{"id":256,"date":"2018-04-09T11:37:02","date_gmt":"2018-04-09T11:37:02","guid":{"rendered":"https:\/\/books.compclassnotes.com\/rothphys110\/?p=256"},"modified":"2021-05-08T23:48:43","modified_gmt":"2021-05-08T23:48:43","slug":"section-7-1","status":"publish","type":"post","link":"https:\/\/books.compclassnotes.com\/rothphys110-2e\/2018\/04\/09\/section-7-1\/","title":{"rendered":"Chapter 7: Linear momentum"},"content":{"rendered":"\n<p>When Isaac Newton was formulating his famous laws of motion, he was not thinking in terms of position, velocity, and acceleration. Instead, he worked with what he called the \u201cquantity of motion.\u201d This was a particular quantity that by itself described the motion of an object. Today we call it <em>linear momentum,<\/em> and usually just refer to it as momentum. An object&#8217;s momentum is its mass times its velocity:<\/p>\n\n\n\n<p align=\"center\"><strong><em>p<\/em><\/strong> = <em>m<\/em><strong><em>v<\/em><\/strong><\/p>\n\n\n\n<p>Since velocity is a vector quantity, and momentum is velocity times mass (which is a scalar), momentum is also a vector quantity. When working with one-dimensional motion (only moving forward or backwards along a single axis), a positive or negative sign is all we need to tell us direction, and we drop the vector notation. When we work with two-dimensional motion, formal vector notation becomes important (see <a rel=\"noreferrer noopener\" href=\"https:\/\/books.compclassnotes.com\/rothphys110-2e\/2017\/10\/22\/section-2-1-2\/\" target=\"_blank\">chapter 2<\/a> to review vectors).<\/p>\n\n\n\n<p>The SI units for momentum are kilograms times meters per second (kg\u00b7m\/s). There is no special name for this unit.<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">Example<\/h4>\n\n\n\n<!-- iframe plugin v.6.0 wordpress.org\/plugins\/iframe\/ -->\n<iframe loading=\"lazy\" src=\"https:\/\/my.compclassnotes.com\/canonical\/PHYS110\/PHYS110_book_ch7_1_EXMP\" width=\"100%\" height=\"1000\" marginwidth=\"0\" marginheight=\"0\" 0=\"scrolling=&quot;yes\u201d\" scrolling=\"yes\" class=\"iframe-class\" frameborder=\"0\"><\/iframe>\n\n\n\n\n<h4 class=\"wp-block-heading\">Practice<\/h4>\n\n\n\n<!-- iframe plugin v.6.0 wordpress.org\/plugins\/iframe\/ -->\n<iframe loading=\"lazy\" src=\"https:\/\/my.compclassnotes.com\/canonical\/PHYS110\/PHYS110_book_ch7v2_prac1\" width=\"100%\" height=\"600\" marginwidth=\"0\" marginheight=\"0\" 0=\"scrolling=&quot;yes\u201d\" scrolling=\"yes\" class=\"iframe-class\" frameborder=\"0\"><\/iframe>\n\n\n\n\n<h2 class=\"wp-block-heading\">7.1 Conservation of Momentum<\/h2>\n\n\n\n<p>Momentum is a conserved quantity. For any system of objects, the total momentum of the system does not change as long as there is no \u201cexternal agent\u201d acting on the system.<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">Example<\/h4>\n\n\n\n<p>A rocket is able to move because of conservation of momentum. Even when there is nothing to push off of, firing a blast still propels the ship forward. The momentum of the system must be constant; since the exhaust gases have momentum in one direction, the ship must have the same momentum in the opposite direction. This means the ship must move forward with some velocity. (Due to it\u2019s much larger mass, the rocket will have a lower velocity than the exhaust gases.)<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">7.1.1 Problem Solving Steps<\/h3>\n\n\n\n<p>Since we\u2019re still dealing with a conservation law, solving momentum problems is very similar to solving energy problems (see <a rel=\"noreferrer noopener\" aria-label=\"section&nbsp;6.5 (opens in a new tab)\" href=\"https:\/\/books.compclassnotes.com\/rothphys110-2e\/2018\/03\/20\/section-6-5\/\" target=\"_blank\">section&nbsp;6.5<\/a>). Conservation of momentum is a useful tool to analyze systems where multiple objects are interacting with each other over a short period of time.