{"id":15893,"date":"2017-11-13T11:07:34","date_gmt":"2017-11-13T11:07:34","guid":{"rendered":"http:\/\/sitepourvtc.com\/?page_id=15893"},"modified":"2022-11-03T07:54:06","modified_gmt":"2022-11-03T07:54:06","slug":"law-of-conservation-of-momentum","status":"publish","type":"page","link":"https:\/\/sitepourvtc.com\/laws-of-conservation\/law-of-conservation-of-momentum\/","title":{"rendered":"Law of Conservation of Momentum"},"content":{"rendered":"<div   id="8n9s8rpp6h"    class=\"su-quote su-quote-style-default\"><div   id="8n9s8rpp6h"    class=\"su-quote-inner su-u-clearfix su-u-trim\"><b> Generally<\/b>, the <strong>law of conservation of momentum<\/strong> or principle of momentum conservation states that the momentum of an isolated system is a <strong>constant<\/strong>.<\/div><\/div>\n<p>The vector sum of the momenta (momentum is equal to the mass of an object multiplied by its velocity) of all the objects of a system cannot be changed by interactions within the system. In classical mechanics, this law is implied by <strong>Newton\u2019s laws<\/strong>. This principle is a direct consequence of Newton\u2019s third law.<\/p>\n<figure id=\"attachment_15722\" aria-describedby=\"caption-attachment-15722\" style=\"width: 290px\" class=\"wp-caption alignright\"><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/10\/conservation-of-energy-pendulum.gif\"><img loading=\"lazy\" class=\" wp-image-15722\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/10\/conservation-of-energy-pendulum.gif\" alt=\"law of conservation of energy - pendulum\" width=\"300\" height=\"225\" \/><\/a><figcaption id=\"caption-attachment-15722\" class=\"wp-caption-text\">Newton\u2019s cradle. A device that demonstrates the Law of Conservation of Mechanical Energy and Momentum.<\/figcaption><\/figure>\n<p>Let us assume the <strong>one-dimensional elastic collision<\/strong> of two billiard balls, ball A and ball B. We assume the net external force on this system of two balls is zero\u2014that is, the only significant forces during the collision are the forces that each ball exerts on the other. According to <strong>Newton\u2019s third law<\/strong>, the two forces are always equal in magnitude and opposite direction. Hence, the impulses that act on the two balls are equal and opposite, and the changes in momentum of the two balls are equal and opposite.<\/p>\n<p>These two balls are moving with velocities <strong>v<\/strong><strong><sub>A<\/sub><\/strong> and<strong> v<\/strong><strong><sub>B<\/sub><\/strong> along the x-axis before the collision. After the collision, their velocities are <b>v\u2019A<\/b>\u00a0and <b>v\u2019B<\/b>. The <strong>conservation of the total momentum<\/strong> demands that the total momentum before the collision is the same as the total momentum after the collision.<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/11\/conservation-of-momentum-formula.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-15895\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/11\/conservation-of-momentum-formula.png\" alt=\"conservation-of-momentum-formula\" width=\"340\" height=\"62\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/11\/conservation-of-momentum-formula.png 340w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/11\/conservation-of-momentum-formula-300x55.png 300w\" sizes=\"(max-width: 340px) 100vw, 340px\" \/><\/a><\/p>\n<p><strong>The law of conservation of linear momentum<\/strong> in words:<\/p>\n<p style=\"text-align: left;\"><em><span style=\"font-size: 20px;\">If no net external force acts on a system of particles, the total linear momentum of the system cannot change.<\/span><\/em><\/p>\n<div   id="8n9s8rpp6h"    class=\"collapsible-container\">\n<h2>Conservation of Momentum and Energy in Collisions<\/h2>\n<p style=\"text-align: left;\">The use of the <strong>conservation laws for momentum and energy<\/strong> is also very important in <strong>particle collisions<\/strong>. This is a very powerful rule because it can allow us to determine the results of a collision without knowing the details of the collision. The law of <strong>conservation of momentum<\/strong> states that <strong>the total momentum is conserved<\/strong> in the collision of two objects such as billiard balls. The assumption of momentum and kinetic energy conservation makes it possible to calculate the <strong>final velocities<\/strong> in two-body collisions. At this point, we have to distinguish between two types of collisions:<\/p>\n<ul>\n<li><a title=\"Elastic Collisions\" href=\"http:\/\/sitepourvtc.com\/laws-of-conservation\/law-of-conservation-of-energy\/elastic-collisions\/\"><strong>Elastic collisions<\/strong><\/a><\/li>\n<li><a title=\"Inelastic Collision\" href=\"http:\/\/sitepourvtc.com\/laws-of-conservation\/law-of-conservation-of-energy\/inelastic-collision\/\"><strong>Inelastic collisions<\/strong><\/a><\/li>\n<\/ul>\n<h2>Elastic Collisions<\/h2>\n<p>A <strong>perfectly elastic collision<\/strong> is defined as one in which there is <strong>no net conversion of kinetic energy<\/strong> into other forms (such as heat or noise). For the brief moment during which the two objects are in contact, some (or all) of the energy is stored momentarily in the form of <strong>elastic potential energy<\/strong>. But if we compare the total kinetic energy just before the collision with the total kinetic energy just after the collision, they are found to be the same. We say that the <strong>total kinetic energy is conserved<\/strong>.<\/p>\n<canvas id=\"the_canvas\" width=\"250\" height=\"250\"><\/canvas>\r\n<script>\r\n\r\nvar __VERSION__ = \"0.3.5\"\r\nvar inspector = {};\r\n\r\nvar Physics = {\r\n    \r\n    collide_balls: function(ball1, ball2) {\r\n        \/\/ calculate restitution\r\n        \/\/ TODO: add elasticity attr to circles\r\n        var ball1_elasticity = 1.0;\r\n        var ball2_elasticity = 1.0;\r\n        var restitution = ball1_elasticity * ball2_elasticity;\r\n        \r\n        \/\/ invert masses\r\n        var im1 = 1.0 \/ ball1.m;\r\n        var im2 = 1.0 \/ ball2.m;\r\n        \r\n        \/\/ get minimum translation distance\r\n        var delta = ball1.point().subtract(ball2.point());\r\n        var d = delta.get_length();\r\n        var mtd = delta.multiply(ball1.r + ball2.r - d).multiply(1.0 \/ d);\r\n        \r\n        \/\/ push-pull apart based on mass\r\n        var ball1_pt = ball1.point().add(mtd.multiply(im1 \/ (im1 + im2)));\r\n        var ball2_pt = ball2.point().subtract(mtd.multiply(im2 \/ (im1 + im2)));\r\n        ball1.set_point(ball1_pt.x, ball1_pt.y);\r\n        ball2.set_point(ball2_pt.x, ball2_pt.y);\r\n        \r\n        \/\/ calculate impact speed\r\n        var v = ball1.velocity().subtract(ball2.velocity());\r\n        var vn = v.dot(mtd.normalized());\r\n        \r\n        \/\/ positive vn means balls are already moving away for one another\r\n        if ( vn > 0 ) {\r\n            return;\r\n        }\r\n        \r\n        \/\/ collision impulse\r\n        var i = (-1.0 * (1.0 + restitution) * vn) \/ (im1 + im2);\r\n        var impulse = mtd.normalized().multiply(i);\r\n        \r\n        \/\/ transfer momentum\r\n        var v1 = ball1.velocity().add(impulse.multiply(im1));\r\n        var v2 = ball2.velocity().subtract(impulse.multiply(im2));\r\n        \r\n        \/\/ set new velocities\r\n        ball1.set_velocity(v1.x, v1.y);\r\n        ball2.set_velocity(v2.x, v2.y);\r\n    },\r\n    \r\n    bounce_ball_off_world: function(ball, world) {\r\n        \/\/ translate ball back inside world\r\n        var contact_pt = this.find_contact_point(world, ball);\r\n        ball.set_point(contact_pt.x, contact_pt.y);\r\n    \r\n        \/\/ reflect velocity of ball\r\n        var new_ball_vel = this.reflect_ball(world, ball);\r\n        ball.set_velocity(new_ball_vel.x, new_ball_vel.y);\r\n    },\r\n    \r\n    find_contact_point: function(world, ball) {\r\n        \/\/ see http:\/\/gamedev.stackexchange.com\/a\/29658\r\n        var A = world.point();\r\n        var B = ball.last_point();\r\n        var C = ball.point();\r\n        var R = world.r;\r\n        var r = ball.r;\r\n        \r\n        var AB = B.subtract(A);\r\n        var BC = C.subtract(B);\r\n        var AB_len = AB.get_length();\r\n        var BC_len = BC.get_length();\r\n        \r\n        \/\/ Avoid divide-by-zero erros (this should never really happen)\r\n        if ( ! BC_len ) {\r\n            return C;\r\n        }\r\n\r\n        var b = AB.dot(BC) \/ Math.pow(BC_len, 2) * -1;\r\n        var c = (Math.pow(AB_len, 2) - Math.pow(R - r, 2)) \/ Math.pow(BC_len, 2);\r\n        var d = b * b - c;\r\n        var k = b - Math.sqrt(d);\r\n        \r\n        if ( k < 0 ) {\r\n            k = b + Math.sqrt(d);\r\n        }\r\n        \r\n        var BD = C.subtract(B);\r\n        var BD_len = BC_len * k;\r\n        BD.set_length(BD_len);\r\n        \r\n        var D = B.add(BD);\r\n        return D;\r\n    },\r\n    \r\n    reflect_ball: function(world, ball) {\r\n        \/\/ see http:\/\/stackoverflow.com\/a\/573206\/1093087\r\n        var world_pt = world.point();\r\n        var ball_pt = ball.point();\r\n        var v = ball.velocity();\r\n        var n = ball_pt.subtract(world_pt).normalized();\r\n        \r\n        \/\/ assume perfect elasticity for now\r\n        \/\/ TODO: add elasticity attr to balls\r\n        var ball_elasticity = 1.0;\r\n        var world_elasticity = 1.0;\r\n        var restitution = ball_elasticity * world_elasticity;\r\n        \r\n        \/\/ solve reflection\r\n        var u = n.multiply(v.dot(n));\r\n        var w = v.subtract(u);\r\n        var v_after = w.subtract(u);\r\n        var reflection = v_after.subtract(v).multiply(restitution);\r\n        \r\n        \/\/ return new