{"id":11297,"date":"2026-07-09T14:28:03","date_gmt":"2026-07-09T14:28:03","guid":{"rendered":"https:\/\/www.myengineeringbuddy.com\/blog\/the-law-of-conservation-of-linear-momentum\/"},"modified":"2026-07-12T04:24:29","modified_gmt":"2026-07-12T04:24:29","slug":"the-law-of-conservation-of-linear-momentum","status":"publish","type":"post","link":"https:\/\/www.myengineeringbuddy.com\/blog\/the-law-of-conservation-of-linear-momentum\/","title":{"rendered":"The Law of Conservation of Linear Momentum: Explained with Examples"},"content":{"rendered":"\n<div style=\"background-color:#f8f8f8; border-left:4px solid #d0d0d0; padding:12px 16px; margin-bottom:20px;\"><strong>Key Takeaways<\/strong>\n<ul>\n<li>Linear momentum is conserved when the net external force on a system is zero.<\/li>\n<li>Newton&#8217;s second law directly proves the conservation of linear momentum.<\/li>\n<li>Gun recoil, collisions, and explosions all demonstrate this conservation law.<\/li>\n<li>In elastic collisions between equal masses, the first object stops completely.<\/li>\n<li>Center of mass position stays fixed when no external horizontal force acts.<\/li>\n<\/ul><\/div>\n\n<p>The law of conservation of momentum is divided into two parts:<\/p>\n<ol>\n<li>The law of conservation of linear momentum<\/li>\n<li>The law of conservation of angular momentum<\/li>\n<\/ol>\n<p>In this article, we will learn about the law of conservation of LINEAR momentum.<\/p>\n<p>If you want to learn Physics, a <a href=\"https:\/\/www.myengineeringbuddy.com\/online-tutoring\/online-physics-tutor\/\">online Physics tutor<\/a> can help \u2014 feel free to contact us at <a href=\"mailto:meb@myengineeringbuddy.com\">meb@myengineeringbuddy.com<\/a> or on <a href=\"https:\/\/api.whatsapp.com\/send?phone=918971383660&#038;text=Hi.%20I%20am%20not%20sure%20how%20to%20proceed%20on%20your%20website%20MEB.%20Need%20help.%20Please%20assist.\" target=\"_blank\" rel=\"noopener\">WhatsApp<\/a>.<\/p>\n\n<p>Students who enjoy working through physics problems often find that strong programming skills complement their analytical thinking. If you are also studying coding alongside physics, you might benefit from working with a <a href=\"https:\/\/www.myengineeringbuddy.com\/subject\/java\/\">Java tutor<\/a> to strengthen your computational problem-solving.<\/p>\n\n<h2>The Law of Conservation of Linear Momentum<\/h2>\n<p><strong>The law of conservation of linear momentum says that when the resultant external force acting on a system is zero, the system&#8217;s total momentum (vector sum) remains constant.<\/strong><\/p>\n\n<h3>Examples of Linear Momentum Conservation<\/h3>\n<p>Explosions, the disintegration of nuclei, recoil of a gun, collisions, etc., can be explained using the law of conservation of linear momentum.<\/p>\n<p><strong>The recoil of the gun is one of the best daily life examples.<\/strong><\/p>\n<p>In this example, when a shot is fired from the gun, while the shot moves forwards, the gun moves backward. This motion of the gun is called recoil of the gun. When a gun of mass M fires a bullet of mass m with a muzzle velocity v, then the gun recoils with a velocity V given by $V = frac{{mv}}{M}$.<\/p>\n<ul>\n<li>$F = frac{{dP}}{{dt}}$, If $F = 0$, then $P$ remains constant or conserved.