{"id":52,"date":"2023-01-31T11:56:58","date_gmt":"2023-01-31T16:56:58","guid":{"rendered":"https:\/\/opentextbc.ca\/foundationsofphysics\/?post_type=chapter&#038;p=52"},"modified":"2023-12-21T15:30:22","modified_gmt":"2023-12-21T20:30:22","slug":"conservation-of-momentum","status":"publish","type":"chapter","link":"https:\/\/opentextbc.ca\/foundationsofphysics\/chapter\/conservation-of-momentum\/","title":{"raw":"Conservation of Momentum","rendered":"Conservation of Momentum"},"content":{"raw":"<div class=\"textbox textbox--key-takeaways\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Resources<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n<ul>\r\n \t<li><span class=\"TextRun BlobObject DragDrop SCXW20133551 BCX0\" lang=\"EN-CA\" xml:lang=\"EN-CA\" data-contrast=\"none\"><span class=\"Superscript SCXW20133551 BCX0\" data-fontsize=\"10\">Video to Watch: <\/span><\/span><a href=\"https:\/\/www.youtube.com\/watch?v=B44InZz-_pE\"><span class=\"TextRun SCXW20133551 BCX0\" lang=\"EN-US\" xml:lang=\"EN-US\" data-contrast=\"none\"><span class=\"NormalTextRun SCXW20133551 BCX0\" data-ccp-parastyle=\"footnote text\">Mechanical Universe - Episode 15 - Conservation of Momentum<\/span><\/span><\/a><\/li>\r\n \t<li><span class=\"TextRun BlobObject DragDrop SCXW239971195 BCX0\" lang=\"EN-CA\" xml:lang=\"EN-CA\" data-contrast=\"none\"><span class=\"Superscript SCXW239971195 BCX0\" data-fontsize=\"10\">Extra Help: <\/span><\/span><a href=\"http:\/\/www.a-levelphysicstutor.com\/index-mech.php\"><span class=\"TextRun SCXW239971195 BCX0\" lang=\"EN-US\" xml:lang=\"EN-US\" data-contrast=\"none\"><span class=\"NormalTextRun SCXW239971195 BCX0\" data-ccp-parastyle=\"footnote text\">A-Level Physics Tutor<\/span><\/span><\/a><\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<span class=\"TextRun MacChromeBold SCXW150514885 BCX0\" lang=\"EN-US\" xml:lang=\"EN-US\" data-contrast=\"none\"><span class=\"NormalTextRun SCXW150514885 BCX0\" data-ccp-parastyle=\"Default\" data-ccp-parastyle-defn=\"{&quot;ObjectId&quot;:&quot;0874aa0a-d072-44e5-be38-606a8e1758c2|30&quot;,&quot;ClassId&quot;:1073872969,&quot;Properties&quot;:[469775450,&quot;Default&quot;,201340122,&quot;2&quot;,134233614,&quot;true&quot;,469778129,&quot;Default&quot;,335572020,&quot;1&quot;,469777841,&quot;Helvetica&quot;,469777842,&quot;Arial Unicode MS&quot;,469777843,&quot;Arial Unicode MS&quot;,469777844,&quot;Helvetica&quot;,469769226,&quot;Helvetica,Arial Unicode MS&quot;,335551500,&quot;0&quot;,268442635,&quot;24&quot;,335551547,&quot;1033&quot;]}\">Equations Introduced and Used in this Topic:<\/span><\/span><span class=\"TextRun SCXW150514885 BCX0\" lang=\"EN-US\" xml:lang=\"EN-US\" data-contrast=\"none\"><span class=\"NormalTextRun SCXW150514885 BCX0\" data-ccp-parastyle=\"Default\"> (All equations can be written and solved as both <\/span><\/span><span class=\"TextRun MacChromeBold SCXW150514885 BCX0\" lang=\"IT-IT\" xml:lang=\"IT-IT\" data-contrast=\"none\"><span class=\"NormalTextRun SCXW150514885 BCX0\" data-ccp-parastyle=\"Default\">scalar<\/span><\/span><span class=\"TextRun SCXW150514885 BCX0\" lang=\"EN-US\" xml:lang=\"EN-US\" data-contrast=\"none\"><span class=\"NormalTextRun SCXW150514885 BCX0\" data-ccp-parastyle=\"Default\"> and <\/span><\/span><span class=\"TextRun MacChromeBold SCXW150514885 BCX0\" lang=\"FR-FR\" xml:lang=\"FR-FR\" data-contrast=\"none\"><span class=\"NormalTextRun SpellingErrorV2Themed SCXW150514885 BCX0\" data-ccp-parastyle=\"Default\">vector<\/span> <\/span><span class=\"TextRun SCXW150514885 BCX0\" lang=\"EN-US\" xml:lang=\"EN-US\" data-contrast=\"none\"><span class=\"NormalTextRun SCXW150514885 BCX0\" data-ccp-parastyle=\"Default\">and<\/span><\/span><span class=\"TextRun MacChromeBold SCXW150514885 BCX0\" lang=\"EN-US\" xml:lang=\"EN-US\" data-contrast=\"none\"> <span class=\"NormalTextRun SCXW150514885 BCX0\" data-ccp-parastyle=\"Default\">a<\/span><span class=\"NormalTextRun SCXW150514885 BCX0\" data-ccp-parastyle=\"Default\">ll equations<\/span><\/span><span class=\"TextRun SCXW150514885 BCX0\" lang=\"EN-US\" xml:lang=\"EN-US\" data-contrast=\"none\"><span class=\"NormalTextRun SCXW150514885 BCX0\" data-ccp-parastyle=\"Default\"> are generally solved as <\/span><\/span><span class=\"TextRun MacChromeBold SCXW150514885 BCX0\" lang=\"EN-US\" xml:lang=\"EN-US\" data-contrast=\"none\"><span class=\"NormalTextRun SCXW150514885 BCX0\" data-ccp-parastyle=\"Default\">vectors)<\/span><\/span>\r\n<p style=\"text-align: center;\">In an Isolated System...<\/p>\r\n<p style=\"text-align: center;\">[latex]\\sum \\vec{p} = \\text{ constant}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">or ...<\/p>\r\n<p style=\"text-align: center;\">[latex]\\vec{p}_{\\text{before}} = \\vec{p}_{\\text{after}}[\/latex] (Vector Equation)<\/p>\r\n<p style=\"text-align: center;\">Where...<\/p>\r\n<p style=\"text-align: center;\">[latex]\\vec{p} = m\\vec{y}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">For two or more bodies:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\vec{p}_{1i(x,y,z)} + \\vec{p}_{2i(x, y z)} + ... = \\vec{p}_{1f(x,y,z)} + \\vec{p}_{2f(x,y,z)} + ...[\/latex]<\/p>\r\nWhere...\r\n<ul>\r\n \t<li>[latex]\\vec{p}[\/latex] is <strong>momentum<\/strong>, the product of mass times velocity and is measured in either newton-seconds (Ns) or kilogram-metres per second (kg m\/s)<\/li>\r\n \t<li><span style=\"orphans: 1; text-align: initial; font-size: 14pt;\">[latex]m[\/latex] is <strong>mass<\/strong>, commonly measured in kilograms (Kg), grams (g) or tonnes (t)<\/span><\/li>\r\n \t<li>[latex]\\vec{v}[\/latex] is the <strong>velocity<\/strong><span style=\"orphans: 1; text-align: initial; font-size: 14pt;\">, commonly measured in metres\/second (m\/s) or kilometres\/hour (km\/h) and includes a direction<\/span><\/li>\r\n<\/ul>\r\n<div class=\"textbox textbox--sidebar\">Extra Help:\u00a0 <a href=\"https:\/\/www.physicsclassroom.com\/class\/momentum\/Lesson-2\/The-Law-of-Action-Reaction-(Revisited)\">The Law of Momentum Conservation<\/a><\/div>\r\n<strong>The Law of the Conservation of Momentum <\/strong>is one of the most powerful and useful laws in the study of mechanics. It belongs to a family of four conservation laws: Conservation of Mass Energy, Conservation of Electric Charge, Conservation of Linear Momentum and the Conservation of Angular Momentum [footnote]There are other physics conservation laws that exist in the study of particle physics.[\/footnote].\r\n\r\nThe law of conservation of momentum mainly is derived from Newton\u2019s third law and can be analyzed as in the following description excerpted from the Physics Classroom:\r\n<p style=\"padding-left: 40px;\"><strong><em>In every interaction, there is a pair of forces acting on the two interacting objects. The size of the force on the first object <\/em><em>equals<\/em><em> the size of the force on the second object. The direction of the force on the first object is <\/em><em>opposite<\/em><em> to the direction of the force on the second object. Forces always come in pairs - equal and opposite action-reaction force pairs.<\/em><\/strong><\/p>\r\n\r\n<h1>9.1 Total Momentum of a System<\/h1>\r\n<p style=\"text-align: center;\"><strong>Total Momentum of a System: The Resolutions of multiple Momentums acting during an event or interaction<\/strong><\/p>\r\n<strong>Momentum are Vectors... <\/strong>They are added, subtracted and combined according to the direction they have.\u00a0 Quite often, vectors in different directions, such as North and South, East and West, and up and down, are left as the sum of the three different vector directions and not combined any further.\r\n\r\nVectors in the same directions are added and vectors in the opposite direction are subtracted.\r\n\r\n<strong>For instance:\u00a0 <\/strong>\r\n<p style=\"text-align: center;\"><strong>100 Ns East + 50 Ns East = 150 Ns East <\/strong><\/p>\r\n(The vectors are in the same direction so they are added.)\r\n<p style=\"text-align: center;\"><strong>100 Ns North + 50 Ns South = 50 Ns North<\/strong><\/p>\r\n(The vectors are in the opposite direction so they are subtracted.)\r\n\r\nVectors in different directions are generally left as the vector sum in each of those directions.\r\n\r\n<img class=\"aligncenter wp-image-58 \" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/Picture-w.png\" alt=\"\" width=\"341\" height=\"176\" \/>\r\n\r\n<strong>For instance:\u00a0 <\/strong>\r\n<p style=\"text-align: center;\"><strong>100 Ns North + 50 Ns East = 100 Ns North + 50 Ns East<\/strong><\/p>\r\n(The vectors are in the different direction so they are left alone.)\r\n\r\nVectors that are angled between two directions are first broken up into the various components\r\n\r\nfor those directions before they are added or subtracted.\r\n\r\n<strong>For instance: <\/strong>Using right angle trigonometry, this 150 Ns vector\u00a0 would be broken up into the force directed upwards and the force to the left.\r\n\r\nTo find the <strong>Momentum upwards<\/strong> use the Sine function:\r\n<p style=\"text-align: center;\">[latex]\\sin 42\u00b0 = \\dfrac{\\text{Momentum up}}{150 \\text{ Ns}}[\/latex]<\/p>\r\nTo find the <strong>Momentum right<\/strong> use the Cosine function:\r\n<p style=\"text-align: center;\">[latex]\\cos 42\u00b0 = \\dfrac{\\text{Momentum right}}{150 \\text{ Ns}}[\/latex]<\/p>\r\n\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 9.1.1<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\n<img class=\"wp-image-59 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-a.png\" alt=\"\" width=\"208\" height=\"257\" \/>\r\n\r\nResolve the Following Multiple Momentum Vectors:\r\n<ol type=\"a\">\r\n \t<li><img class=\"alignnone wp-image-60 size-full\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-a.1.png\" alt=\"\" width=\"990\" height=\"110\" \/><strong>Solution<\/strong>\r\n[latex]\\vec{p} = 10 \\text{ Ns West } + 10 \\text{ Ns East or } 0 Ns[\/latex]<\/li>\r\n \t<li><img class=\"alignnone wp-image-61 size-full\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-b.png\" alt=\"\" width=\"962\" height=\"119\" \/><strong>Solution<\/strong>\r\n[latex]\\vec{p} = 10 \\text{ Ns East } + 10 \\text{ Ns East or } 20 \\text{ Ns East}[\/latex]<\/li>\r\n \t<li><img class=\"alignnone wp-image-62 size-full\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-c.png\" alt=\"\" width=\"1448\" height=\"128\" \/><strong>Solution<\/strong>\r\n[latex]\\vec{p} = 10 \\text{ Ns East } + 10 \\text{ Ns East } + 10 \\text{ Ns East or } 30 \\text{ Ns East}[\/latex]<\/li>\r\n \t<li><img class=\"alignnone wp-image-63 size-full\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-d.png\" alt=\"\" width=\"1425\" height=\"132\" \/><strong>Solution<\/strong>\r\n[latex]\\vec{p} = 10 \\text{ Ns East } + 10 \\text{ Ns West } + 10 \\text{ Ns West or } 10 \\text{ Ns}[\/latex]<\/li>\r\n \t<li><img class=\"alignnone wp-image-64 size-full\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-e-e1675186230959.png\" alt=\"\" width=\"1456\" height=\"145\" \/>\r\n<strong>Solution<\/strong>\r\n[latex]\\vec{p} = 10 \\text{ Ns East } + 6.0 \\text{ Ns West } + 6.0 \\text{ Ns West or } 2.0 \\text{ Ns West}[\/latex]<\/li>\r\n \t<li><img class=\"alignnone wp-image-65 \" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-f-e1675186255470.png\" alt=\"\" width=\"311\" height=\"238\" \/>\r\n<strong>Solution<\/strong>\r\n[latex]\\vec{p} = 10 \\text{ Ns East } + 5.0 \\text{ Ns South (left like this)}[\/latex]<\/li>\r\n \t<li><img class=\"alignnone wp-image-66 \" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-g.png\" alt=\"\" width=\"334\" height=\"294\" \/>\r\n<strong>Solution<\/strong>\r\nBreaking the 15 Ns up, we get:\r\n<ul>\r\n \t<li>[latex]\\vec{p_{\\text{North}}} = 15 \\text{ Ns } \\sin 40^{\\circ}[\/latex] or 9.6 Ns North<\/li>\r\n \t<li>[latex]\\vec{p_{\\text{East}}} = 15 \\text{ Ns } \\cos 40^{\\circ}[\/latex] or 11.5 Ns East<\/li>\r\n<\/ul>\r\n[latex]\\vec{p} = 11.5 \\text{ Ns East } + 9.6 \\text{ Ns North } + 5.0 \\text{ Ns South...}[\/latex]\r\nResultant is [latex]11.5 \\text{ Ns East } + 4.6 \\text{ Ns North}[\/latex]\r\n[latex](\\approx 12 \\text{ Ns East } + 4.6 \\text{ Ns North}[\/latex])<\/li>\r\n \t<li>\u00a0<img class=\"alignnone wp-image-68 \" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-H-1.png\" alt=\"\" width=\"280\" height=\"252\" \/>\r\n<strong>Solution<\/strong>\r\nBreaking the 15 Ns up, we get:\r\n[latex]\\begin{array}\r\n\\vec{p}_{north} = 15 Ns \\sin{40}^{\\circ} \\text{ or } 9.6 Ns \\text{ North} \\\\\r\n\\vec{p}_{north} = 15 Ns \\cos{40}^{\\circ} \\text{ or } 11.5 Ns \\text{ East}\r\n\\end{array}[\/latex]Breaking the 12 Ns up, we get:\r\n[latex]\\vec{p}_{north} = 12 Ns \\sin{50}^{\\circ} \\text{ or } 9.2 Ns \\text{ North}[\/latex]\r\n[latex]\\vec{p}_{north} = 12 Ns \\cos{50}^{\\circ} \\text{ or } 17.7 Ns \\text{ East}[\/latex][latex]\\vec{p}_{north} = 9.6 Ns \\text{ North } + 9.2 Ns \\text{ North } + 5 Ns \\text{ South or } 13.8 Ns \\text{ North}[\/latex]\r\n[latex]\\vec{p}_{north} = 11.5 Ns \\text{ East} + 7.7 Ns \\text{ West or } 4.8 Ns \\text{ East}[\/latex]Resultant is [latex]\\vec{p} = 13.8 Ns \\text{ North} 3.8 Ns \\text{ East}[\/latex], (\u2248 [latex]14 Ns \\text{ North } + 3.8 Ns { East}[\/latex])<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<h1><img class=\"wp-image-72 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/Picture-4.jpg\" alt=\"\" width=\"497\" height=\"349\" \/>\r\n9.2 Linear Collisions<\/h1>\r\n<div class=\"textbox\">\r\n<ul>\r\n \t<li>Extra help: <a href=\"https:\/\/www.physicsclassroom.com\/class\/momentum\/Lesson-1\/Real-World-Applications\">Real World Applications<\/a><\/li>\r\n \t<li>Extra help: \u00a0<a href=\"https:\/\/www.physicsclassroom.com\/class\/momentum\/Lesson-2\/Momentum-Conservation-Principle\">The Law of Momentum Conservation<\/a><\/li>\r\n \t<li>Research News: <a href=\"https:\/\/phys.org\/news\/2019-05-neutron-stars-collided-solar-billions.html\">Two neutron stars collided near the solar system billions of years ago<\/a><\/li>\r\n<\/ul>\r\n<\/div>\r\n<div style=\"text-align: left;\">Equations Introduced or Used for this Section:<\/div>\r\n<div style=\"text-align: center;\">In an Isolated System...