velocity\r\n        var new_ball_vel = v.add(reflection);\r\n        return new_ball_vel\r\n    },\r\n    \r\n    polar_to_vector: function(distance, angle) {\r\n        var radians = angle * (Math.PI \/ 180);\r\n        return new Vec2d(distance * Math.cos(radians),\r\n            distance * Math.sin(radians));\r\n    },\r\n};\r\n\r\n\r\nvar Graphics = {\r\n\r\n    init: function(canvas) {\r\n        this.ctx = canvas.getContext('2d');\r\n        this.center_x = canvas.width \/ 2;\r\n        this.center_y = canvas.height \/ 2;\r\n    },\r\n\r\n    draw_circle: function(x, y, r, color, line_width, stroke_color) {\r\n        var screen_x = this.world_to_screen_x(x);\r\n        var screen_y = this.world_to_screen_y(y);\r\n\r\n        this.ctx.beginPath();\r\n        this.ctx.arc(screen_x, screen_y, r, 0, Math.PI * 2, false);\r\n        this.ctx.fillStyle = color;\r\n        this.ctx.fill();\r\n\r\n        if (line_width) {\r\n            this.ctx.lineWidth = line_width;\r\n            this.ctx.strokeStyle = stroke_color;\r\n            this.ctx.stroke();\r\n        }\r\n\r\n        this.ctx.closePath();\r\n    },\r\n\r\n    world_to_screen_x: function(x) {\r\n        return this.center_x + x;\r\n    },\r\n\r\n    world_to_screen_y: function(y) {\r\n        return this.center_y - y;\r\n    },\r\n};\r\n\r\n\r\nvar Random = {\r\n    integer: function(min, max) {\r\n        return min + Math.floor(Math.random() * (max - min + 1));\r\n    }\r\n};\r\n\r\n\r\nfunction Vec2d(x, y) {\r\n    this.x = x;\r\n    this.y = y;\r\n    \r\n    this.add = function(v2) {\r\n        return new Vec2d(this.x + v2.x, this.y + v2.y);\r\n    };\r\n    \r\n    this.subtract = function(v2) {\r\n        return new Vec2d(this.x - v2.x, this.y - v2.y);\r\n    };\r\n    \r\n    this.multiply = function(v2) {\r\n        if ( ! v2.length ) {\r\n            \/\/ v2 is a scalar\r\n            return new Vec2d(this.x * v2, this.y * v2);\r\n        }\r\n        else {\r\n            return new Vec2d(this.x * v2.x, this.y * v2.y);\r\n        }\r\n    };\r\n    \r\n    this.get_length = function() {\r\n        return Math.sqrt((this.x * this.x) + (this.y * this.y));\r\n    };\r\n    \r\n    this.set_length = function(new_length) {\r\n        var old_length = this.get_length();\r\n        this.x *= new_length\/old_length;\r\n        this.y *= new_length\/old_length;\r\n        return this;\r\n    };\r\n    \r\n    this.distance = function(v2) {\r\n        return Math.sqrt(Math.pow(this.x - v2.x, 2) + Math.pow(\r\n            this.y - v2.y, 2));\r\n    };\r\n    \r\n    this.dot = function(v2) {\r\n        return (this.x * v2.x) + (this.y * v2.y);\r\n    };\r\n    \r\n    this.normalized = function() {\r\n        var length = this.get_length();\r\n        \r\n        if ( length != 0 ) {\r\n            return new Vec2d(this.x \/ length, this.y \/ length);\r\n        }\r\n        else {\r\n            return new Vec2d(this.x, this.y);\r\n        }\r\n    };\r\n}\r\n\r\n\r\nfunction Circle(x, y, r) {    \r\n    this.init = function(x, y, r) {\r\n        this.set_point(x, y);\r\n        this.r = r;\r\n    };\r\n    \r\n    this.point = function() {\r\n        return new Vec2d(this.x, this.y);\r\n    };\r\n    \r\n    this.set_point = function(x, y) {\r\n        this.x = x;\r\n        this.y = y;\r\n        return this;\r\n    };\r\n    \r\n    this.contains_circle = function(c2) {\r\n        var d = this.point().distance(c2.point());\r\n        return d + c2.r - this.r < 0;\r\n    };\r\n    \r\n    this.overlaps_circle = function(c2) {\r\n        \/\/ overlap if combined length is less than distance (use squares for\r\n        \/\/ quicker calculations)\r\n        var dsq = Math.pow(c2.y - this.y, 2) + Math.pow(c2.x - this.x, 2);\r\n        var lsq = Math.pow(this.r + c2.r, 2);\r\n        return dsq < lsq;\r\n    };\r\n    \r\n    this.init(x, y, r);\r\n}\r\n\r\n\r\nfunction World() {\r\n    \/\/ parasitic inheritance\r\n    \/\/ see http:\/\/www.crockford.com\/javascript\/inheritance.html\r\n    var self = new Circle(0, 0, 100);\r\n    \r\n    self.balls = [];\r\n    \r\n    self.color = 'white';\r\n    self.line = 4;\r\n    self.stroke = 'lightgray';\r\n    \r\n    self.update = function() {\r\n        self.balls.map(function(ball) {\r\n            ball.update();\r\n        });\r\n        self.resolve_collisions();\r\n        self.enforce_boundaries();\r\n        return self;\r\n    };\r\n\r\n    self.draw = function () {\r\n        Graphics.draw_circle(self.x, self.y, self.r, self.color, self.line,\r\n            self.stroke);\r\n        self.balls.map(function(ball) {\r\n            ball.draw();\r\n        });\r\n        return self;\r\n    };\r\n    \r\n    self.enforce_boundaries = function() {\r\n        self.balls.map(function(ball) {\r\n            if (! self.contains_circle(ball)) {\r\n                Physics.bounce_ball_off_world(ball, self);\r\n            }\r\n        });\r\n        return self;\r\n    };\r\n    \r\n    self.resolve_collisions = function() {\r\n        for (var i=0; i<self.balls.length; i++) {\r\n            for (var j=i+1; j<self.balls.length; j++) {\r\n                if ( self.balls[i].collides_with(self.balls[j]) ) {\r\n                    self.balls[i].bounce_off(self.balls[j]);\r\n                }\r\n            }\r\n        }\r\n        return self;\r\n    };\r\n    \r\n    self.add_ball = function(ball) {\r\n        self.balls.push(ball);\r\n        return self;\r\n    };\r\n\r\n    return self;\r\n}\r\n\r\n\r\nfunction Ball(x, y, r, color) {\r\n    var self = new Circle(x, y, r);\r\n    self.color = color;\r\n    \r\n    self.m = 10;\r\n    self.vx = 0;\r\n    self.vy = 0;\r\n    \r\n    \/\/ last point: this is needed to resolve collisions with world's wall\r\n    self.lx = undefined;\r\n    self.ly = undefined;\r\n    \r\n    self.max_speed = r \/ 2.0;\r\n    \r\n    self.update = function() {\r\n        self.move();\r\n    };\r\n\r\n    self.move = function() {\r\n        self.lx = self.x;\r\n        self.ly = self.y;\r\n        self.x += self.vx;\r\n        self.y += self.vy;\r\n        return self;\r\n    };\r\n    \r\n    self.last_point = function() {\r\n        return new Vec2d(self.lx, self.ly);\r\n    };\r\n    \r\n    self.set_velocity = function(vx, vy) {\r\n        self.vx = vx;\r\n        self.vy = vy;\r\n        return self;\r\n    };\r\n    \r\n    self.velocity = function() {\r\n        return new Vec2d(self.vx, self.vy);\r\n    };\r\n    \r\n    self.clamp_speed = function () {\r\n        var v = self.velocity();\r\n        var speed = v.get_length();\r\n        if (speed > self.max_speed) {\r\n            v.set_length(self.max_speed);\r\n            self.set_velocity(v.x, v.y);\r\n        }\r\n        return self;\r\n    };\r\n\r\n    self.draw = function () {\r\n        Graphics.draw_circle(self.x, self.y, self.r, self.color, 0);\r\n        return self;\r\n    };\r\n    \r\n    self.collides_with = function(ball) {\r\n        return self.overlaps_circle(ball);\r\n    };\r\n    \r\n    self.bounce_off = function(ball) {\r\n        Physics.collide_balls(self, ball);\r\n        return self;\r\n    };\r\n    \r\n    self.set_direction = function(degrees) {\r\n        var d = self.velocity().get_length();\r\n        var v = Physics.polar_to_vector(d, degrees);\r\n        self.set_velocity(v.x, v.y);\r\n        return self;        \r\n    };\r\n    \r\n    self.set_random_direction = function() {\r\n        return self.set_direction(Random.integer(0,360));\r\n    };\r\n    \r\n    return self;\r\n}\r\n\r\n\r\nfunction main() {\r\n    console.log('launching version: ' + __VERSION__);\r\n    \r\n    \/\/ load canvas and Graphics handler\r\n    var canvas = document.getElementById(\"the_canvas\");\r\n    Graphics.init(canvas);\r\n    \r\n    \/\/ set animation handler\r\n    var fps = 30;\r\n    var mspf = 1000 \/ fps;    \r\n    window.animate = (function () {\r\n        return window.requestAnimationFrame ||\r\n            window.webkitRequestAnimationFrame ||\r\n            window.mozRequestAnimationFrame ||\r\n            window.oRequestAnimationFrame ||\r\n            window.msRequestAnimationFrame ||\r\n            function (callback, element) {\r\n                window.setTimeout(callback, mspf);\r\n            };\r\n    })();\r\n    \r\n    \/\/ init balls\r\n    var red_ball = new Ball(0, 25, 10, 'red');\r\n    red_ball.set_velocity(4, 0).set_random_direction();\r\n    var green_ball = new Ball(-17, -17, 10, 'green');\r\n    green_ball.set_velocity(4, 0).set_random_direction();\r\n    var blue_ball = new Ball(17, -17, 10, 'blue');\r\n    blue_ball.set_velocity(4, 0).set_random_direction();\r\n    \r\n    \/\/ init world\r\n    var world = new World();\r\n    world.add_ball(red_ball);\r\n    world.add_ball(green_ball);\r\n    world.add_ball(blue_ball);\r\n    \r\n    \/\/ attach world to window so that we can inspect\r\n    inspector.world = world;\r\n    \r\n    \/\/ animate\r\n    function anim_loop() {\r\n        world.update();\r\n        world.draw();\r\n        window.animate(anim_loop);\r\n    };\r\n    anim_loop();\r\n}\r\n\r\nmain();\r\n<\/script>\n<ul>\n<li>Large-scale interactions between satellites and planets are <strong>perfectly elastic<\/strong>, like the <strong>slingshot type gravitational interactions<\/strong> (also known as a planetary swing-by or a gravity-assist maneuver).<\/li>\n<li>Collisions between <strong>very hard spheres<\/strong> may be <strong>nearly elastic<\/strong>, so it is useful to calculate the limiting case of an elastic collision.<\/li>\n<li>Collisions in<strong> ideal gases<\/strong> approach perfectly elastic collisions, as scattering interactions of sub-atomic particles deflect by the electromagnetic force.<\/li>\n<li><strong>Rutherford scattering<\/strong> is the elastic scattering of charged particles also by the electromagnetic force.