{"id":161,"date":"2023-02-13T12:18:14","date_gmt":"2023-02-13T17:18:14","guid":{"rendered":"https:\/\/opentextbc.ca\/foundationsofphysics\/?post_type=chapter&#038;p=161"},"modified":"2023-12-21T17:03:52","modified_gmt":"2023-12-21T22:03:52","slug":"conservation-of-mechanical-energy","status":"publish","type":"chapter","link":"https:\/\/opentextbc.ca\/foundationsofphysics\/chapter\/conservation-of-mechanical-energy\/","title":{"raw":"Conservation of Mechanical Energy","rendered":"Conservation of Mechanical Energy"},"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>Video to Watch: <a href=\"https:\/\/www.youtube.com\/watch?v=8BDHN2bY1bg\">Mechanical Universe - Episode 13 - Conservation of Energy<\/a><\/li>\r\n \t<li>Extra Help: <a href=\"http:\/\/www.a-levelphysicstutor.com\/index-mech.php\">A-Level Physics Tutor<\/a><\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\nEquations Introduced and Used in this Topic: (All solutions are scalars)\r\n<p style=\"text-align: center;\">In an Isolated System, Mechanical Energy can be Conserved:<\/p>\r\n<p style=\"text-align: center;\">Energy<sub>Before<\/sub> = Energy<sub>After<\/sub><\/p>\r\nFor Mechanical Energy this means:\r\n<p style=\"text-align: center;\">[latex]E_{k i} + E_{p i} = E_{k f} + E_{p f}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]\\frac{1}{2} m v_i^2 + m g h_i = \\frac{1}{2} m v_f^2 + m g h_f[\/latex]<\/p>\r\n<strong>Since external forces can change the energy of a system<\/strong> the only forces we are looking at in these problems is gravitational force and assuming that frictional forces do not come in to play.\r\n\r\nWhere...\r\n<ul>\r\n \t<li>[latex]W[\/latex] is <strong>work<\/strong>, measured in joules (J)<\/li>\r\n \t<li>[latex]E_p[\/latex] or [latex]\\Delta E_p[\/latex] is <strong>gravitational potential energy<\/strong> or <strong>change in gravitational potential energy<\/strong>, measured in joules (J)<\/li>\r\n \t<li>[latex]E_{p i}[\/latex] is the <strong>initial gravitational potential energy<\/strong>, measured in joules (J)<\/li>\r\n \t<li>[latex]E_{p f}[\/latex] is the <strong>final gravitational potential energy<\/strong>, measured in joules (J)<\/li>\r\n \t<li>[latex]E_{k i}[\/latex]is the <strong>initial kinetic energy<\/strong>, measured in joules (J)<\/li>\r\n \t<li>[latex]E_{k f}[\/latex] is the <strong>final kinetic energy<\/strong>, measured in joules (J)<\/li>\r\n \t<li>[latex]E_k[\/latex] or [latex]\\Delta E_k[\/latex] is <strong>kinetic energy<\/strong> or <strong>change in kinetic energy<\/strong>, measured in joules (J)<\/li>\r\n \t<li>[latex]V[\/latex] or [latex]V_g[\/latex] is the <strong>gravitational potential<\/strong>, measured in newtons per kilogram (N\/kg)<\/li>\r\n \t<li>[latex]m[\/latex] is the <strong>mass of the object<\/strong>, measured in kilograms (kg)<\/li>\r\n \t<li>[latex]g[\/latex] or [latex]a_g[\/latex] is the <strong>gravitational field strength<\/strong>, measured in metres per second squared (m\/s<sup>2<\/sup>)<\/li>\r\n \t<li>[latex]v[\/latex] is the <strong>speed of the object<\/strong>, measured in metres per second (m\/s)<\/li>\r\n \t<li>[latex]v[\/latex] or [latex]\\Delta v[\/latex] is the <strong>speed<\/strong> or <strong>change in speed<\/strong> of the object, measured in metres per second (m\/s)<\/li>\r\n \t<li>[latex]v_f[\/latex] &amp; [latex]v_i[\/latex] is the <strong>final and initial speed of the object<\/strong>, measured in metres per second (m\/s)<\/li>\r\n \t<li>[latex]\\vec{F_{\\text{net}}}[\/latex] is the <strong>net force or the vector sum of the forces<\/strong> acting on a body or a system, measured in newtons (N)<\/li>\r\n \t<li>\u00f8 is the <strong>angle<\/strong> between the vectors of the net force and displacement, measure in degrees (\u00b0)<\/li>\r\n \t<li>[latex]\\vec{d}[\/latex]\u00a0is the <strong>displacement through which the net force acts on a body or a system<\/strong>, measured in metres (m)<\/li>\r\n \t<li>[latex]h[\/latex] or [latex]\\Delta h[\/latex] is the <strong>height<\/strong> or <strong>change in vertical height<\/strong>, measured in metres (m)<\/li>\r\n \t<li>[latex]h_f[\/latex] &amp; [latex]h_i[\/latex] is the<strong> final and initial height<\/strong> of the object measured in metres (m)<\/li>\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<\/ul>\r\n<strong>Notes:<\/strong>\r\n\r\nThe <strong>Dot Product <\/strong><strong>\u201c[latex]\\cdot[\/latex]\u201c<\/strong> represents the cosine of the angle between two vectors. For Work it represents the angle between the direction of the Net Force and the displacement change.\r\n\r\nThe <strong>Cross Product <\/strong><strong>\u201c<\/strong><strong>[latex]\\times[\/latex]<\/strong><strong>\u201c<\/strong> represents the sine of the angle between two vectors. Equations using the cross product will be given later in your study of physics, such as the measure of Torque or the force on a charged particle moving in a magnetic field.\r\n\r\n<strong>Work<\/strong> and <strong>Energy<\/strong> are measured in joules (J) or newton-metres (Nm)\r\n\r\n<strong>For Example: <\/strong>\u00a0Consider the following multiplication of these two vectors shown below\r\n\r\n<img class=\"aligncenter wp-image-724 \" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/10-dot-and-cross-product-1.png\" alt=\"\" width=\"486\" height=\"414\" \/>\r\n\r\n<strong>The Dot Product for these two <\/strong><strong>vectors is:<\/strong><strong>\u00a0<\/strong>\r\n<ul>\r\n \t<li>[latex]\\vec{A} \\cdot \\vec{B} = \\vec{A}\\vec{B} \\cos \u00f8[\/latex]<\/li>\r\n \t<li>[latex]\\vec{A} \\cdot \\vec{B} = (20)(18) \\cos 42^{\\circ}[\/latex]<\/li>\r\n \t<li>[latex]\\vec{A} \\cdot \\vec{B} = 267.5[\/latex] or 270<\/li>\r\n<\/ul>\r\n<strong>The Cross Product for these <\/strong><strong>two vectors is:<\/strong>\r\n<ul>\r\n \t<li>[latex]\\vec{A} \\times \\vec{B} = \\vec{A}\\vec{B} \\sin \u00f8[\/latex]<\/li>\r\n \t<li>[latex]\\vec{A} \\times \\vec{B} = (20)(18) \\sin 42^{\\circ}[\/latex]<\/li>\r\n \t<li>[latex]\\vec{A} \\times \\vec{B} = 240.9[\/latex] or 240<\/li>\r\n<\/ul>\r\n<strong>For Example: <\/strong>\u00a0Consider the following multiplication of these two vectors shown below.\r\n<div class=\"textbox shaded\">\r\n<h2><strong>Kinetic Energy and Collisions<\/strong>: (WHO (2008) Speed Management Manual)<\/h2>\r\n<p style=\"padding-left: 40px;\"><em>According to research, harmful injury is the result of <\/em><em>\u2018<\/em><em>energy interchange<\/em><em>\u2019<\/em><em>. During a collision, injury results from the transfer of energy to the human body in amounts and at rates that damage cellular structure, tissues, blood vessels and other bodily structures. This includes kinetic energy, for example when a motor vehicle user\u2019s head strikes the windshield during a crash. Of the various forms of energy <\/em><em>\u2013 <\/em><em>kinetic, thermal, chemical, electrical and radiation <\/em><em>\u2013 <\/em><em>kinetic energy transfer is the biggest contributor to injury. It is useful for road traffic injury prevention researchers and practitioners to understand the biomechanics of kinetic energy injuries. This will help them develop measures that will limit the generation, distribution, transfer and effect of this energy during a road traffic collision. <\/em><\/p>\r\n(<a href=\"https:\/\/cdn.who.int\/media\/docs\/default-source\/documents\/health-topics\/road-traffic-injuries\/speed-management-manual.pdf\">WHO (2008) Speed Management Manual<\/a>)\r\n\r\n<\/div>\r\n<h1>11.1 Conservation of Mechanical Energy<\/h1>\r\n<div class=\"textbox\">\r\n<ul>\r\n \t<li>Research in the News: <a href=\"https:\/\/anu.prezly.com\/anu-finds-530000-potential-pumped-hydro-sites-worldwide-223526#\">ANU finds 530,000 potential pumped-hydro sites worldwide<\/a><\/li>\r\n \t<li>Extra Help: <a href=\"https:\/\/www.physicsclassroom.com\/class\/energy\/Lesson-2\/Analysis-of-Situations-Involving-External-Forces\">Analysis of Situations Involving External Forces<\/a><\/li>\r\n \t<li>Extra Help: <a href=\"https:\/\/www.physicsclassroom.com\/class\/energy\/Lesson-2\/Analysis-of-Situations-in-Which-Mechanical-Energy\">Analysis of Situations in Which Mechanical Energy is Conserved<\/a><\/li>\r\n \t<li>Extra Help: <a href=\"https:\/\/www.physicsclassroom.com\/class\/energy\/Lesson-2\/Application-and-Practice-Questions\">Application and Practice Questions<\/a><\/li>\r\n<\/ul>\r\n<\/div>\r\nEquations Introduced or Used for this Section:\r\n<ul class=\"twocolumn\" style=\"list-style-type: none;\">\r\n \t<li style=\"text-align: left;\">Energy<sub>Before<\/sub> = Energy<sub>After<\/sub><\/li>\r\n \t<li style=\"text-align: left;\">[latex]E_{k i} + E_{p i} = E_{k f} + E_{p f}[\/latex]<\/li>\r\n \t<li style=\"text-align: left;\">[latex]E_p = m g h[\/latex]<\/li>\r\n \t<li>[latex]\\Delta E_p = m g \\Delta h[\/latex]<\/li>\r\n \t<li>[latex]E_k = \\frac{1}{2} m v^2[\/latex]<\/li>\r\n \t<li>[latex]\\Delta E_k = \\frac{1}{2} m \\Delta v^2[\/latex]<\/li>\r\n \t<li>[latex]\\Delta h = h_f - h_i[\/latex]<\/li>\r\n \t<li>[latex]\\Delta v = v_f - v_i[\/latex]<\/li>\r\n<\/ul>\r\nIn physics, mechanical energy refers only to gravitational potential and kinetic energy.\u00a0 In a system where an object is isolated from external, non-conservative forces, the mechanical energy will remain constant.\u00a0 This can be written as a conservation law where the sum of the kinetic and potential; energies after some event and using conservative forces will remain constant.\r\n\r\nWritten in an equation, the conservation of Mechanical Energy looks like:\r\n<p style=\"text-align: center;\">[latex]E_{k i} + E_{p i} = E_{k f} + E_{p f}[\/latex]<\/p>\r\nThis equation works for multiple bodies in a system. In real life, most forces acting on a system are non-conservative with the classic example being friction. One of the main byproducts of non-conservative forces such as friction is heat and is a chapter covered later in this text\/workbook.\r\n<h2>Conceptual Origins<\/h2>\r\n[caption id=\"attachment_1213\" align=\"aligncenter\" width=\"450\"]<img class=\"wp-image-1213 size-full\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/Galileo-pendulum.gif\" alt=\"\" width=\"450\" height=\"256\" \/> Galileo\u2019s Interrupted Pendulum[\/caption]\r\n\r\nThe history of the conservation of mechanical energy is credited as far back as Thales of Miletus (c. 550 BCE) and Empedocles (490\u2013430 BCE), who explored ideas relating to the conservation of mass-energy. Transitioning from the Greek philosophers, Simon Stevinus (1548\u20131620) used the principle that stated perpetual motion machines were impossible to create (non-conservative forces are always present and act to remove mechanical energy from a system). Galileo\u2019s (1564-1642) interrupted pendulum was the first known classic example of conservation of mechanical energy. In this demonstration, where a pendulum is interrupted by a peg of some sort that shortens the length of the pendulum, the pendulum will rise to nearly the same height from which it was released. This pendulum example is generally introduced in detail in the next physics course, which explains this motion in terms of kinetic and potential energy, momentum and impulse, force, torques and free body diagrams.\r\n\r\nThe development of the conservation of mechanical energy is in reality the product of multiple scientists that followed, which extended energy conservation to heat, light and eventually Albert Einstein\u2019s most famous: [latex]\\Delta E = m c^2[\/latex].\u00a0 The inclusion of heat and the <strong>First Law of Thermodynamics<\/strong> (energy cannot be created or destroyed, only transformed) are covered in following topics.\r\n\r\nExamples of systems that are isolated and mainly experiencing conservative forces are pendulums and orbiting satellites, planets and comets where the main force that is acting is gravity. Examples and questions explored in this section will assume that non-conservative forces acting on a system will be non-existent or negligible.\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 11.1.1<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nIf you were to shoot a crossbow bolt with a velocity of 75 m\/s straight upwards, how high above its launch would the bolt travel?\r\n\r\n<strong>Solution<\/strong>\r\n<ul>\r\n \t<li>[latex]E_{k i} + E_{p i} = E_{k f} + E_{p f}[\/latex]<\/li>\r\n \t<li>[latex]\\frac{1}{2} m v_i^2 + m g h_i = \\frac{1}{2} m v_f^2 + m g h_f[\/latex]<\/li>\r\n \t<li>[latex]\\frac{1}{2} m (75 \\text{ m\/s})^2 + m (9.8 \\text{ m\/s}^2)(0 \\text{ m}) = \\frac{1}{2} m (0 \\text{ m\/s})^2 + m (9.8 \\text{ m\/s}^2)h_f[\/latex]<\/li>\r\n \t<li>Mass is common... [latex]\\frac{1}{2} (75 \\text{ m\/s})^2 + 0 \\text{ J } = 0 \\text{ J } + (9.8 \\text{ m\/s}^2) h_f[\/latex]<\/li>\r\n \t<li>[latex]2812.5 \\text{ J } = (9.8 \\text{ m\/s}^2) h_f[\/latex]<\/li>\r\n \t<li>[latex]h_f = 2812.5 \\text{ J } \\div 9.8 \\text{ m\/s}^2[\/latex]<\/li>\r\n \t<li>[latex]h_f = 287 \\text{ m } (\\approx 290 \\text{ m})[\/latex] (In the absence of air resistance)<\/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 11.1.2<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nChildren take turns dropping from a treehouse that is 2.0 m above the ground. What would be the vertical velocity they would impact the ground with?