<\/div>\r\n<div>\r\n<p style=\"text-align: center;\">[latex]\\sum \\vec{p} = constant[\/latex]<\/p>\r\n<p style=\"text-align: center;\">or ...<\/p>\r\n<p style=\"text-align: center;\">[latex]\\vec{p}_{before} = \\vec{p}_{after} (Vector Equation)[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Where...<\/p>\r\n\r\n<ul>\r\n \t<li style=\"text-align: left;\">[latex]\\vec{p} = m\\vec{y}[\/latex]<\/li>\r\n<\/ul>\r\nFor two or more bodies:\r\n<p style=\"text-align: center;\">[latex]\\vec{p}_{1i \\text{(x, y, z)}} + \\vec{p}_{2i \\text{(x, y, z)}} + ... = \\vec{p}_{1f \\text{(x, y, z)}} + \\vec{p}_{2f \\text{(x, y, z)}} + ...[\/latex]<\/p>\r\n\r\n<\/div>\r\nThe <strong>Law of Momentum Conservation<\/strong> is one of the foundational laws in Physics. The law is stated as follows:\r\n\r\nIn the absence of external forces that can change the momentum of objects inside a system from the outside, the vector sum of the momentum of all objects moving within the system will equal the vector sum of momentum of these same objects after an interaction has occurred, be it a collision or explosion.\r\n<div class=\"textbox textbox--sidebar\">Extra help: <a href=\"https:\/\/www.physicsclassroom.com\/Class\/momentum\/u4l2c.cfm\">Isolated Systems<\/a><\/div>\r\nThis absence of external forces acting on the object means that the system being looked at is termed an <strong>isolated system<\/strong>. Isolated systems require that no external forces outside of the forces acting between the objects within a system can be acting.\u00a0 This means that the forces acting on the objects are in fact being balanced between the objects and not by outside means.\r\n\r\nHow this works for two bodies that collide is explained as follows:\u00a0 If two objects collide, the forces acting between the two objects are equal and opposite as defined by Newton\u2019s third law.\r\n<p style=\"text-align: center;\">[latex]\\vec{\\text{Force}}_{\\text{object 1}} = - \\vec{\\text{Force}}_{\\text{object 2}}[\/latex]<\/p>\r\n<strong>These forces are equal in strength (magnitude) but act in opposite directions.<\/strong>\r\n\r\nThese forces can only act on each other for exactly the same amount of time, meaning:\r\n<p style=\"text-align: center;\">[latex]\\text{time}_{\\text{object 1}}=\\text{time}_{\\text{object 2}}[\/latex]<\/p>\r\nYou should recognize that we are talking about equal and opposite <strong>Impulses<\/strong> (<a href=\"https:\/\/opentextbc.ca\/foundationsofphysics\/chapter\/equilibrium-newtons-first-law\/\">Chapter 7<\/a>) that are acting on the two colliding bodies. This means that:\r\n<p style=\"text-align: center;\">[latex](\\vec{\\text{Force}}_{\\text{object 1}})(\\text{time}_{\\text{object 1}})=- (\\vec{\\text{Force}}_{\\text{object 2}})(\\text{time}_{\\text{object 2}}) [\/latex]<\/p>\r\nYou can conclude from this that we are talking about equal and opposite <strong>changes in momentum<\/strong> (<a href=\"https:\/\/opentextbc.ca\/foundationsofphysics\/chapter\/equilibrium-newtons-first-law\/\">Chapter 7<\/a>) that will result from the equal and opposite impulses that are acting on the two colliding bodies. This means that:\r\n<p style=\"text-align: center;\">[latex](\\text{mass}_{\\text{object 1}})(\\Delta \\vec{\\text{velocity}}_{\\text{object 1}}) = - (\\text{mass}_{\\text{object 2}})(\\Delta \\vec{\\text{velocity}}_{\\text{object 2}})[\/latex]<\/p>\r\n<strong>The changes in momentum that occur for either body are both equal and opposite.<\/strong>\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 9.2.1<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nA circus cannon on wheels (500 kg) launches a clown of mass 65 kg with a horizontal velocity of 4.2 m\/s, calculate:\r\n<ol type=\"i\">\r\n \t<li>\u00a0Momentum of the clown (before and after)<\/li>\r\n \t<li>The combined momentum of the clown and circus cannon before and after firing<\/li>\r\n \t<li>The speed that this circus cannon recoils at if originally at rest<\/li>\r\n<\/ol>\r\n<strong>Solution<\/strong>\r\n\r\nData:\r\n<ul>\r\n \t<li>[latex]m_{cannon} = 500 \\text{ kg}[\/latex]<\/li>\r\n \t<li>[latex]\\vec{v}_{i\\ cannon} = 0 \\text{ m\/s}[\/latex]<\/li>\r\n \t<li>[latex]\\vec{f}_{f\\ cannon} = \\text{ Unknown}[\/latex]<\/li>\r\n \t<li>[latex]m_{clown} = 65 \\text{ kg}[\/latex]<\/li>\r\n \t<li>[latex]\\vec{v}_{i\\ clown} = 0 \\text{ m\/s}[\/latex]<\/li>\r\n \t<li>[latex]\\vec{v}_{f\\ clown} = 4.2 \\text{ m\/s}[\/latex]<\/li>\r\n<\/ul>\r\nQuestion (i):\r\n<ul>\r\n \t<li>[latex]\\vec{p}_{\\text{i clown}} = (65 \\text{ kg})(0 \\text{ m\/s}) \\text{ or } 0 \\text{ Ns}[\/latex]<\/li>\r\n \t<li>[latex]\\vec{p}_{\\text{f clown}} = (65 \\text{ kg})(4.2 \\text{ m\/s}) \\text{ or } 273 \\text{ Ns }(\\approx 270 \\text{ Ns}[\/latex])<\/li>\r\n<\/ul>\r\nQuestion (ii):\r\n<ul>\r\n \t<li>[latex]\\vec{p}_{\\text{i clown}} + \\vec{p}_{\\text{i cannon}} = 0 \\text{ Ns}[\/latex]<\/li>\r\n \t<li>Therefore, [latex]\\vec{p}_{\\text{f clown}} + \\vec{p}_{\\text{f cannon}} = 0 \\text{ Ns}[\/latex]<\/li>\r\n<\/ul>\r\nQuestion (iii):\r\n<ul>\r\n \t<li>[latex]\\vec{p}_{\\text{i clown}} + \\vec{p}_{\\text{i cannon}} = \\vec{p}_{\\text{f clown}} + \\vec{p}_{\\text{f cannon}}[\/latex]<\/li>\r\n \t<li>[latex](m_{\\text{clown}})(\\vec{v}_{\\text{i clown}}) + (m_{\\text{cannon}})(\\vec{v}_{\\text{i cannon}}) = (m_{\\text{clown}})(\\vec{v}_{\\text{f clown}}) + (m_{\\text{cannon}})(\\vec{v}_{\\text{f cannon}})[\/latex]<\/li>\r\n \t<li>[latex](65 \\text{ kg})(0 \\text{ m\/s}) + (500 \\text{ kg})(0 \\text{ m\/s}) = (65 \\text{ kg})(4.2 \\text{ m\/s}) + (500 \\text{ kg})(\\vec{v}_{\\text{f cannon}})[\/latex]<\/li>\r\n \t<li>[latex]0 \\text{ Ns } + 0 \\text{ Ns } = 273 \\text{ Ns } + (500 \\text{ kg})(\\vec{v}_{\\text{f cannon}})[\/latex]<\/li>\r\n \t<li>[latex]\\vec{v}_{\\text{f cannon}} = -273 \\text{ Ns } \\div 500 \\text{ kg or } -0.55 \\text{m\/s (cannon recoils backwards)}[\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 9.2.2<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\n<span class=\"TextRun SCXW240874856 BCX0\" lang=\"EN-US\" xml:lang=\"EN-US\" data-contrast=\"none\"><span class=\"NormalTextRun SCXW240874856 BCX0\" data-ccp-parastyle=\"Default\" data-ccp-parastyle-defn=\"{&quot;ObjectId&quot;:&quot;91c66a2c-ba54-4445-bd05-2c763565bfc2|30&quot;,&quot;ClassId&quot;:1073872969,&quot;Properties&quot;:[469775450,&quot;Default&quot;,201340122,&quot;2&quot;,134233614,&quot;true&quot;,469778129,&quot;Default&quot;,335572020,&quot;1&quot;,469777841,&quot;Helvetica&quot;,469777842,&quot;Arial Unicode MS&quot;,469777843,&quot;Arial Unicode MS&quot;,469777844,&quot;Helvetica&quot;,469769226,&quot;Helvetica,Arial Unicode MS&quot;,335551500,&quot;0&quot;,268442635,&quot;24&quot;,335551547,&quot;1033&quot;]}\">A 4.0 kg curling rock moving 0.5 m\/s West collides with a stationary 4.0 kg curling rock at rest. If the original curling rock comes to a full stop upon collision what velocity should the second curling rock be moving at?<\/span><\/span>\r\n\r\n<strong>Solution<\/strong>\r\n\r\nData:\r\n<ul>\r\n \t<li>[latex]m_{\\text{rock 1}} = 4.0 \\text{ kg}[\/latex]<\/li>\r\n \t<li>[latex]\\vec{v}_{\\text{i rock 1}} = 0.5 \\text{ m\/s West}[\/latex]<\/li>\r\n \t<li>[latex]\\vec{v}_{\\text{f rock 1}} = 0 \\text{ m\/s}[\/latex]<\/li>\r\n<\/ul>\r\n&nbsp;\r\n<ul>\r\n \t<li>[latex]m_{\\text{rock 2}} = 4.0 \\text{ kg}[\/latex]<\/li>\r\n \t<li>[latex]\\vec{v}_{\\text{i rock 2}} = 0.5 \\text{ m\/s West}[\/latex]<\/li>\r\n \t<li>[latex]\\vec{v}_{\\text{f rock 2}} = 0 \\text{ m\/s}[\/latex]<\/li>\r\n<\/ul>\r\nSolution:\r\n<ul>\r\n \t<li>[latex]\\vec{p}_{\\text{i rock 1}} + \\vec{p}_{\\text{i rock 2}} = \\vec{p}_{\\text{f rock 1}} + \\vec{p}_{\\text{f rock 1}}[\/latex]<\/li>\r\n \t<li>[latex](m_{\\text{rock 1}})(\\vec{v}_{\\text{i rock 1}}) + (m_{\\text{rock 2}})(\\vec{v}_{\\text{i rock 2}}) = (m_{\\text{rock 1}})(\\vec{v}_{\\text{f rock 1}}) + (m_{\\text{rock 2}})(\\vec{v}_{\\text{f rock 2}})[\/latex]<\/li>\r\n \t<li>[latex](4.0 \\text{ kg})(0.5 \\text{ m\/s West}) + (4.0 \\text{ kg})(0 \\text{ m\/s}) = (4.0 \\text{ kg})(0 \\text{ m\/s}) + (4.0 \\text{ kg})(\\vec{v}_{\\text{f rock 2}})[\/latex]<\/li>\r\n \t<li>[latex]2.0 \\text{ Ns West } + 0 \\text{ Ns } = 0 \\text{ Ns } + (4.0 \\text{ kg})(\\vec{v}_{\\text{f rock 2}})[\/latex]<\/li>\r\n \t<li>[latex](\\vec{v}_{\\text{f rock 2}}) = 2.0 \\text{ Ns West } \\div 4.0 \\text{ kg or } 0.5 \\text{ m\/s West}[\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 9.2.3<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\n<span class=\"NormalTextRun SCXW94985447 BCX0\" data-ccp-parastyle=\"Default\">During a tackle in a football game<\/span><span class=\"NormalTextRun SCXW94985447 BCX0\" data-ccp-parastyle=\"Default\">,<\/span><span class=\"NormalTextRun SCXW94985447 BCX0\" data-ccp-parastyle=\"Default\"> two players of mass 90 kg and 70 kg were moving towards each other<\/span><span class=\"NormalTextRun SCXW94985447 BCX0\" data-ccp-parastyle=\"Default\">,<\/span><span class=\"NormalTextRun SCXW94985447 BCX0\" data-ccp-parastyle=\"Default\"> and after collision both c<\/span><span class=\"NormalTextRun SCXW94985447 BCX0\" data-ccp-parastyle=\"Default\">a<\/span><span class=\"NormalTextRun SCXW94985447 BCX0\" data-ccp-parastyle=\"Default\">me to a full stop.\u00a0 If the velocity of the larger player before collision was 7.0 m\/s what was the velocity of the smaller player?<\/span>\r\n\r\n<strong>Solution<\/strong>\r\n\r\nAll we have for the direction in this problem is that both players were running towards each other.\r\n\r\nData:\r\n<ul>\r\n \t<li>[latex]m_{90} = 90 \\text{ kg}[\/latex]<\/li>\r\n \t<li>[latex]\\vec{v}_{\\text{i 90}} = 7.0 \\text{ m\/s}[\/latex]<\/li>\r\n \t<li>[latex]\\vec{v}_{\\text{f 90}} = 0 \\text{ m\/s}[\/latex]<\/li>\r\n<\/ul>\r\n&nbsp;\r\n<ul>\r\n \t<li>[latex]m_{70} = 70 \\text{ kg}[\/latex]<\/li>\r\n \t<li>[latex]\\vec{v}_{\\text{i 70}} = \\text{ Find}[\/latex]<\/li>\r\n \t<li>[latex]\\vec{v}_{\\text{f 70}} = 0 \\text{ m\/s}[\/latex]<\/li>\r\n<\/ul>\r\nSolution:\r\n<ul>\r\n \t<li>[latex]\\vec{p}_{\\text{i 90}} + \\vec{p}_{\\text{i 70}} = \\vec{p}_{\\text{f 90}} + \\vec{p}_{\\text{f 70}}[\/latex]<\/li>\r\n \t<li>[latex](m_{\\text{90}})(\\vec{v}_{\\text{i 90}}) + (m_{\\text{70}})(\\vec{v}_{\\text{i 70}}) = (m_{\\text{90}})(\\vec{v}_{\\text{f 90}}) + (m_{\\text{70}})(\\vec{v}_{\\text{f 70}})[\/latex]<\/li>\r\n \t<li>[latex](90 \\text{ kg})(7.0 \\text{ m\/s}) + (70 \\text{ kg})(\\vec{v}_{\\text{i 70}}) = (90 \\text{ kg})(0 \\text{ m\/s}) + (70 \\text{ kg})(0 \\text{ m\/s})[\/latex]<\/li>\r\n \t<li>[latex]630 \\text{ Ns } + (70 \\text{ kg})(\\vec{v}_{\\text{i 70}}) = 0 \\text{ Ns } + 0 \\text{ Ns}[\/latex]<\/li>\r\n \t<li>[latex](\\vec{v}_{\\text{i 70}}) = - 630 \\text{ Ns } \\div 70 \\text{ kg or } - 9.0 \\text{ m\/s}[\/latex]<\/li>\r\n<\/ul>\r\nNote:\u00a0 The [latex]- 9.0[\/latex] m\/s means that the smaller player was running in the opposite direction (towards) as the larger player.\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 9.2.4<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\n<span class=\"NormalTextRun SCXW11241910 BCX0\" data-ccp-parastyle=\"Default\">If a 16 kg medicine ball <\/span><span class=\"NormalTextRun SCXW11241910 BCX0\" data-ccp-parastyle=\"Default\">is <\/span><span class=\"NormalTextRun SCXW11241910 BCX0\" data-ccp-parastyle=\"Default\">thrown horizontally to a stationary 65 kg skater on ice (consider this to be frictionless) and causes the skater who is now holding the ball to move off with a horizontal velocity of 0.50 m\/s, what was the velocity the ball was thrown at?