<\/li>\n<li>A <strong>neutron-nucleus scattering reaction<\/strong> may also be elastic, but in this case, the neutron is deflected by a strong nuclear force.<\/li>\n<\/ul>\n<div   id="8n9s8rpp6h"    class=\"su-accordion su-u-trim\"><div   id="8n9s8rpp6h"    class=\"su-spoiler su-spoiler-style-default su-spoiler-icon-plus su-spoiler-closed\" data-scroll-offset=\"0\"><div   id="8n9s8rpp6h"    class=\"su-spoiler-title\" tabindex=\"0\" role=\"button\"><span class=\"su-spoiler-icon\"><\/span>Equations of Conservation of Momentum and Energy<\/div><div   id="8n9s8rpp6h"    class=\"su-spoiler-content su-u-clearfix su-u-trim\">Let us assume the <strong>one-dimensional elastic collision<\/strong> of two objects, object A and object B. These two objects are moving with velocities <strong>v<sub>A<\/sub><\/strong> and<strong> v<sub>B<\/sub> <\/strong>along the x-axis before the collision. After the collision, their velocities are <b>v\u2019A<\/b>\u00a0and <b>v\u2019B<\/b>. The <strong>conservation of the total momentum<\/strong> demands that the total momentum before the collision is the same as the total momentum after the collision. Likewise, the <strong>conservation of the total kinetic energy<\/strong> demands that the total kinetic energy of both objects before the collision is the same as the total kinetic energy after the collision. Both laws may be expressed in equations as:<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/11\/equations-of-conservation-of-momentum-energy.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-15741\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/11\/equations-of-conservation-of-momentum-energy.png\" alt=\"equations-of-conservation-of-momentum-energy\" width=\"371\" height=\"149\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/11\/equations-of-conservation-of-momentum-energy.png 371w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/11\/equations-of-conservation-of-momentum-energy-300x120.png 300w\" sizes=\"(max-width: 371px) 100vw, 371px\" \/><\/a><\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/11\/solution-conservation-of-momentum-energy.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-15742\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/11\/solution-conservation-of-momentum-energy.png\" alt=\"solution-conservation-of-momentum-energy\" width=\"583\" height=\"340\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/11\/solution-conservation-of-momentum-energy.png 583w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/11\/solution-conservation-of-momentum-energy-300x175.png 300w\" sizes=\"(max-width: 583px) 100vw, 583px\" \/><\/a><\/p>\n<p><strong>The relative speed of the two objects<\/strong> after the collision has <strong>the same magnitude<\/strong> (but opposite direction) as before the collision, <strong>no matter what the masses are<\/strong>.<\/div><\/div><\/div><div   id="8n9s8rpp6h"    class=\"su-divider su-divider-style-dotted\" style=\"margin:15px 0;border-width:2px;border-color:#999999\"><\/div>\n<h2>Inelastic Collisions<\/h2>\n<p><strong>An inelastic collision<\/strong> is when part of the <strong>kinetic energy is changed<\/strong> to some other form of energy in the collision. Any macroscopic collision between objects will convert some kinetic energy into<strong> internal energy<\/strong> and other forms of energy, so <strong>no large-scale impacts are perfectly elastic<\/strong>. For example, in collisions of common bodies, such as two cars, some energy is always transferred from <strong>kinetic energy<\/strong> to other forms of energy, such as <strong>thermal energy<\/strong> or <strong>energy of sound<\/strong>. The inelastic collision of two bodies always involves a loss in the kinetic energy of the system. The greatest loss occurs if the bodies stick together, in which case the collision is called a <strong>completely inelastic collision<\/strong>. Thus, the <strong>kinetic energy<\/strong> of the system is <strong>not conserved<\/strong>, while the <strong>total energy is conserved<\/strong> as required by the general principle of conservation of energy. <strong>Momentum is conserved in inelastic collisions<\/strong>, but one cannot track the kinetic energy through the collision since some of it is converted to other forms of energy.<\/p>\n<p><strong>In nuclear physics<\/strong>, an inelastic collision is when the incoming particle causes the nucleus to strike to become excited or break up. Deep inelastic scattering is a method of probing the structure of subatomic particles in much the same way as Rutherford probed the inside of the atom (see Rutherford scattering).<\/p>\n<p><strong>In nuclear reactors<\/strong>, inelastic collisions are of importance in the <strong>neutron moderation<\/strong> process. An<a title=\"Neutron Inelastic Scattering\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/nuclear-engineering-fundamentals\/neutron-nuclear-reactions\/neutron-inelastic-scattering\/\"> inelastic scattering<\/a> plays an important role in slowing down neutrons, especially <strong>at high energies and by heavy nuclei<\/strong>. Inelastic scattering occurs above <strong>threshold energy<\/strong>. This threshold energy is higher than the energy of the first excited state of the target nucleus (due to the laws of conservation), and it is given by the following formula:<\/p>\n<p style=\"text-align: center;\"><strong>E<\/strong><strong><sub>t<\/sub><\/strong><strong> = ((A+1)\/A)* \u03b5<\/strong><strong><sub>1<\/sub><\/strong><\/p>\n<p>where <strong>E<\/strong><strong><sub>t<\/sub><\/strong> is the<strong> inelastic threshold energy,<\/strong> and <strong>\u03b5<\/strong><strong><sub>1<\/sub><\/strong> is the energy of the first excited state. Therefore, especially scattering data of <strong><sup>238<\/sup><\/strong><strong>U<\/strong>, a major component of <a href=\"http:\/\/sitepourvtc.com\/nuclear-power-plant\/nuclear-fuel\/\">nuclear fuel<\/a> in commercial power reactors, is one of the most important data in the neutron transport calculations in the <a href=\"http:\/\/sitepourvtc.com\/nuclear-power-plant\/nuclear-reactor-core\/\">reactor core<\/a>.<\/p>\n<div   id="8n9s8rpp6h"    class=\"su-accordion su-u-trim\"><div   id="8n9s8rpp6h"    class=\"su-spoiler su-spoiler-style-default su-spoiler-icon-plus su-spoiler-closed\" data-scroll-offset=\"0\"><div   id="8n9s8rpp6h"    class=\"su-spoiler-title\" tabindex=\"0\" role=\"button\"><span class=\"su-spoiler-icon\"><\/span>Example: Ballistic Pendulum<\/div><div   id="8n9s8rpp6h"    class=\"su-spoiler-content su-u-clearfix su-u-trim\">\n<figure id=\"attachment_15745\" aria-describedby=\"caption-attachment-15745\" style=\"width: 390px\" class=\"wp-caption alignright\"><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/11\/ballistic-pendulum-example.png\"><img loading=\"lazy\" class=\" wp-image-15745\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/11\/ballistic-pendulum-example.png\" alt=\"Ballistic Pendulum\" width=\"400\" height=\"294\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/11\/ballistic-pendulum-example.png 590w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/11\/ballistic-pendulum-example-300x221.png 300w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/11\/ballistic-pendulum-example-220x161.png 220w\" sizes=\"(max-width: 400px) 100vw, 400px\" \/><\/a><figcaption id=\"caption-attachment-15745\" class=\"wp-caption-text\">The ballistic pendulum is a kind of \u201ctransformer,\u201d exchanging the high speed of a light object (the bullet) for the low speed of a massive object.<\/figcaption><\/figure>\n<p><strong>A ballistic pendulum<\/strong> is a device for measuring the velocity of a projectile, such as a <strong>bullet<\/strong>. The ballistic pendulum is a kind of \u201ctransformer,\u201d exchanging the high speed of a light object (the bullet) for the low speed of a massive object (the block). When a bullet is fired into the block, its momentum is transferred to the block. The bullet\u2019s momentum can be determined from the<strong> amplitude of the pendulum swing<\/strong>.<\/p>\n<p>When the bullet is embedding itself in the block, it occurs so quickly that the block does not move appreciably. The supporting strings remain nearly vertical, so negligible external horizontal force acts on the bullet\u2013block system, and the<strong> horizontal<\/strong> component of <strong>momentum is conserved<\/strong>. However, <strong>mechanical\u00a0energy is not conserved<\/strong> during this stage because a <strong>nonconservative force<\/strong> does work (the force of friction between bullet and block).<\/p>\n<p>In the second stage, the bullet and block move together. The only forces acting on this system are gravity (a conservative force) and string tensions (which do not work). Thus, as the block swings, <strong>mechanical energy is conserved<\/strong>. <strong>Momentum is not conserved during this stage<\/strong> because there is a net external force (the forces of gravity and string tension don\u2019t cancel when the strings are inclined).<\/p>\n<h2><strong><b>Equations governing the ballistic pendulum<\/b><\/strong><\/h2>\n<p>In the first stage, <strong>momentum is conserved,<\/strong> and therefore:<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/11\/conservation-of-momentum-inelastic.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-15746\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/11\/conservation-of-momentum-inelastic.png\" alt=\"conservation-of-momentum-inelastic\" width=\"231\" height=\"121\" \/><\/a><\/p>\n<p>where<strong> v<\/strong> is the initial velocity of the projectile of mass <strong>m<sub>P<\/sub><\/strong>. <strong>v\u2019<\/strong> is the velocity of the block and embedded projectile (both of mass <strong>m<sub>P<\/sub> + m<sub>B<\/sub><\/strong>) just after the collision, before they have moved significantly.