v\u2082 = [2(2.0)\/(3.0)](3.0) = (4\/3)(3.0) = 4.0 m\/s<br>\nStep 4: Check: momentum before = 6.0, after = (2.0)(1.0)+(1.0)(4.0) = 6.0 \u2713; KE before = 9.0 J, after = 1.0 + 8.0 = 9.0 J \u2713<br>\n<strong>Answer: v\u2081 = 1.0 m\/s and v\u2082 = 4.0 m\/s, both forward<\/strong>\n<\/div><\/details><\/div>\n\n<h2>Frequently Asked Questions<\/h2>\n\n<details class=\"pf-faq-item\"><summary>What is the law of conservation of momentum in simple words?<\/summary><div class=\"pf-faq-item-answer\">\nIt means the total momentum of a closed group of objects never changes. Momentum (mass \u00d7 velocity) can move from one object to another during a collision, but the grand total before equals the grand total after, as long as no outside force acts on the system.\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>Is momentum conserved in an inelastic collision?<\/summary><div class=\"pf-faq-item-answer\">\nYes. Momentum is conserved in every collision, elastic or inelastic, provided the system is isolated. What changes in an inelastic collision is kinetic energy: some is converted into heat, sound, or deformation. Only momentum is guaranteed to stay constant \u2014 kinetic energy is not.\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>Is momentum always conserved?<\/summary><div class=\"pf-faq-item-answer\">\nOnly for an isolated system with no net external force. Internal forces between objects cancel by Newton&#8217;s third law, so they never change the total. External forces such as friction, gravity, or a wall can add or remove momentum, so in those cases the system&#8217;s total momentum is not conserved.\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>What is the difference between momentum and kinetic energy?<\/summary><div class=\"pf-faq-item-answer\">\nMomentum is mass \u00d7 velocity (mv), a vector with direction, and it is conserved in all collisions. Kinetic energy is \u00bdmv\u00b2, a scalar with no direction, and it is only conserved in elastic collisions. An object can keep its momentum while losing kinetic energy to heat or sound.\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>Does conservation of momentum apply to explosions and recoil?<\/summary><div class=\"pf-faq-item-answer\">\nYes. If an object starts at rest, its total momentum is zero, so the fragments must fly apart with momenta that cancel to zero. This is exactly why a gun recoils and a rocket accelerates: the forward momentum of one part is balanced by the backward momentum of another.\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>Why is momentum conserved?<\/summary><div class=\"pf-faq-item-answer\">\nAt a basic level, it follows from Newton&#8217;s third law: the forces two objects exert on each other are equal and opposite, so the momentum one gains the other loses. More deeply, it reflects a symmetry of nature \u2014 the laws of physics are the same at every point in space.\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>What units is momentum measured in?<\/summary><div class=\"pf-faq-item-answer\">\nIn SI units, momentum is measured in kilogram-metres per second (kg\u00b7m\/s). This comes directly from its definition, p = mv, where mass is in kilograms and velocity is in metres per second. The same unit, written as the newton-second (N\u00b7s), is used for impulse.\n<\/div><\/details>\n","protected":false},"excerpt":{"rendered":"<p>Conservation of momentum explained simply: the formula, elastic and inelastic collisions, eight worked examples, common misconceptions, and an interactive collision lab for students.<\/p>\n","protected":false},"author":1,"featured_media":240,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[2],"tags":[111,110,108,109,112],"class_list":["post-239","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-mechanics","tag-collisions","tag-conservation-of-momentum","tag-impulse","tag-momentum","tag-newtons-third-law"],"_links":{"self":[{"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts\/239","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/comments?post=239"}],"version-history":[{"count":1,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts\/239\/revisions"}],"predecessor-version":[{"id":242,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts\/239\/revisions\/242"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/media\/240"}],"wp:attachment":[{"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/media?parent=239"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/categories?post=239"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/tags?post=239"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}