\r\n\r\n<strong>Solution<\/strong>\r\n<ul>\r\n \t<li>[latex]E_{k i} + E_{p i} = E_{k f} + E_{p f}[\/latex]<\/li>\r\n \t<li>[latex]\\frac{1}{2} m v_i^2 + m g h_i = \\frac{1}{2} m v_f^2 + m g h_f[\/latex]<\/li>\r\n \t<li>[latex]\\frac{1}{2} m (0 \\text{ m\/s})^2 + m (9.8 \\text{ m\/s}^2)(2.0 \\text{ m}) = \\frac{1}{2} m v_f^2 + m (9.8 \\text{ m\/s}^2)(0 \\text{ m})[\/latex]<\/li>\r\n \t<li>Mass is common... [latex]0 \\text{ J } + (9.8 \\text{ m\/s}^2)(2.0 \\text{ m}) = \\frac{1}{2} v_f^2 + 0 \\text{ J} [\/latex]<\/li>\r\n \t<li>[latex]\\frac{1}{2} v_f^2 = 19.6[\/latex] m<sup>2<\/sup>\/s<sup>2<\/sup><\/li>\r\n \t<li>[latex]v_f^2 = 39.2[\/latex]\u00a0m<sup>2<\/sup>\/s<sup>2<\/sup><\/li>\r\n \t<li>[latex]v_f = \\pm 6.26 \\text{ m\/s}[\/latex]<\/li>\r\n<\/ul>\r\n<strong>Since velocity is asked for, the impact velocity is <\/strong>[latex]v_f = - 6.26 \\text{ m\/s}\u00a0 \u00a0(\\approx - 6.3 \\text{ m\/s})[\/latex]\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 11.1.3<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nA stuntman runs at 8.0 m\/s off a 4.0 m vertical drop.\u00a0 At what speed does he impact the cushions stopping his fall at the bottom of the drop?\r\n\r\n<strong>Solution<\/strong>\r\n<ul>\r\n \t<li>[latex]E_{k i} + E_{p i} = E_{k f} + E_{p f}[\/latex]<\/li>\r\n \t<li>[latex]\\frac{1}{2} m v_i^2 + m g h_i = \\frac{1}{2} m v_f^2 + m g h_f[\/latex]<\/li>\r\n \t<li>[latex]\\frac{1}{2} m (8.0 \\text{ m\/s})^2 + m (9.8 \\text{ m\/s}^2)(4.0 \\text{ m}) = \\frac{1}{2} m v_f^2 + m (9.8 \\text{ m\/s}^2)(0 \\text{ m})[\/latex]<\/li>\r\n \t<li>Mass is common... [latex]32 \\text{m}^2\\text{\/s}^2 + 39.2 \\text{m}^2\\text{\/s}^2 = \\frac{1}{2} v_f^2 + 0 \\text{ J}[\/latex]<\/li>\r\n \t<li>[latex]71.2 \\text{ m}^2\\text{\/s}^2 = \\frac{1}{2}v_f^2[\/latex]<\/li>\r\n \t<li>[latex]v_f^2 = 142.4 \\text{ m}^2\\text{\/s}^2[\/latex]<\/li>\r\n \t<li>[latex]v_f = 11.9 \\text{ m\/s } (\\approx 12 \\text{ m\/s})[\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--exercises\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Exercise 11.1<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n<ol>\r\n \t<li>The Steamboat Geyser in Yellowstone is the world\u2019s tallest currently active geyser. What is the ejected velocity of the water if during a major eruption, the geyser can reach heights in excess of 91 m?<\/li>\r\n \t<li>A paint bucket tipped over on a worker\u2019s scaffold falls through a potential difference of 118 N\/kg or (m<sup>2<\/sup>\/s<sup>2<\/sup> ... 12 m drop) to strike a patio table below. At what speed does this paint strike the table?<\/li>\r\n \t<li>It is estimated that the medieval longbow could launch an arrow with a speed of 60 m\/s. If we ignore the effects of air resistance, how high would these arrows travel if fired straight upwards?<\/li>\r\n \t<li>A child climbs a tree house and drops her pet guinea pig to see if it could fly. If we assume the guinea pig fell about 2.5 m to the grass what impact speed would it strike the ground with? (Its impact speed was far less than this since she had a cape tied to her \u201cSuper Hammy\u201d)<\/li>\r\n \t<li>A golf ball is launched with a speed of 125 km\/h.\u00a0 What speed is it traveling when it has risen 40 m above its starting position?<\/li>\r\n \t<li>Water flowing at 8.0 m\/s flows off a 12 m vertical waterfall. At what speed does it impact the base of the falls?<\/li>\r\n \t<li>A 0.50 kg mass is projected horizontally with speed 8.0 m\/s from the top of the cliff 45 m high.\r\n<ol type=\"i\">\r\n \t<li>What is the initial kinetic energy of the body?<\/li>\r\n \t<li>What is its kinetic energy just before striking the ground?<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>A car drives off a 50 m high cliff at 90 km\/h. At what speed should it reach before impacting the base of the cliff?<\/li>\r\n \t<li>A 2.2 kg projectile is fired upwards at an angle at 33 m\/s. It gains a maximum height of 30.0 m above the level from which it was fired.\u00a0 Calculate:\r\n<ol type=\"i\">\r\n \t<li>its initial kinetic energy<\/li>\r\n \t<li>its horizontal speed at maximum height.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>A 17.34 kg mass has a total mechanical energy of 93<span style=\"margin-left: 0.25em;\">500<\/span> J when moving at a height of 360 m. What is its speed?<\/li>\r\n \t<li>If a roller coaster is traveling at 75 km\/h when it goes over a drop and falls through a potential difference of 420 N\/kg, what speed can it reach? Assume no external friction acts to slow it down.<\/li>\r\n \t<li>\u00a0If we assume that a cyclist is at the top of the Eshima Ohashi bridge in Japan traveling at 10 m\/s, what speed could be reached if this cyclist rolls to the bottom through a potential difference of 430 N\/kg? (ignore all friction)\r\n\r\n[caption id=\"attachment_795\" align=\"aligncenter\" width=\"500\"]<img class=\"wp-image-795\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/Eshima_oohashi_04_14781816175.jpg\" alt=\"\" width=\"500\" height=\"331\" \/> Eshima Ohashi Bridge[\/caption]\r\n\r\n[caption id=\"attachment_796\" align=\"aligncenter\" width=\"500\"]<img class=\"wp-image-796\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/Eshima_bidge.jpg\" alt=\"\" width=\"500\" height=\"333\" \/> Eshima Ohashi Bridge Side View[\/caption]<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<h1>11.2 Non-Horizontally Launched Projectiles<\/h1>\r\nEquations Introduced or Used for this Section:\r\n<ul class=\"twocolumn\" style=\"list-style-type: none;\">\r\n \t<li style=\"text-align: left;\">Energy<sub>Before<\/sub> = Energy<sub>After<\/sub><\/li>\r\n \t<li style=\"text-align: left;\">[latex]E_{k i} + E_{p i} = E_{k f} + E_{p f}[\/latex]<\/li>\r\n \t<li style=\"text-align: left;\">[latex]E_p = m g h[\/latex]<\/li>\r\n \t<li>[latex]\\Delta E_p = m g \\Delta h[\/latex]<\/li>\r\n \t<li>[latex]E_k = \\frac{1}{2} m v^2[\/latex]<\/li>\r\n \t<li>[latex]\\Delta E_k = \\frac{1}{2} m \\Delta v^2[\/latex]<\/li>\r\n \t<li>[latex]\\Delta h = h_f - h_i[\/latex]<\/li>\r\n \t<li>[latex]\\Delta v = v_f - v_i[\/latex]<\/li>\r\n<\/ul>\r\nIn some problems, projectiles launched at an angle can be solved with the conservation of mechanical energy instead of using trigonometry to break the motion into vector quantities of vertical and horizontal components to solve. Using mechanical energy greatly simplifies the solutions to these problems. In all of these problems, air resistance is not accounted for.\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 11.2.1<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nConsider a bottle of water thrown at 18 m\/s at 30\u00b0 above the horizontal from the edge of one balcony to be caught by a friend on another balcony 12 m below. At what speed would the friend catch the bottle of water?\r\n\r\n<strong>Solution<\/strong>\r\n\r\nSince we are using energy to solve this, we can skip using vectors and trigonometry.\r\n<ul>\r\n \t<li>[latex]E_{k i} + E_{p i} = E_{k f} + E_{p f}[\/latex]<\/li>\r\n \t<li>[latex]\\frac{1}{2} m v_i^2 + m g h_i = \\frac{1}{2} m v_f^2 + m g h_f[\/latex]<\/li>\r\n \t<li>[latex]\\frac{1}{2} (m)(18 \\text{ m\/s})^2 + m (9.8 \\text{ m\/s}^2)(12 \\text{ m}) = \\frac{1}{2} m v_f^2 + 0 \\text{ J}[\/latex]<\/li>\r\n \t<li>Mass is common... [latex]\\frac{1}{2} (324 \\text{ m}^2\\text{\/s}^2) + (117.6 \\text{m}^2\\text{\/s}^2) = \\frac{1}{2} v_f^2 + 0 \\text{ J}[\/latex]<\/li>\r\n \t<li>[latex]162 \\text{ m}^2\\text{\/s}^2 + 117.6 \\text{ m}^2\\text{\/s}^2 = \\frac{1}{2}v_f^2 + 0 \\text{ J}[\/latex]<\/li>\r\n \t<li>[latex]279.6 \\text{ m}^2\\text{\/s}^2 = \\frac{1}{2}v_f^2[\/latex]<\/li>\r\n \t<li>[latex]v_f^2 = 559.2 \\text{ m}^2\\text{\/s}^2[\/latex]<\/li>\r\n \t<li>[latex]v_f = 23.6 \\text{ m\/s } (\\approx 24 \\text{ m\/s})[\/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 11.2.2<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nA competitive archer shoots an arrow at 48 m\/s angled at 35\u00b0 above the horizontal from a castle rampart 25 m above the ground below. At what speed would we expect this arrow to strike the ground below if we ignore air resistance?\r\n\r\n<strong>Solution<\/strong>\r\n<ul>\r\n \t<li>[latex]E_{k i} + E_{p i} = E_{k f} + E_{p f}[\/latex]<\/li>\r\n \t<li>[latex]\\frac{1}{2} m v_i^2 + m g h_i = \\frac{1}{2} m v_f^2 + m g h_f[\/latex]<\/li>\r\n \t<li>[latex]\\frac{1}{2}(m)(48 \\text{ m\/s})^2 + (m)(9.8 \\text{ m\/s}^2)(25 \\text{ m}) = \\frac{1}{2} (m)v_f^2 + (m)(9.8 \\text{ m\/s}^2)(0 \\text{ m})[\/latex]<\/li>\r\n \t<li>Mass is common... [latex](1152 \\text{m}^2\\text{\/s}^2) + (245 \\text{m}^2\\text{\/s}^2) = \\frac{1}{2} v_f^2 + 0 \\text{ J}[\/latex]<\/li>\r\n \t<li>[latex]139 \\text{ m}^2\\text{\/s}^2 = \\frac{1}{2} v_f^2[\/latex]<\/li>\r\n \t<li>[latex]v_f^2 = 2794 \\text{ m}^2\\text{\/s}^2[\/latex]<\/li>\r\n \t<li>[latex]v_f = 52.9 \\text{ m\/s } (\\approx 53 \\text{ m\/s})[\/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 11.2.3<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nA baseball player throws a ball into the stands with a velocity of 90 km\/h angled at 40\u00b0 above the horizontal. If this ball is caught by a fan that is 25 m above the height from which it was thrown, at what speed was it traveling?\r\n\r\n<strong>Solution<\/strong>\r\n<ul>\r\n \t<li>[latex]E_{k i} + E_{p i} = E_{k f} + E_{p f}[\/latex]<\/li>\r\n \t<li>[latex]\\frac{1}{2} m v_i^2 + m g h_i = \\frac{1}{2} m v_f^2 + m g h_f[\/latex]<\/li>\r\n \t<li>[latex]\\frac{1}{2} (m)(25 \\text{ m\/s})^2 + (m)(9.8 \\text{ m\/s}^2)(0 \\text{ m}) = \\frac{1}{2} (m) v_f^2 + (m)(245 \\text{ m}^2\\text{\/s}^2)[\/latex]<\/li>\r\n \t<li>Mass is common... [latex]312.5 \\text{ m}^2\\text{\/s}^2 - 245 \\text{ m}^2\\text{\/s}^2 = \\frac{1}{2} v_f^2[\/latex]<\/li>\r\n \t<li>[latex]v_f^2 = 135 \\text{ m}^2\\text{\/s}^2[\/latex]<\/li>\r\n \t<li>[latex]v_f = 11.6 \\text{ m\/s } (\\approx 12 \\text{ m\/s})[\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--exercises\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Exercise 11.2<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n<ol>\r\n \t<li>Firefighters are directing a water stream of 53 m\/s upwards at 32\u00b0 above the horizontal. What is the speed of this water 16 m above where it is launched?<img class=\"wp-image-765 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/11.2.1-exercise.jpeg\" alt=\"\" width=\"504\" height=\"359\" \/><\/li>\r\n \t<li>If someone throws a first aid kit at 8.2 m\/s downwards at 28\u00b0 below the horizontal, what should its speed be if caught by a person 9.0 m below where it was thrown?<\/li>\r\n \t<li>An archer shoots an arrow at 60 m\/s at an angle of 45\u00b0 above the horizontal from the top of a cliff. At what speed should the arrow strike the field 48 m below where the arrow was launched?<\/li>\r\n \t<li>A football player throws a football into the stands at a velocity of 21 m\/s angled at 50\u00b0 above the horizontal. If this football is caught by a fan that is 18 m above the height from where it was thrown, at what speed was it traveling?<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<h1>11.3 Elastic and Inelastic Collisions<\/h1>\r\n<div class=\"textbox\">Video to Watch: <a href=\"https:\/\/www.youtube.com\/watch?v=8ko3qy9vgLQ\">Elastic and Inelastic Collisions<\/a><\/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: center;\">Where...<\/p>\r\n<p style=\"text-align: center;\">[latex]\\vec{p} = m \\vec{v}[\/latex]<\/p>\r\n\r\n<ul class=\"twocolumn\" style=\"list-style-type: none;\">\r\n \t<li style=\"text-align: left;\">Energy<sub>Before<\/sub> = Energy<sub>After<\/sub><\/li>\r\n \t<li style=\"text-align: left;\">[latex]E_{k i} + E_{p i} = E_{k f} + E_{p f}[\/latex]<\/li>\r\n \t<li style=\"text-align: left;\">[latex]E_p = m g h[\/latex]<\/li>\r\n \t<li>[latex]\\Delta E_p = m g \\Delta h[\/latex]<\/li>\r\n \t<li>[latex]E_k = \\frac{1}{2} m v^2[\/latex]<\/li>\r\n \t<li>[latex]\\Delta E_k = \\frac{1}{2} m \\Delta v^2[\/latex]<\/li>\r\n \t<li>[latex]\\Delta h = h_f - h_i[\/latex]<\/li>\r\n \t<li>[latex]\\Delta v = v_f - v_i[\/latex]<\/li>\r\n<\/ul>\r\n<h2>Concepts used in this Section:<\/h2>\r\nCollisions must always obey the <strong>Law of Conservation of Momentum<\/strong>. That is, the vector sum of momenta prior to a collision equals the vector sum after a collision carrying a requirement that no external forces are acting on the colliding objects.\u00a0 However, collisions are classified due to kinetic energy being or not being conserved in the collision.\r\n\r\nCollisions where some of the initial kinetic energy is turned into heat or some other form of energy are inelastic. Collisions where kinetic energy is not lost are classified as elastic.