<\/p>\n<p>In the second stage, <strong>mechanical energy is conserved<\/strong>. We choose y = 0 when the pendulum hangs vertically and then y = h when the block and embedded projectile system reach maximum height. The system swings up and comes to rest for an instant at a height y, where its kinetic energy is zero, and the potential energy is <strong>(m<sub>P<\/sub> + m<sub>B<\/sub>)gh<\/strong>. Thus we write the law of conservation of energy:<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/11\/solution-inelastic-collision.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-15747\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/11\/solution-inelastic-collision.png\" alt=\"solution-inelastic-collision\" width=\"486\" height=\"243\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/11\/solution-inelastic-collision.png 486w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/11\/solution-inelastic-collision-300x150.png 300w\" sizes=\"(max-width: 486px) 100vw, 486px\" \/><\/a><\/p>\n<p>which is the initial velocity of the projectile and our final result.<\/p>\n<p><strong>When we use some realistic numbers:<\/strong><\/p>\n<ul>\n<li>m<sub>P<\/sub> = 5 g<\/li>\n<li>m<sub>B<\/sub> = 2 kg<\/li>\n<li>h = 3 cm<\/li>\n<li>v = ?<\/li>\n<\/ul>\n<p><strong>then we have:<\/strong><\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/11\/solution-example.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-15748\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/11\/solution-example.png\" alt=\"solution-example\" width=\"320\" height=\"52\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/11\/solution-example.png 320w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/11\/solution-example-300x49.png 300w\" sizes=\"(max-width: 320px) 100vw, 320px\" \/><\/a><\/div><\/div><\/div><div   id="8n9s8rpp6h"    class=\"su-divider su-divider-style-dotted\" style=\"margin:15px 0;border-width:2px;border-color:#999999\"><\/div>\n<h2>Conservation Laws in Nuclear Reactions<\/h2>\n<p>A <a title=\"Nuclear Reactions\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/nuclear-reactions\/\"><strong>nuclear reaction<\/strong><\/a> is considered to be the process in which two nuclear particles (two nuclei or a nucleus and a nucleon) interact to produce two or more nuclear particles or \u02e0-rays (<a href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/atomic-nuclear-physics\/fundamental-particles\/photon\/gamma-ray\/\">gamma rays<\/a>). Thus, a <strong>nuclear reaction<\/strong> must cause a transformation of at least one nuclide to another. Sometimes if a nucleus interacts with another nucleus or particle without changing the nature of any nuclide, the process is referred to as a <a href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/nuclear-engineering-fundamentals\/neutron-nuclear-reactions\/neutron-elastic-scattering\/elastic-scattering-cross-section\/\"><strong>nuclear scattering<\/strong><\/a>\u00a0rather than a nuclear reaction.<\/p>\n<p>In analyzing nuclear reactions, we apply the <strong>many conservation laws<\/strong>. <strong>Nuclear reactions<\/strong> are subject to classical <strong>conservation laws for a charge, momentum, angular momentum, and energy<\/strong> (including rest energies). \u00a0Not anticipated by classical physics, other conservation laws are <strong>electric charge<\/strong>, <strong>lepton number, and baryon number<\/strong>. Certain of these laws are obeyed under all circumstances, and others are not. We have accepted the conservation of energy and momentum. In all the examples given, we assume that the number of protons and the number of neutrons is separately conserved. We shall find circumstances and conditions in which this rule is not true. Where we are considering non-relativistic nuclear reactions, it is essentially true. However, we shall find that these principles must be extended when we consider relativistic nuclear energies or those involving weak interactions.<\/p>\n<p>Some conservation principles have arisen from theoretical considerations, and others are just empirical relationships. Notwithstanding, any reaction not expressly forbidden by the conservation laws will generally occur, if perhaps at a slow rate. This expectation is based on quantum mechanics. Unless the barrier between the initial and final states is infinitely high, there is always a <strong>non-zero probability<\/strong> that a system will make the transition between them.<\/p>\n<p>For the purposes of this article, it is sufficient to note four of the fundamental laws governing these reactions.<\/p>\n<ol>\n<li><strong>Conservation of nucleons<\/strong>. The total number of nucleons before and after a reaction are the same.<\/li>\n<li><strong>Conservation of charge<\/strong>. The sum of the charges on all the particles before and after a reaction are the same.<\/li>\n<li><strong>Conservation of momentum<\/strong>. The total momentum of the interacting particles before and after a reaction is the same.<\/li>\n<li><strong>Conservation of energy<\/strong>. Energy, including rest mass energy, is conserved in nuclear reactions.<\/li>\n<\/ol>\n<div   id="8n9s8rpp6h"    class=\"su-accordion su-u-trim\"><div   id="8n9s8rpp6h"    class=\"su-spoiler su-spoiler-style-default su-spoiler-icon-plus su-spoiler-closed\" data-scroll-offset=\"0\"><div   id="8n9s8rpp6h"    class=\"su-spoiler-title\" tabindex=\"0\" role=\"button\"><span class=\"su-spoiler-icon\"><\/span>Types of Nuclear Reactions<\/div><div   id="8n9s8rpp6h"    class=\"su-spoiler-content su-u-clearfix su-u-trim\">Although the number of possible\u00a0<a title=\"Nuclear Reactions\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/nuclear-reactions\/\"><strong>nuclear reactions<\/strong><\/a> is enormous, nuclear reactions can be sorted by types. Most nuclear reactions are accompanied by gamma emissions. Some examples are:<\/p>\n<ul>\n<li><a href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/nuclear-engineering-fundamentals\/neutron-nuclear-reactions\/neutron-elastic-scattering\/\"><strong>Elastic scattering<\/strong><\/a>. In this process, the kinetic energy of a particle is conserved in the center-of-mass frame, but its direction of propagation is modified (by interaction with other particles and\/or potentials) meaning the two particles in the collision do not lose energy.<\/li>\n<\/ul>\n<p><strong>\u00a0208Pb (n, n) 208Pb<\/strong><\/p>\n<ul>\n<li><strong>\u00a0<a href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/nuclear-engineering-fundamentals\/neutron-nuclear-reactions\/neutron-inelastic-scattering\/\">Inelastic scattering<\/a><\/strong>. Occurs when energy is transferred. The difference of kinetic energies is saved in an excited nuclide.<\/li>\n<\/ul>\n<p><strong>\u00a040Ca (\u03b1, \u03b1\u2019) 40mCa<\/strong><\/p>\n<ul>\n<li><a href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/nuclear-engineering-fundamentals\/neutron-nuclear-reactions\/neutron-capture-radiative-capture\/\"><strong>Capture reactions<\/strong><\/a>. Nuclei can capture both charged and neutral particles. This is accompanied by the emission of <a href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/atomic-nuclear-physics\/fundamental-particles\/photon\/gamma-ray\/\">\u02e0-rays<\/a>. Neutron capture reaction produces radioactive nuclides (induced radioactivity).<\/li>\n<\/ul>\n<p><strong>\u00a0238U (n, \u02e0) 239U<\/strong><\/p>\n<ul>\n<li><strong>Transfer\u00a0Reactions<\/strong>. The absorption of a particle accompanied by the emission of one or more particles is called the\u00a0transfer\u00a0reaction.<\/li>\n<\/ul>\n<p><strong>4He (\u03b1, p) 7Li<\/strong><\/p>\n<ul>\n<li><a href=\"http:\/\/sitepourvtc.com\/nuclear-power\/fission\/\"><strong>Fission reactions<\/strong><\/a>. Nuclear fission is a nuclear reaction in which the nucleus of an atom splits into smaller parts (lighter nuclei). The fission process often produces\u00a0<a href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/atomic-nuclear-physics\/fundamental-particles\/neutron\/free-neutron\/\">free <\/a><a href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/atomic-nuclear-physics\/fundamental-particles\/neutron\/free-neutron\/\">neutrons<\/a> and photons (in the form of gamma rays) and releases a large amount of energy.<\/li>\n<\/ul>\n<p><strong><a href=\"http:\/\/sitepourvtc.com\/nuclear-power-plant\/nuclear-fuel\/uranium\/uranium-235\/\">235U<\/a>\u00a0(n, 3 n)\u00a0<em>fission products<\/em><\/strong><\/p>\n<ul>\n<li><strong>Fusion reactions<\/strong>. \u00a0Occur when two or more atomic nuclei collide at a very high speed and join to form a new type of atomic nucleus. The fusion reaction of deuterium and tritium is particularly interesting because of its potential of providing energy for the future.<\/li>\n<\/ul>\n<p><strong>3T (d, n) 4He<\/strong><\/p>\n<ul>\n<li><strong>Spallation reactions<\/strong>. Occur occurs when a particle hits a nucleus with sufficient energy and momentum to knock out several small fragments or smash them into many fragments.<\/li>\n<\/ul>\n<ul>\n<li><strong>Nuclear decay<\/strong>\u00a0(<strong>Radioactive decay<\/strong>). Occurs when an unstable atom loses energy by emitting\u00a0<a href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/atomic-nuclear-physics\/radiation\/\">ionizing radiation<\/a>. Radioactive decay is a random process at the level of single atoms, in that, according to quantum theory, it is impossible to predict when a particular atom will decay. There are many types of radioactive decay:\n<ul>\n<li><strong>Alpha radioactivity<\/strong>. <a title=\"Alpha Particle\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/atomic-nuclear-physics\/fundamental-particles\/alpha-particle\/\">Alpha particles<\/a> consist of two protons and two neutrons bound together into a particle identical to a helium nucleus. Because of its very large mass (more than 7000 times the mass of the beta particle) and its charge, it heavily ionizes material and has a very short range.