{"id":87860,"date":"2025-06-01T06:55:43","date_gmt":"2025-06-01T06:55:43","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=87860"},"modified":"2025-06-01T06:55:43","modified_gmt":"2025-06-01T06:55:43","slug":"a-boy-of-mass-52-kg-jumps-with-a-horizontal-velocity-of-2-m-s-onto-a-s","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/a-boy-of-mass-52-kg-jumps-with-a-horizontal-velocity-of-2-m-s-onto-a-s\/","title":{"rendered":"A boy of mass 52 kg jumps with a horizontal velocity of 2 m\/s onto a s"},"content":{"rendered":"

A boy of mass 52 kg jumps with a horizontal velocity of 2 m\/s onto a stationary cart of mass 3 kg. The cart is fixed with frictionless wheels. Which one of the following would be the speed of the cart?<\/p>\n

[amp_mcq option1=”2.15 m\/s” option2=”1.89 m\/s” option3=”1.51 m\/s” option4=”2.51 m\/s” correct=”option2″]<\/p>\n

\n
\n
This question was previously asked in<\/div>\n
UPSC NDA-1 – 2022<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nThe correct answer is 1.89 m\/s. This is calculated using the principle of conservation of linear momentum for the inelastic collision between the boy and the cart.
\n<\/section>\n
\n– The system consists of the boy and the stationary cart.
\n– Before the jump, the boy has momentum (mass \u00d7 velocity) and the cart has zero momentum.
\n– After the boy jumps onto the cart, they move together as a single system with a common velocity.
\n– According to the law of conservation of momentum, the total momentum of the system before the collision is equal to the total momentum after the collision, provided no external forces act on the system (frictionless wheels are mentioned).
\n– Initial momentum = (mass of boy \u00d7 velocity of boy) + (mass of cart \u00d7 velocity of cart) = (52 kg \u00d7 2 m\/s) + (3 kg \u00d7 0 m\/s) = 104 kg m\/s.
\n– Let the final velocity of the combined system (boy + cart) be ‘v’. The combined mass is 52 kg + 3 kg = 55 kg.
\n– Final momentum = (combined mass \u00d7 final velocity) = 55 kg \u00d7 v.
\n– By conservation of momentum: 104 kg m\/s = 55 kg \u00d7 v.
\n– v = 104 \/ 55 \u2248 1.89 m\/s.
\n<\/section>\n
\nThis is an example of a perfectly inelastic collision because the two objects stick together after colliding. Kinetic energy is not conserved in an inelastic collision, but momentum is always conserved in the absence of external forces.
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

A boy of mass 52 kg jumps with a horizontal velocity of 2 m\/s onto a stationary cart of mass 3 kg. The cart is fixed with frictionless wheels. Which one of the following would be the speed of the cart? [amp_mcq option1=”2.15 m\/s” option2=”1.89 m\/s” option3=”1.51 m\/s” option4=”2.51 m\/s” correct=”option2″] This question was previously … <\/p>\n