\r\n\r\nCollisions between atoms[footnote]One general exception to collisions between atoms being inelastic is where energy is released during the collision as electromagnetic radiation such as a burst of light energy.[\/footnote] in ideal gases are generally elastic since no kinetic energy is lost during the collision. Other examples of elastic collisions are the scattering of sub-atomic particles when deflected by electromagnetic forces or the gravitational interactions between planets and satellites[footnote]<a href=\"http:\/\/hyperphysics.phy-astr.gsu.edu\/hbase\/elacol.html\">http:\/\/hyperphysics.phy-astr.gsu.edu\/hbase\/elacol.html<\/a>[\/footnote]. Inelastic collisions are further differentiated if they lose all kinetic energy during the collision or only part of the initial kinetic energy. In collisions where all kinetic energy is lost the collision is classified as perfectly inelastic. For collisions where only some of the kinetic energy is lost, the collision is defined as partially elastic or inelastic.\r\n<div>\r\n<div><\/div>\r\n<div>\r\n\r\n<strong>Summing this up:<\/strong>\r\n<ul>\r\n \t<li>Perfectly elastic collisions conserve both momentum and kinetic energy.<\/li>\r\n<\/ul>\r\n<ul>\r\n \t<li>Partially elastic collisions or inelastic conserve momentum and some kinetic energy.<\/li>\r\n<\/ul>\r\n<ul>\r\n \t<li>Perfectly inelastic collisions conserve only momentum and lose all kinetic energy.<\/li>\r\n<\/ul>\r\n<img class=\"alignleft wp-image-767 \" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/11.3-newtons-cradle-scaled.jpg\" alt=\"\" width=\"292\" height=\"219\" \/>Generally, macroscopic collisions are partially elastic or perfectly inelastic. Newton\u2019s cradle is one apparatus that comes close to a perfectly elastic collision and is used for demonstrations in introductory physics classes.<strong>\u00a0<\/strong>\r\n\r\nObservers watching a demonstration of Newton\u2019s cradle can hear the sound of the metal balls colliding which acts to extract energy out of the collision, so it becomes quite obvious that the collision is not perfectly elastic.\r\n<div>\r\n<div>\r\n\r\nThe following examples explore the elasticity of collisions.\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 11.3.1<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nTwo objects collide on a frictionless surface. The first object having a mass of 4.0 kg has a velocity of 2.0 m\/s prior to the collision and \u2212 5.2 m\/s after.\u00a0 The second object having a mass of 6.0 kg has a velocity of \u2212 4.0 m\/s prior to the collision and 0.8 m\/s after. Verify that this collision could exist and identify what type of a collision it is. (Perfectly Elastic, Partially Elastic or Inelastic, or Perfectly Inelastic).\r\n\r\n<strong>Solution<\/strong>\r\n\r\nData:\r\n<ul>\r\n \t<li>[latex]m_1 = 4.0 \\text{ kg}[\/latex]<\/li>\r\n \t<li>[latex]\\vec{v}_{1 i} = 2.0 \\text{ m\/s}[\/latex]<\/li>\r\n \t<li>[latex]\\vec{v}_{1 f} = - 5.2 \\text{ m\/s}[\/latex]<\/li>\r\n \t<li>[latex]m_2 = 6.0 \\text{ kg}[\/latex]<\/li>\r\n \t<li>[latex]\\vec{v}_{2 i} = - 4.0 \\text{ m\/s}[\/latex]<\/li>\r\n \t<li>[latex]\\vec{v}_{2 f} = 0.8 \\text{ m\/s}[\/latex]<\/li>\r\n<\/ul>\r\nSolution:\r\n\r\nFirst solve for the velocity of mass 2 ([latex]\\vec{v}_{2 f}[\/latex]) before the collision.\r\n<p style=\"text-align: center;\">[latex]\\vec{P}_{\\text{before}} = \\vec{P}_{\\text{after}}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">or: [latex]m_1 \\vec{v}_{1 i} + m_2 \\vec{v}_{2 i} = m_1 \\vec{v}_{1 f} + m_2 \\vec{v}_{2 f}[\/latex]<\/p>\r\nInputing variables yields:\r\n<ul>\r\n \t<li>[latex](4.0 \\text{ kg})(2.0 \\text{ m\/s}) + (6.0 \\text{ kg})(- 4.0 \\text{ m\/s}) = (4.0 \\text{ kg})(- 5.2 \\text{ m\/s}) + (6.0 \\text{ kg})(0.8 \\text{ m\/s})[\/latex]<\/li>\r\n \t<li>[latex]8.0 \\text{ Ns } - 24 \\text{ Ns } = - 20.8 \\text{ Ns }+ 4.8 \\text{ Ns}[\/latex]<\/li>\r\n \t<li>[latex]16 \\text{ Ns } = 16 \\text{ Ns}[\/latex]. Momentum is conserved.<\/li>\r\n<\/ul>\r\nNow to check to see if any Kinetic Energy was lost during the collision:\r\n<p style=\"text-align: center;\">[latex]E_{ k i} = E_{k f}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">or: [latex]\\frac{1}{2} m_1 \\vec{v}^2_{1 i} + \\frac{1}{2} m_2 \\vec{v}^2_{2 i} = \\frac{1}{2} m_1 \\vec{v}^2_{1 f} + \\frac{1}{2} m_2 \\vec{v}^2_{2 f}[\/latex]<\/p>\r\nInputting variables yields:\r\n<ul>\r\n \t<li>[latex]\\frac{1}{2}(4.0 \\text{ kg})(2.0 \\text{ m\/s})^2 + \\frac{1}{2}(6.0 \\text{ kg})(- 4.0 \\text{ m\/s})^2 = \\frac{1}{2}(4.0 \\text{ kg})(- 5.2 \\text{ m\/s})^2 + \\frac{1}{2}(6.0 \\text{ kg})(0.8 \\text{ m\/s})^2[\/latex]<\/li>\r\n \t<li>[latex]8.0 \\text{ J } + 48 \\text{ J } = 54.08 \\text{ J } + 1.92 \\text{ J}[\/latex]<\/li>\r\n \t<li>[latex]56 \\text{ J } = 56 \\text{ J}[\/latex]. Kinetic energy is conserved.<\/li>\r\n<\/ul>\r\nBecause Kinetic Energy was conserved this collision is an example of a perfectly elastic collision.\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 11.3.2<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nTwo objects collide on a frictionless surface. The first object having a mass of 0.42 kg has a velocity of \u22120.32 m\/s prior to the collision and 0.26 m\/s after.\u00a0 The second object has a mass of 0.88 kg at some unknown velocity prior to the collision and at rest after.\u00a0 Find the velocity and identify what type of a collision it is. (Perfectly Elastic, Partially Elastic or Inelastic, or Perfectly Inelastic).\r\n\r\n<strong>Solution<\/strong>\r\n\r\nData:\r\n<ul>\r\n \t<li>[latex]m_1 = 0.42 \\text{ kg}[\/latex]<\/li>\r\n \t<li>[latex]\\vec{v}_{1 i} = - 0.32 \\text{ m\/s}[\/latex]<\/li>\r\n \t<li>[latex]\\vec{v}_{1 f} = 0.26 \\text{ m\/s}[\/latex]<\/li>\r\n \t<li>[latex]m_2 = 0.88 \\text{ kg}[\/latex]<\/li>\r\n \t<li>[latex]\\vec{v}_{2 i} = 0 \\text{ m\/s}[\/latex]<\/li>\r\n \t<li>[latex]\\vec{v}_{2 f} = \\text{ Unknown}[\/latex]<\/li>\r\n<\/ul>\r\nSolution:\r\n\r\nFirst solve for the velocity of mass 2 ([latex]\\vec{v}_{2 f}[\/latex]) before the collision.\r\n<p style=\"text-align: center;\">[latex]\\vec{P}_{\\text{before}} = \\vec{P}_{\\text{after}}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">or: [latex]m_1 \\vec{v}_{1 i} + m_2 \\vec{v}_{2 i} = m_1 \\vec{v}_{1 f} + m_2 \\vec{v}_{2 f}[\/latex]<\/p>\r\nInputting variables yields:\r\n<ul>\r\n \t<li>[latex](0.42 \\text{ kg})(- 0.32 \\text{ m\/s}) + (0.88 \\text{ kg})(\\vec{v}_{2 f}) = (0.42 \\text{ kg})(0.26 \\text{ m\/s}) + (0.88 \\text{ kg})(0 \\text{ m\/s})[\/latex]<\/li>\r\n \t<li>[latex]-0.1344 \\text{ Ns } + (0.88 \\text{ kg})(\\vec{v}_{2 f}) = 0.1092 \\text{ Ns } + 0 \\text{ Ns}[\/latex]<\/li>\r\n \t<li>[latex](0.88 \\text{ kg})(\\vec{v}_{2f}) = 0.2436 \\text{ Ns}[\/latex]<\/li>\r\n \t<li>[latex]\\vec{v}_{2 f} = \\dfrac{0.2436 \\text{ Ns}}{0.88 \\text{ kg}}[\/latex]<\/li>\r\n \t<li>[latex]\\vec{v}_{2 f} = 0.277 \\text{ m\/s}[\/latex]<\/li>\r\n<\/ul>\r\nNow to check to see if any Kinetic Energy was lost during the collision:\r\n<p style=\"text-align: center;\">[latex]E_{ k i} = E_{k f}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">or: [latex]\\frac{1}{2} m_1 \\vec{v}^2_{1 i} + \\frac{1}{2} m_2 \\vec{v}^2_{2 i} = \\frac{1}{2} m_1 \\vec{v}^2_{1 f} + \\frac{1}{2} m_2 \\vec{v}^2_{2 f}[\/latex]<\/p>\r\nInputting variables yields:\r\n<ul>\r\n \t<li>[latex]\\frac{1}{2}(0.42 \\text{ kg})(- 0.32 \\text{ m\/s})^2 + \\frac{1}{2}(0.88 \\text{ kg})(0.277 \\text{ m\/s})^2 = \\frac{1}{2}(0.42 \\text{ kg})(0.26 \\text{ m\/s})^2 + 0 \\text{ Nm}[\/latex]<\/li>\r\n \t<li>[latex]0.0215 \\text{ J } + 0.0338 \\text{ J } = 0.0142 \\text{ J } + 0 \\text{ J}[\/latex]<\/li>\r\n \t<li>[latex]0.0553 \\text{ N J m} = 0.0142 \\text{ J}[\/latex]. Kinetic Energy is lost.<\/li>\r\n<\/ul>\r\nBecause Kinetic Energy was not conserved this collision is an example of an inelastic collision.\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 11.3<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n<ol>\r\n \t<li>A 0.50 kg billiard ball traveling at 1.5 m\/s strikes a 0.45 kg billiard ball at rest.\u00a0 If the 0.50 kg ball slows to 0.75 m\/s, at what speed is the 0.45 kg ball moving at?\u00a0 What type of elastic collision is this?<\/li>\r\n \t<li>Two balls of putty sliding towards each other collide and come to a complete stop. If the small ball had a mass of 150 g and a velocity of 12.0 cm\/s North, prior to the collision, what was the velocity of the 200 g larger ball prior to the collision?\u00a0 What type of elastic collision was this?<\/li>\r\n \t<li>Two balls of identical mass (0.500 kg) each having a kinetic energy of 3.0 joules are directed toward each other. During the collision each experience equal and opposite impulses of 2.0 Nm.\r\n<ol type=\"i\">\r\n \t<li>What were their speeds prior to the collision?<\/li>\r\n \t<li>What are their speeds after the collision?<\/li>\r\n \t<li>Was this collision elastic or inelastic?<\/li>\r\n \t<li>What impulse would be needed to make this a perfectly elastic collision?<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<h1>11.4 Work, Force &amp; Change in Energy Problems - Launching and Braking Devices<\/h1>\r\n<div class=\"textbox\">\r\n<ul>\r\n \t<li>Extra Help: <a href=\"https:\/\/www.physicsclassroom.com\/class\/energy\/Lesson-2\/Analysis-of-Situations-Involving-External-Forces\">Analysis of Situations Involving External Forces<\/a><\/li>\r\n \t<li>Extra Help: <a href=\"https:\/\/www.physicsclassroom.com\/class\/energy\/Lesson-2\/Analysis-of-Situations-in-Which-Mechanical-Energy\">Analysis of Situations in Which Mechanical Energy is Conserved<\/a><\/li>\r\n<\/ul>\r\n<\/div>\r\nEquations Introduced or Used for this Section:\r\n<ul class=\"twocolumn\" style=\"list-style-type: none;\">\r\n \t<li>Work [latex](W) = \\vec{F}_{\\text{net}} \\cdot \\vec{d}[\/latex] or [latex]\\vec{F}_{\\text{net}} \\vec{d} \\cos \u00f8[\/latex]<\/li>\r\n \t<li style=\"text-align: left;\">[latex]E_p = m g h[\/latex]<\/li>\r\n \t<li>[latex]\\Delta E_p = m g \\Delta h[\/latex]<\/li>\r\n \t<li>[latex]E_k = \\frac{1}{2} m v^2[\/latex]<\/li>\r\n \t<li>[latex]\\vec{W} = \\Delta \\text{ Mechanical Energy}[\/latex]<\/li>\r\n \t<li>[latex]\\Delta E_k = \\frac{1}{2} m \\Delta v^2[\/latex]<\/li>\r\n \t<li>[latex]\\Delta h = h_f - h_i[\/latex]<\/li>\r\n \t<li>[latex]\\Delta v = v_f - v_i[\/latex]<\/li>\r\n \t<li>[latex]V[\/latex] or [latex]V_g = g h[\/latex]<\/li>\r\n<\/ul>\r\n<h2>Concepts used in this Section:<\/h2>\r\nWhen one is using the <strong>work-energy theorem<\/strong>, in situations where only conservative forces are acting, then [latex]W = \\Delta E_k[\/latex] or [latex]W = (E_{k f} - E_{k i})[\/latex].\u00a0 Under some conditions, the work done on a system can result in a change in both potential and kinetic energy [latex]W = \\Delta E_k + \\Delta E_p[\/latex]. This equation is not defined by the <strong>work-energy theorem <\/strong>but is a combination of the <strong>work-energy theorem <\/strong>and the<strong> conservation of mechanical energy<\/strong>. In these cases work can be looked at as the function of adding or subtracting energy of the system using the conservation of mechanical energy as defining the total energy of the system which could be converted to a kinetic energy.\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example 11.4.1<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nA 1400 kg car is traveling at 45 km\/h at the top of a 42 m high hill.\r\n<ol type=\"i\">\r\n \t<li>What amount of work is needed to stop this car at the bottom of the hill?<\/li>\r\n \t<li>What average braking force is needed to do this if the car stops in 200 m at the bottom of this hill?<\/li>\r\n<\/ol>\r\n<strong>Solution<\/strong>\r\n\r\nSolution, Question (i):\r\n<ul>\r\n \t<li>[latex]W = \\Delta E_k + \\Delta E_p[\/latex]<\/li>\r\n \t<li>[latex]W = (E_{k f} - E_{k i}) + (E_{p f} - E_{p i})[\/latex]<\/li>\r\n \t<li>[latex]W = \\frac{1}{2} m v_v^2-\\frac{1}{2} m v_i^2 + m g h_f- m g h_i[\/latex]<\/li>\r\n \t<li>[latex]W = \\frac{1}{2} m v_f^2 - \\frac{1}{2} m v_i^2 + m g h_f - m g h_i[\/latex]<\/li>\r\n \t<li>[latex]W = \\frac{1}{2}(1400 \\text{ kg})(0 \\text{ m\/s})^2 - \\frac{1}{2}(1400 \\text{ kg})(12.5 \\text{ m\/s})^2 + (1400 \\text{ kg})(9.8 \\text{ m\/s}^2)(0 \\text{ m}) - (1400 \\text{ kg})(9.8 \\text{ m\/s}^2)(42 \\text{ m}) [\/latex]<\/li>\r\n \t<li>[latex]W = 0 \\text{ J} - 109\\;375 \\text{ J } + 0 \\text{ J } - 576\\;240 \\text{ J}[\/latex]<\/li>\r\n \t<li>[latex]W = - 686\\;000 \\text{ J } (\\approx -690\\;000 \\text{ J})[\/latex]<\/li>\r\n<\/ul>\r\nSolution, Question (ii):\r\n<ul>\r\n \t<li>[latex]W = \\Delta \\text{ Energy}[\/latex]<\/li>\r\n \t<li>For this car, [latex]W = -686\\;000 \\text{ J}[\/latex]<\/li>\r\n \t<li>Now, [latex]\\vec{F} \\vec{d} \\cos \u00f8 = -686\\;000 \\text{ J}[\/latex]<\/li>\r\n \t<li>[latex]\\vec{F} (200 \\text{ m}) \\cos 0^{\\circ} = -686\\;000 \\text{ J}[\/latex]<\/li>\r\n \t<li>[latex]\\vec{F} = -686\\;000 \\text{ J } \\div 200 \\text{ m}[\/latex]<\/li>\r\n \t<li>[latex]\\vec{F} = - 3430 \\text{ N } (\\approx 3400 \\text{ N})[\/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 11.4.2<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nA co-worker tosses a 4.0 kg brick to his friend 2.0 m below at 1.5 m\/s. What amount of force is needed to catch and stop this brick in a horizontal distance of 0.50 m?