<\/span>\r\n\r\n<strong>Solution<\/strong>\r\n\r\nData:\r\n<ul>\r\n \t<li>[latex]m_{16} = 16 \\text{ kg}[\/latex]<\/li>\r\n \t<li>[latex]\\vec{v}_{\\text{i 16}} = \\text{ Find}[\/latex]<\/li>\r\n \t<li>[latex]\\vec{v}_{\\text{f 16}} = 0.50 \\text{ m\/s}[\/latex]<\/li>\r\n<\/ul>\r\n&nbsp;\r\n<ul>\r\n \t<li>[latex]m_{65} = 70 \\text{ kg}[\/latex]<\/li>\r\n \t<li>[latex]\\vec{v}_{\\text{i 65}} = 0 \\text{ m\/s}[\/latex]<\/li>\r\n \t<li>[latex]\\vec{v}_{\\text{f 65}} = 0.50 \\text{ m\/s}[\/latex]<\/li>\r\n<\/ul>\r\nSolution:\r\n<ul>\r\n \t<li>[latex]\\vec{p}_{\\text{i 16}} + \\vec{p}_{\\text{i 65}} = \\vec{p}_{\\text{f 16}} + \\vec{p}_{\\text{f 65}}[\/latex]<\/li>\r\n \t<li>[latex](m_{\\text{16}})(\\vec{v}_{\\text{i 16}}) + (m_{\\text{65}})(\\vec{v}_{\\text{i 65}}) = (m_{\\text{16}})(\\vec{v}_{\\text{f 16}}) + (m_{\\text{65}})(\\vec{v}_{\\text{f 65}})[\/latex]<\/li>\r\n \t<li>[latex](16 \\text{ kg})(\\vec{v}_{\\text{i 16}}) + (65 \\text{ kg})(0 \\text{ m\/s}) = (16 \\text{ kg } + 65 \\text{ kg})(0.5 \\text{ m\/s})[\/latex]<\/li>\r\n \t<li>[latex](16 \\text{ kg})(\\vec{v}_{\\text{i 16}}) + (0 \\text{ Ns}) = (40.5 Ns)[\/latex]<\/li>\r\n \t<li>[latex](\\vec{v}_{\\text{i 16}}) = 40.5 Ns \\div 16 kg[\/latex] \u2248 2.5 m\/s horizontally<\/li>\r\n<\/ul>\r\nNote:\u00a0 The \u2248 + 2.5 m\/s means that both the ball and the skater continue moving in the same horizontal direction as the medicine ball was thrown.\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--exercises\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Exercise 9.2<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n<ol>\r\n \t<li>A cannon on wheels (2750 kg) shoots a cannon ball of mass 14.7 kg.\u00a0 If this cannon ball leaves the muzzle with a horizontal velocity of 487 m\/s, calculate:\r\n<ol type=\"i\">\r\n \t<li>Momentum of this cannon ball (before and after)<\/li>\r\n \t<li>The combined momentum of the cannon ball and cannon before and after firing<\/li>\r\n \t<li>The recoil velocity of this cannon<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>A 0.40 kg rock moving 0.50 m\/s West collides with a stationary 0.50 kg rubber ball at rest. If the 0.40 kg rock comes to a full stop upon collision, what velocity should this ball have?<\/li>\r\n \t<li>During a dynamics experiment, two trolley cars of mass 9.0 kg and 6.0 kg were moving towards each other, and after collision both come to a full stop. If the velocity of the larger trolley car before collision was 0.025 m\/s, what was the velocity of the smaller trolley car?<\/li>\r\n \t<li>In a bakery fight, a 2.0 kg mass of bread dough moving at 16.0 m\/s North collides with a stationary molasses container. The two objects coalesce and move off at 7.50 m\/s North.\u00a0 What was the mass of the stationary molasses container?<\/li>\r\n \t<li>An 800 kg asteroid in space has a velocity of 5.00 m\/s. An explosion causes the rear section (mass 240 kg) of this asteroid to be sent in the opposite direction at 1.20 m\/s.\u00a0 At what velocity did the remaining part of this asteroid continue to move?<\/li>\r\n \t<li>If a 5.40 g bullet is fired horizontally into a suspended 10.0 kg block of wood and causes the block with bullet embedded to move off with a horizontal velocity of 0.30 m\/s, what was the impact velocity of this bullet?<\/li>\r\n \t<li>Brian\u2019s 10 gram toy train car is traveling at 5.0 cm\/s when it collides with a 15 gram tank car originally at rest. If the two cars lock together, at what speed should they travel off?<\/li>\r\n \t<li>A 95 kg hockey player is skating at 12 m\/s when he runs into a larger 120 kg player that is only skating at 5.0 m\/s in the same direction. If the two skaters become entangled, at what speed do they slide off?<\/li>\r\n \t<li>A 105 kg fullback is running straight towards the end zone at 9.5 m\/s. A 95 kg defensive back is running straight towards the fullback, tackling the fullback with both dropping to the ground with a velocity of 0 m\/s.\u00a0 What was the velocity of this defensive back prior to the tackle?<\/li>\r\n \t<li>Kyle\u2019s small toy train boxcar of mass of 60 g is rolling at 4.0 cm\/s towards a flatcar of mass 45 g that is stationary. If the two cars lock together, at what speed do they continue to move down the track?<\/li>\r\n \t<li>In the movies, an object such as a football is thrown at a running person, striking and stopping them. If we consider a 70 kg person running toward you at 8.0 m\/s, what speed must a 2.0 kg football be thrown at to stop this person?<\/li>\r\n \t<li>A medicine ball is thrown at 8.0 m\/s towards a stationary 55 kg skater on ice. After she catches it, both the skater and the ball move off at 0.8 m\/s.\u00a0 What is the mass of this medicine ball?<\/li>\r\n \t<li>A large railway wagon, mass 4200 kg, rolls down a track at 2.5 m\/s and collides with a smaller wagon, mass 2750 kg, moving at 1.0 m\/s in the same direction. If the smaller wagon has a velocity of 3.0 m\/s after the collision, what is the final velocity of the large wagon?<\/li>\r\n \t<li>A small particle A of mass [latex]1.41 \\times 10^{-5}[\/latex] g moving at [latex]3.33 \\times 10^3[\/latex] m\/s is hit by particle B moving at 5.46 x 104 m\/s in the same direction. As a result, particle A increased its velocity to [latex]6.66 \\times 10^3[\/latex] m\/s and particle B was slowed to [latex]4.60 \\times 10^3[\/latex] m\/s in the original direction.\u00a0 What was the mass of particle B?<\/li>\r\n \t<li>A 0.85 kg spring toy (at rest) \u201cpops\u201d into 3 pieces. The largest piece of mass 0.45 kg moves at 5.0 cm\/s to the right.\u00a0 A smaller piece of mass, of mass 0.25 kg, moves at 8.0 cm\/s to the left.\u00a0 At what speed does the remaining piece move off?<\/li>\r\n \t<li>A small 65 kg hockey player takes a run at a larger 115 kg player at rest. If the smaller player was traveling at 16 m\/s towards the larger before the collision and rebounded at 4 m\/s after, at what speed did the larger player move off?<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox\">\r\n<ul>\r\n \t<li>Article to Read: <a href=\"https:\/\/phys.org\/news\/2018-12-pedestrians-cm-comfort-zone-collisions.html\">Eindhoven University of Technology (2018) Pedestrians keep a 75 cm comfort zone to prevent collisions<\/a><\/li>\r\n \t<li>Extra Help: <a href=\"https:\/\/www.physicsclassroom.com\/class\/momentum\/Lesson-2\/Using-Equations-as-a-Recipe-for-Algebraic-Problem\">Using Equations as a Recipe for Algebraic Problem-Solving<\/a><\/li>\r\n \t<li>Extra Help: <a href=\"https:\/\/www.physicsclassroom.com\/class\/momentum\/Lesson-2\/Using-Equations-as-a-Guide-to-Thinking\">Using Equations as a Guide to Thinking<\/a><\/li>\r\n<\/ul>\r\n<\/div>\r\n<h1><img class=\"wp-image-696 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.3-two-dimensional-collisions.jpg\" alt=\"\" width=\"528\" height=\"384\" \/><\/h1>\r\n<h1>9.3 Two Dimensional Collisions<\/h1>\r\n<div class=\"textbox\">\r\n<ul>\r\n \t<li>Article to Read:\u00a0 <a href=\"https:\/\/phys.org\/news\/2018-12-big-space-uranus-lopsided.html\">A big space crash likely made Uranus lopsided<\/a><\/li>\r\n \t<li>Article to Read: <a href=\"http:\/\/www.iflscience.com\/space\/uranus-experienced-a-colossal-impact-that-knocked-it-on-its-side\/\">Uranus has experienced a colossal pounding<\/a><\/li>\r\n<\/ul>\r\n<\/div>\r\nEquations Introduced or Used for this Section:\r\n<p style=\"text-align: center;\">In an Isolated System...<\/p>\r\n<p style=\"text-align: center;\">[latex]\\sum \\vec{p} = \\text{ constant}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">or ...<\/p>\r\n<p style=\"text-align: center;\">[latex]\\vec{p}_{\\text{before}} = \\vec{p}_{\\text{after}}[\/latex] (Vector Equation)<\/p>\r\n<p style=\"text-align: left;\">Where...<\/p>\r\n\r\n<ul>\r\n \t<li style=\"text-align: left;\">[latex]\\vec{p} = m\\vec{y}[\/latex]<\/li>\r\n<\/ul>\r\n<p style=\"text-align: center;\">For two or more bodies:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\vec{p}_{1i(x,y,z)} + \\vec{p}_{2i(x, y z)} + ... = \\vec{p}_{1f(x,y,z)} + \\vec{p}_{2f(x,y,z)} + ...[\/latex]<\/p>\r\n&nbsp;\r\n\r\nThe<strong> Law of the Conservation of Momentum<\/strong> is a law that can be expanded from linear one-dimensional to the multi-dimensional universe of 2 or 3 dimensions.\u00a0 This is due to the fact that momentum must be conserved in its original sum as long as no external forces act to change the momentum of the system.\r\n\r\nThe grid shown to the right is the common (x, y, z) representation of 3 dimensional space. All that is missing from this would be the time that identifies the spot where an object would be at a specific time.\r\n\r\nIt is also common to see vector momenta sums to be identified according to the directions defined by <strong>(x, y, z)<\/strong>.\r\n\r\n<img class=\"wp-image-697 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.3-3D-coordinates.png\" alt=\"\" width=\"553\" height=\"563\" \/>\r\n\r\nAs such, you could see the momentum of an object defined by:\r\n<p style=\"text-align: center;\">[latex]\\vec{p}_{\\text{i object}} = \\vec{p}_{i\\ x} + \\vec{p}_{i\\ y} + \\vec{p}_{i\\ z}[\/latex]<\/p>\r\nWhen solving two or three dimensional problems using the Law of Conservation of Momentum, you solve as you did before with the one dimensional problems, except now you need to make sure that the momentum in each of the three directions is conserved both before and after an event occurs.\r\n\r\nThis means that:\r\n<p style=\"text-align: center;\">[latex]\\vec{p}_{\\text{before (x, y, z)}} = \\vec{p}_{\\text{after (x, y, z)}}[\/latex]<\/p>\r\nAs an example of this, if the vector sum of the momentum of two or more interacting objects is:\r\n<p style=\"text-align: center;\">[latex]\\vec{p}_{\\text{before (x, y, z)}} = 25 \\text{ Ns}_x - 30 \\text{ Ns}_y + 10 \\text{ Ns}_z[\/latex]<\/p>\r\nThen the vector sum of the momentum of two or more interacting objects afterwards is:\r\n<p style=\"text-align: center;\">[latex]\\vec{p}_{\\text{after (x, y, z)}} = 25 \\text{ Ns}_x - 30 \\text{ Ns}_y + 10 \\text{ Ns}_z[\/latex]<\/p>\r\nIn solving these two or three dimensional problems, you will be using trigonometry and the Pythagorean Theorem extensively, as you can see in the following two examples.\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 9.3.1<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nA 120 kg lineman is running at 8.5 m\/s North and tackles a 85 kg halfback running at 9.6 m\/s East.\u00a0 At what velocity do these two move after the tackle occurs?\r\n\r\n<img class=\"wp-image-704 aligncenter\" style=\"font-size: 0.9em; word-spacing: normal;\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.3.1-example.png\" alt=\"\" width=\"482\" height=\"309\" \/>\r\n\r\n<strong>Solution<\/strong>\r\n\r\nFirst, find the momenta of the two football players heading into the tackle.\r\n\r\nThe 85 kg Halfback:\r\n<ul>\r\n \t<li>[latex]\\vec{p}_{85} = m_{85} \\vec{v}_{\\text{i 85}}[\/latex]<\/li>\r\n \t<li>[latex]p_{85} = (85 \\text{ kg})(9.6 \\text{ m\/s East})[\/latex] or 816 Ns East<\/li>\r\n<\/ul>\r\nThe 120 kg Lineman:\r\n<ul>\r\n \t<li>[latex]\\vec{p}_{120} = m_{120} \\vec{v}_{\\text{ i 120}}[\/latex]<\/li>\r\n \t<li>[latex]p_{120} = (120 \\text{ kg})(8.5 \\text{ m\/s North})[\/latex] or 1020 Ns North<\/li>\r\n<\/ul>\r\nThe total momentum before the tackle is 816 Ns East + 1020 Ns North.\r\n\r\nNow, in the absence of any other forces acting during this collision, the total momentum after the tackle for these two players must also be 816 Ns East + 1020 Ns North.\r\n\r\n<img class=\"wp-image-703 aligncenter\" style=\"font-size: 0.9em; word-spacing: normal;\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.3.1-example-2.png\" alt=\"\" width=\"497\" height=\"215\" \/>\r\n\r\nAt this point, you will use trigonometry and the Pythagorean theorem to further analyze this collision.\r\n\r\n<img class=\"wp-image-702 aligncenter\" style=\"word-spacing: normal; text-align: initial; font-size: 0.9em;\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.3.1-example-3.png\" alt=\"\" width=\"453\" height=\"202\" \/>\r\n\r\nNow: The total by using the momentum can be found using the Pythagorean Theorem.\r\n<ul>\r\n \t<li>[latex](\\vec{p}_{\\text{total}})^2 = (816 \\text{ Ns})^2 + (1020 \\text{ Ns})^2[\/latex]<\/li>\r\n \t<li>[latex](\\vec{p}_{\\text{total}})^2 = 665856 \\text{ N}^2 \\text{S}^2 + 1040400 \\text{ N}^2 \\text{S}^2[\/latex]<\/li>\r\n \t<li>[latex](\\vec{p}_{\\text{total}})^2 = 1706256 \\text{ N}^2 \\text{S}^2[\/latex]<\/li>\r\n \t<li>[latex]\\vec{p}_{\\text{total}} = 1306 \\text{ Ns}[\/latex]<\/li>\r\n<\/ul>\r\nThe angle at\u00a0 which these players move off is now found by using trigonometry, specifically the Tangent function:\r\n<ul>\r\n \t<li>[latex]\\tan \u00f8 = \\dfrac{\\text{Opposite Side}}{\\text{Adjacent Side}}[\/latex]<\/li>\r\n \t<li>[latex]\\tan \u00f8 = 1.245[\/latex]<\/li>\r\n \t<li>[latex]\u00f8 = \\tan ^{-1} 1.245[\/latex]<\/li>\r\n \t<li>\u00f8 = 51.3\u00b0, specifically \u2248 51\u00b0 North from East<\/li>\r\n<\/ul>\r\nThe velocity at which these players move off can be found by using the equation for momentum:\r\n<ul>\r\n \t<li>[latex]\\vec{p} = m \\vec{v}[\/latex]<\/li>\r\n \t<li>[latex]1306 \\text{ Ns} = (120 \\text{ kg } + 85 \\text{ kg}) \\vec{v}_{\\text{f (120 + 85)}}[\/latex]<\/li>\r\n \t<li>[latex]\\vec{v}_{\\text{f (120 + 85)}} = 1306 \\text{ Ns } \\div 205 \\text{ kg}[\/latex]<\/li>\r\n \t<li>[latex]\\vec{v}_{\\text{f (120 + 85)}} = 64. \\text{ m\/s}[\/latex] 51\u00b0 North from East<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 9.3.2<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nAccident investigators are analyzing a car crash to see if excessive speed was involved. One of the cars had a mass of 1250 kg and was traveling at 40 km\/h East when it was struck by a 1600 kg pickup truck traveling South that ran a red light. If the wreckage moved at <strong>23.5<\/strong><strong>\u00b0 East from South<\/strong>, should the truck driver also be charged for driving above the 60 km\/h speed limit?