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<ul>\n<li style=\"list-style-type: none;\">\n<ul>\n<li><strong>Beta radioactivity<\/strong>. <a title=\"Beta Particle\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/atomic-nuclear-physics\/fundamental-particles\/beta-particle\/\">Beta particles<\/a> are high-energy, high-speed electrons or positrons emitted by certain types of radioactive nuclei such as potassium-40. The beta particles have a greater range of penetration than alpha particles but still much less than gamma rays. The beta particles emitted are a form of ionizing radiation, also known as beta rays. The production of beta particles is termed beta decay.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<ul>\n<li style=\"list-style-type: none;\">\n<ul>\n<li><strong>Gamma radioactivity<\/strong>. <a title=\"Gamma Rays \/ Gamma Radiation\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/atomic-nuclear-physics\/fundamental-particles\/photon\/gamma-ray\/\">Gamma rays<\/a> are electromagnetic radiation of a very high frequency and are therefore high-energy photons. Nuclei decay produces them as they transition from a high-energy state to a lower state known as gamma decay. Most nuclear reactions are accompanied by gamma emissions.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<ul>\n<li style=\"list-style-type: none;\">\n<ul>\n<li><strong><a href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/nuclear-engineering-fundamentals\/neutron-nuclear-reactions\/neutron-emission\/\">Neutron emission<\/a>.\u00a0<\/strong>Neutron emission is a type of radioactive decay of nuclei containing excess neutrons (especially fission products), in which a neutron is simply ejected from the nucleus. This type of radiation plays a key role in nuclear reactor control because these neutrons are <a href=\"http:\/\/sitepourvtc.com\/nuclear-power\/fission\/delayed-neutrons\/\">delayed neutrons<\/a>.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/div><\/div><\/div>\n<h2>Conservation of Momentum in Nuclear Reactions &#8211; Neutron Moderation<\/h2>\n<p>See also: <a title=\"Neutron Elastic Scattering\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/nuclear-engineering-fundamentals\/neutron-nuclear-reactions\/neutron-elastic-scattering\/\">Neutron Elastic Scattering<\/a><\/p>\n<p>See also: <a title=\"Neutron Inelastic Scattering\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/nuclear-engineering-fundamentals\/neutron-nuclear-reactions\/neutron-inelastic-scattering\/\">Neutron Inelastic Scattering<\/a><\/p>\n<p>See also: <a title=\"Neutron Moderator\" href=\"http:\/\/sitepourvtc.com\/neutron-moderator\/\">Neutron Moderators<\/a><\/p>\n<p>It is known the fission neutrons are of importance in any <a title=\"Nuclear Fission Chain Reaction\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/nuclear-fission-chain-reaction\/\">chain-reacting system<\/a>. All <a title=\"Neutron\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/atomic-nuclear-physics\/fundamental-particles\/neutron\/\">neutrons<\/a> produced by fission are born as <strong>fast neutrons<\/strong> with high kinetic energy. Before such neutrons can efficiently cause additional fissions, they must be slowed down by collisions with nuclei in the reactor\u2019s moderator. The probability of the fission <a href=\"http:\/\/sitepourvtc.com\/nuclear-power-plant\/nuclear-fuel\/uranium\/uranium-235\/\">U-235<\/a> becomes very large <strong>at the thermal energies<\/strong> of slow neutrons. This fact implies the increase of the <a title=\"Six Factor Formula \u2013 Effective Multiplication Factor\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/nuclear-fission-chain-reaction\/six-factor-formula-effective-multiplication-factor\/\">multiplication factor<\/a> of the reactor (i.e., lower fuel enrichment is needed to sustain a chain reaction).<\/p>\n<p>The neutrons released during fission with an average energy of <strong>2 MeV<\/strong> in a reactor on average undergo <strong>many\u00a0collisions<\/strong> (<a href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/nuclear-engineering-fundamentals\/neutron-nuclear-reactions\/neutron-elastic-scattering\/\">elastic<\/a> or <a href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/nuclear-engineering-fundamentals\/neutron-nuclear-reactions\/neutron-inelastic-scattering\/\">inelastic<\/a>) before they are absorbed. During the scattering reaction, a fraction of the neutron\u2019s kinetic energy <strong>is transferred to the nucleus<\/strong>. It is possible to derive the following equation for the mass of target or <a href=\"http:\/\/sitepourvtc.com\/neutron-moderator\/\">moderator<\/a> nucleus (M), the energy of incident <a href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/atomic-nuclear-physics\/fundamental-particles\/neutron\/\">neutron<\/a> (E<sub>i<\/sub>), and the energy of scattered neutron (E<sub>s<\/sub>) using the laws of <strong>conservation of momentum and energy<\/strong> and the analogy of collisions of billiard balls for elastic scattering.<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/11\/equation-momentum-energy.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-13057\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/11\/equation-momentum-energy.png\" alt=\"equation momentum energy\" width=\"493\" height=\"98\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/11\/equation-momentum-energy.png 493w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/11\/equation-momentum-energy-300x60.png 300w\" sizes=\"(max-width: 493px) 100vw, 493px\" \/><\/a><\/p>\n<p>where A is the atomic mass number, in case of the <strong>hydrogen (A = 1)<\/strong> as the target nucleus, the incident neutron <strong>can be completely stopped<\/strong>. But this works when the direction of the neutron is completely reversed (i.e., scattered at 180\u00b0). In reality, the direction of scattering ranges from 0 to 180 \u00b0, and the energy transferred also ranges from 0% to maximum. Therefore, the average energy of the scattered neutron is taken as the average of energies with scattering angles 0 and 180\u00b0.<\/p>\n<p>Moreover, it is useful to work <strong>with logarithmic quantities<\/strong>. Therefore\u00a0one defines <strong><a title=\"Average Logarithmic Energy Decrement\" href=\"http:\/\/sitepourvtc.com\/glossary\/neutron-moderatoraverage-logarithmic-energy-decrement\/\">the logarithmic energy decrement per collision<\/a> (\u03be)<\/strong> as a key material constant describing energy transfers during a neutron slowing down. \u03be is not dependent on energy, only on A and is defined as follows:<a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/11\/logarithmic-energy-decrement-equation.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-13059\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/11\/logarithmic-energy-decrement-equation.png\" alt=\"logarithmic energy decrement - equation\" width=\"742\" height=\"272\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/11\/logarithmic-energy-decrement-equation.png 742w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/11\/logarithmic-energy-decrement-equation-300x110.png 300w\" sizes=\"(max-width: 742px) 100vw, 742px\" \/><\/a>For heavy target nuclei,<strong> \u03be<\/strong> may be approximated by the following formula:<a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/11\/logarithmic-energy-decrement-heavy-nuclei.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-13060\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/11\/logarithmic-energy-decrement-heavy-nuclei.png\" alt=\"the logarithmic energy decrement per collision\" width=\"684\" height=\"663\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/11\/logarithmic-energy-decrement-heavy-nuclei.png 684w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/11\/logarithmic-energy-decrement-heavy-nuclei-300x291.png 300w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/11\/logarithmic-energy-decrement-heavy-nuclei-52x50.png 52w\" sizes=\"(max-width: 684px) 100vw, 684px\" \/><\/a>From these equations, it is easy to determine the number of collisions required to slow down a neutron from, for example from <strong>2 MeV to 1 eV<\/strong>.<\/p>\n<p>Example: Determine the number of collisions required for thermalization for the 2 MeV neutrons in the carbon.<\/p>\n<p>\u03be<sub>CARBON<\/sub> = 0.158<\/p>\n<p>N(<strong>2MeV \u2192 1eV<\/strong>) = ln 2\u22c510<sup>6<\/sup>\/\u03be =14.5\/0.158 = <strong>92<\/strong><\/p>\n<figure id=\"attachment_13062\" aria-describedby=\"caption-attachment-13062\" style=\"width: 630px\" class=\"wp-caption aligncenter\"><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/11\/logarithmic-energy-decrement-table.png\"><img loading=\"lazy\" class=\"wp-image-13062 size-full\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/11\/logarithmic-energy-decrement-table.png\" alt=\"Table of average logarithmic energy decrement for some elements\" width=\"640\" height=\"538\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/11\/logarithmic-energy-decrement-table.png 640w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/11\/logarithmic-energy-decrement-table-300x252.png 300w\" sizes=\"(max-width: 640px) 100vw, 640px\" \/><\/a><figcaption id=\"caption-attachment-13062\" class=\"wp-caption-text\">Table of average logarithmic energy decrement for some elements.