\r\n\r\n<strong>Solution<\/strong>\r\n\r\nFirst, find out how much energy needs to be removed from the brick when catching it.\r\n<ul>\r\n \t<li>[latex]W = \\Delta \\text{ Energy}[\/latex]<\/li>\r\n \t<li>[latex]W = \\Delta E_k + \\Delta E_p[\/latex]<\/li>\r\n \t<li>[latex]W = (E_{k f} - E_{k i}) + (E_{p f} - E_{p i})[\/latex]<\/li>\r\n \t<li>[latex]W = \\frac{1}{2} m v_v^2 - \\frac{1}{2} m v_i^2 + m g h_f- m g h_i[\/latex]<\/li>\r\n \t<li>[latex]W = \\frac{1}{2}(4.0 \\text{ kg})(0 \\text{ m\/s})^2 - \\frac{1}{2}(4.0 \\text{ kg})(1.5 \\text{ m\/s})^2 + (4.0 \\text{ kg})(9.8 \\text{ m\/s}^2)(0 \\text{ m}) - (4.0 \\text{ kg})(9.8 \\text{ m\/s}^2)(2.0 \\text{ m})[\/latex]<\/li>\r\n \t<li>[latex] W = 0 \\text{ J } - 4.5 \\text{ J } + 0 \\text{ J } - 78.4 \\text{ J}[\/latex]<\/li>\r\n<\/ul>\r\nSecond, find out the average braking force needed to do this in 0.5 m.\r\n<ul>\r\n \t<li>For the brick, [latex]W = - 82.9 \\text{ J}[\/latex]<\/li>\r\n \t<li>Now, [latex]\\vec{F} \\vec{d} \\cos \u00f8 = -82.9 \\text{ J}[\/latex]<\/li>\r\n \t<li>[latex]\\vec{F} (0.5 \\text{ m}) \\cos 0^{\\circ} = - 82.9 \\text{ J}[\/latex]<\/li>\r\n \t<li>[latex]\\vec{F} = -82.9 \\text{ J } \\div 0.5 \\text{ m}[\/latex]<\/li>\r\n \t<li>[latex]\\vec{F} = - 166 \\text{ N } (\\approx - 170 \\text{ N})[\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--exercises\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Exercise 11.4<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n<ol>\r\n \t<li>A 35 kg bag of sand is tied to the end of a 3.2 m long rope.\u00a0 It is drawn back from its vertical position to the horizontal then released.\r\n<ol type=\"i\">\r\n \t<li>What speed does it have at the bottom of its swing?<\/li>\r\n \t<li>A mechanical catching device brings it to a full stop (in 0.10 m) at the bottom of the swing. How much work is done by the device?<\/li>\r\n \t<li>What braking force is needed to stop the sand bag?<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>What is the average force that a circus cannon exerts on a 75 kg stunt-person over the 2.0 m they are launched with, so that the stunt-person is traveling at 3.0 m\/s when 5.0 m above the launched height?<\/li>\r\n \t<li>A 4200 kg truck is rolling along at 80 km\/h at the top of a 102 m high hill. If this truck is allowed to freely accelerate to the bottom of the hill before the brakes are applied, what average braking force would be needed to stop this truck in half a kilometre?<\/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 11.5.1: A Trebuchet Catapult<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nA Trebuchet Catapult[footnote]A trebuchet or trebucket is a siege engine that was employed in the Middle Ages either to smash masonry walls or to throw projectiles over them. It is sometimes called a \"counterweight trebuchet\" or \"counterpoise trebuchet\" in order to distinguish it from an earlier weapon that has come to be called the \"traction trebuchet\", the original version with pulling men instead of a counterweight. The counterweight trebuchet appeared in both Christian and Muslim lands around the Mediterranean in the twelfth century. It could fling three-hundred-pound (140 kg ... largest appear to be 1500 kg) projectiles at high speeds into enemy fortifications. On occasion, disease-infected corpses were flung into cities in an attempt to infect the people under siege a medieval variant of biological warfare. Traction trebuchets appeared in China in about the 4th century BCE and in Europe in the 6th century CE, and did not become obsolete until the 16th century, well after the introduction of gunpowder. Trebuchets were far more accurate than other medieval catapults. Adapted from <a href=\"http:\/\/en.wikipedia.org\/wiki\/Trebuchet\">http:\/\/en.wikipedia.org\/wiki\/Trebuchet<\/a>[\/footnote]\u00a0is capable of launching 140 kg stones with such energy they will still be traveling at 80 km\/h when 60 m above its original launch height.\r\n\r\n<strong>Question:<\/strong> At what speed are these stones released?\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 11.5.2: Kingda Ka... Six Flags Great Adventure<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nOn May 21, 2005, Six Flags Great Adventure introduced Kingda Ka the tallest and fastest roller coaster on the planet to the public.\u00a0 This Swiss designed \"rocket coaster\" uses hydraulic motors to launch the cars along a horizontal section of track from zero to 205 km\/h in an impressive 3.5 seconds.\u00a0 The cars then begin a vertical ascent up a steel tower that peaks at 139 m (\u2248 45 stories).\u00a0 Crossing over the apex the train enters a vertical descent plunging through a 270\u00b0 spiral twist again reaching speeds more than 160 km\/h.\u00a0 One final surprise comes before the brake run a 39 m tall camelback hill that offers plenty of negative g's also known fondly as \"airtime\".\r\n\r\n<strong>Question:<\/strong> With its launch speed of 205 km\/h how fast could this roller coaster be moving after it has climbed a vertical height of 139 m?\u00a0 (It will be less than this due to friction.)\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 11.5.3: The Ballistic Pendulum<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nThe Ballistic Pendulum was invented in 1742 by English mathematician Benjamin Robins (1707\u20131751) and published in his book New Principles of Gunnery, which revolutionized the science of ballistics as it provided the first way to accurately measure the velocity of a bullet.\u00a0 Robins used the ballistic pendulum to measure projectile velocity in two ways.\u00a0 The first was to attach the gun to the pendulum and measure the recoil.\u00a0 Since the momentum of the gun is equal to the momentum of the ejecta and since the projectile was (in those experiments) the large majority of the mass of the ejecta the velocity of the bullet could be approximated.\u00a0 The second and more accurate method was to directly measure the bullet momentum by firing it into the pendulum and measuring the change in potential energy of the pendulum + embedded bullet.\r\n\r\n<strong>Question<\/strong>: Suppose one wanted to set up a ballistics pendulum to check the muzzle velocity of a bullet fired from an 1894 Winchester (30-30) as shown below.\u00a0 Specs show that it should shoot an 8.0 g (130 grain) bullet at 759 m\/s.\u00a0 What mass of wooden block would be needed so that it would rise 10.0 cm above its rest position after being struck by this bullet?\r\n\r\n<\/div>\r\n<\/div>\r\n<h1>Exercise Answers<\/h1>\r\n<h2>11.1 Conservation of Mechanical Energy<\/h2>\r\n<ol class=\"twocolumn\">\r\n \t<li>42 m<\/li>\r\n \t<li>15.4 m\/s<\/li>\r\n \t<li>184 m<\/li>\r\n \t<li>7.0 m\/s<\/li>\r\n \t<li>21 m\/s<\/li>\r\n \t<li>17 m\/s\r\n<ol class=\"twocolumn\" type=\"i\">\r\n \t<li>16 J<\/li>\r\n \t<li>237 J<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>40 m\/s\r\n<ol class=\"twocolumn\" type=\"i\">\r\n \t<li>1200 J<\/li>\r\n \t<li>22 m\/s<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>61 m\/s<\/li>\r\n \t<li>35 m\/s<\/li>\r\n \t<li>31 m\/s<\/li>\r\n<\/ol>\r\n<h2>11.2 Non-Horizontally Launched Projectiles made easy using Conservation of Energy<\/h2>\r\n<ol class=\"twocolumn\">\r\n \t<li>50 m\/s<\/li>\r\n \t<li>15.6 m\/s<\/li>\r\n \t<li>67 m\/s<\/li>\r\n \t<li>31 m\/s<\/li>\r\n<\/ol>\r\n<h2>11.3 Elastic and Inelastic collisions<\/h2>\r\n<ol>\r\n \t<li>Inelastic Collision ([latex]E_{ki}[\/latex] = 0.5625 J,\u00a0 [latex]E_{kf}[\/latex] = 0.3437 J)<\/li>\r\n \t<li>0.0135 m\/s South\r\n<ol class=\"twocolumn\" type=\"i\">\r\n \t<li>3.46 m\/s<\/li>\r\n \t<li>\u2212 0.54 m\/s<\/li>\r\n \t<li>Inelastic Collision<\/li>\r\n \t<li>3.46 Ns<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n<h2>11.4 Work, Force &amp; Change in Energy Problems - Launching and Braking Devices<\/h2>\r\n<ol>\r\n \t<li>\r\n<ol class=\"threecolumn\" type=\"i\">\r\n \t<li>7.9 m\/s<\/li>\r\n \t<li>\u2212 1100 J<\/li>\r\n \t<li>\u2212 11<span style=\"margin-left: 0.25em;\">000<\/span> N<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>2000 N<\/li>\r\n \t<li>\u2212 10<span style=\"margin-left: 0.25em;\">500<\/span> N<\/li>\r\n<\/ol>\r\n<h2>11.5.1 A Trebuchet Catapult<\/h2>\r\n<ol>\r\n \t<li>[latex]v_i[\/latex] = 41 m\/s<\/li>\r\n<\/ol>\r\n<h2>11.5.2 Kingda Ka ... Six Flags Great Adventure<\/h2>\r\n<ol>\r\n \t<li>[latex]v_f[\/latex] = 23 m\/s<\/li>\r\n<\/ol>\r\n<h3>Media Attributions<\/h3>\r\n<ul>\r\n \t<li>\"Galileo's Interrupted Pendulum\" by Galileo Galilei is in the <a href=\"https:\/\/en.wikipedia.org\/wiki\/Public_domain\">public domain<\/a>.\r\n<div class=\"hHq9Z\">\r\n<div data-hveid=\"CA8QAA\" data-ved=\"2ahUKEwjqsYjXh6uBAxXSMDQIHeeJBU4QlcAGegQIDxAA\">\r\n<div class=\"KsRP6\">\r\n<div class=\"MDY31c\">\r\n<div class=\"QpPSMb\">\r\n<div class=\"DoxwDb\"><\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div><\/li>\r\n \t<li>\"<a href=\"https:\/\/commons.wikimedia.org\/wiki\/File:Eshima_oohashi_04_(14781816175).jpg\">Eshima oohashi 04 (14781816175)<\/a>\" by mstk is licensed under a <a href=\"https:\/\/creativecommons.org\/licenses\/by\/2.0\/deed.en\">CC BY 2.0 licence<\/a>.<\/li>\r\n \t<li>\"<a href=\"https:\/\/commons.wikimedia.org\/wiki\/File:Eshima_bidge.jpg\">Eshima bidge<\/a>\" by \u5199\u771f\u6295\u7a3f\u7528 is licensed under a <a href=\"https:\/\/creativecommons.org\/licenses\/by-sa\/4.0\/deed.en\">CC BY-SA 4.0 licence<\/a>.<\/li>\r\n \t<li>\"<a href=\"https:\/\/commons.wikimedia.org\/wiki\/File:Fire_inside_an_abandoned_convent_in_Massueville,_Quebec,_Canada.jpg\">Fire inside an abandoned convent in Massueville, Quebec, Canada<\/a>\" by Sylvain Pedneault is licensed under a <a href=\"https:\/\/creativecommons.org\/licenses\/by-sa\/3.0\/\">CC BY-SA 3.0 Unported licence<\/a>.<\/li>\r\n \t<li>\"<a href=\"https:\/\/pixabay.com\/photos\/newtons-cradle-physics-pendulum-2891856\/\">Newton's cradle<\/a>\" by <a href=\"https:\/\/pixabay.com\/users\/quincecreative-1031690\/\">QuinceCreative<\/a> is licensed under a <a href=\"https:\/\/pixabay.com\/service\/terms\/\">Pixabay licence<\/a>.<\/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>Video to Watch: <a href=\"https:\/\/www.youtube.com\/watch?v=8BDHN2bY1bg\">Mechanical Universe &#8211; Episode 13 &#8211; Conservation of Energy<\/a><\/li>\n<li>Extra Help: <a href=\"http:\/\/www.a-levelphysicstutor.com\/index-mech.php\">A-Level Physics Tutor<\/a><\/li>\n<\/ul>\n<\/div>\n<\/div>\n<p>Equations Introduced and Used in this Topic: (All solutions are scalars)<\/p>\n<p style=\"text-align: center;\">In an Isolated System, Mechanical Energy can be Conserved:<\/p>\n<p style=\"text-align: center;\">Energy<sub>Before<\/sub> = Energy<sub>After<\/sub><\/p>\n<p>For Mechanical Energy this means:<\/p>\n<p style=\"text-align: center;\">[latex]E_{k i} + E_{p i} = E_{k f} + E_{p f}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{1}{2} m v_i^2 + m g h_i = \\frac{1}{2} m v_f^2 + m g h_f[\/latex]<\/p>\n<p><strong>Since external forces can change the energy of a system<\/strong> the only forces we are looking at in these problems is gravitational force and assuming that frictional forces do not come in to play.<\/p>\n<p>Where&#8230;<\/p>\n<ul>\n<li>[latex]W[\/latex] is <strong>work<\/strong>, measured in joules (J)<\/li>\n<li>[latex]E_p[\/latex] or [latex]\\Delta E_p[\/latex] is <strong>gravitational potential energy<\/strong> or <strong>change in gravitational potential energy<\/strong>, measured in joules (J)<\/li>\n<li>[latex]E_{p i}[\/latex] is the <strong>initial gravitational potential energy<\/strong>, measured in joules (J)<\/li>\n<li>[latex]E_{p f}[\/latex] is the <strong>final gravitational potential energy<\/strong>, measured in joules (J)<\/li>\n<li>[latex]E_{k i}[\/latex]is the <strong>initial kinetic energy<\/strong>, measured in joules (J)<\/li>\n<li>[latex]E_{k f}[\/latex] is the <strong>final kinetic energy<\/strong>, measured in joules (J)<\/li>\n<li>[latex]E_k[\/latex] or [latex]\\Delta E_k[\/latex] is <strong>kinetic energy<\/strong> or <strong>change in kinetic energy<\/strong>, measured in joules (J)<\/li>\n<li>[latex]V[\/latex] or [latex]V_g[\/latex] is the <strong>gravitational potential<\/strong>, measured in newtons per kilogram (N\/kg)<\/li>\n<li>[latex]m[\/latex] is the <strong>mass of the object<\/strong>, measured in kilograms (kg)<\/li>\n<li>[latex]g[\/latex] or [latex]a_g[\/latex] is the <strong>gravitational field strength<\/strong>, measured in metres per second squared (m\/s<sup>2<\/sup>)<\/li>\n<li>[latex]v[\/latex] is the <strong>speed of the object<\/strong>, measured in metres per second (m\/s)<\/li>\n<li>[latex]v[\/latex] or [latex]\\Delta v[\/latex] is the <strong>speed<\/strong> or <strong>change in speed<\/strong> of the object, measured in metres per second (m\/s)<\/li>\n<li>[latex]v_f[\/latex] &amp; [latex]v_i[\/latex] is the <strong>final and initial speed of the object<\/strong>, measured in metres per second (m\/s)<\/li>\n<li>[latex]\\vec{F_{\\text{net}}}[\/latex] is the <strong>net force or the vector sum of the forces<\/strong> acting on a body or a system, measured in newtons (N)<\/li>\n<li>\u00f8 is the <strong>angle<\/strong> between the vectors of the net force and displacement, measure in degrees (\u00b0)<\/li>\n<li>[latex]\\vec{d}[\/latex]\u00a0is the <strong>displacement through which the net force acts on a body or a system<\/strong>, measured in metres (m)<\/li>\n<li>[latex]h[\/latex] or [latex]\\Delta h[\/latex] is the <strong>height<\/strong> or <strong>change in vertical height<\/strong>, measured in metres (m)<\/li>\n<li>[latex]h_f[\/latex] &amp; [latex]h_i[\/latex] is the<strong> final and initial height<\/strong> of the object measured in metres (m)<\/li>\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<\/ul>\n<p><strong>Notes:<\/strong><\/p>\n<p>The <strong>Dot Product <\/strong><strong>\u201c[latex]\\cdot[\/latex]\u201c<\/strong> represents the cosine of the angle between two vectors. For Work it represents the angle between the direction of the Net Force and the displacement change.