\r\n\r\n<strong>Solution<\/strong>\r\n\r\nFirst, draw a sketch of the two vehicles:\r\n\r\n<img class=\"alignnone wp-image-701 \" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.3.1-example-4.png\" alt=\"\" width=\"536\" height=\"513\" \/>\r\n\r\nAll that is known are the mass of the two colliding vehicles, original velocity of the car and the angle at which the wreck moved off.\r\n\r\nCombining this data, we can calculate the momentum of the car, and by using trigonometry we can calculate the momentum of the truck.\u00a0 The momentum of the truck allows us to estimate the initial velocity that it was traveling at.\r\n\r\nFirst, the momentum of the car:\r\n\r\n<img class=\"wp-image-717 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.3.2-example-2.png\" alt=\"\" width=\"360\" height=\"398\" \/>\r\n<ul>\r\n \t<li>[latex]\\vec{p} = m \\vec{v}[\/latex]<\/li>\r\n \t<li>[latex]\\vec{p} = (1250 \\text{ kg})(11.1 \\text{ m\/s East})[\/latex]<\/li>\r\n \t<li>[latex]\\vec{p} = 13\\;900 \\text{ Ns East}[\/latex]<\/li>\r\n<\/ul>\r\nNow, using the Tangent function:\r\n<ul>\r\n \t<li>[latex]\\tan 23.5^{\\circ} = \\dfrac{13\\;900 \\text{ Ns}}{\\vec{p}_{\\text{truck}}}[\/latex]<\/li>\r\n \t<li>[latex]\\vec{p}_{\\text{truck}} = \\dfrac{13\\;900 \\text{ Ns}}{\\tan 23.5^{\\circ}}[\/latex]<\/li>\r\n \t<li>[latex]\\vec{p}_{\\text{truck}} = 32\\;000 \\text{ Ns South}[\/latex]<\/li>\r\n<\/ul>\r\nFinally, the velocity of the truck is:\r\n<ul>\r\n \t<li>[latex]\\vec{p} = m \\vec{v}[\/latex]<\/li>\r\n \t<li>[latex]\\vec{v} = (32\\;000 \\text{ Ns South}) \\div (1600 \\text{ kg})[\/latex]<\/li>\r\n \t<li>[latex]\\vec{v} = 20[\/latex] m\/s South or 72 km\/h<\/li>\r\n<\/ul>\r\n<strong>The truck was speeding by 12 km\/h and could also be charged with speeding.<\/strong>\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--exercises\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Exercise 9.3<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n<ol>\r\n \t<li>A 85 kg fullback is running at 9.0 m\/s South and is tackled by a 70 kg halfback running at 8.0 m\/s West. At what velocity do these two move after the tackle occurs?<\/li>\r\n \t<li>Two hockey players collide mid-ice in a race for a puck. If the smaller player (80 kg) is racing at 12 m\/s North and the larger player (90 kg) is racing at 15 m\/s at 40\u00b0 West from North, at what angle do the colliding players move off?<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--exercises\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Exercise 9.4.1: Survival Rates of Seat Areas in an Airplane 1971-2007<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\n<strong>Question:<\/strong> Why is the greatest survival rate at the rear of the airplane?\r\n\r\n&nbsp;\r\n\r\n[caption id=\"attachment_1211\" align=\"aligncenter\" width=\"389\"]<img class=\"wp-image-1211\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.4.1-airplane-1.png\" alt=\"\" width=\"389\" height=\"350\" \/> <strong>Survival rates for various parts of the passenger cabin<\/strong>. First\/Business class: 49%. Ahead of the wing: 56%. Over wing: 56%. Rear cabin 69%.[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--exercises\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Exercise 9.4.2: Bullet Proof Vests<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nBullet proof vests were first developed by Casimir Zeglen, a Catholic priest who in 1893 created a vest made from silks that was capable of stopping bullets. The principle behind these vests was to stop the bullet, causing it to deform, which then spread the bullet\u2019s point of impact over a much larger area on the vest. Zeglen\u2019s choice of silk for the vest derived from his reading of how a physician described how a man had been shot but was saved by a silk handkerchief in his breast pocket.\u00a0 Soon tests were done by folding layers of silk some 10 to 30 times. Zeglen\u2019s original vests came from using multiple layers of silk that had been combined with an unknown chemical that he never disclosed. His later vests included blends of Angora and silk threads for handgun protection and even a steel sheet insert that worked to stop rifle bullets. His finished product was a cloth that was a little over 0.3 cm thick.\r\n\r\n<strong>\u00a0<\/strong>The photo to the left is taken from the May 1924 Popular Mechanics article showing a test subject being fired upon with 38 caliber (largest mass of 9.53 g, speed of 270 m\/s) and 45 caliber (largest mass of 15 g, speed of 255 m\/s) handguns.\r\n\r\n<strong>Question:<\/strong> For the man in this photo, if we assume a mass of 68 kg (150 lb man), what should be his recoil velocity in stopping a 45 caliber bullet?\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox\">\r\n\r\nResearch in the News: <a href=\"https:\/\/phys.org\/news\/2019-06-metal-foam-caliber-rounds-steel.html?\">Metal Foam stops .50 caliber rounds as well as steel - at less than half the weight<\/a>\r\n\r\n<\/div>\r\n<h1>Exercise Answers<\/h1>\r\n<h2>9.2 Simple Collisions<\/h2>\r\n<ol class=\"twocolumn\">\r\n \t<li>\r\n<ol type=\"i\">\r\n \t<li>0 Ns (before), 7160 Ns (after)<\/li>\r\n \t<li>\u2212\u00a02.60 m\/s (recoils at 2.60 m\/s)<\/li>\r\n \t<li>0 Ns in both cases<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>0.4 m\/s West<\/li>\r\n \t<li>0.0375 m\/s opposite direction<\/li>\r\n \t<li>2.27 kg<\/li>\r\n \t<li>7.66 m\/s<\/li>\r\n \t<li>556 m\/s<\/li>\r\n \t<li>2.0 cm\/s in original direction<\/li>\r\n \t<li>8.1 m\/s<\/li>\r\n \t<li>10.5 m\/s away from the end-zone<\/li>\r\n \t<li>0.0229 m\/s or 2.3 cm\/s<\/li>\r\n \t<li>280 m\/s (roughly 1000 km\/h)<\/li>\r\n \t<li>6.1 kg<\/li>\r\n \t<li>1.19 m\/s same direction<\/li>\r\n \t<li>[latex]2.82 \\times 10^{-7}[\/latex] g<\/li>\r\n \t<li>0.17 m\/s left<\/li>\r\n \t<li>11.3 m\/s<\/li>\r\n<\/ol>\r\n<h2>9.3 Two Dimensional Collisions<\/h2>\r\n<ol>\r\n \t<li>6.1 m\/s, 36\u00b0 West of South<\/li>\r\n \t<li>23.5\u00b0 West of North<\/li>\r\n<\/ol>\r\n<h2>9.4.1 Survival Rates of Seat Areas in an Airplane<\/h2>\r\n<ul>\r\n \t<li>The passengers in the back would have a longer distance to come to a stop, therefore the impulse is stretched over a longer time.<\/li>\r\n \t<li>The front of the aircraft acts like a crumple zone for the rest of the aircraft. Similar to a front end collision of a vehicle.<\/li>\r\n \t<li>Other possible reasons?<\/li>\r\n<\/ul>\r\n<h2>9.4.2 Bullet Proof Vests<\/h2>\r\n<ol>\r\n \t<li>0.050 m\/s recoil<\/li>\r\n<\/ol>\r\n<h3>Media Attributions<\/h3>\r\n<ul>\r\n \t<li>\"<a href=\"https:\/\/pixabay.com\/fr\/photos\/berceau-de-newton-des-balles-balle-256213\/\">black and gray newton's cradle<\/a>\" by <a href=\"https:\/\/pixabay.com\/fr\/users\/jarmoluk-143740\/\">jarmoluk<\/a> licensed under a <a href=\"https:\/\/pixabay.com\/service\/license\/\">Pixabay licence<\/a>.<\/li>\r\n \t<li>\"<a href=\"https:\/\/www.flickr.com\/photos\/8623220@N02\/2477625961\">Auto polo (LOC)<\/a>\" from <a href=\"https:\/\/www.flickr.com\/photos\/library_of_congress\/\">The Library of Congress<\/a> is in the <a href=\"https:\/\/creativecommons.org\/share-your-work\/public-domain\/\">public domain<\/a>.<\/li>\r\n \t<li>\"Safest seats on a plane\" by Gil Ahn. Image courtesy of seatguru.com<\/li>\r\n<\/ul>","rendered":"<div class=\"textbox textbox--key-takeaways\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Resources<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<ul>\n<li><span class=\"TextRun BlobObject DragDrop SCXW20133551 BCX0\" lang=\"EN-CA\" xml:lang=\"EN-CA\" data-contrast=\"none\"><span class=\"Superscript SCXW20133551 BCX0\" data-fontsize=\"10\">Video to Watch: <\/span><\/span><a href=\"https:\/\/www.youtube.com\/watch?v=B44InZz-_pE\"><span class=\"TextRun SCXW20133551 BCX0\" lang=\"EN-US\" xml:lang=\"EN-US\" data-contrast=\"none\"><span class=\"NormalTextRun SCXW20133551 BCX0\" data-ccp-parastyle=\"footnote text\">Mechanical Universe &#8211; Episode 15 &#8211; Conservation of Momentum<\/span><\/span><\/a><\/li>\n<li><span class=\"TextRun BlobObject DragDrop SCXW239971195 BCX0\" lang=\"EN-CA\" xml:lang=\"EN-CA\" data-contrast=\"none\"><span class=\"Superscript SCXW239971195 BCX0\" data-fontsize=\"10\">Extra Help: <\/span><\/span><a href=\"http:\/\/www.a-levelphysicstutor.com\/index-mech.php\"><span class=\"TextRun SCXW239971195 BCX0\" lang=\"EN-US\" xml:lang=\"EN-US\" data-contrast=\"none\"><span class=\"NormalTextRun SCXW239971195 BCX0\" data-ccp-parastyle=\"footnote text\">A-Level Physics Tutor<\/span><\/span><\/a><\/li>\n<\/ul>\n<\/div>\n<\/div>\n<p><span class=\"TextRun MacChromeBold SCXW150514885 BCX0\" lang=\"EN-US\" xml:lang=\"EN-US\" data-contrast=\"none\"><span class=\"NormalTextRun SCXW150514885 BCX0\" data-ccp-parastyle=\"Default\" data-ccp-parastyle-defn=\"{&quot;ObjectId&quot;:&quot;0874aa0a-d072-44e5-be38-606a8e1758c2|30&quot;,&quot;ClassId&quot;:1073872969,&quot;Properties&quot;:[469775450,&quot;Default&quot;,201340122,&quot;2&quot;,134233614,&quot;true&quot;,469778129,&quot;Default&quot;,335572020,&quot;1&quot;,469777841,&quot;Helvetica&quot;,469777842,&quot;Arial Unicode MS&quot;,469777843,&quot;Arial Unicode MS&quot;,469777844,&quot;Helvetica&quot;,469769226,&quot;Helvetica,Arial Unicode MS&quot;,335551500,&quot;0&quot;,268442635,&quot;24&quot;,335551547,&quot;1033&quot;]}\">Equations Introduced and Used in this Topic:<\/span><\/span><span class=\"TextRun SCXW150514885 BCX0\" lang=\"EN-US\" xml:lang=\"EN-US\" data-contrast=\"none\"><span class=\"NormalTextRun SCXW150514885 BCX0\" data-ccp-parastyle=\"Default\"> (All equations can be written and solved as both <\/span><\/span><span class=\"TextRun MacChromeBold SCXW150514885 BCX0\" lang=\"IT-IT\" xml:lang=\"IT-IT\" data-contrast=\"none\"><span class=\"NormalTextRun SCXW150514885 BCX0\" data-ccp-parastyle=\"Default\">scalar<\/span><\/span><span class=\"TextRun SCXW150514885 BCX0\" lang=\"EN-US\" xml:lang=\"EN-US\" data-contrast=\"none\"><span class=\"NormalTextRun SCXW150514885 BCX0\" data-ccp-parastyle=\"Default\"> and <\/span><\/span><span class=\"TextRun MacChromeBold SCXW150514885 BCX0\" lang=\"FR-FR\" xml:lang=\"FR-FR\" data-contrast=\"none\"><span class=\"NormalTextRun SpellingErrorV2Themed SCXW150514885 BCX0\" data-ccp-parastyle=\"Default\">vector<\/span> <\/span><span class=\"TextRun SCXW150514885 BCX0\" lang=\"EN-US\" xml:lang=\"EN-US\" data-contrast=\"none\"><span class=\"NormalTextRun SCXW150514885 BCX0\" data-ccp-parastyle=\"Default\">and<\/span><\/span><span class=\"TextRun MacChromeBold SCXW150514885 BCX0\" lang=\"EN-US\" xml:lang=\"EN-US\" data-contrast=\"none\"> <span class=\"NormalTextRun SCXW150514885 BCX0\" data-ccp-parastyle=\"Default\">a<\/span><span class=\"NormalTextRun SCXW150514885 BCX0\" data-ccp-parastyle=\"Default\">ll equations<\/span><\/span><span class=\"TextRun SCXW150514885 BCX0\" lang=\"EN-US\" xml:lang=\"EN-US\" data-contrast=\"none\"><span class=\"NormalTextRun SCXW150514885 BCX0\" data-ccp-parastyle=\"Default\"> are generally solved as <\/span><\/span><span class=\"TextRun MacChromeBold SCXW150514885 BCX0\" lang=\"EN-US\" xml:lang=\"EN-US\" data-contrast=\"none\"><span class=\"NormalTextRun SCXW150514885 BCX0\" data-ccp-parastyle=\"Default\">vectors)<\/span><\/span><\/p>\n<p style=\"text-align: center;\">In an Isolated System&#8230;<\/p>\n<p style=\"text-align: center;\">[latex]\\sum \\vec{p} = \\text{ constant}[\/latex]<\/p>\n<p style=\"text-align: center;\">or &#8230;<\/p>\n<p style=\"text-align: center;\">[latex]\\vec{p}_{\\text{before}} = \\vec{p}_{\\text{after}}[\/latex] (Vector Equation)<\/p>\n<p style=\"text-align: center;\">Where&#8230;<\/p>\n<p style=\"text-align: center;\">[latex]\\vec{p} = m\\vec{y}[\/latex]<\/p>\n<p style=\"text-align: center;\">For two or more bodies:<\/p>\n<p style=\"text-align: center;\">[latex]\\vec{p}_{1i(x,y,z)} + \\vec{p}_{2i(x, y z)} + ... = \\vec{p}_{1f(x,y,z)} + \\vec{p}_{2f(x,y,z)} + ...[\/latex]<\/p>\n<p>Where&#8230;<\/p>\n<ul>\n<li>[latex]\\vec{p}[\/latex] is <strong>momentum<\/strong>, the product of mass times velocity and is measured in either newton-seconds (Ns) or kilogram-metres per second (kg m\/s)<\/li>\n<li><span style=\"orphans: 1; text-align: initial; font-size: 14pt;\">[latex]m[\/latex] is <strong>mass<\/strong>, commonly measured in kilograms (Kg), grams (g) or tonnes (t)<\/span><\/li>\n<li>[latex]\\vec{v}[\/latex] is the <strong>velocity<\/strong><span style=\"orphans: 1; text-align: initial; font-size: 14pt;\">, commonly measured in metres\/second (m\/s) or kilometres\/hour (km\/h) and includes a direction<\/span><\/li>\n<\/ul>\n<div class=\"textbox textbox--sidebar\">Extra Help:\u00a0 <a href=\"https:\/\/www.physicsclassroom.com\/class\/momentum\/Lesson-2\/The-Law-of-Action-Reaction-(Revisited)\">The Law of Momentum Conservation<\/a><\/div>\n<p><strong>The Law of the Conservation of Momentum <\/strong>is one of the most powerful and useful laws in the study of mechanics. It belongs to a family of four conservation laws: Conservation of Mass Energy, Conservation of Electric Charge, Conservation of Linear Momentum and the Conservation of Angular Momentum <a class=\"footnote\" title=\"There are other physics conservation laws that exist in the study of particle physics.