<\/li>\n<\/ul>\n\n<p>For students exploring simulation-based approaches to mechanics problems, working with an <a href=\"https:\/\/www.myengineeringbuddy.com\/subject\/abaqus\/\">Abaqus expert<\/a> can provide useful exposure to finite element analysis of physical systems.<\/p>\n\n<h3>Proof of the Linear Momentum Conservation Law<\/h3>\n<p>As we know, from Newton&#8217;s 2<sup>nd<\/sup> law of motion, the rate of change of momentum of a body is directly proportional to the resultant (or) net external force acting on the body and taking place along the direction of the force.<\/p>\n<p>Mathematically we can write, ${F_{net}} = frac{{dP}}{{dt}}$<\/p>\n<p>If ${F_{net}} = 0$; this means, $frac{{dP}}{{dt}} = 0$.<\/p>\n<p>Hence, the system&#8217;s total momentum will not change with respect to time. This will remain constant.<\/p>\n<p>If $sum {F_{ext}} = 0,;,{left( {sum {P_{sy}}} right)_i} = {left( {sum {P_{sy}}} right)_f}$<\/p>\n<ul>\n<li>If ${F_{net}} = frac{{dL}}{{dt}} = 0,;$ then L remains constant or conserved.<\/li>\n<\/ul>\n<p>As we know, the angular momentum of a body is defined as,<\/p>\n<p>$L = sum left( {{r_i} times {P_i}} right)$<\/p>\n<p>Now, differentiating with respect to time,<\/p>\n<p>$frac{{dL}}{{dt}} = frac{d}{{dt}}sum left( {{r_i} times {P_i}} right) = sumlimits_i^{} {left[ {frac{{d{r_i}}}{{dt}} times {P_i} + r times frac{{d{P_i}}}{{dt}}} right]} $<\/p>\n<p>$frac{{dL}}{{dt}} = sumlimits_i^{} {left[ {{v_i} times m{v_i} + {r_i} times {F_i}} right]} $<\/p>\n<p>Here, ${v_i} times m{v_i} = 0$, so,<\/p>\n<p>$frac{{dL}}{{dt}} = sumlimits_i^{} {{r_i} times {F_i}} = {Gamma _{total}}$<\/p>\n<p>Here, ${F_i}$ is the total force acting on the i<sup>th<\/sup> particle. This includes any external force as well as the forces on the i<sup>th<\/sup> particle by all the other particles. Taking summation of both sides, internal torque adds to zero.<\/p>\n<p>$sumlimits_{}^{} {{Gamma ^{ext}}} = frac{{dL}}{{dt}}$<\/p>\n<p>Here, ${Gamma ^{ext}}$ is the total torque due to all the external forces acting on the system.<\/p>\n<p>If the torque of external forces about any particular point is equal to zero, then $frac{{dL}}{{dt}} = 0$. This means that the system&#8217;s total angular momentum about that particular point will remain constant.<\/p>\n<p>${left( {sumlimits_{}^{} {{L_{sys,,0}}} } right)_i} = {left( {sumlimits_{}^{} {{L_{sys,,0}}} } right)_f}$ if $Gamma _0^{ext} = 0$.<\/p>\n\n<p>Understanding the mathematical structure behind conservation laws can also sharpen your skills in <a href=\"https:\/\/www.myengineeringbuddy.com\/subject\/c-plus-plus-programming\/\">C++ programming<\/a>, where similar logical frameworks govern algorithm design.<\/p>\n\n<p>If you are curious about how structured learning platforms compare when studying topics like this, our <a href=\"https:\/\/www.myengineeringbuddy.com\/blog\/codecademy-reviews-alternatives-pricing-offerings\/\">Codecademy review covering alternatives, pricing, and offerings<\/a> may help you choose the right resource.<\/p>\n\n<h2>Conservation of Linear Momentum: Solved Problems<\/h2>\n\n<h3>Problem 1: A Person Jumping on a Moving Car<\/h3>\n<p><strong>A car of mass M is moving with uniform velocity v on a horizontal road when a person drops himself on it from above. Taking mass of person to be m; What will be the car&#8217;s velocity after the event?<\/strong><\/p>\n<p><strong>Ans.