<\/figcaption><\/figure>\n<p>For a mixture of isotopes:<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/11\/logarithmic-energy-decrement-mixtures.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-13061\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/11\/logarithmic-energy-decrement-mixtures.png\" alt=\"the logarithmic energy decrement for mixtures\" width=\"230\" height=\"94\" \/><\/a><\/p>\n<div   id="8n9s8rpp6h"    class=\"su-accordion su-u-trim\"><div   id="8n9s8rpp6h"    class=\"su-spoiler su-spoiler-style-default su-spoiler-icon-plus su-spoiler-closed\" data-scroll-offset=\"0\"><div   id="8n9s8rpp6h"    class=\"su-spoiler-title\" tabindex=\"0\" role=\"button\"><span class=\"su-spoiler-icon\"><\/span>Example: Elastic Nuclear Collision<\/div><div   id="8n9s8rpp6h"    class=\"su-spoiler-content su-u-clearfix su-u-trim\"><strong>A neutron (n)<\/strong> of mass <strong>1.01 u<\/strong> traveling with a speed of <strong>3.60 x\u00a010<sup>4<\/sup>m\/s<\/strong> interacts with a <strong>carbon (C)<\/strong> nucleus (<strong>m<sub>C<\/sub> = 12.00 u<\/strong>) initially at rest in an <strong>elastic head-on collision<\/strong>.<\/p>\n<p>What are the velocities of the neutron and carbon nucleus after the collision?<\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>This is an <strong>elastic head-on collision<\/strong> of two objects with <strong>unequal masses<\/strong>. We have to use the conservation laws of momentum and kinetic energy and apply them to our system of two particles.<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/11\/conservation-laws-elastic-collisions.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-15743\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/11\/conservation-laws-elastic-collisions.png\" alt=\"conservation-laws-elastic-collisions\" width=\"337\" height=\"259\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/11\/conservation-laws-elastic-collisions.png 337w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/11\/conservation-laws-elastic-collisions-300x231.png 300w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/11\/conservation-laws-elastic-collisions-180x138.png 180w\" sizes=\"(max-width: 337px) 100vw, 337px\" \/><\/a><\/p>\n<p>We can solve this system of equations or use the equation derived in the previous section. This equation stated that the relative speed of the two objects after the collision has the same magnitude (but opposite direction) as before the collision, no matter what the masses are.<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/11\/solution-elastic-collision.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-15744\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/11\/solution-elastic-collision.png\" alt=\"solution-elastic-collision\" width=\"446\" height=\"357\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/11\/solution-elastic-collision.png 446w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/11\/solution-elastic-collision-300x240.png 300w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/11\/solution-elastic-collision-177x142.png 177w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/11\/solution-elastic-collision-147x118.png 147w\" sizes=\"(max-width: 446px) 100vw, 446px\" \/><\/a><\/p>\n<p>The minus sign for v\u2019 tells us that the <strong>neutron scatters back<\/strong> of the carbon nucleus because the carbon nucleus is significantly heavier. On the other hand, <strong>its speed is less<\/strong> than its initial speed. This process is known as <strong>neutron moderation,<\/strong> and it significantly depends on the mass of moderator nuclei.<\/div><\/div><\/div><div   id="8n9s8rpp6h"    class=\"su-divider su-divider-style-dotted\" style=\"margin:15px 0;border-width:2px;border-color:#999999\"><\/div>\n<h2>The momentum of a Photon<\/h2>\n<p><strong>A<a title=\"Photon \u2013 Fundamental Particle\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/atomic-nuclear-physics\/fundamental-particles\/photon\/\"> photon<\/a>, the quantum of electromagnetic radiation<\/strong>, \u00a0is an <a title=\"Fundamental Particles\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/atomic-nuclear-physics\/fundamental-particles\/\">elementary particle<\/a>, the force carrier of the electromagnetic force. The modern photon concept was developed (1905) by <strong>Albert Einstein<\/strong> to explain the <a title=\"Photoelectric Effect\" href=\"http:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/interaction-radiation-matter\/interaction-gamma-radiation-matter\/photoelectric-effect\/\">photoelectric effect<\/a>, in which he proposed the existence of discrete energy packets during the transmission of light.<\/p>\n<p>In 1916, Einstein extended his concept of light quanta (photons) by proposing that a <strong>quantum of light has linear momentum<\/strong>. Although a photon is <strong>massless<\/strong>, it has momentum, which is related to its energy E, frequency f, and wavelength by:<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/11\/momentum-photon-equation.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-15901\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/11\/momentum-photon-equation.png\" alt=\"momentum-photon-equation\" width=\"101\" height=\"48\" \/><\/a><\/p>\n<p>Thus, when a photon interacts with another object, <strong>energy and momentum are transferred<\/strong>, as if there were a collision between the photon and matter in the classical sense.<\/p>\n<div   id="8n9s8rpp6h"    class=\"su-accordion su-u-trim\"><div   id="8n9s8rpp6h"    class=\"su-spoiler su-spoiler-style-default su-spoiler-icon-plus su-spoiler-closed\" data-scroll-offset=\"0\"><div   id="8n9s8rpp6h"    class=\"su-spoiler-title\" tabindex=\"0\" role=\"button\"><span class=\"su-spoiler-icon\"><\/span>Example: Momentum and Force of a Photon <\/div><div   id="8n9s8rpp6h"    class=\"su-spoiler-content su-u-clearfix su-u-trim\">Suppose the <strong>10<sup>19<\/sup> photons<\/strong> (let \u03bb = 650 nm) emitted per second from the<strong> 100 W lightbulb<\/strong>. \u00a0Suppose all photons are focused onto a piece of black paper and absorbed. Assume that the momentum of a photon changes <strong>from p = h\/\u03bb to zero<\/strong>.<\/p>\n<p><strong>Calculate the momentum<\/strong> of one photon and calculate the <strong>force<\/strong> all these photons could exert on the paper.<\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>We use the formula of momentum of a single photon:<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/11\/momentum-photon-example.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-15902\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/11\/momentum-photon-example.png\" alt=\"momentum-photon-example\" width=\"368\" height=\"153\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/11\/momentum-photon-example.png 368w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/11\/momentum-photon-example-300x125.png 300w\" sizes=\"(max-width: 368px) 100vw, 368px\" \/><\/a><\/div><\/div><\/div>\n<h2>The momentum of a Photon &#8211; Compton Scattering<\/h2>\n<figure id=\"attachment_11831\" aria-describedby=\"caption-attachment-11831\" style=\"width: 334px\" class=\"wp-caption alignright\"><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/04\/compton-scattering.gif\"><img loading=\"lazy\" class=\" wp-image-11831\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/04\/compton-scattering.gif\" alt=\"Compton Scattering\" width=\"344\" height=\"248\" \/><\/a><figcaption id=\"caption-attachment-11831\" class=\"wp-caption-text\">In Compton scattering, the incident gamma-ray photon is deflected through an angle \u0398 with respect to its original direction. This deflection results in a decrease in the photon\u2019s frequency\u2019s energy (decrease in photon\u2019s frequency) and is called the Compton effect.<br \/>Source: hyperphysics.phy-astr.gsu.edu<\/figcaption><\/figure>\n<p>The Compton formula was published in 1923 in the Physical Review. Compton explained that the X-ray shift is caused by the particle-like momentum of photons. Compton scattering formula is the mathematical relationship between the shift in wavelength and the scattering angle of the X-rays. In the case of Compton scattering, the photon of frequency <em>f<\/em> collides with an electron at rest. The photon bounces off the electron upon collision, giving up some of its initial energy (given by Planck\u2019s formula E=hf). While the electron gains momentum (mass x velocity), the <strong>photon cannot lower its velocity<\/strong>. As a result of momentum conservation law, the photon must lower its momentum given by:<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/04\/Photon-momentum-formula.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-11869\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/04\/Photon-momentum-formula.png\" alt=\"As a result of momentum conservetion law, the photon must lower its momentum given by this formula.\" width=\"177\" height=\"59\" \/><\/a><\/p>\n<p>So the decrease in photon\u2019s momentum must be translated into a <strong>decrease in frequency<\/strong>\u00a0(increase in wavelength \u0394<strong>\u03bb = \u03bb\u2019 \u2013 \u03bb<\/strong>). The shift of the wavelength increased with scattering angle according to\u00a0<strong>the Compton formula<\/strong>:<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/04\/Compton-Scattering-Formula.