\" id=\"return-footnote-52-1\" href=\"#footnote-52-1\" aria-label=\"Footnote 1\"><sup class=\"footnote\">[1]<\/sup><\/a>.<\/p>\n<p>The law of conservation of momentum mainly is derived from Newton\u2019s third law and can be analyzed as in the following description excerpted from the Physics Classroom:<\/p>\n<p style=\"padding-left: 40px;\"><strong><em>In every interaction, there is a pair of forces acting on the two interacting objects. The size of the force on the first object <\/em><em>equals<\/em><em> the size of the force on the second object. The direction of the force on the first object is <\/em><em>opposite<\/em><em> to the direction of the force on the second object. Forces always come in pairs &#8211; equal and opposite action-reaction force pairs.<\/em><\/strong><\/p>\n<h1>9.1 Total Momentum of a System<\/h1>\n<p style=\"text-align: center;\"><strong>Total Momentum of a System: The Resolutions of multiple Momentums acting during an event or interaction<\/strong><\/p>\n<p><strong>Momentum are Vectors&#8230; <\/strong>They are added, subtracted and combined according to the direction they have.\u00a0 Quite often, vectors in different directions, such as North and South, East and West, and up and down, are left as the sum of the three different vector directions and not combined any further.<\/p>\n<p>Vectors in the same directions are added and vectors in the opposite direction are subtracted.<\/p>\n<p><strong>For instance:\u00a0 <\/strong><\/p>\n<p style=\"text-align: center;\"><strong>100 Ns East + 50 Ns East = 150 Ns East <\/strong><\/p>\n<p>(The vectors are in the same direction so they are added.)<\/p>\n<p style=\"text-align: center;\"><strong>100 Ns North + 50 Ns South = 50 Ns North<\/strong><\/p>\n<p>(The vectors are in the opposite direction so they are subtracted.)<\/p>\n<p>Vectors in different directions are generally left as the vector sum in each of those directions.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-58\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/Picture-w.png\" alt=\"\" width=\"341\" height=\"176\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/Picture-w.png 682w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/Picture-w-300x155.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/Picture-w-65x34.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/Picture-w-225x116.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/Picture-w-350x181.png 350w\" sizes=\"auto, (max-width: 341px) 100vw, 341px\" \/><\/p>\n<p><strong>For instance:\u00a0 <\/strong><\/p>\n<p style=\"text-align: center;\"><strong>100 Ns North + 50 Ns East = 100 Ns North + 50 Ns East<\/strong><\/p>\n<p>(The vectors are in the different direction so they are left alone.)<\/p>\n<p>Vectors that are angled between two directions are first broken up into the various components<\/p>\n<p>for those directions before they are added or subtracted.<\/p>\n<p><strong>For instance: <\/strong>Using right angle trigonometry, this 150 Ns vector\u00a0 would be broken up into the force directed upwards and the force to the left.<\/p>\n<p>To find the <strong>Momentum upwards<\/strong> use the Sine function:<\/p>\n<p style=\"text-align: center;\">[latex]\\sin 42\u00b0 = \\dfrac{\\text{Momentum up}}{150 \\text{ Ns}}[\/latex]<\/p>\n<p>To find the <strong>Momentum right<\/strong> use the Cosine function:<\/p>\n<p style=\"text-align: center;\">[latex]\\cos 42\u00b0 = \\dfrac{\\text{Momentum right}}{150 \\text{ Ns}}[\/latex]<\/p>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 9.1.1<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-59 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-a.png\" alt=\"\" width=\"208\" height=\"257\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-a.png 371w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-a-243x300.png 243w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-a-65x80.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-a-225x278.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-a-350x432.png 350w\" sizes=\"auto, (max-width: 208px) 100vw, 208px\" \/><\/p>\n<p>Resolve the Following Multiple Momentum Vectors:<\/p>\n<ol type=\"a\">\n<li><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-60 size-full\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-a.1.png\" alt=\"\" width=\"990\" height=\"110\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-a.1.png 990w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-a.1-300x33.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-a.1-768x85.png 768w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-a.1-65x7.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-a.1-225x25.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-a.1-350x39.png 350w\" sizes=\"auto, (max-width: 990px) 100vw, 990px\" \/><strong>Solution<\/strong><br \/>\n[latex]\\vec{p} = 10 \\text{ Ns West } + 10 \\text{ Ns East or } 0 Ns[\/latex]<\/li>\n<li><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-61 size-full\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-b.png\" alt=\"\" width=\"962\" height=\"119\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-b.png 962w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-b-300x37.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-b-768x95.png 768w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-b-65x8.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-b-225x28.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-b-350x43.png 350w\" sizes=\"auto, (max-width: 962px) 100vw, 962px\" \/><strong>Solution<\/strong><br \/>\n[latex]\\vec{p} = 10 \\text{ Ns East } + 10 \\text{ Ns East or } 20 \\text{ Ns East}[\/latex]<\/li>\n<li><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-62 size-full\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-c.png\" alt=\"\" width=\"1448\" height=\"128\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-c.png 1448w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-c-300x27.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-c-1024x91.png 1024w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-c-768x68.png 768w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-c-65x6.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-c-225x20.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-c-350x31.png 350w\" sizes=\"auto, (max-width: 1448px) 100vw, 1448px\" \/><strong>Solution<\/strong><br \/>\n[latex]\\vec{p} = 10 \\text{ Ns East } + 10 \\text{ Ns East } + 10 \\text{ Ns East or } 30 \\text{ Ns East}[\/latex]<\/li>\n<li><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-63 size-full\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-d.png\" alt=\"\" width=\"1425\" height=\"132\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-d.png 1425w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-d-300x28.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-d-1024x95.png 1024w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-d-768x71.png 768w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-d-65x6.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-d-225x21.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-d-350x32.png 350w\" sizes=\"auto, (max-width: 1425px) 100vw, 1425px\" \/><strong>Solution<\/strong><br \/>\n[latex]\\vec{p} = 10 \\text{ Ns East } + 10 \\text{ Ns West } + 10 \\text{ Ns West or } 10 \\text{ Ns}[\/latex]<\/li>\n<li><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-64 size-full\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-e-e1675186230959.png\" alt=\"\" width=\"1456\" height=\"145\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-e-e1675186230959.png 1456w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-e-e1675186230959-300x30.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-e-e1675186230959-1024x102.png 1024w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-e-e1675186230959-768x76.png 768w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-e-e1675186230959-65x6.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-e-e1675186230959-225x22.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-e-e1675186230959-350x35.png 350w\" sizes=\"auto, (max-width: 1456px) 100vw, 1456px\" \/><br \/>\n<strong>Solution<\/strong><br \/>\n[latex]\\vec{p} = 10 \\text{ Ns East } + 6.0 \\text{ Ns West } + 6.0 \\text{ Ns West or } 2.0 \\text{ Ns West}[\/latex]<\/li>\n<li><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-65\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-f-e1675186255470.png\" alt=\"\" width=\"311\" height=\"238\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-f-e1675186255470.png 592w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-f-e1675186255470-300x229.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-f-e1675186255470-65x50.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-f-e1675186255470-225x172.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-f-e1675186255470-350x267.png 350w\" sizes=\"auto, (max-width: 311px) 100vw, 311px\" \/><br \/>\n<strong>Solution<\/strong><br \/>\n[latex]\\vec{p} = 10 \\text{ Ns East } + 5.0 \\text{ Ns South (left like this)}[\/latex]<\/li>\n<li><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-66\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-g.png\" alt=\"\" width=\"334\" height=\"294\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-g.png 682w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-g-300x264.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-g-65x57.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-g-225x198.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-g-350x308.png 350w\" sizes=\"auto, (max-width: 334px) 100vw, 334px\" \/><br \/>\n<strong>Solution<\/strong><br \/>\nBreaking the 15 Ns up, we get:<\/p>\n<ul>\n<li>[latex]\\vec{p_{\\text{North}}} = 15 \\text{ Ns } \\sin 40^{\\circ}[\/latex] or 9.6 Ns North<\/li>\n<li>[latex]\\vec{p_{\\text{East}}} = 15 \\text{ Ns } \\cos 40^{\\circ}[\/latex] or 11.5 Ns East<\/li>\n<\/ul>\n<p>[latex]\\vec{p} = 11.5 \\text{ Ns East } + 9.6 \\text{ Ns North } + 5.0 \\text{ Ns South...}[\/latex]<br \/>\nResultant is [latex]11.5 \\text{ Ns East } + 4.6 \\text{ Ns North}[\/latex]<br \/>\n[latex](\\approx 12 \\text{ Ns East } + 4.6 \\text{ Ns North}[\/latex])<\/li>\n<li>\u00a0<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-68\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-H-1.png\" alt=\"\" width=\"280\" height=\"252\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-H-1.png 783w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-H-1-300x270.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-H-1-768x691.png 768w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-H-1-65x59.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-H-1-225x203.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.1.1-H-1-350x315.png 350w\" sizes=\"auto, (max-width: 280px) 100vw, 280px\" \/><br \/>\n<strong>Solution<\/strong><br \/>\nBreaking the 15 Ns up, we get:<br \/>\n[latex]\\begin{array}  \\vec{p}_{north} = 15 Ns \\sin{40}^{\\circ} \\text{ or } 9.6 Ns \\text{ North} \\\\  \\vec{p}_{north} = 15 Ns \\cos{40}^{\\circ} \\text{ or } 11.5 Ns \\text{ East}  \\end{array}[\/latex]Breaking the 12 Ns up, we get:<br \/>\n[latex]\\vec{p}_{north} = 12 Ns \\sin{50}^{\\circ} \\text{ or } 9.2 Ns \\text{ North}[\/latex]<br \/>\n[latex]\\vec{p}_{north} = 12 Ns \\cos{50}^{\\circ} \\text{ or } 17.7 Ns \\text{ East}[\/latex][latex]\\vec{p}_{north} = 9.6 Ns \\text{ North } + 9.2 Ns \\text{ North } + 5 Ns \\text{ South or } 13.8 Ns \\text{ North}[\/latex]<br \/>\n[latex]\\vec{p}_{north} = 11.5 Ns \\text{ East} + 7.7 Ns \\text{ West or } 4.8 Ns \\text{ East}[\/latex]Resultant is [latex]\\vec{p} = 13.8 Ns \\text{ North} 3.8 Ns \\text{ East}[\/latex], (\u2248 [latex]14 Ns \\text{ North } + 3.8 Ns { East}[\/latex])<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<h1><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-72 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/Picture-4.jpg\" alt=\"\" width=\"497\" height=\"349\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/Picture-4.jpg 706w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/Picture-4-300x211.jpg 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/Picture-4-65x46.jpg 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/Picture-4-225x158.jpg 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/Picture-4-350x246.jpg 350w\" sizes=\"auto, (max-width: 497px) 100vw, 497px\" \/><br \/>\n9.2 Linear Collisions<\/h1>\n<div class=\"textbox\">\n<ul>\n<li>Extra help: <a href=\"https:\/\/www.physicsclassroom.com\/class\/momentum\/Lesson-1\/Real-World-Applications\">Real World Applications<\/a><\/li>\n<li>Extra help: \u00a0<a href=\"https:\/\/www.physicsclassroom.com\/class\/momentum\/Lesson-2\/Momentum-Conservation-Principle\">The Law of Momentum Conservation<\/a><\/li>\n<li>Research News: <a href=\"https:\/\/phys.org\/news\/2019-05-neutron-stars-collided-solar-billions.html\">Two neutron stars collided near the solar system billions of years ago<\/a><\/li>\n<\/ul>\n<\/div>\n<div style=\"text-align: left;\">Equations Introduced or Used for this Section:<\/div>\n<div style=\"text-align: center;\">In an Isolated System&#8230;<\/div>\n<div>\n<p style=\"text-align: center;\">[latex]\\sum \\vec{p} = constant[\/latex]<\/p>\n<p style=\"text-align: center;\">or &#8230;<\/p>\n<p style=\"text-align: center;\">[latex]\\vec{p}_{before} = \\vec{p}_{after} (Vector Equation)[\/latex]<\/p>\n<p style=\"text-align: left;\">Where&#8230;<\/p>\n<ul>\n<li style=\"text-align: left;\">[latex]\\vec{p} = m\\vec{y}[\/latex]<\/li>\n<\/ul>\n<p>For two or more bodies:<\/p>\n<p style=\"text-align: center;\">[latex]\\vec{p}_{1i \\text{(x, y, z)}} + \\vec{p}_{2i \\text{(x, y, z)}} + ... = \\vec{p}_{1f \\text{(x, y, z)}} + \\vec{p}_{2f \\text{(x, y, z)}} + ...