<\/p>\n<p>The <strong>Cross Product <\/strong><strong>\u201c<\/strong><strong>[latex]\\times[\/latex]<\/strong><strong>\u201c<\/strong> represents the sine of the angle between two vectors. Equations using the cross product will be given later in your study of physics, such as the measure of Torque or the force on a charged particle moving in a magnetic field.<\/p>\n<p><strong>Work<\/strong> and <strong>Energy<\/strong> are measured in joules (J) or newton-metres (Nm)<\/p>\n<p><strong>For Example: <\/strong>\u00a0Consider the following multiplication of these two vectors shown below<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-724\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/10-dot-and-cross-product-1.png\" alt=\"\" width=\"486\" height=\"414\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/10-dot-and-cross-product-1.png 1017w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/10-dot-and-cross-product-1-300x255.png 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/10-dot-and-cross-product-1-768x654.png 768w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/10-dot-and-cross-product-1-65x55.png 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/10-dot-and-cross-product-1-225x192.png 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/10-dot-and-cross-product-1-350x298.png 350w\" sizes=\"auto, (max-width: 486px) 100vw, 486px\" \/><\/p>\n<p><strong>The Dot Product for these two <\/strong><strong>vectors is:<\/strong><strong>\u00a0<\/strong><\/p>\n<ul>\n<li>[latex]\\vec{A} \\cdot \\vec{B} = \\vec{A}\\vec{B} \\cos \u00f8[\/latex]<\/li>\n<li>[latex]\\vec{A} \\cdot \\vec{B} = (20)(18) \\cos 42^{\\circ}[\/latex]<\/li>\n<li>[latex]\\vec{A} \\cdot \\vec{B} = 267.5[\/latex] or 270<\/li>\n<\/ul>\n<p><strong>The Cross Product for these <\/strong><strong>two vectors is:<\/strong><\/p>\n<ul>\n<li>[latex]\\vec{A} \\times \\vec{B} = \\vec{A}\\vec{B} \\sin \u00f8[\/latex]<\/li>\n<li>[latex]\\vec{A} \\times \\vec{B} = (20)(18) \\sin 42^{\\circ}[\/latex]<\/li>\n<li>[latex]\\vec{A} \\times \\vec{B} = 240.9[\/latex] or 240<\/li>\n<\/ul>\n<p><strong>For Example: <\/strong>\u00a0Consider the following multiplication of these two vectors shown below.<\/p>\n<div class=\"textbox shaded\">\n<h2><strong>Kinetic Energy and Collisions<\/strong>: (WHO (2008) Speed Management Manual)<\/h2>\n<p style=\"padding-left: 40px;\"><em>According to research, harmful injury is the result of <\/em><em>\u2018<\/em><em>energy interchange<\/em><em>\u2019<\/em><em>. During a collision, injury results from the transfer of energy to the human body in amounts and at rates that damage cellular structure, tissues, blood vessels and other bodily structures. This includes kinetic energy, for example when a motor vehicle user\u2019s head strikes the windshield during a crash. Of the various forms of energy <\/em><em>\u2013 <\/em><em>kinetic, thermal, chemical, electrical and radiation <\/em><em>\u2013 <\/em><em>kinetic energy transfer is the biggest contributor to injury. It is useful for road traffic injury prevention researchers and practitioners to understand the biomechanics of kinetic energy injuries. This will help them develop measures that will limit the generation, distribution, transfer and effect of this energy during a road traffic collision. <\/em><\/p>\n<p>(<a href=\"https:\/\/cdn.who.int\/media\/docs\/default-source\/documents\/health-topics\/road-traffic-injuries\/speed-management-manual.pdf\">WHO (2008) Speed Management Manual<\/a>)<\/p>\n<\/div>\n<h1>11.1 Conservation of Mechanical Energy<\/h1>\n<div class=\"textbox\">\n<ul>\n<li>Research in the News: <a href=\"https:\/\/anu.prezly.com\/anu-finds-530000-potential-pumped-hydro-sites-worldwide-223526#\">ANU finds 530,000 potential pumped-hydro sites worldwide<\/a><\/li>\n<li>Extra Help: <a href=\"https:\/\/www.physicsclassroom.com\/class\/energy\/Lesson-2\/Analysis-of-Situations-Involving-External-Forces\">Analysis of Situations Involving External Forces<\/a><\/li>\n<li>Extra Help: <a href=\"https:\/\/www.physicsclassroom.com\/class\/energy\/Lesson-2\/Analysis-of-Situations-in-Which-Mechanical-Energy\">Analysis of Situations in Which Mechanical Energy is Conserved<\/a><\/li>\n<li>Extra Help: <a href=\"https:\/\/www.physicsclassroom.com\/class\/energy\/Lesson-2\/Application-and-Practice-Questions\">Application and Practice Questions<\/a><\/li>\n<\/ul>\n<\/div>\n<p>Equations Introduced or Used for this Section:<\/p>\n<ul class=\"twocolumn\" style=\"list-style-type: none;\">\n<li style=\"text-align: left;\">Energy<sub>Before<\/sub> = Energy<sub>After<\/sub><\/li>\n<li style=\"text-align: left;\">[latex]E_{k i} + E_{p i} = E_{k f} + E_{p f}[\/latex]<\/li>\n<li style=\"text-align: left;\">[latex]E_p = m g h[\/latex]<\/li>\n<li>[latex]\\Delta E_p = m g \\Delta h[\/latex]<\/li>\n<li>[latex]E_k = \\frac{1}{2} m v^2[\/latex]<\/li>\n<li>[latex]\\Delta E_k = \\frac{1}{2} m \\Delta v^2[\/latex]<\/li>\n<li>[latex]\\Delta h = h_f - h_i[\/latex]<\/li>\n<li>[latex]\\Delta v = v_f - v_i[\/latex]<\/li>\n<\/ul>\n<p>In physics, mechanical energy refers only to gravitational potential and kinetic energy.\u00a0 In a system where an object is isolated from external, non-conservative forces, the mechanical energy will remain constant.\u00a0 This can be written as a conservation law where the sum of the kinetic and potential; energies after some event and using conservative forces will remain constant.<\/p>\n<p>Written in an equation, the conservation of Mechanical Energy looks like:<\/p>\n<p style=\"text-align: center;\">[latex]E_{k i} + E_{p i} = E_{k f} + E_{p f}[\/latex]<\/p>\n<p>This equation works for multiple bodies in a system. In real life, most forces acting on a system are non-conservative with the classic example being friction. One of the main byproducts of non-conservative forces such as friction is heat and is a chapter covered later in this text\/workbook.<\/p>\n<h2>Conceptual Origins<\/h2>\n<figure id=\"attachment_1213\" aria-describedby=\"caption-attachment-1213\" style=\"width: 450px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1213 size-full\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/Galileo-pendulum.gif\" alt=\"\" width=\"450\" height=\"256\" \/><figcaption id=\"caption-attachment-1213\" class=\"wp-caption-text\">Galileo\u2019s Interrupted Pendulum<\/figcaption><\/figure>\n<p>The history of the conservation of mechanical energy is credited as far back as Thales of Miletus (c. 550 BCE) and Empedocles (490\u2013430 BCE), who explored ideas relating to the conservation of mass-energy. Transitioning from the Greek philosophers, Simon Stevinus (1548\u20131620) used the principle that stated perpetual motion machines were impossible to create (non-conservative forces are always present and act to remove mechanical energy from a system). Galileo\u2019s (1564-1642) interrupted pendulum was the first known classic example of conservation of mechanical energy. In this demonstration, where a pendulum is interrupted by a peg of some sort that shortens the length of the pendulum, the pendulum will rise to nearly the same height from which it was released. This pendulum example is generally introduced in detail in the next physics course, which explains this motion in terms of kinetic and potential energy, momentum and impulse, force, torques and free body diagrams.<\/p>\n<p>The development of the conservation of mechanical energy is in reality the product of multiple scientists that followed, which extended energy conservation to heat, light and eventually Albert Einstein\u2019s most famous: [latex]\\Delta E = m c^2[\/latex].\u00a0 The inclusion of heat and the <strong>First Law of Thermodynamics<\/strong> (energy cannot be created or destroyed, only transformed) are covered in following topics.<\/p>\n<p>Examples of systems that are isolated and mainly experiencing conservative forces are pendulums and orbiting satellites, planets and comets where the main force that is acting is gravity. Examples and questions explored in this section will assume that non-conservative forces acting on a system will be non-existent or negligible.<\/p>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 11.1.1<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>If you were to shoot a crossbow bolt with a velocity of 75 m\/s straight upwards, how high above its launch would the bolt travel?<\/p>\n<p><strong>Solution<\/strong><\/p>\n<ul>\n<li>[latex]E_{k i} + E_{p i} = E_{k f} + E_{p f}[\/latex]<\/li>\n<li>[latex]\\frac{1}{2} m v_i^2 + m g h_i = \\frac{1}{2} m v_f^2 + m g h_f[\/latex]<\/li>\n<li>[latex]\\frac{1}{2} m (75 \\text{ m\/s})^2 + m (9.8 \\text{ m\/s}^2)(0 \\text{ m}) = \\frac{1}{2} m (0 \\text{ m\/s})^2 + m (9.8 \\text{ m\/s}^2)h_f[\/latex]<\/li>\n<li>Mass is common&#8230; [latex]\\frac{1}{2} (75 \\text{ m\/s})^2 + 0 \\text{ J } = 0 \\text{ J } + (9.8 \\text{ m\/s}^2) h_f[\/latex]<\/li>\n<li>[latex]2812.5 \\text{ J } = (9.8 \\text{ m\/s}^2) h_f[\/latex]<\/li>\n<li>[latex]h_f = 2812.5 \\text{ J } \\div 9.8 \\text{ m\/s}^2[\/latex]<\/li>\n<li>[latex]h_f = 287 \\text{ m } (\\approx 290 \\text{ m})[\/latex] (In the absence of air resistance)<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 11.1.2<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>Children take turns dropping from a treehouse that is 2.0 m above the ground. What would be the vertical velocity they would impact the ground with?<\/p>\n<p><strong>Solution<\/strong><\/p>\n<ul>\n<li>[latex]E_{k i} + E_{p i} = E_{k f} + E_{p f}[\/latex]<\/li>\n<li>[latex]\\frac{1}{2} m v_i^2 + m g h_i = \\frac{1}{2} m v_f^2 + m g h_f[\/latex]<\/li>\n<li>[latex]\\frac{1}{2} m (0 \\text{ m\/s})^2 + m (9.8 \\text{ m\/s}^2)(2.0 \\text{ m}) = \\frac{1}{2} m v_f^2 + m (9.8 \\text{ m\/s}^2)(0 \\text{ m})[\/latex]<\/li>\n<li>Mass is common&#8230; [latex]0 \\text{ J } + (9.8 \\text{ m\/s}^2)(2.0 \\text{ m}) = \\frac{1}{2} v_f^2 + 0 \\text{ J}[\/latex]<\/li>\n<li>[latex]\\frac{1}{2} v_f^2 = 19.6[\/latex] m<sup>2<\/sup>\/s<sup>2<\/sup><\/li>\n<li>[latex]v_f^2 = 39.2[\/latex]\u00a0m<sup>2<\/sup>\/s<sup>2<\/sup><\/li>\n<li>[latex]v_f = \\pm 6.26 \\text{ m\/s}[\/latex]<\/li>\n<\/ul>\n<p><strong>Since velocity is asked for, the impact velocity is <\/strong>[latex]v_f = - 6.26 \\text{ m\/s}\u00a0 \u00a0(\\approx - 6.3 \\text{ m\/s})[\/latex]<\/p>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 11.1.3<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>A stuntman runs at 8.0 m\/s off a 4.0 m vertical drop.\u00a0 At what speed does he impact the cushions stopping his fall at the bottom of the drop?<\/p>\n<p><strong>Solution<\/strong><\/p>\n<ul>\n<li>[latex]E_{k i} + E_{p i} = E_{k f} + E_{p f}[\/latex]<\/li>\n<li>[latex]\\frac{1}{2} m v_i^2 + m g h_i = \\frac{1}{2} m v_f^2 + m g h_f[\/latex]<\/li>\n<li>[latex]\\frac{1}{2} m (8.0 \\text{ m\/s})^2 + m (9.8 \\text{ m\/s}^2)(4.0 \\text{ m}) = \\frac{1}{2} m v_f^2 + m (9.8 \\text{ m\/s}^2)(0 \\text{ m})[\/latex]<\/li>\n<li>Mass is common&#8230; [latex]32 \\text{m}^2\\text{\/s}^2 + 39.2 \\text{m}^2\\text{\/s}^2 = \\frac{1}{2} v_f^2 + 0 \\text{ J}[\/latex]<\/li>\n<li>[latex]71.2 \\text{ m}^2\\text{\/s}^2 = \\frac{1}{2}v_f^2[\/latex]<\/li>\n<li>[latex]v_f^2 = 142.4 \\text{ m}^2\\text{\/s}^2[\/latex]<\/li>\n<li>[latex]v_f = 11.9 \\text{ m\/s } (\\approx 12 \\text{ m\/s})[\/latex]<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Exercise 11.1<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<ol>\n<li>The Steamboat Geyser in Yellowstone is the world\u2019s tallest currently active geyser. What is the ejected velocity of the water if during a major eruption, the geyser can reach heights in excess of 91 m?<\/li>\n<li>A paint bucket tipped over on a worker\u2019s scaffold falls through a potential difference of 118 N\/kg or (m<sup>2<\/sup>\/s<sup>2<\/sup> &#8230; 12 m drop) to strike a patio table below. At what speed does this paint strike the table?