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-11870\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/04\/Compton-Scattering-Formula.png\" alt=\"The shift of the wavelength increased with scattering angle according to the Compton formula\" width=\"236\" height=\"64\" \/><\/a><\/p>\n<p>where <strong>\u03bb<\/strong>\u00a0is the initial wavelength of photon <strong>\u03bb\u2019<\/strong>\u00a0is the wavelength after scattering, <strong>h\u00a0<\/strong>is the Planck constant = 6.626 x 10<sup>-34<\/sup>\u00a0J.s, <strong>m<sub>e<\/sub><\/strong>\u00a0is the electron rest mass (0.511 MeV)<strong>c<\/strong>\u00a0is the speed of light <strong>\u0398<\/strong>\u00a0is the scattering angle. The minimum change in wavelength (<em>\u03bb\u2032<\/em>\u00a0\u2212\u00a0<em>\u03bb<\/em>) for the photon occurs when \u0398 = 0\u00b0 (cos(\u0398)=1) and is at least zero. The maximum change in wavelength (<em>\u03bb\u2032<\/em>\u00a0\u2212\u00a0<em>\u03bb<\/em>) for the photon occurs when \u0398 = 180\u00b0 (cos(\u0398)=-1). In this case, the photon transfers to the electron as much momentum as possible. The maximum change in wavelength can be derived from the Compton formula:<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/04\/Compton-length.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-11867\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/04\/Compton-length.png\" alt=\"The maximum change in wavelength can be derived from Compton formula. Compton length\" width=\"561\" height=\"78\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/04\/Compton-length.png 561w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2015\/04\/Compton-length-300x42.png 300w\" sizes=\"(max-width: 561px) 100vw, 561px\" \/><\/a><\/p>\n<p>The quantity h\/m<sub>e<\/sub>c is known as the\u00a0<strong>Compton wavelength<\/strong>\u00a0of the electron and is equal to\u00a0<strong>2.43\u00d710<sup>\u221212<\/sup><\/strong><strong>\u00a0<\/strong><strong>m<\/strong>.<\/p>\n<h2>Conservation of Momentum in Fluid Dynamics<\/h2>\n<p><strong>In general<\/strong>, the <strong>law of conservation of momentum<\/strong> or principle of momentum conservation states that the momentum of an isolated system is a <strong>constant<\/strong>. The vector sum of the momenta (momentum is equal to the mass of an object multiplied by its velocity) of all the objects of a system cannot be changed by interactions within the system. In classical mechanics, this law is implied by <strong>Newton\u2019s laws<\/strong>.<\/p>\n<p><strong>In fluid dynamics<\/strong>, the analysis of motion is performed in the same way as in solid mechanics &#8211; using <strong>Newton\u2019s laws of motion<\/strong>. As can be seen, moving fluids are exerting forces. The <a title=\"Lift Force\" href=\"http:\/\/sitepourvtc.com\/nuclear-engineering\/fluid-dynamics\/bernoullis-equation-bernoullis-principle\/lift-force\/\">lift force <\/a>on an aircraft is exerted by the air moving\u00a0over the wing.<\/p>\n<p>A jet of water from a hose exerts a force on whatever it hits. But here, it is not clear <strong>what mass of moving fluid<\/strong> we should use, and therefore it is necessary to use a different form of the equation.<\/p>\n<p><strong>Newton\u2019s 2nd Law can be written:<\/strong><\/p>\n<div   id="8n9s8rpp6h"    class=\"su-quote su-quote-style-default\"><div   id="8n9s8rpp6h"    class=\"su-quote-inner su-u-clearfix su-u-trim\">The rate of change of momentum of a body is equal to the resultant force acting on the body and takes place in the direction of the force.<\/div><\/div><div   id="8n9s8rpp6h"    class=\"su-divider su-divider-style-dotted\" style=\"margin:15px 0;border-width:2px;border-color:#999999\"><\/div><div   id="8n9s8rpp6h"    class=\"su-spacer\" style=\"height:15px\"><\/div><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/05\/Conservation-of-Momentum-Fluids-min-2.png\"><img loading=\"lazy\" class=\"alignright size-medium wp-image-14358\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/05\/Conservation-of-Momentum-Fluids-min-2-300x210.png\" alt=\"Conservation of Momentum - Fluids\" width=\"300\" height=\"210\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/05\/Conservation-of-Momentum-Fluids-min-2-300x210.png 300w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/05\/Conservation-of-Momentum-Fluids-min-2.png 1021w\" sizes=\"(max-width: 300px) 100vw, 300px\" \/><\/a>We assume fluid to be both <strong>steady<\/strong> and <strong>incompressible<\/strong>. To determine the rate of change of momentum for a fluid, we will consider a stream tube (<strong>control volume<\/strong>) as we did for the <a title=\"Bernoulli\u2019s Equation \u2013 Bernoulli\u2019s Principle\" href=\"http:\/\/sitepourvtc.com\/nuclear-engineering\/fluid-dynamics\/bernoullis-equation-bernoullis-principle\/\">Bernoulli equation<\/a>. In this<strong> control volume,<\/strong> any change in momentum of the fluid within a control volume is due to the action of external forces on the fluid within the volume.<\/p>\n<p>As can be seen from the picture, the <strong>control volume method<\/strong> can analyze the law of conservation of momentum in the fluid. The control volume is an <strong>imaginary surface<\/strong> enclosing a volume of interest. The control volume can be fixed or moving, and it can be rigid or deformable. To determine all forces acting on the surfaces of the control volume, we have to solve the conservation laws in this control volume.<\/p>\n<p>The first conservation equation we have to consider in the control volume is the <a title=\"Continuity Equation\" href=\"http:\/\/sitepourvtc.com\/nuclear-engineering\/fluid-dynamics\/continuity-equation\/\">continuity equation<\/a> (<a title=\"Law of Conservation of Matter\" href=\"http:\/\/sitepourvtc.com\/laws-of-conservation\/law-of-conservation-of-matter\/\">the law of conservation of matter<\/a>). In the simplest form, it is represented by the following equation:<\/p>\n<p style=\"text-align: center;\"><span style=\"font-size: 30px;\">\u2211m\u0307<sub>in<\/sub> = \u2211m\u0307<sub>out<\/sub><\/span><\/p>\n<p style=\"text-align: center;\">Sum of mass flow rates entering per unit time = Sum of mass flow rates leaving per unit time<\/p>\n<p><strong><span style=\"font-weight: 400;\">The second conservation equation we have to consider in the control volume is the<strong> momentum formula<\/strong>.<\/span><\/strong><\/p>\n<h2>Momentum Formula<\/h2>\n<p>In the simplest form, the <strong>momentum formula<\/strong> can be represented by the following equation:<\/p>\n<p>&#8211;<br \/>\n<a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/05\/Momentum-Formula-fluids.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-14362\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/05\/Momentum-Formula-fluids.png\" alt=\"Momentum Formula - fluids\" width=\"559\" height=\"185\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/05\/Momentum-Formula-fluids.png 559w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/05\/Momentum-Formula-fluids-300x99.png 300w\" sizes=\"(max-width: 559px) 100vw, 559px\" \/><\/a><\/p>\n<h2><\/h2>\n<h2>Choosing a Control Volume<\/h2>\n<p><strong><span style=\"font-weight: 400;\">A control volume can be selected as any arbitrary volume through which fluid flows. This volume can be static, moving, and even deforming during flow. To solve any problem, we have to solve basic conservation laws in this volume. It is very important to know all relative flow velocities to the control surface. Therefore it is very important to define exactly the boundaries of the control volume during an analysis.<\/span><\/strong><br \/>\n<div   id="8n9s8rpp6h"    class=\"su-accordion su-u-trim\"><div   id="8n9s8rpp6h"    class=\"su-spoiler su-spoiler-style-default su-spoiler-icon-plus su-spoiler-closed\" data-scroll-offset=\"0\"><div   id="8n9s8rpp6h"    class=\"su-spoiler-title\" tabindex=\"0\" role=\"button\"><span class=\"su-spoiler-icon\"><\/span>Example: The force acting on a deflector elbow<\/div><div   id="8n9s8rpp6h"    class=\"su-spoiler-content su-u-clearfix su-u-trim\"><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/05\/Momentum-equation-control-volume.png\"><img loading=\"lazy\" class=\"alignright size-medium wp-image-14368\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/05\/Momentum-equation-control-volume-300x255.png\" alt=\"Momentum equation - control volume\" width=\"300\" height=\"255\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/05\/Momentum-equation-control-volume-300x255.png 300w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/05\/Momentum-equation-control-volume.png 433w\" sizes=\"(max-width: 300px) 100vw, 300px\" \/><\/a><strong>An elbow<\/strong> (let say of primary piping) is used to deflect water flow at a velocity of <strong>17 m\/s<\/strong>. The piping diameter is equal to<strong> 700 mm<\/strong>. The gauge pressure inside the pipe is about <strong>16 MPa<\/strong> at a temperature of 290\u00b0C. Fluid is of constant density <strong>\u2374 ~ 720 kg\/m<sup>3<\/sup><\/strong> (at 290\u00b0C). The angle of the elbow is <strong>45\u00b0<\/strong>.<\/p>\n<p>Calculate the <strong>force on the wall<\/strong> of a deflector elbow (i.e., calculate vector F<sub>3<\/sub>).<\/p>\n<p><strong>Assumptions:<\/strong><\/p>\n<ul>\n<li>The flow is steady.<\/li>\n<li>The frictional losses are negligible.<\/li>\n<li>The weight of the elbow is negligible.<\/li>\n<li>The weight of the water in the elbow is negligible.<\/li>\n<\/ul>\n<p>We take the elbow as the <strong>control volume<\/strong>. The control volume is shown in the picture. The momentum equation is a vector equation, so it has three components. We take the x- and z- coordinates as shown, and we will solve the problem separately according to these coordinates.