[\/latex]<\/p>\n<\/div>\n<p>The <strong>Law of Momentum Conservation<\/strong> is one of the foundational laws in Physics. The law is stated as follows:<\/p>\n<p>In the absence of external forces that can change the momentum of objects inside a system from the outside, the vector sum of the momentum of all objects moving within the system will equal the vector sum of momentum of these same objects after an interaction has occurred, be it a collision or explosion.<\/p>\n<div class=\"textbox textbox--sidebar\">Extra help: <a href=\"https:\/\/www.physicsclassroom.com\/Class\/momentum\/u4l2c.cfm\">Isolated Systems<\/a><\/div>\n<p>This absence of external forces acting on the object means that the system being looked at is termed an <strong>isolated system<\/strong>. Isolated systems require that no external forces outside of the forces acting between the objects within a system can be acting.\u00a0 This means that the forces acting on the objects are in fact being balanced between the objects and not by outside means.<\/p>\n<p>How this works for two bodies that collide is explained as follows:\u00a0 If two objects collide, the forces acting between the two objects are equal and opposite as defined by Newton\u2019s third law.<\/p>\n<p style=\"text-align: center;\">[latex]\\vec{\\text{Force}}_{\\text{object 1}} = - \\vec{\\text{Force}}_{\\text{object 2}}[\/latex]<\/p>\n<p><strong>These forces are equal in strength (magnitude) but act in opposite directions.<\/strong><\/p>\n<p>These forces can only act on each other for exactly the same amount of time, meaning:<\/p>\n<p style=\"text-align: center;\">[latex]\\text{time}_{\\text{object 1}}=\\text{time}_{\\text{object 2}}[\/latex]<\/p>\n<p>You should recognize that we are talking about equal and opposite <strong>Impulses<\/strong> (<a href=\"https:\/\/opentextbc.ca\/foundationsofphysics\/chapter\/equilibrium-newtons-first-law\/\">Chapter 7<\/a>) that are acting on the two colliding bodies. This means that:<\/p>\n<p style=\"text-align: center;\">[latex](\\vec{\\text{Force}}_{\\text{object 1}})(\\text{time}_{\\text{object 1}})=- (\\vec{\\text{Force}}_{\\text{object 2}})(\\text{time}_{\\text{object 2}})[\/latex]<\/p>\n<p>You can conclude from this that we are talking about equal and opposite <strong>changes in momentum<\/strong> (<a href=\"https:\/\/opentextbc.ca\/foundationsofphysics\/chapter\/equilibrium-newtons-first-law\/\">Chapter 7<\/a>) that will result from the equal and opposite impulses that are acting on the two colliding bodies. This means that:<\/p>\n<p style=\"text-align: center;\">[latex](\\text{mass}_{\\text{object 1}})(\\Delta \\vec{\\text{velocity}}_{\\text{object 1}}) = - (\\text{mass}_{\\text{object 2}})(\\Delta \\vec{\\text{velocity}}_{\\text{object 2}})[\/latex]<\/p>\n<p><strong>The changes in momentum that occur for either body are both equal and opposite.<\/strong><\/p>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 9.2.1<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>A circus cannon on wheels (500 kg) launches a clown of mass 65 kg with a horizontal velocity of 4.2 m\/s, calculate:<\/p>\n<ol type=\"i\">\n<li>\u00a0Momentum of the clown (before and after)<\/li>\n<li>The combined momentum of the clown and circus cannon before and after firing<\/li>\n<li>The speed that this circus cannon recoils at if originally at rest<\/li>\n<\/ol>\n<p><strong>Solution<\/strong><\/p>\n<p>Data:<\/p>\n<ul>\n<li>[latex]m_{cannon} = 500 \\text{ kg}[\/latex]<\/li>\n<li>[latex]\\vec{v}_{i\\ cannon} = 0 \\text{ m\/s}[\/latex]<\/li>\n<li>[latex]\\vec{f}_{f\\ cannon} = \\text{ Unknown}[\/latex]<\/li>\n<li>[latex]m_{clown} = 65 \\text{ kg}[\/latex]<\/li>\n<li>[latex]\\vec{v}_{i\\ clown} = 0 \\text{ m\/s}[\/latex]<\/li>\n<li>[latex]\\vec{v}_{f\\ clown} = 4.2 \\text{ m\/s}[\/latex]<\/li>\n<\/ul>\n<p>Question (i):<\/p>\n<ul>\n<li>[latex]\\vec{p}_{\\text{i clown}} = (65 \\text{ kg})(0 \\text{ m\/s}) \\text{ or } 0 \\text{ Ns}[\/latex]<\/li>\n<li>[latex]\\vec{p}_{\\text{f clown}} = (65 \\text{ kg})(4.2 \\text{ m\/s}) \\text{ or } 273 \\text{ Ns }(\\approx 270 \\text{ Ns}[\/latex])<\/li>\n<\/ul>\n<p>Question (ii):<\/p>\n<ul>\n<li>[latex]\\vec{p}_{\\text{i clown}} + \\vec{p}_{\\text{i cannon}} = 0 \\text{ Ns}[\/latex]<\/li>\n<li>Therefore, [latex]\\vec{p}_{\\text{f clown}} + \\vec{p}_{\\text{f cannon}} = 0 \\text{ Ns}[\/latex]<\/li>\n<\/ul>\n<p>Question (iii):<\/p>\n<ul>\n<li>[latex]\\vec{p}_{\\text{i clown}} + \\vec{p}_{\\text{i cannon}} = \\vec{p}_{\\text{f clown}} + \\vec{p}_{\\text{f cannon}}[\/latex]<\/li>\n<li>[latex](m_{\\text{clown}})(\\vec{v}_{\\text{i clown}}) + (m_{\\text{cannon}})(\\vec{v}_{\\text{i cannon}}) = (m_{\\text{clown}})(\\vec{v}_{\\text{f clown}}) + (m_{\\text{cannon}})(\\vec{v}_{\\text{f cannon}})[\/latex]<\/li>\n<li>[latex](65 \\text{ kg})(0 \\text{ m\/s}) + (500 \\text{ kg})(0 \\text{ m\/s}) = (65 \\text{ kg})(4.2 \\text{ m\/s}) + (500 \\text{ kg})(\\vec{v}_{\\text{f cannon}})[\/latex]<\/li>\n<li>[latex]0 \\text{ Ns } + 0 \\text{ Ns } = 273 \\text{ Ns } + (500 \\text{ kg})(\\vec{v}_{\\text{f cannon}})[\/latex]<\/li>\n<li>[latex]\\vec{v}_{\\text{f cannon}} = -273 \\text{ Ns } \\div 500 \\text{ kg or } -0.55 \\text{m\/s (cannon recoils backwards)}[\/latex]<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 9.2.2<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p><span class=\"TextRun SCXW240874856 BCX0\" lang=\"EN-US\" xml:lang=\"EN-US\" data-contrast=\"none\"><span class=\"NormalTextRun SCXW240874856 BCX0\" data-ccp-parastyle=\"Default\" data-ccp-parastyle-defn=\"{&quot;ObjectId&quot;:&quot;91c66a2c-ba54-4445-bd05-2c763565bfc2|30&quot;,&quot;ClassId&quot;:1073872969,&quot;Properties&quot;:[469775450,&quot;Default&quot;,201340122,&quot;2&quot;,134233614,&quot;true&quot;,469778129,&quot;Default&quot;,335572020,&quot;1&quot;,469777841,&quot;Helvetica&quot;,469777842,&quot;Arial Unicode MS&quot;,469777843,&quot;Arial Unicode MS&quot;,469777844,&quot;Helvetica&quot;,469769226,&quot;Helvetica,Arial Unicode MS&quot;,335551500,&quot;0&quot;,268442635,&quot;24&quot;,335551547,&quot;1033&quot;]}\">A 4.0 kg curling rock moving 0.5 m\/s West collides with a stationary 4.0 kg curling rock at rest. If the original curling rock comes to a full stop upon collision what velocity should the second curling rock be moving at?<\/span><\/span><\/p>\n<p><strong>Solution<\/strong><\/p>\n<p>Data:<\/p>\n<ul>\n<li>[latex]m_{\\text{rock 1}} = 4.0 \\text{ kg}[\/latex]<\/li>\n<li>[latex]\\vec{v}_{\\text{i rock 1}} = 0.5 \\text{ m\/s West}[\/latex]<\/li>\n<li>[latex]\\vec{v}_{\\text{f rock 1}} = 0 \\text{ m\/s}[\/latex]<\/li>\n<\/ul>\n<p>&nbsp;<\/p>\n<ul>\n<li>[latex]m_{\\text{rock 2}} = 4.0 \\text{ kg}[\/latex]<\/li>\n<li>[latex]\\vec{v}_{\\text{i rock 2}} = 0.5 \\text{ m\/s West}[\/latex]<\/li>\n<li>[latex]\\vec{v}_{\\text{f rock 2}} = 0 \\text{ m\/s}[\/latex]<\/li>\n<\/ul>\n<p>Solution:<\/p>\n<ul>\n<li>[latex]\\vec{p}_{\\text{i rock 1}} + \\vec{p}_{\\text{i rock 2}} = \\vec{p}_{\\text{f rock 1}} + \\vec{p}_{\\text{f rock 1}}[\/latex]<\/li>\n<li>[latex](m_{\\text{rock 1}})(\\vec{v}_{\\text{i rock 1}}) + (m_{\\text{rock 2}})(\\vec{v}_{\\text{i rock 2}}) = (m_{\\text{rock 1}})(\\vec{v}_{\\text{f rock 1}}) + (m_{\\text{rock 2}})(\\vec{v}_{\\text{f rock 2}})[\/latex]<\/li>\n<li>[latex](4.0 \\text{ kg})(0.5 \\text{ m\/s West}) + (4.0 \\text{ kg})(0 \\text{ m\/s}) = (4.0 \\text{ kg})(0 \\text{ m\/s}) + (4.0 \\text{ kg})(\\vec{v}_{\\text{f rock 2}})[\/latex]<\/li>\n<li>[latex]2.0 \\text{ Ns West } + 0 \\text{ Ns } = 0 \\text{ Ns } + (4.0 \\text{ kg})(\\vec{v}_{\\text{f rock 2}})[\/latex]<\/li>\n<li>[latex](\\vec{v}_{\\text{f rock 2}}) = 2.0 \\text{ Ns West } \\div 4.0 \\text{ kg or } 0.5 \\text{ m\/s West}[\/latex]<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 9.2.3<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p><span class=\"NormalTextRun SCXW94985447 BCX0\" data-ccp-parastyle=\"Default\">During a tackle in a football game<\/span><span class=\"NormalTextRun SCXW94985447 BCX0\" data-ccp-parastyle=\"Default\">,<\/span><span class=\"NormalTextRun SCXW94985447 BCX0\" data-ccp-parastyle=\"Default\"> two players of mass 90 kg and 70 kg were moving towards each other<\/span><span class=\"NormalTextRun SCXW94985447 BCX0\" data-ccp-parastyle=\"Default\">,<\/span><span class=\"NormalTextRun SCXW94985447 BCX0\" data-ccp-parastyle=\"Default\"> and after collision both c<\/span><span class=\"NormalTextRun SCXW94985447 BCX0\" data-ccp-parastyle=\"Default\">a<\/span><span class=\"NormalTextRun SCXW94985447 BCX0\" data-ccp-parastyle=\"Default\">me to a full stop.\u00a0 If the velocity of the larger player before collision was 7.0 m\/s what was the velocity of the smaller player?<\/span><\/p>\n<p><strong>Solution<\/strong><\/p>\n<p>All we have for the direction in this problem is that both players were running towards each other.<\/p>\n<p>Data:<\/p>\n<ul>\n<li>[latex]m_{90} = 90 \\text{ kg}[\/latex]<\/li>\n<li>[latex]\\vec{v}_{\\text{i 90}} = 7.0 \\text{ m\/s}[\/latex]<\/li>\n<li>[latex]\\vec{v}_{\\text{f 90}} = 0 \\text{ m\/s}[\/latex]<\/li>\n<\/ul>\n<p>&nbsp;<\/p>\n<ul>\n<li>[latex]m_{70} = 70 \\text{ kg}[\/latex]<\/li>\n<li>[latex]\\vec{v}_{\\text{i 70}} = \\text{ Find}[\/latex]<\/li>\n<li>[latex]\\vec{v}_{\\text{f 70}} = 0 \\text{ m\/s}[\/latex]<\/li>\n<\/ul>\n<p>Solution:<\/p>\n<ul>\n<li>[latex]\\vec{p}_{\\text{i 90}} + \\vec{p}_{\\text{i 70}} = \\vec{p}_{\\text{f 90}} + \\vec{p}_{\\text{f 70}}[\/latex]<\/li>\n<li>[latex](m_{\\text{90}})(\\vec{v}_{\\text{i 90}}) + (m_{\\text{70}})(\\vec{v}_{\\text{i 70}}) = (m_{\\text{90}})(\\vec{v}_{\\text{f 90}}) + (m_{\\text{70}})(\\vec{v}_{\\text{f 70}})[\/latex]<\/li>\n<li>[latex](90 \\text{ kg})(7.0 \\text{ m\/s}) + (70 \\text{ kg})(\\vec{v}_{\\text{i 70}}) = (90 \\text{ kg})(0 \\text{ m\/s}) + (70 \\text{ kg})(0 \\text{ m\/s})[\/latex]<\/li>\n<li>[latex]630 \\text{ Ns } + (70 \\text{ kg})(\\vec{v}_{\\text{i 70}}) = 0 \\text{ Ns } + 0 \\text{ Ns}[\/latex]<\/li>\n<li>[latex](\\vec{v}_{\\text{i 70}}) = - 630 \\text{ Ns } \\div 70 \\text{ kg or } - 9.0 \\text{ m\/s}[\/latex]<\/li>\n<\/ul>\n<p>Note:\u00a0 The [latex]- 9.0[\/latex] m\/s means that the smaller player was running in the opposite direction (towards) as the larger player.<\/p>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 9.2.4<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p><span class=\"NormalTextRun SCXW11241910 BCX0\" data-ccp-parastyle=\"Default\">If a 16 kg medicine ball <\/span><span class=\"NormalTextRun SCXW11241910 BCX0\" data-ccp-parastyle=\"Default\">is <\/span><span class=\"NormalTextRun SCXW11241910 BCX0\" data-ccp-parastyle=\"Default\">thrown horizontally to a stationary 65 kg skater on ice (consider this to be frictionless) and causes the skater who is now holding the ball to move off with a horizontal velocity of 0.50 m\/s, what was the velocity the ball was thrown at?<\/span><\/p>\n<p><strong>Solution<\/strong><\/p>\n<p>Data:<\/p>\n<ul>\n<li>[latex]m_{16} = 16 \\text{ kg}[\/latex]<\/li>\n<li>[latex]\\vec{v}_{\\text{i 16}} = \\text{ Find}[\/latex]<\/li>\n<li>[latex]\\vec{v}_{\\text{f 16}} = 0.50 \\text{ m\/s}[\/latex]<\/li>\n<\/ul>\n<p>&nbsp;<\/p>\n<ul>\n<li>[latex]m_{65} = 70 \\text{ kg}[\/latex]<\/li>\n<li>[latex]\\vec{v}_{\\text{i 65}} = 0 \\text{ m\/s}[\/latex]<\/li>\n<li>[latex]\\vec{v}_{\\text{f 65}} = 0.50 \\text{ m\/s}[\/latex]<\/li>\n<\/ul>\n<p>Solution:<\/p>\n<ul>\n<li>[latex]\\vec{p}_{\\text{i 16}} + \\vec{p}_{\\text{i 65}} = \\vec{p}_{\\text{f 16}} + \\vec{p}_{\\text{f 65}}[\/latex]<\/li>\n<li>[latex](m_{\\text{16}})(\\vec{v}_{\\text{i 16}}) + (m_{\\text{65}})(\\vec{v}_{\\text{i 65}}) = (m_{\\text{16}})(\\vec{v}_{\\text{f 16}}) + (m_{\\text{65}})(\\vec{v}_{\\text{f 65}})[\/latex]<\/li>\n<li>[latex](16 \\text{ kg})(\\vec{v}_{\\text{i 16}}) + (65 \\text{ kg})(0 \\text{ m\/s}) = (16 \\text{ kg } + 65 \\text{ kg})(0.5 \\text{ m\/s})[\/latex]<\/li>\n<li>[latex](16 \\text{ kg})(\\vec{v}_{\\text{i 16}}) + (0 \\text{ Ns}) = (40.5 Ns)[\/latex]<\/li>\n<li>[latex](\\vec{v}_{\\text{i 16}}) = 40.5 Ns \\div 16 kg[\/latex] \u2248 2.5 m\/s horizontally<\/li>\n<\/ul>\n<p>Note:\u00a0 The \u2248 + 2.5 m\/s means that both the ball and the skater continue moving in the same horizontal direction as the medicine ball was thrown.