<\/p>\n\n\n\n<p>You\u2019ll usually find it helpful to draw a series of pictures in step 1 of the problem-solving process, showing what is going on&nbsp;just before&nbsp;and&nbsp;just after&nbsp;the interaction. The fundamental law that we\u2019ll use in step 2 of the process is conservation of momentum:<\/p>\n\n\n\n<p align=\"center\"> <em>p<sub>i<\/sub><\/em> = <em>p<sub>f<\/sub><\/em> <\/p>\n\n\n\n<p> To solve mathematically (step 3), we\u2019ll identify which objects have momentum before and after an interaction, and go from there.<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">Example<\/h4>\n\n\n\n<!-- iframe plugin v.6.0 wordpress.org\/plugins\/iframe\/ -->\n<iframe loading=\"lazy\" src=\"https:\/\/my.compclassnotes.com\/canonical\/PHYS110\/PHYS110_book_ch7_3_EXMP\" width=\"100%\" height=\"1000\" marginwidth=\"0\" marginheight=\"0\" 0=\"scrolling=&quot;yes\u201d\" scrolling=\"yes\" class=\"iframe-class\" frameborder=\"0\"><\/iframe>\n\n\n\n\n<h4 class=\"wp-block-heading\">Example<\/h4>\n\n\n\n<!-- iframe plugin v.6.0 wordpress.org\/plugins\/iframe\/ -->\n<iframe loading=\"lazy\" src=\"https:\/\/my.compclassnotes.com\/canonical\/PHYS110\/PHYS110_book_ch7_4_EXMP\" width=\"100%\" height=\"700\" marginwidth=\"0\" marginheight=\"0\" 0=\"scrolling=&quot;yes\u201d\" scrolling=\"yes\" class=\"iframe-class\" frameborder=\"0\"><\/iframe>\n\n\n\n\n<h4 class=\"wp-block-heading\">Practice<\/h4>\n\n\n\n<!-- iframe plugin v.6.0 wordpress.org\/plugins\/iframe\/ -->\n<iframe loading=\"lazy\" src=\"https:\/\/my.compclassnotes.com\/canonical\/PHYS110\/PHYS110_book_ch7_prac_part1_q2\" width=\"100%\" height=\"800\" marginwidth=\"0\" marginheight=\"0\" scrolling=\"yes\" class=\"iframe-class\" frameborder=\"0\"><\/iframe>\n\n","protected":false},"excerpt":{"rendered":"<p>When Isaac Newton was formulating his famous laws of motion, he was not thinking in terms of position, velocity, and acceleration. Instead, he worked with what he called the \u201cquantity of motion.\u201d This was a particular quantity that by itself <span class=\"readmore\"><a href=\"https:\/\/books.compclassnotes.com\/rothphys110-2e\/2018\/04\/09\/section-7-1\/\">Continue Reading<\/a><\/span><\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-256","post","type-post","status-publish","format-standard","hentry","category-uncategorized"],"_links":{"self":[{"href":"https:\/\/books.compclassnotes.com\/rothphys110-2e\/wp-json\/wp\/v2\/posts\/256","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/books.compclassnotes.com\/rothphys110-2e\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/books.compclassnotes.com\/rothphys110-2e\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/books.compclassnotes.com\/rothphys110-2e\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/books.compclassnotes.com\/rothphys110-2e\/wp-json\/wp\/v2\/comments?post=256"}],"version-history":[{"count":8,"href":"https:\/\/books.compclassnotes.com\/rothphys110-2e\/wp-json\/wp\/v2\/posts\/256\/revisions"}],"predecessor-version":[{"id":864,"href":"https:\/\/books.compclassnotes.com\/rothphys110-2e\/wp-json\/wp\/v2\/posts\/256\/revisions\/864"}],"wp:attachment":[{"href":"https:\/\/books.compclassnotes.com\/rothphys110-2e\/wp-json\/wp\/v2\/media?parent=256"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/books.compclassnotes.com\/rothphys110-2e\/wp-json\/wp\/v2\/categories?post=256"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/books.compclassnotes.com\/rothphys110-2e\/wp-json\/wp\/v2\/tags?post=256"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}