<\/strong>: Consider the car plus the person as a system. In the horizontal direction, there is no external force since the momentum of that person is equal to zero.<\/p>\n<p>Initially, the horizontal momentum of the system = MV.<\/p>\n<p>Finally, the person sticks to the car&#8217;s roof, so they move with equal velocity V. Now, final horizontal momentum of system = (M + m)V.<\/p>\n<p>Now, from the principle of conservation of angular momentum,<\/p>\n<p>$Mv = (M + m)V$<\/p>\n<p>$V = frac{{Mv}}{{M + m}}$<\/p>\n\n<h3>Problem 2: Force of Recoil of a Machine Gun<\/h3>\n<p><strong>The hero of a stunt film fires bullets from a machine gun, each at a speed of 1km\/s. If he fires 20 bullets in 4 seconds, what average force does he exert against the machine gun during this period?<\/strong><\/p>\n<p><strong>Ans.<\/strong>: The momentum of each bullet,<\/p>\n<p>$P = (0.05,kg),(1000,m{rm{\/s)}} = 50,{rm{kg &#8211; m\/s}}$<\/p>\n<p>The gun is imparted this much momentum by each bullet fired. So, the rate of change of momentum of the gun (total)<\/p>\n<p>$ = (50,k{rm{g &#8211; m\/s) times }}frac{{({rm{Number of bullets}})}}{{Delta t}}$<\/p>\n<p>$ = (50,k{rm{g &#8211; m\/s) times }}frac{{20}}{4} = 250,N$<\/p>\n\n<p>Analytical problem-solving of this kind also appears in engineering optimization contexts. Our <a href=\"https:\/\/www.myengineeringbuddy.com\/blog\/the-complete-guide-to-operations-research-from-complex-problems-to-career-success\/\">complete guide to operations research<\/a> explores how mathematical methods tackle complex real-world problems.<\/p>\n\n<h3>Problem 3: Two Men Walking on a Boat<\/h3>\n<p><strong>Mr. John (50 kg) and Mr. Alex (60 kg) are sitting at the two extremes of a 4m long boat (40kg) still in the water. They come to the middle of the boat to discuss a mechanical problem. Neglecting friction with water, how far does the boat move on the water during the process?<\/strong><\/p>\n<p><strong>Ans.<\/strong>: (Mr. John + Mr. Alex + Boat) as a system.<\/p>\n<p>Here velocity of the center of the system along the x-axis equals zero. The sum of all external forces acting on the system along the x-axis is equal to zero; hence, the momentum (total) of the system along the x-axis must be conserved.<\/p>\n<p>Since, ${({V_{cm}})_x} = 0quad Rightarrow quad frac{d}{{dt}}{X_{cm}} = 0$<\/p>\n<p>$ Rightarrow quad {({X_{cm}})_i} = {({X_{cm}})_f}$ &nbsp;&nbsp;&nbsp;&nbsp;\u2026(1)<\/p>\n<img class=\"lazyload\" decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAAAAACH5BAEKAAEALAAAAAABAAEAAAICTAEAOw==\" data-orig-src=\"https:\/\/myengineeringbuddy.com\/wp-content\/uploads\/2022\/07\/fig.3-300x220.png\" alt=\"Diagram showing two men walking on a boat on water\">\n<p>Let the boat moves along the \u2013ve x-axis by a distance $ell $ as they approach the middle of the boat.<\/p>\n<p>Now, from (1),<\/p>\n<p>$frac{{50 times d + 40(d + 2) + 60(4 + d)}}{{(60 + 40 + 50)}} = frac{{(50 + 40 + 60)(2)}}{{(60 + 40 + 50)}}$<\/p>\n<p>$ Rightarrow quad 50d + 40d + 80 + 240 + 60d = 150 times 2$<\/p>\n<p>$ Rightarrow quad 150d + 320 = 300quad Rightarrow quad 150d = &#8211; 20$<\/p>\n<p>$ Rightarrow quad d = &#8211; frac{2}{{15}}m = &#8211; frac{2}{{15}} times 100,cm$<\/p>\n<p>Hence, boat will move along +ve x-direction by a distance of $frac{2}{{15}}m$ or $ approx 13,cm$.