<\/p>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Exercise 9.2<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<ol>\n<li>A cannon on wheels (2750 kg) shoots a cannon ball of mass 14.7 kg.\u00a0 If this cannon ball leaves the muzzle with a horizontal velocity of 487 m\/s, calculate:\n<ol type=\"i\">\n<li>Momentum of this cannon ball (before and after)<\/li>\n<li>The combined momentum of the cannon ball and cannon before and after firing<\/li>\n<li>The recoil velocity of this cannon<\/li>\n<\/ol>\n<\/li>\n<li>A 0.40 kg rock moving 0.50 m\/s West collides with a stationary 0.50 kg rubber ball at rest. If the 0.40 kg rock comes to a full stop upon collision, what velocity should this ball have?<\/li>\n<li>During a dynamics experiment, two trolley cars of mass 9.0 kg and 6.0 kg were moving towards each other, and after collision both come to a full stop. If the velocity of the larger trolley car before collision was 0.025 m\/s, what was the velocity of the smaller trolley car?<\/li>\n<li>In a bakery fight, a 2.0 kg mass of bread dough moving at 16.0 m\/s North collides with a stationary molasses container. The two objects coalesce and move off at 7.50 m\/s North.\u00a0 What was the mass of the stationary molasses container?<\/li>\n<li>An 800 kg asteroid in space has a velocity of 5.00 m\/s. An explosion causes the rear section (mass 240 kg) of this asteroid to be sent in the opposite direction at 1.20 m\/s.\u00a0 At what velocity did the remaining part of this asteroid continue to move?<\/li>\n<li>If a 5.40 g bullet is fired horizontally into a suspended 10.0 kg block of wood and causes the block with bullet embedded to move off with a horizontal velocity of 0.30 m\/s, what was the impact velocity of this bullet?<\/li>\n<li>Brian\u2019s 10 gram toy train car is traveling at 5.0 cm\/s when it collides with a 15 gram tank car originally at rest. If the two cars lock together, at what speed should they travel off?<\/li>\n<li>A 95 kg hockey player is skating at 12 m\/s when he runs into a larger 120 kg player that is only skating at 5.0 m\/s in the same direction. If the two skaters become entangled, at what speed do they slide off?<\/li>\n<li>A 105 kg fullback is running straight towards the end zone at 9.5 m\/s. A 95 kg defensive back is running straight towards the fullback, tackling the fullback with both dropping to the ground with a velocity of 0 m\/s.\u00a0 What was the velocity of this defensive back prior to the tackle?<\/li>\n<li>Kyle\u2019s small toy train boxcar of mass of 60 g is rolling at 4.0 cm\/s towards a flatcar of mass 45 g that is stationary. If the two cars lock together, at what speed do they continue to move down the track?<\/li>\n<li>In the movies, an object such as a football is thrown at a running person, striking and stopping them. If we consider a 70 kg person running toward you at 8.0 m\/s, what speed must a 2.0 kg football be thrown at to stop this person?<\/li>\n<li>A medicine ball is thrown at 8.0 m\/s towards a stationary 55 kg skater on ice. After she catches it, both the skater and the ball move off at 0.8 m\/s.\u00a0 What is the mass of this medicine ball?<\/li>\n<li>A large railway wagon, mass 4200 kg, rolls down a track at 2.5 m\/s and collides with a smaller wagon, mass 2750 kg, moving at 1.0 m\/s in the same direction. If the smaller wagon has a velocity of 3.0 m\/s after the collision, what is the final velocity of the large wagon?<\/li>\n<li>A small particle A of mass [latex]1.41 \\times 10^{-5}[\/latex] g moving at [latex]3.33 \\times 10^3[\/latex] m\/s is hit by particle B moving at 5.46 x 104 m\/s in the same direction. As a result, particle A increased its velocity to [latex]6.66 \\times 10^3[\/latex] m\/s and particle B was slowed to [latex]4.60 \\times 10^3[\/latex] m\/s in the original direction.\u00a0 What was the mass of particle B?<\/li>\n<li>A 0.85 kg spring toy (at rest) \u201cpops\u201d into 3 pieces. The largest piece of mass 0.45 kg moves at 5.0 cm\/s to the right.\u00a0 A smaller piece of mass, of mass 0.25 kg, moves at 8.0 cm\/s to the left.\u00a0 At what speed does the remaining piece move off?<\/li>\n<li>A small 65 kg hockey player takes a run at a larger 115 kg player at rest. If the smaller player was traveling at 16 m\/s towards the larger before the collision and rebounded at 4 m\/s after, at what speed did the larger player move off?<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<div class=\"textbox\">\n<ul>\n<li>Article to Read: <a href=\"https:\/\/phys.org\/news\/2018-12-pedestrians-cm-comfort-zone-collisions.html\">Eindhoven University of Technology (2018) Pedestrians keep a 75 cm comfort zone to prevent collisions<\/a><\/li>\n<li>Extra Help: <a href=\"https:\/\/www.physicsclassroom.com\/class\/momentum\/Lesson-2\/Using-Equations-as-a-Recipe-for-Algebraic-Problem\">Using Equations as a Recipe for Algebraic Problem-Solving<\/a><\/li>\n<li>Extra Help: <a href=\"https:\/\/www.physicsclassroom.com\/class\/momentum\/Lesson-2\/Using-Equations-as-a-Guide-to-Thinking\">Using Equations as a Guide to Thinking<\/a><\/li>\n<\/ul>\n<\/div>\n<h1><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-696 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.3-two-dimensional-collisions.jpg\" alt=\"\" width=\"528\" height=\"384\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.3-two-dimensional-collisions.jpg 2497w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.3-two-dimensional-collisions-300x218.jpg 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.3-two-dimensional-collisions-1024x746.jpg 1024w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.3-two-dimensional-collisions-768x559.jpg 768w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.3-two-dimensional-collisions-1536x1118.jpg 1536w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.3-two-dimensional-collisions-2048x1491.jpg 2048w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.3-two-dimensional-collisions-65x47.jpg 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.3-two-dimensional-collisions-225x164.jpg 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.3-two-dimensional-collisions-350x255.jpg 350w\" sizes=\"auto, (max-width: 528px) 100vw, 528px\" \/><\/h1>\n<h1>9.3 Two Dimensional Collisions<\/h1>\n<div class=\"textbox\">\n<ul>\n<li>Article to Read:\u00a0 <a href=\"https:\/\/phys.org\/news\/2018-12-big-space-uranus-lopsided.html\">A big space crash likely made Uranus lopsided<\/a><\/li>\n<li>Article to Read: <a href=\"http:\/\/www.iflscience.com\/space\/uranus-experienced-a-colossal-impact-that-knocked-it-on-its-side\/\">Uranus has experienced a colossal pounding<\/a><\/li>\n<\/ul>\n<\/div>\n<p>Equations Introduced or Used for this Section:<\/p>\n<p style=\"text-align: center;\">In an Isolated System&#8230;<\/p>\n<p style=\"text-align: center;\">[latex]\\sum \\vec{p} = \\text{ constant}[\/latex]<\/p>\n<p style=\"text-align: center;\">or &#8230;<\/p>\n<p style=\"text-align: center;\">[latex]\\vec{p}_{\\text{before}} = \\vec{p}_{\\text{after}}[\/latex] (Vector Equation)<\/p>\n<p style=\"text-align: left;\">Where&#8230;<\/p>\n<ul>\n<li style=\"text-align: left;\">[latex]\\vec{p} = m\\vec{y}[\/latex]<\/li>\n<\/ul>\n<p style=\"text-align: center;\">For two or more bodies:<\/p>\n<p style=\"text-align: center;\">[latex]\\vec{p}_{1i(x,y,z)} + \\vec{p}_{2i(x, y z)} + ... = \\vec{p}_{1f(x,y,z)} + \\vec{p}_{2f(x,y,z)} + ...[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>The<strong> Law of the Conservation of Momentum<\/strong> is a law that can be expanded from linear one-dimensional to the multi-dimensional universe of 2 or 3 dimensions.\u00a0 This is due to the fact that momentum must be conserved in its original sum as long as no external forces act to change the momentum of the system.<\/p>\n<p>The grid shown to the right is the common (x, y, z) representation of 3 dimensional space. All that is missing from this would be the time that identifies the spot where an object would be at a specific time.<\/p>\n<p>It is also common to see vector momenta sums to be identified according to the directions defined by <strong>(x, y, z)<\/strong>.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-697 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.3-3D-coordinates.png\" alt=\"\" width=\"553\" height=\"563\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.3-3D-coordinates.png 1081w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.3-3D-coordinates-295x300.png 295w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.3-3D-coordinates-1006x1024.png 1006w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.3-3D-coordinates-768x781.png 768w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.3-3D-coordinates-65x66.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.3-3D-coordinates-225x229.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.3-3D-coordinates-350x356.png 350w\" sizes=\"auto, (max-width: 553px) 100vw, 553px\" \/><\/p>\n<p>As such, you could see the momentum of an object defined by:<\/p>\n<p style=\"text-align: center;\">[latex]\\vec{p}_{\\text{i object}} = \\vec{p}_{i\\ x} + \\vec{p}_{i\\ y} + \\vec{p}_{i\\ z}[\/latex]<\/p>\n<p>When solving two or three dimensional problems using the Law of Conservation of Momentum, you solve as you did before with the one dimensional problems, except now you need to make sure that the momentum in each of the three directions is conserved both before and after an event occurs.<\/p>\n<p>This means that:<\/p>\n<p style=\"text-align: center;\">[latex]\\vec{p}_{\\text{before (x, y, z)}} = \\vec{p}_{\\text{after (x, y, z)}}[\/latex]<\/p>\n<p>As an example of this, if the vector sum of the momentum of two or more interacting objects is:<\/p>\n<p style=\"text-align: center;\">[latex]\\vec{p}_{\\text{before (x, y, z)}} = 25 \\text{ Ns}_x - 30 \\text{ Ns}_y + 10 \\text{ Ns}_z[\/latex]<\/p>\n<p>Then the vector sum of the momentum of two or more interacting objects afterwards is:<\/p>\n<p style=\"text-align: center;\">[latex]\\vec{p}_{\\text{after (x, y, z)}} = 25 \\text{ Ns}_x - 30 \\text{ Ns}_y + 10 \\text{ Ns}_z[\/latex]<\/p>\n<p>In solving these two or three dimensional problems, you will be using trigonometry and the Pythagorean Theorem extensively, as you can see in the following two examples.<\/p>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 9.3.1<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>A 120 kg lineman is running at 8.5 m\/s North and tackles a 85 kg halfback running at 9.6 m\/s East.\u00a0 At what velocity do these two move after the tackle occurs?<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-704 aligncenter\" style=\"font-size: 0.9em; word-spacing: normal;\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.3.1-example.png\" alt=\"\" width=\"482\" height=\"309\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.3.1-example.png 1576w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.3.1-example-300x193.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.3.1-example-1024x658.png 1024w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.3.1-example-768x493.png 768w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.3.1-example-1536x986.png 1536w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.3.1-example-65x42.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.3.1-example-225x144.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.3.1-example-350x225.png 350w\" sizes=\"auto, (max-width: 482px) 100vw, 482px\" \/><\/p>\n<p><strong>Solution<\/strong><\/p>\n<p>First, find the momenta of the two football players heading into the tackle.<\/p>\n<p>The 85 kg Halfback:<\/p>\n<ul>\n<li>[latex]\\vec{p}_{85} = m_{85} \\vec{v}_{\\text{i 85}}[\/latex]<\/li>\n<li>[latex]p_{85} = (85 \\text{ kg})(9.6 \\text{ m\/s East})[\/latex] or 816 Ns East<\/li>\n<\/ul>\n<p>The 120 kg Lineman:<\/p>\n<ul>\n<li>[latex]\\vec{p}_{120} = m_{120} \\vec{v}_{\\text{ i 120}}[\/latex]<\/li>\n<li>[latex]p_{120} = (120 \\text{ kg})(8.5 \\text{ m\/s North})[\/latex] or 1020 Ns North<\/li>\n<\/ul>\n<p>The total momentum before the tackle is 816 Ns East + 1020 Ns North.<\/p>\n<p>Now, in the absence of any other forces acting during this collision, the total momentum after the tackle for these two players must also be 816 Ns East + 1020 Ns North.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-703 aligncenter\" style=\"font-size: 0.9em; word-spacing: normal;\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.3.1-example-2.png\" alt=\"\" width=\"497\" height=\"215\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.3.1-example-2.png 1370w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.3.1-example-2-300x129.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.3.1-example-2-1024x442.png 1024w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.3.1-example-2-768x331.png 768w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.3.1-example-2-65x28.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.3.1-example-2-225x97.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.3.1-example-2-350x151.png 350w\" sizes=\"auto, (max-width: 497px) 100vw, 497px\" \/><\/p>\n<p>At this point, you will use trigonometry and the Pythagorean theorem to further analyze this collision.