{"id":252881,"date":"2025-07-11T14:26:24","date_gmt":"2025-07-11T14:26:24","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=252881"},"modified":"2025-07-11T14:26:26","modified_gmt":"2025-07-11T14:26:26","slug":"a-billiard-ball-moving-at-5m-s-strikes-a-stationary-ball-of-the-same-mass","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/11\/a-billiard-ball-moving-at-5m-s-strikes-a-stationary-ball-of-the-same-mass\/","title":{"rendered":"A billiard ball moving at 5m\/s strikes a stationary ball of the same mass."},"content":{"rendered":"\n<p>A billiard ball moving at 5m\/s strikes a stationary ball of the same mass. After the collision, the first ball moves at 4.33m\/s at an angle of 30 with respect to the original line of motion. Assuming an elastic collision (and ignoring friction and rotational motion), a) Find the struck ball&#8217;s speed magnitude and direction. b) Is the collision really elastic? Prove that.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/image-399.png\" alt=\"\" class=\"wp-image-252882\"\/><\/figure>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>Here is the solution to the physics problem.<\/p>\n\n\n\n<p>This problem involves a two-dimensional collision between two billiard balls of equal mass. We can solve this by applying the principle of conservation of linear momentum. Since momentum is a vector quantity, we must conserve it independently in both the horizontal (x) and vertical (y) directions.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>a) Find the struck ball&#8217;s speed magnitude and direction.<\/strong><\/h3>\n\n\n\n<p>To find the final velocity of the second ball, we will set up conservation of momentum equations. Let &#8216;m&#8217; be the mass of each ball. Let the initial velocity of the first ball be&nbsp;v1i&nbsp;and the second ball be&nbsp;v2i. Let their final velocities be&nbsp;v1f&nbsp;and&nbsp;v2f, respectively. The problem provides that&nbsp;v1i = 5.00 m\/s&nbsp;along the x-axis,&nbsp;v2i = 0, and&nbsp;v1f&nbsp;has a magnitude of&nbsp;4.33 m\/s&nbsp;at an angle of 30 degrees.<\/p>\n\n\n\n<p><strong>Conservation of Momentum in the x-direction:<\/strong><br>The total initial momentum in the x-direction must equal the total final momentum in the x-direction.<\/p>\n\n\n\n<p>Initial x-momentum =&nbsp;m * v1ix + m * v2ix = m * (5.00) + m * (0) = 5.00m<br>Final x-momentum =&nbsp;m * v1fx + m * v2fx = m * (4.33 * cos(30\u00b0)) + m * v2fx<\/p>\n\n\n\n<p>Setting them equal and canceling &#8216;m&#8217;:<br>5.00 = 4.33 * cos(30\u00b0) + v2fx<br>5.00 = 4.33 * (0.866) + v2fx<br>5.00 = 3.75 + v2fx<br>v2fx = 5.00 &#8211; 3.75 = 1.25 m\/s<\/p>\n\n\n\n<p><strong>Conservation of Momentum in the y-direction:<\/strong><br>The total initial momentum in the y-direction was zero, so the total final momentum in the y-direction must also be zero.<\/p>\n\n\n\n<p>Initial y-momentum =&nbsp;m * v1iy + m * v2iy = m * (0) + m * (0) = 0<br>Final y-momentum =&nbsp;m * v1fy + m * v2fy = m * (4.33 * sin(30\u00b0)) + m * v2fy<\/p>\n\n\n\n<p>Setting them equal and canceling &#8216;m&#8217;:<br>0 = 4.33 * sin(30\u00b0) + v2fy<br>0 = 4.33 * (0.5) + v2fy<br>0 = 2.165 + v2fy<br>v2fy = -2.165 m\/s<\/p>\n\n\n\n<p><strong>Finding the Struck Ball&#8217;s Final Speed and Direction:<\/strong><br>Now that we have the x and y components of the second ball&#8217;s final velocity (v2fx&nbsp;and&nbsp;v2fy), we can find its speed (magnitude) using the Pythagorean theorem and its direction using trigonometry.<\/p>\n\n\n\n<p>Speed&nbsp;v2f = sqrt(v2fx\u00b2 + v2fy\u00b2) = sqrt((1.25)\u00b2 + (-2.165)\u00b2) = sqrt(1.5625 + 4.687225) = sqrt(6.25) = 2.50 m\/s<\/p>\n\n\n\n<p>Direction&nbsp;\u03b8 = arctan(v2fy \/ v2fx) = arctan(-2.165 \/ 1.25) = arctan(-1.732) = -60\u00b0<\/p>\n\n\n\n<p><strong>Answer for (a): The struck ball moves with a speed of 2.50 m\/s at an angle of -60\u00b0 (or 60\u00b0 below the original line of motion).<\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>b) Is the collision really elastic? Prove that.<\/strong><\/h3>\n\n\n\n<p>A collision is elastic if the total kinetic energy of the system is conserved, meaning the total kinetic energy before the collision is equal to the total kinetic energy after the collision. We will calculate both values to verify this.<\/p>\n\n\n\n<p><strong>Initial Kinetic Energy (KEi):<\/strong><br>Only the first ball is moving initially.<br>KEi = (1\/2)m * v1i\u00b2 + (1\/2)m * v2i\u00b2<br>KEi = (1\/2)m * (5.00)\u00b2 + (1\/2)m * (0)\u00b2<br>KEi = (1\/2)m * 25 = 12.5m<\/p>\n\n\n\n<p><strong>Final Kinetic Energy (KEf):<\/strong><br>Both balls are moving after the collision.<br>KEf = (1\/2)m * v1f\u00b2 + (1\/2)m * v2f\u00b2<br>KEf = (1\/2)m * (4.33)\u00b2 + (1\/2)m * (2.50)\u00b2<br>KEf = (1\/2)m * (18.7489 + 6.25)<br>KEf = (1\/2)m * (24.9989)<br>KEf \u2248 12.5m<\/p>\n\n\n\n<p><strong>Proof:<\/strong><br>Comparing the initial and final kinetic energies, we find that&nbsp;KEi = 12.5m&nbsp;and&nbsp;KEf \u2248 12.5m. The very small difference is due to the rounding of the value 4.33 m\/s in the problem statement (the exact value would be&nbsp;5 * cos(30\u00b0) \u2248 4.3301 m\/s). Within the precision of the given numbers, the kinetic energy is conserved.<\/p>\n\n\n\n<p><strong>Answer for (b): Yes, the collision is elastic because the total kinetic energy before the collision (12.5m) is equal to the total kinetic energy after the collision (approximately 12.5m).<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner6-288.jpeg\" alt=\"\" class=\"wp-image-252883\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>A billiard ball moving at 5m\/s strikes a stationary ball of the same mass. After the collision, the first ball moves at 4.33m\/s at an angle of 30 with respect to the original line of motion. Assuming an elastic collision (and ignoring friction and rotational motion), a) Find the struck ball&#8217;s speed magnitude and direction. [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-252881","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/252881","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=252881"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/252881\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=252881"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=252881"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=252881"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}