<\/li>\n<li>It is estimated that the medieval longbow could launch an arrow with a speed of 60 m\/s. If we ignore the effects of air resistance, how high would these arrows travel if fired straight upwards?<\/li>\n<li>A child climbs a tree house and drops her pet guinea pig to see if it could fly. If we assume the guinea pig fell about 2.5 m to the grass what impact speed would it strike the ground with? (Its impact speed was far less than this since she had a cape tied to her \u201cSuper Hammy\u201d)<\/li>\n<li>A golf ball is launched with a speed of 125 km\/h.\u00a0 What speed is it traveling when it has risen 40 m above its starting position?<\/li>\n<li>Water flowing at 8.0 m\/s flows off a 12 m vertical waterfall. At what speed does it impact the base of the falls?<\/li>\n<li>A 0.50 kg mass is projected horizontally with speed 8.0 m\/s from the top of the cliff 45 m high.\n<ol type=\"i\">\n<li>What is the initial kinetic energy of the body?<\/li>\n<li>What is its kinetic energy just before striking the ground?<\/li>\n<\/ol>\n<\/li>\n<li>A car drives off a 50 m high cliff at 90 km\/h. At what speed should it reach before impacting the base of the cliff?<\/li>\n<li>A 2.2 kg projectile is fired upwards at an angle at 33 m\/s. It gains a maximum height of 30.0 m above the level from which it was fired.\u00a0 Calculate:\n<ol type=\"i\">\n<li>its initial kinetic energy<\/li>\n<li>its horizontal speed at maximum height.<\/li>\n<\/ol>\n<\/li>\n<li>A 17.34 kg mass has a total mechanical energy of 93<span style=\"margin-left: 0.25em;\">500<\/span> J when moving at a height of 360 m. What is its speed?<\/li>\n<li>If a roller coaster is traveling at 75 km\/h when it goes over a drop and falls through a potential difference of 420 N\/kg, what speed can it reach? Assume no external friction acts to slow it down.<\/li>\n<li>\u00a0If we assume that a cyclist is at the top of the Eshima Ohashi bridge in Japan traveling at 10 m\/s, what speed could be reached if this cyclist rolls to the bottom through a potential difference of 430 N\/kg? (ignore all friction)<br \/>\n<figure id=\"attachment_795\" aria-describedby=\"caption-attachment-795\" style=\"width: 500px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-795\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/Eshima_oohashi_04_14781816175.jpg\" alt=\"\" width=\"500\" height=\"331\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/Eshima_oohashi_04_14781816175.jpg 640w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/Eshima_oohashi_04_14781816175-300x199.jpg 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/Eshima_oohashi_04_14781816175-65x43.jpg 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/Eshima_oohashi_04_14781816175-225x149.jpg 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/Eshima_oohashi_04_14781816175-350x232.jpg 350w\" sizes=\"auto, (max-width: 500px) 100vw, 500px\" \/><figcaption id=\"caption-attachment-795\" class=\"wp-caption-text\">Eshima Ohashi Bridge<\/figcaption><\/figure>\n<figure id=\"attachment_796\" aria-describedby=\"caption-attachment-796\" style=\"width: 500px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-796\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/Eshima_bidge.jpg\" alt=\"\" width=\"500\" height=\"333\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/Eshima_bidge.jpg 640w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/Eshima_bidge-300x200.jpg 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/Eshima_bidge-65x43.jpg 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/Eshima_bidge-225x150.jpg 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/Eshima_bidge-350x233.jpg 350w\" sizes=\"auto, (max-width: 500px) 100vw, 500px\" \/><figcaption id=\"caption-attachment-796\" class=\"wp-caption-text\">Eshima Ohashi Bridge Side View<\/figcaption><\/figure>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<h1>11.2 Non-Horizontally Launched Projectiles<\/h1>\n<p>Equations Introduced or Used for this Section:<\/p>\n<ul class=\"twocolumn\" style=\"list-style-type: none;\">\n<li style=\"text-align: left;\">Energy<sub>Before<\/sub> = Energy<sub>After<\/sub><\/li>\n<li style=\"text-align: left;\">[latex]E_{k i} + E_{p i} = E_{k f} + E_{p f}[\/latex]<\/li>\n<li style=\"text-align: left;\">[latex]E_p = m g h[\/latex]<\/li>\n<li>[latex]\\Delta E_p = m g \\Delta h[\/latex]<\/li>\n<li>[latex]E_k = \\frac{1}{2} m v^2[\/latex]<\/li>\n<li>[latex]\\Delta E_k = \\frac{1}{2} m \\Delta v^2[\/latex]<\/li>\n<li>[latex]\\Delta h = h_f - h_i[\/latex]<\/li>\n<li>[latex]\\Delta v = v_f - v_i[\/latex]<\/li>\n<\/ul>\n<p>In some problems, projectiles launched at an angle can be solved with the conservation of mechanical energy instead of using trigonometry to break the motion into vector quantities of vertical and horizontal components to solve. Using mechanical energy greatly simplifies the solutions to these problems. In all of these problems, air resistance is not accounted for.<\/p>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 11.2.1<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>Consider a bottle of water thrown at 18 m\/s at 30\u00b0 above the horizontal from the edge of one balcony to be caught by a friend on another balcony 12 m below. At what speed would the friend catch the bottle of water?<\/p>\n<p><strong>Solution<\/strong><\/p>\n<p>Since we are using energy to solve this, we can skip using vectors and trigonometry.<\/p>\n<ul>\n<li>[latex]E_{k i} + E_{p i} = E_{k f} + E_{p f}[\/latex]<\/li>\n<li>[latex]\\frac{1}{2} m v_i^2 + m g h_i = \\frac{1}{2} m v_f^2 + m g h_f[\/latex]<\/li>\n<li>[latex]\\frac{1}{2} (m)(18 \\text{ m\/s})^2 + m (9.8 \\text{ m\/s}^2)(12 \\text{ m}) = \\frac{1}{2} m v_f^2 + 0 \\text{ J}[\/latex]<\/li>\n<li>Mass is common&#8230; [latex]\\frac{1}{2} (324 \\text{ m}^2\\text{\/s}^2) + (117.6 \\text{m}^2\\text{\/s}^2) = \\frac{1}{2} v_f^2 + 0 \\text{ J}[\/latex]<\/li>\n<li>[latex]162 \\text{ m}^2\\text{\/s}^2 + 117.6 \\text{ m}^2\\text{\/s}^2 = \\frac{1}{2}v_f^2 + 0 \\text{ J}[\/latex]<\/li>\n<li>[latex]279.6 \\text{ m}^2\\text{\/s}^2 = \\frac{1}{2}v_f^2[\/latex]<\/li>\n<li>[latex]v_f^2 = 559.2 \\text{ m}^2\\text{\/s}^2[\/latex]<\/li>\n<li>[latex]v_f = 23.6 \\text{ m\/s } (\\approx 24 \\text{ m\/s})[\/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 11.2.2<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>A competitive archer shoots an arrow at 48 m\/s angled at 35\u00b0 above the horizontal from a castle rampart 25 m above the ground below. At what speed would we expect this arrow to strike the ground below if we ignore air resistance?<\/p>\n<p><strong>Solution<\/strong><\/p>\n<ul>\n<li>[latex]E_{k i} + E_{p i} = E_{k f} + E_{p f}[\/latex]<\/li>\n<li>[latex]\\frac{1}{2} m v_i^2 + m g h_i = \\frac{1}{2} m v_f^2 + m g h_f[\/latex]<\/li>\n<li>[latex]\\frac{1}{2}(m)(48 \\text{ m\/s})^2 + (m)(9.8 \\text{ m\/s}^2)(25 \\text{ m}) = \\frac{1}{2} (m)v_f^2 + (m)(9.8 \\text{ m\/s}^2)(0 \\text{ m})[\/latex]<\/li>\n<li>Mass is common&#8230; [latex](1152 \\text{m}^2\\text{\/s}^2) + (245 \\text{m}^2\\text{\/s}^2) = \\frac{1}{2} v_f^2 + 0 \\text{ J}[\/latex]<\/li>\n<li>[latex]139 \\text{ m}^2\\text{\/s}^2 = \\frac{1}{2} v_f^2[\/latex]<\/li>\n<li>[latex]v_f^2 = 2794 \\text{ m}^2\\text{\/s}^2[\/latex]<\/li>\n<li>[latex]v_f = 52.9 \\text{ m\/s } (\\approx 53 \\text{ m\/s})[\/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 11.2.3<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>A baseball player throws a ball into the stands with a velocity of 90 km\/h angled at 40\u00b0 above the horizontal. If this ball is caught by a fan that is 25 m above the height from which it was thrown, at what speed was it traveling?<\/p>\n<p><strong>Solution<\/strong><\/p>\n<ul>\n<li>[latex]E_{k i} + E_{p i} = E_{k f} + E_{p f}[\/latex]<\/li>\n<li>[latex]\\frac{1}{2} m v_i^2 + m g h_i = \\frac{1}{2} m v_f^2 + m g h_f[\/latex]<\/li>\n<li>[latex]\\frac{1}{2} (m)(25 \\text{ m\/s})^2 + (m)(9.8 \\text{ m\/s}^2)(0 \\text{ m}) = \\frac{1}{2} (m) v_f^2 + (m)(245 \\text{ m}^2\\text{\/s}^2)[\/latex]<\/li>\n<li>Mass is common&#8230; [latex]312.5 \\text{ m}^2\\text{\/s}^2 - 245 \\text{ m}^2\\text{\/s}^2 = \\frac{1}{2} v_f^2[\/latex]<\/li>\n<li>[latex]v_f^2 = 135 \\text{ m}^2\\text{\/s}^2[\/latex]<\/li>\n<li>[latex]v_f = 11.6 \\text{ m\/s } (\\approx 12 \\text{ m\/s})[\/latex]<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Exercise 11.2<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<ol>\n<li>Firefighters are directing a water stream of 53 m\/s upwards at 32\u00b0 above the horizontal. What is the speed of this water 16 m above where it is launched?<img loading=\"lazy\" decoding=\"async\" class=\"wp-image-765 aligncenter\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/11.2.1-exercise.jpeg\" alt=\"\" width=\"504\" height=\"359\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/11.2.1-exercise.jpeg 1600w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/11.2.1-exercise-300x214.jpeg 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/11.2.1-exercise-1024x729.jpeg 1024w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/11.2.1-exercise-768x547.jpeg 768w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/11.2.1-exercise-1536x1093.jpeg 1536w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/11.2.1-exercise-65x46.jpeg 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/11.2.1-exercise-225x160.jpeg 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/11.2.1-exercise-350x249.jpeg 350w\" sizes=\"auto, (max-width: 504px) 100vw, 504px\" \/><\/li>\n<li>If someone throws a first aid kit at 8.2 m\/s downwards at 28\u00b0 below the horizontal, what should its speed be if caught by a person 9.0 m below where it was thrown?<\/li>\n<li>An archer shoots an arrow at 60 m\/s at an angle of 45\u00b0 above the horizontal from the top of a cliff. At what speed should the arrow strike the field 48 m below where the arrow was launched?<\/li>\n<li>A football player throws a football into the stands at a velocity of 21 m\/s angled at 50\u00b0 above the horizontal. If this football is caught by a fan that is 18 m above the height from where it was thrown, at what speed was it traveling?<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<h1>11.3 Elastic and Inelastic Collisions<\/h1>\n<div class=\"textbox\">Video to Watch: <a href=\"https:\/\/www.youtube.com\/watch?v=8ko3qy9vgLQ\">Elastic and Inelastic Collisions<\/a><\/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: center;\">Where&#8230;<\/p>\n<p style=\"text-align: center;\">[latex]\\vec{p} = m \\vec{v}[\/latex]<\/p>\n<ul class=\"twocolumn\" style=\"list-style-type: none;\">\n<li style=\"text-align: left;\">Energy<sub>Before<\/sub> = Energy<sub>After<\/sub><\/li>\n<li style=\"text-align: left;\">[latex]E_{k i} + E_{p i} = E_{k f} + E_{p f}[\/latex]<\/li>\n<li style=\"text-align: left;\">[latex]E_p = m g h[\/latex]<\/li>\n<li>[latex]\\Delta E_p = m g \\Delta h[\/latex]<\/li>\n<li>[latex]E_k = \\frac{1}{2} m v^2[\/latex]<\/li>\n<li>[latex]\\Delta E_k = \\frac{1}{2} m \\Delta v^2[\/latex]<\/li>\n<li>[latex]\\Delta h = h_f - h_i[\/latex]<\/li>\n<li>[latex]\\Delta v = v_f - v_i[\/latex]<\/li>\n<\/ul>\n<h2>Concepts used in this Section:<\/h2>\n<p>Collisions must always obey the <strong>Law of Conservation of Momentum<\/strong>. That is, the vector sum of momenta prior to a collision equals the vector sum after a collision carrying a requirement that no external forces are acting on the colliding objects.\u00a0 However, collisions are classified due to kinetic energy being or not being conserved in the collision.<\/p>\n<p>Collisions where some of the initial kinetic energy is turned into heat or some other form of energy are inelastic. Collisions where kinetic energy is not lost are classified as elastic.<\/p>\n<p>Collisions between atoms<a class=\"footnote\" title=\"One general exception to collisions between atoms being inelastic is where energy is released during the collision as electromagnetic radiation such as a burst of light energy.\" id=\"return-footnote-161-1\" href=\"#footnote-161-1\" aria-label=\"Footnote 1\"><sup class=\"footnote\">[1]<\/sup><\/a> in ideal gases are generally elastic since no kinetic energy is lost during the collision. Other examples of elastic collisions are the scattering of sub-atomic particles when deflected by electromagnetic forces or the gravitational interactions between planets and satellites<a class=\"footnote\" title=\"http:\/\/hyperphysics.phy-astr.gsu.edu\/hbase\/elacol.html\" id=\"return-footnote-161-2\" href=\"#footnote-161-2\" aria-label=\"Footnote 2\"><sup class=\"footnote\">[2]<\/sup><\/a>. Inelastic collisions are further differentiated if they lose all kinetic energy during the collision or only part of the initial kinetic energy. In collisions where all kinetic energy is lost the collision is classified as perfectly inelastic. For collisions where only some of the kinetic energy is lost, the collision is defined as partially elastic or inelastic.