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-702 aligncenter\" style=\"word-spacing: normal; text-align: initial; font-size: 0.9em;\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.3.1-example-3.png\" alt=\"\" width=\"453\" height=\"202\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.3.1-example-3.png 1856w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.3.1-example-3-300x133.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.3.1-example-3-1024x455.png 1024w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.3.1-example-3-768x341.png 768w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.3.1-example-3-1536x683.png 1536w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.3.1-example-3-65x29.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.3.1-example-3-225x100.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.3.1-example-3-350x156.png 350w\" sizes=\"auto, (max-width: 453px) 100vw, 453px\" \/><\/p>\n<p>Now: The total by using the momentum can be found using the Pythagorean Theorem.<\/p>\n<ul>\n<li>[latex](\\vec{p}_{\\text{total}})^2 = (816 \\text{ Ns})^2 + (1020 \\text{ Ns})^2[\/latex]<\/li>\n<li>[latex](\\vec{p}_{\\text{total}})^2 = 665856 \\text{ N}^2 \\text{S}^2 + 1040400 \\text{ N}^2 \\text{S}^2[\/latex]<\/li>\n<li>[latex](\\vec{p}_{\\text{total}})^2 = 1706256 \\text{ N}^2 \\text{S}^2[\/latex]<\/li>\n<li>[latex]\\vec{p}_{\\text{total}} = 1306 \\text{ Ns}[\/latex]<\/li>\n<\/ul>\n<p>The angle at\u00a0 which these players move off is now found by using trigonometry, specifically the Tangent function:<\/p>\n<ul>\n<li>[latex]\\tan \u00f8 = \\dfrac{\\text{Opposite Side}}{\\text{Adjacent Side}}[\/latex]<\/li>\n<li>[latex]\\tan \u00f8 = 1.245[\/latex]<\/li>\n<li>[latex]\u00f8 = \\tan ^{-1} 1.245[\/latex]<\/li>\n<li>\u00f8 = 51.3\u00b0, specifically \u2248 51\u00b0 North from East<\/li>\n<\/ul>\n<p>The velocity at which these players move off can be found by using the equation for momentum:<\/p>\n<ul>\n<li>[latex]\\vec{p} = m \\vec{v}[\/latex]<\/li>\n<li>[latex]1306 \\text{ Ns} = (120 \\text{ kg } + 85 \\text{ kg}) \\vec{v}_{\\text{f (120 + 85)}}[\/latex]<\/li>\n<li>[latex]\\vec{v}_{\\text{f (120 + 85)}} = 1306 \\text{ Ns } \\div 205 \\text{ kg}[\/latex]<\/li>\n<li>[latex]\\vec{v}_{\\text{f (120 + 85)}} = 64. \\text{ m\/s}[\/latex] 51\u00b0 North from East<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 9.3.2<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>Accident investigators are analyzing a car crash to see if excessive speed was involved. One of the cars had a mass of 1250 kg and was traveling at 40 km\/h East when it was struck by a 1600 kg pickup truck traveling South that ran a red light. If the wreckage moved at <strong>23.5<\/strong><strong>\u00b0 East from South<\/strong>, should the truck driver also be charged for driving above the 60 km\/h speed limit?<\/p>\n<p><strong>Solution<\/strong><\/p>\n<p>First, draw a sketch of the two vehicles:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-701\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.3.1-example-4.png\" alt=\"\" width=\"536\" height=\"513\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.3.1-example-4.png 1581w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.3.1-example-4-300x287.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.3.1-example-4-1024x979.png 1024w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.3.1-example-4-768x734.png 768w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.3.1-example-4-1536x1469.png 1536w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.3.1-example-4-65x62.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.3.1-example-4-225x215.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.3.1-example-4-350x335.png 350w\" sizes=\"auto, (max-width: 536px) 100vw, 536px\" \/><\/p>\n<p>All that is known are the mass of the two colliding vehicles, original velocity of the car and the angle at which the wreck moved off.<\/p>\n<p>Combining this data, we can calculate the momentum of the car, and by using trigonometry we can calculate the momentum of the truck.\u00a0 The momentum of the truck allows us to estimate the initial velocity that it was traveling at.<\/p>\n<p>First, the momentum of the car:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-717 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.3.2-example-2.png\" alt=\"\" width=\"360\" height=\"398\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.3.2-example-2.png 770w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.3.2-example-2-271x300.png 271w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.3.2-example-2-768x850.png 768w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.3.2-example-2-65x72.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.3.2-example-2-225x249.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.3.2-example-2-350x387.png 350w\" sizes=\"auto, (max-width: 360px) 100vw, 360px\" \/><\/p>\n<ul>\n<li>[latex]\\vec{p} = m \\vec{v}[\/latex]<\/li>\n<li>[latex]\\vec{p} = (1250 \\text{ kg})(11.1 \\text{ m\/s East})[\/latex]<\/li>\n<li>[latex]\\vec{p} = 13\\;900 \\text{ Ns East}[\/latex]<\/li>\n<\/ul>\n<p>Now, using the Tangent function:<\/p>\n<ul>\n<li>[latex]\\tan 23.5^{\\circ} = \\dfrac{13\\;900 \\text{ Ns}}{\\vec{p}_{\\text{truck}}}[\/latex]<\/li>\n<li>[latex]\\vec{p}_{\\text{truck}} = \\dfrac{13\\;900 \\text{ Ns}}{\\tan 23.5^{\\circ}}[\/latex]<\/li>\n<li>[latex]\\vec{p}_{\\text{truck}} = 32\\;000 \\text{ Ns South}[\/latex]<\/li>\n<\/ul>\n<p>Finally, the velocity of the truck is:<\/p>\n<ul>\n<li>[latex]\\vec{p} = m \\vec{v}[\/latex]<\/li>\n<li>[latex]\\vec{v} = (32\\;000 \\text{ Ns South}) \\div (1600 \\text{ kg})[\/latex]<\/li>\n<li>[latex]\\vec{v} = 20[\/latex] m\/s South or 72 km\/h<\/li>\n<\/ul>\n<p><strong>The truck was speeding by 12 km\/h and could also be charged with speeding.<\/strong><\/p>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Exercise 9.3<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<ol>\n<li>A 85 kg fullback is running at 9.0 m\/s South and is tackled by a 70 kg halfback running at 8.0 m\/s West. At what velocity do these two move after the tackle occurs?<\/li>\n<li>Two hockey players collide mid-ice in a race for a puck. If the smaller player (80 kg) is racing at 12 m\/s North and the larger player (90 kg) is racing at 15 m\/s at 40\u00b0 West from North, at what angle do the colliding players move off?<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Exercise 9.4.1: Survival Rates of Seat Areas in an Airplane 1971-2007<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p><strong>Question:<\/strong> Why is the greatest survival rate at the rear of the airplane?<\/p>\n<p>&nbsp;<\/p>\n<figure id=\"attachment_1211\" aria-describedby=\"caption-attachment-1211\" style=\"width: 389px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1211\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.4.1-airplane-1.png\" alt=\"\" width=\"389\" height=\"350\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.4.1-airplane-1.png 676w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.4.1-airplane-1-300x270.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.4.1-airplane-1-65x58.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.4.1-airplane-1-225x202.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/01\/9.4.1-airplane-1-350x315.png 350w\" sizes=\"auto, (max-width: 389px) 100vw, 389px\" \/><figcaption id=\"caption-attachment-1211\" class=\"wp-caption-text\"><strong>Survival rates for various parts of the passenger cabin<\/strong>. First\/Business class: 49%. Ahead of the wing: 56%. Over wing: 56%. Rear cabin 69%.<\/figcaption><\/figure>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Exercise 9.4.2: Bullet Proof Vests<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>Bullet proof vests were first developed by Casimir Zeglen, a Catholic priest who in 1893 created a vest made from silks that was capable of stopping bullets. The principle behind these vests was to stop the bullet, causing it to deform, which then spread the bullet\u2019s point of impact over a much larger area on the vest. Zeglen\u2019s choice of silk for the vest derived from his reading of how a physician described how a man had been shot but was saved by a silk handkerchief in his breast pocket.\u00a0 Soon tests were done by folding layers of silk some 10 to 30 times. Zeglen\u2019s original vests came from using multiple layers of silk that had been combined with an unknown chemical that he never disclosed. His later vests included blends of Angora and silk threads for handgun protection and even a steel sheet insert that worked to stop rifle bullets. His finished product was a cloth that was a little over 0.3 cm thick.<\/p>\n<p><strong>\u00a0<\/strong>The photo to the left is taken from the May 1924 Popular Mechanics article showing a test subject being fired upon with 38 caliber (largest mass of 9.53 g, speed of 270 m\/s) and 45 caliber (largest mass of 15 g, speed of 255 m\/s) handguns.<\/p>\n<p><strong>Question:<\/strong> For the man in this photo, if we assume a mass of 68 kg (150 lb man), what should be his recoil velocity in stopping a 45 caliber bullet?<\/p>\n<\/div>\n<\/div>\n<div class=\"textbox\">\n<p>Research in the News: <a href=\"https:\/\/phys.org\/news\/2019-06-metal-foam-caliber-rounds-steel.html?\">Metal Foam stops .50 caliber rounds as well as steel &#8211; at less than half the weight<\/a><\/p>\n<\/div>\n<h1>Exercise Answers<\/h1>\n<h2>9.2 Simple Collisions<\/h2>\n<ol class=\"twocolumn\">\n<li>\n<ol type=\"i\">\n<li>0 Ns (before), 7160 Ns (after)<\/li>\n<li>\u2212\u00a02.60 m\/s (recoils at 2.60 m\/s)<\/li>\n<li>0 Ns in both cases<\/li>\n<\/ol>\n<\/li>\n<li>0.4 m\/s West<\/li>\n<li>0.0375 m\/s opposite direction<\/li>\n<li>2.27 kg<\/li>\n<li>7.66 m\/s<\/li>\n<li>556 m\/s<\/li>\n<li>2.0 cm\/s in original direction<\/li>\n<li>8.1 m\/s<\/li>\n<li>10.5 m\/s away from the end-zone<\/li>\n<li>0.0229 m\/s or 2.3 cm\/s<\/li>\n<li>280 m\/s (roughly 1000 km\/h)<\/li>\n<li>6.1 kg<\/li>\n<li>1.19 m\/s same direction<\/li>\n<li>[latex]2.82 \\times 10^{-7}[\/latex] g<\/li>\n<li>0.17 m\/s left<\/li>\n<li>11.3 m\/s<\/li>\n<\/ol>\n<h2>9.3 Two Dimensional Collisions<\/h2>\n<ol>\n<li>6.1 m\/s, 36\u00b0 West of South<\/li>\n<li>23.5\u00b0 West of North<\/li>\n<\/ol>\n<h2>9.4.1 Survival Rates of Seat Areas in an Airplane<\/h2>\n<ul>\n<li>The passengers in the back would have a longer distance to come to a stop, therefore the impulse is stretched over a longer time.<\/li>\n<li>The front of the aircraft acts like a crumple zone for the rest of the aircraft. Similar to a front end collision of a vehicle.<\/li>\n<li>Other possible reasons?<\/li>\n<\/ul>\n<h2>9.4.2 Bullet Proof Vests<\/h2>\n<ol>\n<li>0.050 m\/s recoil<\/li>\n<\/ol>\n<h3>Media Attributions<\/h3>\n<ul>\n<li>&#8220;<a href=\"https:\/\/pixabay.com\/fr\/photos\/berceau-de-newton-des-balles-balle-256213\/\">black and gray newton&#8217;s cradle<\/a>&#8221; by <a href=\"https:\/\/pixabay.com\/fr\/users\/jarmoluk-143740\/\">jarmoluk<\/a> licensed under a <a href=\"https:\/\/pixabay.com\/service\/license\/\">Pixabay licence<\/a>.<\/li>\n<li>&#8220;<a href=\"https:\/\/www.flickr.com\/photos\/8623220@N02\/2477625961\">Auto polo (LOC)<\/a>&#8221; from <a href=\"https:\/\/www.flickr.com\/photos\/library_of_congress\/\">The Library of Congress<\/a> is in the <a href=\"https:\/\/creativecommons.org\/share-your-work\/public-domain\/\">public domain<\/a>.<\/li>\n<li>&#8220;Safest seats on a plane&#8221; by Gil Ahn. Image courtesy of seatguru.com<\/li>\n<\/ul>\n<hr class=\"before-footnotes clear\" \/><div class=\"footnotes\"><ol><li id=\"footnote-52-1\">There are other physics conservation laws that exist in the study of particle physics. <a href=\"#return-footnote-52-1\" class=\"return-footnote\" aria-label=\"Return to footnote 1\">&crarr;<\/a><\/li><\/ol><\/div>","protected":false},"author":125,"menu_order":9,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-52","chapter","type-chapter","status-publish","hentry"],"part":3,"_links":{"self":[{"href":"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-json\/pressbooks\/v2\/chapters\/52","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-json\/wp\/v2\/users\/125"}],"version-history":[{"count":25,"href":"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-json\/pressbooks\/v2\/chapters\/52\/revisions"}],"predecessor-version":[{"id":1238,"href":"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-json\/pressbooks\/v2\/chapters\/52\/revisions\/1238"}],"part":[{"href":"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-json\/pressbooks\/v2\/parts\/3"}],"metadata":[{"href":"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-json\/pressbooks\/v2\/chapters\/52\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-json\/wp\/v2\/media?parent=52"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-json\/pressbooks\/v2\/chapter-type?post=52"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-json\/wp\/v2\/contributor?post=52"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-json\/wp\/v2\/license?post=52"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}