A proton (mass m = 1.67 x 10:27 kg) is being accelerated along a straight line at 4.60 x 1010 m\/s\u00b2 in a machine. If the proton has an initial speed of 4.00 x 10 m\/s and travels 1.10 cm, what then is (a) its speed and (b) the increase in its kinetic energy?
(a) Number
Units
(b) Number
Units<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n We can solve the problem using basic kinematic and work-energy principles. Here’s a step-by-step solution:<\/p>\n\n\n\n We will use the kinematic equation that relates the initial velocity, final velocity, acceleration, and distance traveled:<\/p>\n\n\n\n [ Where:<\/p>\n\n\n\n Substituting the known values:<\/p>\n\n\n\n [ [ [ [ Now, taking the square root:<\/p>\n\n\n\n [ [ So, the final speed is approximately:<\/p>\n\n\n\n [ The increase in kinetic energy is given by:<\/p>\n\n\n\n [ Substituting the known values:<\/p>\n\n\n\n [ First, calculate the squares:<\/p>\n\n\n\n [ Now, substitute:<\/p>\n\n\n\n [ [ [ So, the increase in kinetic energy is:<\/p>\n\n\n\n [ (a) Speed<\/strong> = 1.01 \u00d7 10\u2075 m\/s A proton (mass m = 1.67 x 10:27 kg) is being accelerated along a straight line at 4.60 x 1010 m\/s\u00b2 in a machine. If the proton has an initial speed of 4.00 x 10 m\/s and travels 1.10 cm, what then is (a) its speed and (b) the increase in its kinetic energy?(a) NumberUnits(b) […]<\/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-193142","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/193142","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=193142"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/193142\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=193142"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=193142"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=193142"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Given:<\/h3>\n\n\n\n
\n
Part (a): Finding Final Speed (v)<\/h3>\n\n\n\n
v^2 = v_0^2 + 2ad
]<\/p>\n\n\n\n\n
v^2 = (4.00 \\times 10^0)^2 + 2 \\times (4.60 \\times 10^{10}) \\times (0.011)
]<\/p>\n\n\n\n
v^2 = 16 + 1.012 \\times 10^{10}
]<\/p>\n\n\n\n
v^2 = 1.012 \\times 10^{10} + 16
]<\/p>\n\n\n\n
v^2 \\approx 1.012 \\times 10^{10} \\, \\text{m}^2\/\\text{s}^2
]<\/p>\n\n\n\n
v \\approx \\sqrt{1.012 \\times 10^{10}} \\, \\text{m\/s}
]<\/p>\n\n\n\n
v \\approx 1.006 \\times 10^5 \\, \\text{m\/s}
]<\/p>\n\n\n\n
v \\approx 1.01 \\times 10^5 \\, \\text{m\/s}
]<\/p>\n\n\n\nPart (b): Increase in Kinetic Energy (\u0394K)<\/h3>\n\n\n\n
\\Delta K = \\frac{1}{2} m (v^2 – v_0^2)
]<\/p>\n\n\n\n
\\Delta K = \\frac{1}{2} \\times (1.67 \\times 10^{-27}) \\times \\left( (1.01 \\times 10^5)^2 – (4.00 \\times 10^0)^2 \\right)
]<\/p>\n\n\n\n
(1.01 \\times 10^5)^2 = 1.0201 \\times 10^{10}
]
[
(4.00 \\times 10^0)^2 = 16
]<\/p>\n\n\n\n
\\Delta K = \\frac{1}{2} \\times (1.67 \\times 10^{-27}) \\times (1.0201 \\times 10^{10} – 16)
]<\/p>\n\n\n\n
\\Delta K = \\frac{1}{2} \\times (1.67 \\times 10^{-27}) \\times 1.0201 \\times 10^{10}
]<\/p>\n\n\n\n
\\Delta K = 8.53 \\times 10^{-18} \\, \\text{J}
]<\/p>\n\n\n\n
\\Delta K \\approx 8.53 \\times 10^{-18} \\, \\text{J}
]<\/p>\n\n\n\nFinal Answer:<\/h3>\n\n\n\n
(b) Increase in Kinetic Energy<\/strong> = 8.53 \u00d7 10\u207b\u00b9\u2078 J<\/p>\n\n\n\nExplanation:<\/h3>\n\n\n\n
\n