<\/p>\n\n<h3>Problem 4: Head-on Collision of Two Perfectly Elastic Balls<\/h3>\n<p>A marble going at a speed of 2 m\/s hits another marble of equal mass at rest. If the collision is perfectly elastic, then find the velocity of the first marble after the collision.<\/p>\n<p><strong>Ans.<\/strong>: Assuming both marbles as a system,<\/p>\n<img class=\"lazyload\" decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAAAAACH5BAEKAAEALAAAAAABAAEAAAICTAEAOw==\" data-orig-src=\"https:\/\/myengineeringbuddy.com\/wp-content\/uploads\/2022\/07\/fig.2-300x149.png\" alt=\"Diagram showing head-on collision of two elastic marbles\">\n<p>Let speeds of marbles become ${v_1}&#8217;$ and ${v_2}&#8217;$ after the collision.<\/p>\n<p>Here $left( {sumlimits_{}^{} {{F_{ext}}} } right)$ along x-axis equals to zero.<\/p>\n<p>${left( {{P_{sys}}} right)_i} = {left( {{P_{sys}}} right)_f}$<\/p>\n<p>$ Rightarrow quad {m_A}{v_1} + {m_B}{v_2} = {m_A}{v_1}&#8217; + {m_B}{v_2}&#8217;$<\/p>\n<p>Since ${m_A} = {m_B}$ and ${v_2} = 0$,<\/p>\n<p>$ Rightarrow quad {v_1} + 0 = {v_1}&#8217; + {v_2}&#8217;$<\/p>\n<p>$ Rightarrow quad 2,{rm{m\/s}},hat i = {v_1}&#8217; + {v_2}&#8217;$ &nbsp;&nbsp;&nbsp;&nbsp;\u2026(1)<\/p>\n<p>Also, in this case, we can write, the velocity of approach = velocity of separation:<\/p>\n<p>${v_1} &#8211; {v_2} = {v_2}&#8217; &#8211; {v_1}&#8217;$<\/p>\n<p>$ Rightarrow quad 2,{rm{m\/s}},hat i &#8211; 0 = {v_2}&#8217; &#8211; {v_1}&#8217;$ &nbsp;&nbsp;&nbsp;&nbsp;\u2026(2)<\/p>\n<p>Subtracting (2) from (1),<\/p>\n<p>${v_1}&#8217; + {v_2}&#8217; &#8211; ({v_2}&#8217; &#8211; {v_1}&#8217;) = 0$<\/p>\n<p>$2{v_1}&#8217; = 0quad Rightarrow quad {v_1}&#8217; = 0$<\/p>\n<p>So, the final velocity of the first marble will be zero.<\/p>\n\n<h3>Problem 5: Falling off a Vertical Pen on a Frictionless Surface<\/h3>\n<p>A uniform pen is kept vertically on a smooth horizontal surface at a point O. If it is rotated slightly and released, it falls on the horizontal surface. The lower end will remain:<\/p>\n<p>(a) &nbsp;&nbsp;at point O<\/p>\n<p>(b) &nbsp;&nbsp;at a distance, less than $frac{ell }{2}$ from O<\/p>\n<p>(c) &nbsp;&nbsp;at a distance, $frac{ell }{2}$ from O<\/p>\n<p>(d) &nbsp;&nbsp;at a distance larger than $frac{ell }{2}$ from O<\/p>\n<p><strong>Ans.<\/strong>: Since the given pen is slightly rotated, the velocity of the center of mass along the x-axis is equal to zero. Hence, position of centre of mass of the pen $({X_{cm}})$ will be constant.<\/p>\n<p>So, ${({X_{cm}})_i} = {({X_{cm}})_f}$.<\/p>\n<img class=\"lazyload\" decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAAAAACH5BAEKAAEALAAAAAABAAEAAAICTAEAOw==\" data-orig-src=\"https:\/\/myengineeringbuddy.com\/wp-content\/uploads\/2022\/07\/fig.1-234x300.png\" alt=\"Diagram showing a vertical pen falling on a frictionless horizontal surface\">\n<p>Hence, the lower end will be at a distance $frac{ell }{2}$ left with respect to the point O when the pen becomes horizontal on the surface.<\/p>\n\n<p>For students who want to explore how academic writing services compare when preparing physics lab reports, our <a href=\"https:\/\/www.myengineeringbuddy.com\/blog\/5staressays-reviews-alternatives-pricing-offerings\/\">5StarEssays review covering alternatives, pricing, and offerings<\/a> provides a useful breakdown.