<\/p>\n<div>\n<div><\/div>\n<div>\n<p><strong>Summing this up:<\/strong><\/p>\n<ul>\n<li>Perfectly elastic collisions conserve both momentum and kinetic energy.<\/li>\n<\/ul>\n<ul>\n<li>Partially elastic collisions or inelastic conserve momentum and some kinetic energy.<\/li>\n<\/ul>\n<ul>\n<li>Perfectly inelastic collisions conserve only momentum and lose all kinetic energy.<\/li>\n<\/ul>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignleft wp-image-767\" src=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/11.3-newtons-cradle-scaled.jpg\" alt=\"\" width=\"292\" height=\"219\" srcset=\"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/11.3-newtons-cradle-scaled.jpg 2560w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/11.3-newtons-cradle-300x225.jpg 300w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/11.3-newtons-cradle-1024x768.jpg 1024w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/11.3-newtons-cradle-768x576.jpg 768w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/11.3-newtons-cradle-1536x1152.jpg 1536w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/11.3-newtons-cradle-2048x1536.jpg 2048w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/11.3-newtons-cradle-65x49.jpg 65w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/11.3-newtons-cradle-225x169.jpg 225w, https:\/\/opentextbc.ca\/foundationsofphysics\/wp-content\/uploads\/sites\/427\/2023\/02\/11.3-newtons-cradle-350x263.jpg 350w\" sizes=\"auto, (max-width: 292px) 100vw, 292px\" \/>Generally, macroscopic collisions are partially elastic or perfectly inelastic. Newton\u2019s cradle is one apparatus that comes close to a perfectly elastic collision and is used for demonstrations in introductory physics classes.<strong>\u00a0<\/strong><\/p>\n<p>Observers watching a demonstration of Newton\u2019s cradle can hear the sound of the metal balls colliding which acts to extract energy out of the collision, so it becomes quite obvious that the collision is not perfectly elastic.<\/p>\n<div>\n<div>\n<p>The following examples explore the elasticity of collisions.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 11.3.1<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>Two objects collide on a frictionless surface. The first object having a mass of 4.0 kg has a velocity of 2.0 m\/s prior to the collision and \u2212 5.2 m\/s after.\u00a0 The second object having a mass of 6.0 kg has a velocity of \u2212 4.0 m\/s prior to the collision and 0.8 m\/s after. Verify that this collision could exist and identify what type of a collision it is. (Perfectly Elastic, Partially Elastic or Inelastic, or Perfectly Inelastic).<\/p>\n<p><strong>Solution<\/strong><\/p>\n<p>Data:<\/p>\n<ul>\n<li>[latex]m_1 = 4.0 \\text{ kg}[\/latex]<\/li>\n<li>[latex]\\vec{v}_{1 i} = 2.0 \\text{ m\/s}[\/latex]<\/li>\n<li>[latex]\\vec{v}_{1 f} = - 5.2 \\text{ m\/s}[\/latex]<\/li>\n<li>[latex]m_2 = 6.0 \\text{ kg}[\/latex]<\/li>\n<li>[latex]\\vec{v}_{2 i} = - 4.0 \\text{ m\/s}[\/latex]<\/li>\n<li>[latex]\\vec{v}_{2 f} = 0.8 \\text{ m\/s}[\/latex]<\/li>\n<\/ul>\n<p>Solution:<\/p>\n<p>First solve for the velocity of mass 2 ([latex]\\vec{v}_{2 f}[\/latex]) before the collision.<\/p>\n<p style=\"text-align: center;\">[latex]\\vec{P}_{\\text{before}} = \\vec{P}_{\\text{after}}[\/latex]<\/p>\n<p style=\"text-align: center;\">or: [latex]m_1 \\vec{v}_{1 i} + m_2 \\vec{v}_{2 i} = m_1 \\vec{v}_{1 f} + m_2 \\vec{v}_{2 f}[\/latex]<\/p>\n<p>Inputing variables yields:<\/p>\n<ul>\n<li>[latex](4.0 \\text{ kg})(2.0 \\text{ m\/s}) + (6.0 \\text{ kg})(- 4.0 \\text{ m\/s}) = (4.0 \\text{ kg})(- 5.2 \\text{ m\/s}) + (6.0 \\text{ kg})(0.8 \\text{ m\/s})[\/latex]<\/li>\n<li>[latex]8.0 \\text{ Ns } - 24 \\text{ Ns } = - 20.8 \\text{ Ns }+ 4.8 \\text{ Ns}[\/latex]<\/li>\n<li>[latex]16 \\text{ Ns } = 16 \\text{ Ns}[\/latex]. Momentum is conserved.<\/li>\n<\/ul>\n<p>Now to check to see if any Kinetic Energy was lost during the collision:<\/p>\n<p style=\"text-align: center;\">[latex]E_{ k i} = E_{k f}[\/latex]<\/p>\n<p style=\"text-align: center;\">or: [latex]\\frac{1}{2} m_1 \\vec{v}^2_{1 i} + \\frac{1}{2} m_2 \\vec{v}^2_{2 i} = \\frac{1}{2} m_1 \\vec{v}^2_{1 f} + \\frac{1}{2} m_2 \\vec{v}^2_{2 f}[\/latex]<\/p>\n<p>Inputting variables yields:<\/p>\n<ul>\n<li>[latex]\\frac{1}{2}(4.0 \\text{ kg})(2.0 \\text{ m\/s})^2 + \\frac{1}{2}(6.0 \\text{ kg})(- 4.0 \\text{ m\/s})^2 = \\frac{1}{2}(4.0 \\text{ kg})(- 5.2 \\text{ m\/s})^2 + \\frac{1}{2}(6.0 \\text{ kg})(0.8 \\text{ m\/s})^2[\/latex]<\/li>\n<li>[latex]8.0 \\text{ J } + 48 \\text{ J } = 54.08 \\text{ J } + 1.92 \\text{ J}[\/latex]<\/li>\n<li>[latex]56 \\text{ J } = 56 \\text{ J}[\/latex]. Kinetic energy is conserved.<\/li>\n<\/ul>\n<p>Because Kinetic Energy was conserved this collision is an example of a perfectly elastic collision.<\/p>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 11.3.2<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>Two objects collide on a frictionless surface. The first object having a mass of 0.42 kg has a velocity of \u22120.32 m\/s prior to the collision and 0.26 m\/s after.\u00a0 The second object has a mass of 0.88 kg at some unknown velocity prior to the collision and at rest after.\u00a0 Find the velocity and identify what type of a collision it is. (Perfectly Elastic, Partially Elastic or Inelastic, or Perfectly Inelastic).<\/p>\n<p><strong>Solution<\/strong><\/p>\n<p>Data:<\/p>\n<ul>\n<li>[latex]m_1 = 0.42 \\text{ kg}[\/latex]<\/li>\n<li>[latex]\\vec{v}_{1 i} = - 0.32 \\text{ m\/s}[\/latex]<\/li>\n<li>[latex]\\vec{v}_{1 f} = 0.26 \\text{ m\/s}[\/latex]<\/li>\n<li>[latex]m_2 = 0.88 \\text{ kg}[\/latex]<\/li>\n<li>[latex]\\vec{v}_{2 i} = 0 \\text{ m\/s}[\/latex]<\/li>\n<li>[latex]\\vec{v}_{2 f} = \\text{ Unknown}[\/latex]<\/li>\n<\/ul>\n<p>Solution:<\/p>\n<p>First solve for the velocity of mass 2 ([latex]\\vec{v}_{2 f}[\/latex]) before the collision.<\/p>\n<p style=\"text-align: center;\">[latex]\\vec{P}_{\\text{before}} = \\vec{P}_{\\text{after}}[\/latex]<\/p>\n<p style=\"text-align: center;\">or: [latex]m_1 \\vec{v}_{1 i} + m_2 \\vec{v}_{2 i} = m_1 \\vec{v}_{1 f} + m_2 \\vec{v}_{2 f}[\/latex]<\/p>\n<p>Inputting variables yields:<\/p>\n<ul>\n<li>[latex](0.42 \\text{ kg})(- 0.32 \\text{ m\/s}) + (0.88 \\text{ kg})(\\vec{v}_{2 f}) = (0.42 \\text{ kg})(0.26 \\text{ m\/s}) + (0.88 \\text{ kg})(0 \\text{ m\/s})[\/latex]<\/li>\n<li>[latex]-0.1344 \\text{ Ns } + (0.88 \\text{ kg})(\\vec{v}_{2 f}) = 0.1092 \\text{ Ns } + 0 \\text{ Ns}[\/latex]<\/li>\n<li>[latex](0.88 \\text{ kg})(\\vec{v}_{2f}) = 0.2436 \\text{ Ns}[\/latex]<\/li>\n<li>[latex]\\vec{v}_{2 f} = \\dfrac{0.2436 \\text{ Ns}}{0.88 \\text{ kg}}[\/latex]<\/li>\n<li>[latex]\\vec{v}_{2 f} = 0.277 \\text{ m\/s}[\/latex]<\/li>\n<\/ul>\n<p>Now to check to see if any Kinetic Energy was lost during the collision:<\/p>\n<p style=\"text-align: center;\">[latex]E_{ k i} = E_{k f}[\/latex]<\/p>\n<p style=\"text-align: center;\">or: [latex]\\frac{1}{2} m_1 \\vec{v}^2_{1 i} + \\frac{1}{2} m_2 \\vec{v}^2_{2 i} = \\frac{1}{2} m_1 \\vec{v}^2_{1 f} + \\frac{1}{2} m_2 \\vec{v}^2_{2 f}[\/latex]<\/p>\n<p>Inputting variables yields:<\/p>\n<ul>\n<li>[latex]\\frac{1}{2}(0.42 \\text{ kg})(- 0.32 \\text{ m\/s})^2 + \\frac{1}{2}(0.88 \\text{ kg})(0.277 \\text{ m\/s})^2 = \\frac{1}{2}(0.42 \\text{ kg})(0.26 \\text{ m\/s})^2 + 0 \\text{ Nm}[\/latex]<\/li>\n<li>[latex]0.0215 \\text{ J } + 0.0338 \\text{ J } = 0.0142 \\text{ J } + 0 \\text{ J}[\/latex]<\/li>\n<li>[latex]0.0553 \\text{ N J m} = 0.0142 \\text{ J}[\/latex]. Kinetic Energy is lost.<\/li>\n<\/ul>\n<p>Because Kinetic Energy was not conserved this collision is an example of an inelastic collision.<\/p>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Exercise 11.3<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<ol>\n<li>A 0.50 kg billiard ball traveling at 1.5 m\/s strikes a 0.45 kg billiard ball at rest.\u00a0 If the 0.50 kg ball slows to 0.75 m\/s, at what speed is the 0.45 kg ball moving at?\u00a0 What type of elastic collision is this?<\/li>\n<li>Two balls of putty sliding towards each other collide and come to a complete stop. If the small ball had a mass of 150 g and a velocity of 12.0 cm\/s North, prior to the collision, what was the velocity of the 200 g larger ball prior to the collision?\u00a0 What type of elastic collision was this?<\/li>\n<li>Two balls of identical mass (0.500 kg) each having a kinetic energy of 3.0 joules are directed toward each other. During the collision each experience equal and opposite impulses of 2.0 Nm.\n<ol type=\"i\">\n<li>What were their speeds prior to the collision?<\/li>\n<li>What are their speeds after the collision?<\/li>\n<li>Was this collision elastic or inelastic?<\/li>\n<li>What impulse would be needed to make this a perfectly elastic collision?<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<h1>11.4 Work, Force &amp; Change in Energy Problems &#8211; Launching and Braking Devices<\/h1>\n<div class=\"textbox\">\n<ul>\n<li>Extra Help: <a href=\"https:\/\/www.physicsclassroom.com\/class\/energy\/Lesson-2\/Analysis-of-Situations-Involving-External-Forces\">Analysis of Situations Involving External Forces<\/a><\/li>\n<li>Extra Help: <a href=\"https:\/\/www.physicsclassroom.com\/class\/energy\/Lesson-2\/Analysis-of-Situations-in-Which-Mechanical-Energy\">Analysis of Situations in Which Mechanical Energy is Conserved<\/a><\/li>\n<\/ul>\n<\/div>\n<p>Equations Introduced or Used for this Section:<\/p>\n<ul class=\"twocolumn\" style=\"list-style-type: none;\">\n<li>Work [latex](W) = \\vec{F}_{\\text{net}} \\cdot \\vec{d}[\/latex] or [latex]\\vec{F}_{\\text{net}} \\vec{d} \\cos \u00f8[\/latex]<\/li>\n<li style=\"text-align: left;\">[latex]E_p = m g h[\/latex]<\/li>\n<li>[latex]\\Delta E_p = m g \\Delta h[\/latex]<\/li>\n<li>[latex]E_k = \\frac{1}{2} m v^2[\/latex]<\/li>\n<li>[latex]\\vec{W} = \\Delta \\text{ Mechanical Energy}[\/latex]<\/li>\n<li>[latex]\\Delta E_k = \\frac{1}{2} m \\Delta v^2[\/latex]<\/li>\n<li>[latex]\\Delta h = h_f - h_i[\/latex]<\/li>\n<li>[latex]\\Delta v = v_f - v_i[\/latex]<\/li>\n<li>[latex]V[\/latex] or [latex]V_g = g h[\/latex]<\/li>\n<\/ul>\n<h2>Concepts used in this Section:<\/h2>\n<p>When one is using the <strong>work-energy theorem<\/strong>, in situations where only conservative forces are acting, then [latex]W = \\Delta E_k[\/latex] or [latex]W = (E_{k f} - E_{k i})[\/latex].\u00a0 Under some conditions, the work done on a system can result in a change in both potential and kinetic energy [latex]W = \\Delta E_k + \\Delta E_p[\/latex]. This equation is not defined by the <strong>work-energy theorem <\/strong>but is a combination of the <strong>work-energy theorem <\/strong>and the<strong> conservation of mechanical energy<\/strong>. In these cases work can be looked at as the function of adding or subtracting energy of the system using the conservation of mechanical energy as defining the total energy of the system which could be converted to a kinetic energy.<\/p>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example 11.4.1<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>A 1400 kg car is traveling at 45 km\/h at the top of a 42 m high hill.<\/p>\n<ol type=\"i\">\n<li>What amount of work is needed to stop this car at the bottom of the hill?<\/li>\n<li>What average braking force is needed to do this if the car stops in 200 m at the bottom of this hill?<\/li>\n<\/ol>\n<p><strong>Solution<\/strong><\/p>\n<p>Solution, Question (i):<\/p>\n<ul>\n<li>[latex]W = \\Delta E_k + \\Delta E_p[\/latex]<\/li>\n<li>[latex]W = (E_{k f} - E_{k i}) + (E_{p f} - E_{p i})[\/latex]<\/li>\n<li>[latex]W = \\frac{1}{2} m v_v^2-\\frac{1}{2} m v_i^2 + m g h_f- m g h_i[\/latex]<\/li>\n<li>[latex]W = \\frac{1}{2} m v_f^2 - \\frac{1}{2} m v_i^2 + m g h_f - m g h_i[\/latex]<\/li>\n<li>[latex]W = \\frac{1}{2}(1400 \\text{ kg})(0 \\text{ m\/s})^2 - \\frac{1}{2}(1400 \\text{ kg})(12.5 \\text{ m\/s})^2 + (1400 \\text{ kg})(9.8 \\text{ m\/s}^2)(0 \\text{ m}) - (1400 \\text{ kg})(9.8 \\text{ m\/s}^2)(42 \\text{ m})[\/latex]<\/li>\n<li>[latex]W = 0 \\text{ J} - 109\\;375 \\text{ J } + 0 \\text{ J } - 576\\;240 \\text{ J}[\/latex]<\/li>\n<li>[latex]W = - 686\\;000 \\text{ J } (\\approx -690\\;000 \\text{ J})[\/latex]<\/li>\n<\/ul>\n<p>Solution, Question (ii):<\/p>\n<ul>\n<li>[latex]W = \\Delta \\text{ Energy}[\/latex]<\/li>\n<li>For this car, [latex]W = -686\\;000 \\text{ J}[\/latex]<\/li>\n<li>Now, [latex]\\vec{F} \\vec{d} \\cos \u00f8 = -686\\;000 \\text{ J}[\/latex]<\/li>\n<li>[latex]\\vec{F} (200 \\text{ m}) \\cos 0^{\\circ} = -686\\;000 \\text{ J}[\/latex]<\/li>\n<li>[latex]\\vec{F} = -686\\;000 \\text{ J } \\div 200 \\text{ m}[\/latex]<\/li>\n<li>[latex]\\vec{F} = - 3430 \\text{ N } (\\approx 3400 \\text{ N})[\/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 11.4.2<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>A co-worker tosses a 4.0 kg brick to his friend 2.0 m below at 1.5 m\/s. What amount of force is needed to catch and stop this brick in a horizontal distance of 0.50 m?<\/p>\n<p><strong>Solution<\/strong><\/p>\n<p>First, find out how much energy needs to be removed from the brick when catching it.