<\/p>\n<p>First, let us consider the component in the <strong>x-coordinate<\/strong>. The conservation of linear momentum equation becomes:<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/05\/Momentum-equation-x-coordinate.png\"><img loading=\"lazy\" class=\" size-full wp-image-14367 alignleft\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/05\/Momentum-equation-x-coordinate.png\" alt=\"Momentum equation - x-coordinate\" width=\"603\" height=\"371\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/05\/Momentum-equation-x-coordinate.png 603w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/05\/Momentum-equation-x-coordinate-300x185.png 300w\" sizes=\"(max-width: 603px) 100vw, 603px\" \/><\/a><\/p>\n<p>Second, let us consider the component in the <strong>y-coordinate<\/strong>. The conservation of linear momentum equation becomes:<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/05\/Momentum-equation-y-coordinate.png\"><img loading=\"lazy\" class=\" size-full wp-image-14365 alignleft\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/05\/Momentum-equation-y-coordinate.png\" alt=\"Momentum equation - y-coordinate\" width=\"566\" height=\"383\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/05\/Momentum-equation-y-coordinate.png 566w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/05\/Momentum-equation-y-coordinate-300x203.png 300w\" sizes=\"(max-width: 566px) 100vw, 566px\" \/><\/a><\/p>\n<p>The final force acting on the wall of a deflector elbow will be:<\/p>\n<p><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/05\/Momentum-equation-resulting-force.png\"><img loading=\"lazy\" class=\" size-full wp-image-14366 alignleft\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/05\/Momentum-equation-resulting-force.png\" alt=\"Momentum equation - resulting force\" width=\"294\" height=\"51\" \/><\/a><\/div><\/div><div   id="8n9s8rpp6h"    class=\"su-spoiler su-spoiler-style-default su-spoiler-icon-plus su-spoiler-closed\" data-scroll-offset=\"0\"><div   id="8n9s8rpp6h"    class=\"su-spoiler-title\" tabindex=\"0\" role=\"button\"><span class=\"su-spoiler-icon\"><\/span>Example: Water Jet Striking a Stationary Plate<\/div><div   id="8n9s8rpp6h"    class=\"su-spoiler-content su-u-clearfix su-u-trim\"><strong><a href=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/05\/Momentum-Equation-Water-Jet.png\"><img loading=\"lazy\" class=\"alignright size-medium wp-image-14373\" src=\"http:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/05\/Momentum-Equation-Water-Jet-300x218.png\" alt=\"Momentum Equation - Water Jet\" width=\"300\" height=\"218\" srcset=\"https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/05\/Momentum-Equation-Water-Jet-300x218.png 300w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/05\/Momentum-Equation-Water-Jet-220x161.png 220w, https:\/\/sitepourvtc.com\/wp-content\/uploads\/2016\/05\/Momentum-Equation-Water-Jet.png 356w\" sizes=\"(max-width: 300px) 100vw, 300px\" \/><\/a>A stationary plate<\/strong> (e.g.,, blade of a watermill) is used to deflect water flow at a velocity of <strong>1 m\/s<\/strong> and an angle of <strong>90\u00b0<\/strong>. It occurs at atmospheric pressure, and the mass flow rate is equal to <strong>Q =1 m<sup>3<\/sup>\/s<\/strong>.<\/p>\n<ol>\n<li><strong>Calculate the pressure force.<\/strong><\/li>\n<li><strong>Calculate the body force.<\/strong><\/li>\n<li><strong>Calculate the total force.<\/strong><\/li>\n<li><strong>Calculate the resultant force.<\/strong><\/li>\n<\/ol>\n<p><strong>Solution<\/strong><\/p>\n<ol>\n<li>The <strong>pressure force<\/strong> is zero as the pressure at the inlet, and the outlets to the control volume are atmospheric.<\/li>\n<li>As the control volume is small, we can ignore the <strong>body force<\/strong> due to the weight of gravity.<\/li>\n<li><strong>F<sub>x<\/sub><\/strong> = \u03c1.Q.(w<sub>1x<\/sub> &#8211; w<sub>2x<\/sub>) = 1000 . 1 . (1 &#8211; 0) = <strong>1000 N<\/strong><br \/>\n<strong>F<sub>y<\/sub><\/strong> = <strong>0<\/strong><br \/>\n<strong>F<\/strong> = <strong>(1000, 0)<\/strong><\/li>\n<li>The <strong>resultant force<\/strong> on the plane is the same magnitude but in the opposite direction as the total force <strong>F<\/strong> (friction and weight\u00a0are\u00a0neglected).<\/li>\n<\/ol>\n<p>The water jet exerts on the plate the force of 1000 N in the x-direction.<\/div><\/div><\/div>\n<\/div>\n<div   id="8n9s8rpp6h"    class=\"su-divider su-divider-style-dotted\" style=\"margin:20px 0;border-width:2px;border-color:#999999\"><\/div>\n<div   id="8n9s8rpp6h"     class=\"lgc-column lgc-grid-parent lgc-grid-100 lgc-tablet-grid-100 lgc-mobile-grid-100 lgc-equal-heights \"><div   id="8n9s8rpp6h"     class=\"inside-grid-column\">\u00a0 <div   id="8n9s8rpp6h"    class=\"su-accordion su-u-trim\"><div   id="8n9s8rpp6h"    class=\"su-spoiler su-spoiler-style-default su-spoiler-icon-plus su-spoiler-closed\" data-scroll-offset=\"0\"><div   id="8n9s8rpp6h"    class=\"su-spoiler-title\" tabindex=\"0\" role=\"button\"><span class=\"su-spoiler-icon\"><\/span>References:<\/div><div   id="8n9s8rpp6h"    class=\"su-spoiler-content su-u-clearfix su-u-trim\"><strong>Nuclear and Reactor Physics:<\/strong>\n<ol>\n<li>J. R. Lamarsh, Introduction to Nuclear Reactor Theory, 2nd ed., Addison-Wesley, Reading,\u00a0MA (1983).<\/li>\n<li>J. R. Lamarsh, A. J. Baratta, Introduction to Nuclear Engineering, 3d ed., Prentice-Hall, 2001, ISBN: 0-201-82498-1.<\/li>\n<li>W. M. Stacey, Nuclear Reactor Physics, John Wiley &amp; Sons, 2001, ISBN: 0- 471-39127-1.<\/li>\n<li>Glasstone, Sesonske. Nuclear Reactor Engineering: Reactor Systems Engineering,\u00a0Springer; 4th edition, 1994, ISBN:\u00a0978-0412985317<\/li>\n<li>W.S.C. Williams. Nuclear and Particle Physics.\u00a0Clarendon Press; 1 edition, 1991, ISBN:\u00a0978-0198520467<\/li>\n<li>Kenneth S. Krane. Introductory Nuclear Physics, 3rd Edition, Wiley, 1987,\u00a0ISBN:\u00a0978-0471805533<\/li>\n<li>G.R.Keepin. Physics of Nuclear Kinetics.\u00a0Addison-Wesley Pub. Co; 1st edition, 1965<\/li>\n<li>Robert Reed Burn, Introduction to Nuclear Reactor Operation, 1988.<\/li>\n<li>U.S. Department of Energy, Nuclear Physics and Reactor Theory.\u00a0DOE Fundamentals Handbook,\u00a0Volume 1 and 2.\u00a0January\u00a01993.<\/li>\n<\/ol>\n<p><strong><\/strong><strong>Advanced Reactor Physics:<\/strong><\/p>\n<ol>\n<li>K. O. Ott, W. A. Bezella, Introductory Nuclear Reactor Statics, American Nuclear Society, Revised edition (1989), 1989, ISBN: 0-894-48033-2.<\/li>\n<li>K. O. Ott, R. J. Neuhold, Introductory Nuclear Reactor Dynamics, American Nuclear Society, 1985, ISBN: 0-894-48029-4.<\/li>\n<li><span style=\"font-weight: 400;\">D. L. Hetrick, Dynamics of Nuclear Reactors, American Nuclear Society, 1993, ISBN: 0-894-48453-2.\u00a0<\/span><\/li>\n<li>E. E. Lewis, W. F. Miller, Computational Methods of Neutron Transport, American Nuclear Society, 1993, ISBN: 0-894-48452-4.<\/li>\n<\/ol>\n<\/div><\/div><\/div><div   id="8n9s8rpp6h"    class=\"su-divider su-divider-style-dotted\" style=\"margin:15px 0;border-width:2px;border-color:#999999\"><\/div><\/div><\/div><div   id="8n9s8rpp6h"    class=\"su-divider su-divider-style-default\" style=\"margin:15px 0;border-width:2px;border-color:#999999\"><\/div><div   id="8n9s8rpp6h"     class=\"lgc-column lgc-grid-parent lgc-grid-33 lgc-tablet-grid-33 lgc-mobile-grid-100 lgc-equal-heights \"><div   id="8n9s8rpp6h"     class=\"inside-grid-column\"><\/div><\/div><div   id="8n9s8rpp6h"     class=\"lgc-column lgc-grid-parent lgc-grid-33 lgc-tablet-grid-33 lgc-mobile-grid-100 lgc-equal-heights \"><div   id="8n9s8rpp6h"     class=\"inside-grid-column\">\n<h2>See above:<\/h2>\n<p>Laws of Conservation<a href=\"http:\/\/sitepourvtc.com\/laws-of-conservation\/\" class=\"su-button su-button-style-flat\" style=\"color:#606060;background-color:#ffffff;border-color:#cccccc;border-radius:10px;-moz-border-radius:10px;-webkit-border-radius:10px\" target=\"_self\"><span style=\"color:#606060;padding:7px 20px;font-size:16px;line-height:24px;border-color:#ffffff;border-radius:10px;-moz-border-radius:10px;-webkit-border-radius:10px;text-shadow:0px 0px 0px #000000;-moz-text-shadow:0px 0px 0px #000000;-webkit-text-shadow:0px 0px 0px #000000\"><i class=\"sui sui-link\" style=\"font-size:16px;color:#5d5d5d\"><\/i> <\/span><\/a><\/p><\/div><\/div><div   id="8n9s8rpp6h"     class=\"lgc-column lgc-grid-parent lgc-grid-33 lgc-tablet-grid-33 lgc-mobile-grid-100 lgc-equal-heights \"><div   id="8n9s8rpp6h"     class=\"inside-grid-column\"><\/div><\/div>\n","protected":false},"excerpt":{"rendered":"<p>The vector sum of the momenta (momentum is equal to the mass of an object multiplied by its velocity) of all the objects of a system cannot be changed by interactions within the system. In classical mechanics, this law is implied by Newton\u2019s laws. This principle is a direct consequence of Newton\u2019s third law. Let &#8230; <a title=\"Law of Conservation of Momentum\" class=\"read-more\" href=\"https:\/\/sitepourvtc.com\/laws-of-conservation\/law-of-conservation-of-momentum\/\" aria-label=\"More on Law of Conservation of Momentum\">Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"parent":14137,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"generate_page_header":""},"_links":{"self":[{"href":"https:\/\/sitepourvtc.com\/wp-json\/wp\/v2\/pages\/15893"}],"collection":[{"href":"https:\/\/sitepourvtc.com\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/sitepourvtc.com\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/sitepourvtc.com\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/sitepourvtc.com\/wp-json\/wp\/v2\/comments?post=15893"}],"version-history":[{"count":4,"href":"https:\/\/sitepourvtc.com\/wp-json\/wp\/v2\/pages\/15893\/revisions"}],"predecessor-version":[{"id":34345,"href":"https:\/\/sitepourvtc.com\/wp-json\/wp\/v2\/pages\/15893\/revisions\/34345"}],"up":[{"embeddable":true,"href":"https:\/\/sitepourvtc.com\/wp-json\/wp\/v2\/pages\/14137"}],"wp:attachment":[{"href":"https:\/\/sitepourvtc.com\/wp-json\/wp\/v2\/media?parent=15893"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}