<\/p>\n\n<p>Physics concepts like momentum conservation also appear in interdisciplinary contexts. Our overview of <a href=\"https:\/\/www.myengineeringbuddy.com\/blog\/religious-studies-tutoring-online\/\">religious studies tutoring online<\/a> shows how structured academic support applies across very different subject areas.<\/p>\n\n<p>For those also studying hardware design alongside physics, connecting with a <a href=\"https:\/\/www.myengineeringbuddy.com\/subject\/vlsi-design\/\">VLSI design tutor<\/a> can help bridge theoretical mechanics with electronic systems engineering.<\/p>\n\n<p>Learners who want to reinforce their physics problem-solving with programming practice may also find it useful to work with a <a href=\"https:\/\/www.myengineeringbuddy.com\/subject\/c-programming\/\">C programming tutor<\/a> to build computational skills alongside their physics studies.<\/p>\n\n<p>Related: <a href=\"https:\/\/www.myengineeringbuddy.com\/the-law-of-conservation-of-angular-momentum\/\">The law of conservation of angular momentum<\/a> | <a href=\"https:\/\/www.myengineeringbuddy.com\/the-law-of-conservation-of-mechanical-energy\/\">The law of conservation of mechanical energy<\/a> | <a href=\"https:\/\/www.myengineeringbuddy.com\/homework-help\/physics-homework-help\/\">Physics Homework Help<\/a><\/p>\n\n<h2>Related Reading<\/h2>\n<ul>\n<li><a href=\"https:\/\/www.myengineeringbuddy.com\/blog\/skillshare-reviews-alternatives-pricing-offerings\/\">Skillshare Reviews: Alternatives, Pricing, and Offerings<\/a><\/li>\n<li><a href=\"https:\/\/www.myengineeringbuddy.com\/blog\/project-management-tutoring-online-boost-grades-skills-and-career-readiness\/\">Project Management Tutoring Online: Boost Grades, Skills, and Career Readiness<\/a><\/li>\n<li><a href=\"https:\/\/www.myengineeringbuddy.com\/blog\/complete-guide-to-physical-education-tutoring-online-find-expert-pe-tutors-homework-help\/\">Complete Guide to Physical Education Tutoring Online<\/a><\/li>\n<li><a href=\"https:\/\/www.myengineeringbuddy.com\/blog\/cymath-reviews-alternatives\/\">Cymath Reviews and Alternatives<\/a><\/li>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"<p>Key Takeaways Linear momentum is conserved when the net external  [&#8230;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-11297","post","type-post","status-publish","format-standard","hentry","category-uncategorized"],"_links":{"self":[{"href":"https:\/\/www.myengineeringbuddy.com\/blog\/wp-json\/wp\/v2\/posts\/11297","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.myengineeringbuddy.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.myengineeringbuddy.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.myengineeringbuddy.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.myengineeringbuddy.com\/blog\/wp-json\/wp\/v2\/comments?post=11297"}],"version-history":[{"count":1,"href":"https:\/\/www.myengineeringbuddy.com\/blog\/wp-json\/wp\/v2\/posts\/11297\/revisions"}],"predecessor-version":[{"id":12173,"href":"https:\/\/www.myengineeringbuddy.com\/blog\/wp-json\/wp\/v2\/posts\/11297\/revisions\/12173"}],"wp:attachment":[{"href":"https:\/\/www.myengineeringbuddy.com\/blog\/wp-json\/wp\/v2\/media?parent=11297"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.myengineeringbuddy.com\/blog\/wp-json\/wp\/v2\/categories?post=11297"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.myengineeringbuddy.com\/blog\/wp-json\/wp\/v2\/tags?post=11297"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}