<\/p>\n<ul>\n<li>[latex]W = \\Delta \\text{ Energy}[\/latex]<\/li>\n<li>[latex]W = \\Delta E_k + \\Delta E_p[\/latex]<\/li>\n<li>[latex]W = (E_{k f} - E_{k i}) + (E_{p f} - E_{p i})[\/latex]<\/li>\n<li>[latex]W = \\frac{1}{2} m v_v^2 - \\frac{1}{2} m v_i^2 + m g h_f- m g h_i[\/latex]<\/li>\n<li>[latex]W = \\frac{1}{2}(4.0 \\text{ kg})(0 \\text{ m\/s})^2 - \\frac{1}{2}(4.0 \\text{ kg})(1.5 \\text{ m\/s})^2 + (4.0 \\text{ kg})(9.8 \\text{ m\/s}^2)(0 \\text{ m}) - (4.0 \\text{ kg})(9.8 \\text{ m\/s}^2)(2.0 \\text{ m})[\/latex]<\/li>\n<li>[latex]W = 0 \\text{ J } - 4.5 \\text{ J } + 0 \\text{ J } - 78.4 \\text{ J}[\/latex]<\/li>\n<\/ul>\n<p>Second, find out the average braking force needed to do this in 0.5 m.<\/p>\n<ul>\n<li>For the brick, [latex]W = - 82.9 \\text{ J}[\/latex]<\/li>\n<li>Now, [latex]\\vec{F} \\vec{d} \\cos \u00f8 = -82.9 \\text{ J}[\/latex]<\/li>\n<li>[latex]\\vec{F} (0.5 \\text{ m}) \\cos 0^{\\circ} = - 82.9 \\text{ J}[\/latex]<\/li>\n<li>[latex]\\vec{F} = -82.9 \\text{ J } \\div 0.5 \\text{ m}[\/latex]<\/li>\n<li>[latex]\\vec{F} = - 166 \\text{ N } (\\approx - 170 \\text{ N})[\/latex]<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Exercise 11.4<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<ol>\n<li>A 35 kg bag of sand is tied to the end of a 3.2 m long rope.\u00a0 It is drawn back from its vertical position to the horizontal then released.\n<ol type=\"i\">\n<li>What speed does it have at the bottom of its swing?<\/li>\n<li>A mechanical catching device brings it to a full stop (in 0.10 m) at the bottom of the swing. How much work is done by the device?<\/li>\n<li>What braking force is needed to stop the sand bag?<\/li>\n<\/ol>\n<\/li>\n<li>What is the average force that a circus cannon exerts on a 75 kg stunt-person over the 2.0 m they are launched with, so that the stunt-person is traveling at 3.0 m\/s when 5.0 m above the launched height?<\/li>\n<li>A 4200 kg truck is rolling along at 80 km\/h at the top of a 102 m high hill. If this truck is allowed to freely accelerate to the bottom of the hill before the brakes are applied, what average braking force would be needed to stop this truck in half a kilometre?<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Exercise 11.5.1: A Trebuchet Catapult<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>A Trebuchet Catapult<a class=\"footnote\" title=\"A trebuchet or trebucket is a siege engine that was employed in the Middle Ages either to smash masonry walls or to throw projectiles over them. It is sometimes called a &quot;counterweight trebuchet&quot; or &quot;counterpoise trebuchet&quot; in order to distinguish it from an earlier weapon that has come to be called the &quot;traction trebuchet&quot;, the original version with pulling men instead of a counterweight. The counterweight trebuchet appeared in both Christian and Muslim lands around the Mediterranean in the twelfth century. It could fling three-hundred-pound (140 kg ... largest appear to be 1500 kg) projectiles at high speeds into enemy fortifications. On occasion, disease-infected corpses were flung into cities in an attempt to infect the people under siege a medieval variant of biological warfare. Traction trebuchets appeared in China in about the 4th century BCE and in Europe in the 6th century CE, and did not become obsolete until the 16th century, well after the introduction of gunpowder. Trebuchets were far more accurate than other medieval catapults. Adapted from http:\/\/en.wikipedia.org\/wiki\/Trebuchet\" id=\"return-footnote-161-3\" href=\"#footnote-161-3\" aria-label=\"Footnote 3\"><sup class=\"footnote\">[3]<\/sup><\/a>\u00a0is capable of launching 140 kg stones with such energy they will still be traveling at 80 km\/h when 60 m above its original launch height.<\/p>\n<p><strong>Question:<\/strong> At what speed are these stones released?<\/p>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Exercise 11.5.2: Kingda Ka&#8230; Six Flags Great Adventure<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>On May 21, 2005, Six Flags Great Adventure introduced Kingda Ka the tallest and fastest roller coaster on the planet to the public.\u00a0 This Swiss designed &#8220;rocket coaster&#8221; uses hydraulic motors to launch the cars along a horizontal section of track from zero to 205 km\/h in an impressive 3.5 seconds.\u00a0 The cars then begin a vertical ascent up a steel tower that peaks at 139 m (\u2248 45 stories).\u00a0 Crossing over the apex the train enters a vertical descent plunging through a 270\u00b0 spiral twist again reaching speeds more than 160 km\/h.\u00a0 One final surprise comes before the brake run a 39 m tall camelback hill that offers plenty of negative g&#8217;s also known fondly as &#8220;airtime&#8221;.<\/p>\n<p><strong>Question:<\/strong> With its launch speed of 205 km\/h how fast could this roller coaster be moving after it has climbed a vertical height of 139 m?\u00a0 (It will be less than this due to friction.)<\/p>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Exercise 11.5.3: The Ballistic Pendulum<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>The Ballistic Pendulum was invented in 1742 by English mathematician Benjamin Robins (1707\u20131751) and published in his book New Principles of Gunnery, which revolutionized the science of ballistics as it provided the first way to accurately measure the velocity of a bullet.\u00a0 Robins used the ballistic pendulum to measure projectile velocity in two ways.\u00a0 The first was to attach the gun to the pendulum and measure the recoil.\u00a0 Since the momentum of the gun is equal to the momentum of the ejecta and since the projectile was (in those experiments) the large majority of the mass of the ejecta the velocity of the bullet could be approximated.\u00a0 The second and more accurate method was to directly measure the bullet momentum by firing it into the pendulum and measuring the change in potential energy of the pendulum + embedded bullet.<\/p>\n<p><strong>Question<\/strong>: Suppose one wanted to set up a ballistics pendulum to check the muzzle velocity of a bullet fired from an 1894 Winchester (30-30) as shown below.\u00a0 Specs show that it should shoot an 8.0 g (130 grain) bullet at 759 m\/s.\u00a0 What mass of wooden block would be needed so that it would rise 10.0 cm above its rest position after being struck by this bullet?<\/p>\n<\/div>\n<\/div>\n<h1>Exercise Answers<\/h1>\n<h2>11.1 Conservation of Mechanical Energy<\/h2>\n<ol class=\"twocolumn\">\n<li>42 m<\/li>\n<li>15.4 m\/s<\/li>\n<li>184 m<\/li>\n<li>7.0 m\/s<\/li>\n<li>21 m\/s<\/li>\n<li>17 m\/s\n<ol class=\"twocolumn\" type=\"i\">\n<li>16 J<\/li>\n<li>237 J<\/li>\n<\/ol>\n<\/li>\n<li>40 m\/s\n<ol class=\"twocolumn\" type=\"i\">\n<li>1200 J<\/li>\n<li>22 m\/s<\/li>\n<\/ol>\n<\/li>\n<li>61 m\/s<\/li>\n<li>35 m\/s<\/li>\n<li>31 m\/s<\/li>\n<\/ol>\n<h2>11.2 Non-Horizontally Launched Projectiles made easy using Conservation of Energy<\/h2>\n<ol class=\"twocolumn\">\n<li>50 m\/s<\/li>\n<li>15.6 m\/s<\/li>\n<li>67 m\/s<\/li>\n<li>31 m\/s<\/li>\n<\/ol>\n<h2>11.3 Elastic and Inelastic collisions<\/h2>\n<ol>\n<li>Inelastic Collision ([latex]E_{ki}[\/latex] = 0.5625 J,\u00a0 [latex]E_{kf}[\/latex] = 0.3437 J)<\/li>\n<li>0.0135 m\/s South\n<ol class=\"twocolumn\" type=\"i\">\n<li>3.46 m\/s<\/li>\n<li>\u2212 0.54 m\/s<\/li>\n<li>Inelastic Collision<\/li>\n<li>3.46 Ns<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<h2>11.4 Work, Force &amp; Change in Energy Problems &#8211; Launching and Braking Devices<\/h2>\n<ol>\n<li>\n<ol class=\"threecolumn\" type=\"i\">\n<li>7.9 m\/s<\/li>\n<li>\u2212 1100 J<\/li>\n<li>\u2212 11<span style=\"margin-left: 0.25em;\">000<\/span> N<\/li>\n<\/ol>\n<\/li>\n<li>2000 N<\/li>\n<li>\u2212 10<span style=\"margin-left: 0.25em;\">500<\/span> N<\/li>\n<\/ol>\n<h2>11.5.1 A Trebuchet Catapult<\/h2>\n<ol>\n<li>[latex]v_i[\/latex] = 41 m\/s<\/li>\n<\/ol>\n<h2>11.5.2 Kingda Ka &#8230; Six Flags Great Adventure<\/h2>\n<ol>\n<li>[latex]v_f[\/latex] = 23 m\/s<\/li>\n<\/ol>\n<h3>Media Attributions<\/h3>\n<ul>\n<li>&#8220;Galileo&#8217;s Interrupted Pendulum&#8221; by Galileo Galilei is in the <a href=\"https:\/\/en.wikipedia.org\/wiki\/Public_domain\">public domain<\/a>.\n<div class=\"hHq9Z\">\n<div data-hveid=\"CA8QAA\" data-ved=\"2ahUKEwjqsYjXh6uBAxXSMDQIHeeJBU4QlcAGegQIDxAA\">\n<div class=\"KsRP6\">\n<div class=\"MDY31c\">\n<div class=\"QpPSMb\">\n<div class=\"DoxwDb\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/li>\n<li>&#8220;<a href=\"https:\/\/commons.wikimedia.org\/wiki\/File:Eshima_oohashi_04_(14781816175).jpg\">Eshima oohashi 04 (14781816175)<\/a>&#8221; by mstk is licensed under a <a href=\"https:\/\/creativecommons.org\/licenses\/by\/2.0\/deed.en\">CC BY 2.0 licence<\/a>.<\/li>\n<li>&#8220;<a href=\"https:\/\/commons.wikimedia.org\/wiki\/File:Eshima_bidge.jpg\">Eshima bidge<\/a>&#8221; by \u5199\u771f\u6295\u7a3f\u7528 is licensed under a <a href=\"https:\/\/creativecommons.org\/licenses\/by-sa\/4.0\/deed.en\">CC BY-SA 4.0 licence<\/a>.<\/li>\n<li>&#8220;<a href=\"https:\/\/commons.wikimedia.org\/wiki\/File:Fire_inside_an_abandoned_convent_in_Massueville,_Quebec,_Canada.jpg\">Fire inside an abandoned convent in Massueville, Quebec, Canada<\/a>&#8221; by Sylvain Pedneault is licensed under a <a href=\"https:\/\/creativecommons.org\/licenses\/by-sa\/3.0\/\">CC BY-SA 3.0 Unported licence<\/a>.<\/li>\n<li>&#8220;<a href=\"https:\/\/pixabay.com\/photos\/newtons-cradle-physics-pendulum-2891856\/\">Newton&#8217;s cradle<\/a>&#8221; by <a href=\"https:\/\/pixabay.com\/users\/quincecreative-1031690\/\">QuinceCreative<\/a> is licensed under a <a href=\"https:\/\/pixabay.com\/service\/terms\/\">Pixabay licence<\/a>.<\/li>\n<\/ul>\n<hr class=\"before-footnotes clear\" \/><div class=\"footnotes\"><ol><li id=\"footnote-161-1\">One general exception to collisions between atoms being inelastic is where energy is released during the collision as electromagnetic radiation such as a burst of light energy. <a href=\"#return-footnote-161-1\" class=\"return-footnote\" aria-label=\"Return to footnote 1\">&crarr;<\/a><\/li><li id=\"footnote-161-2\"><a href=\"http:\/\/hyperphysics.phy-astr.gsu.edu\/hbase\/elacol.html\">http:\/\/hyperphysics.phy-astr.gsu.edu\/hbase\/elacol.html<\/a> <a href=\"#return-footnote-161-2\" class=\"return-footnote\" aria-label=\"Return to footnote 2\">&crarr;<\/a><\/li><li id=\"footnote-161-3\">A trebuchet or trebucket is a siege engine that was employed in the Middle Ages either to smash masonry walls or to throw projectiles over them. It is sometimes called a \"counterweight trebuchet\" or \"counterpoise trebuchet\" in order to distinguish it from an earlier weapon that has come to be called the \"traction trebuchet\", the original version with pulling men instead of a counterweight. The counterweight trebuchet appeared in both Christian and Muslim lands around the Mediterranean in the twelfth century. It could fling three-hundred-pound (140 kg ... largest appear to be 1500 kg) projectiles at high speeds into enemy fortifications. On occasion, disease-infected corpses were flung into cities in an attempt to infect the people under siege a medieval variant of biological warfare. Traction trebuchets appeared in China in about the 4th century BCE and in Europe in the 6th century CE, and did not become obsolete until the 16th century, well after the introduction of gunpowder. Trebuchets were far more accurate than other medieval catapults. Adapted from <a href=\"http:\/\/en.wikipedia.org\/wiki\/Trebuchet\">http:\/\/en.wikipedia.org\/wiki\/Trebuchet<\/a> <a href=\"#return-footnote-161-3\" class=\"return-footnote\" aria-label=\"Return to footnote 3\">&crarr;<\/a><\/li><\/ol><\/div>","protected":false},"author":125,"menu_order":11,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-161","chapter","type-chapter","status-publish","hentry"],"part":3,"_links":{"self":[{"href":"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-json\/pressbooks\/v2\/chapters\/161","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":24,"href":"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-json\/pressbooks\/v2\/chapters\/161\/revisions"}],"predecessor-version":[{"id":1243,"href":"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-json\/pressbooks\/v2\/chapters\/161\/revisions\/1243"}],"part":[{"href":"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-json\/pressbooks\/v2\/parts\/3"}],"metadata":[{"href":"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-json\/pressbooks\/v2\/chapters\/161\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-json\/wp\/v2\/media?parent=161"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-json\/pressbooks\/v2\/chapter-type?post=161"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-json\/wp\/v2\/contributor?post=161"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/opentextbc.ca\/foundationsofphysics\/wp-json\/wp\/v2\/license?post=161"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}