Detailed SolutionA boy of mass 52 kg jumps with a horizontal velocity of 2 m\/s onto a s<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1093],"tags":[1108,1129,1128],"class_list":["post-87860","post","type-post","status-publish","format-standard","hentry","category-upsc-nda-1","tag-1108","tag-mechanics","tag-physics","no-featured-image-padding"],"yoast_head":"\nA boy of mass 52 kg jumps with a horizontal velocity of 2 m\/s onto a s<\/title>\n<meta name=\"description\" content=\"The correct answer is 1.89 m\/s. This is calculated using the principle of conservation of linear momentum for the inelastic collision between the boy and the cart. - The system consists of the boy and the stationary cart. - Before the jump, the boy has momentum (mass \u00d7 velocity) and the cart has zero momentum. - After the boy jumps onto the cart, they move together as a single system with a common velocity. - According to the law of conservation of momentum, the total momentum of the system before the collision is equal to the total momentum after the collision, provided no external forces act on the system (frictionless wheels are mentioned). - Initial momentum = (mass of boy \u00d7 velocity of boy) + (mass of cart \u00d7 velocity of cart) = (52 kg \u00d7 2 m\/s) + (3 kg \u00d7 0 m\/s) = 104 kg m\/s. - Let the final velocity of the combined system (boy + cart) be 'v'. The combined mass is 52 kg + 3 kg = 55 kg. - Final momentum = (combined mass \u00d7 final velocity) = 55 kg \u00d7 v. - By conservation of momentum: 104 kg m\/s = 55 kg \u00d7 v. - v = 104 \/ 55 \u2248 1.89 m\/s.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/exam.pscnotes.com\/mcq\/a-boy-of-mass-52-kg-jumps-with-a-horizontal-velocity-of-2-m-s-onto-a-s\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"A boy of mass 52 kg jumps with a horizontal velocity of 2 m\/s onto a s\" \/>\n<meta property=\"og:description\" content=\"The correct answer is 1.89 m\/s. This is calculated using the principle of conservation of linear momentum for the inelastic collision between the boy and the cart. - The system consists of the boy and the stationary cart. - Before the jump, the boy has momentum (mass \u00d7 velocity) and the cart has zero momentum. - After the boy jumps onto the cart, they move together as a single system with a common velocity. - According to the law of conservation of momentum, the total momentum of the system before the collision is equal to the total momentum after the collision, provided no external forces act on the system (frictionless wheels are mentioned). - Initial momentum = (mass of boy \u00d7 velocity of boy) + (mass of cart \u00d7 velocity of cart) = (52 kg \u00d7 2 m\/s) + (3 kg \u00d7 0 m\/s) = 104 kg m\/s. - Let the final velocity of the combined system (boy + cart) be 'v'. The combined mass is 52 kg + 3 kg = 55 kg. - Final momentum = (combined mass \u00d7 final velocity) = 55 kg \u00d7 v. - By conservation of momentum: 104 kg m\/s = 55 kg \u00d7 v. - v = 104 \/ 55 \u2248 1.89 m\/s.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/a-boy-of-mass-52-kg-jumps-with-a-horizontal-velocity-of-2-m-s-onto-a-s\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T06:55:43+00:00\" \/>\n<meta name=\"author\" content=\"rawan239\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"rawan239\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"1 minute\" \/>\n<!-- \/ Yoast SEO Premium plugin. -->","yoast_head_json":{"title":"A boy of mass 52 kg jumps with a horizontal velocity of 2 m\/s onto a s","description":"The correct answer is 1.89 m\/s. This is calculated using the principle of conservation of linear momentum for the inelastic collision between the boy and the cart. - The system consists of the boy and the stationary cart. - Before the jump, the boy has momentum (mass \u00d7 velocity) and the cart has zero momentum. - After the boy jumps onto the cart, they move together as a single system with a common velocity. - According to the law of conservation of momentum, the total momentum of the system before the collision is equal to the total momentum after the collision, provided no external forces act on the system (frictionless wheels are mentioned). - Initial momentum = (mass of boy \u00d7 velocity of boy) + (mass of cart \u00d7 velocity of cart) = (52 kg \u00d7 2 m\/s) + (3 kg \u00d7 0 m\/s) = 104 kg m\/s. - Let the final velocity of the combined system (boy + cart) be 'v'. The combined mass is 52 kg + 3 kg = 55 kg. - Final momentum = (combined mass \u00d7 final velocity) = 55 kg \u00d7 v. - By conservation of momentum: 104 kg m\/s = 55 kg \u00d7 v. - v = 104 \/ 55 \u2248 1.89 m\/s.","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/exam.pscnotes.com\/mcq\/a-boy-of-mass-52-kg-jumps-with-a-horizontal-velocity-of-2-m-s-onto-a-s\/","og_locale":"en_US","og_type":"article","og_title":"A boy of mass 52 kg jumps with a horizontal velocity of 2 m\/s onto a s","og_description":"The correct answer is 1.89 m\/s. This is calculated using the principle of conservation of linear momentum for the inelastic collision between the boy and the cart. - The system consists of the boy and the stationary cart. - Before the jump, the boy has momentum (mass \u00d7 velocity) and the cart has zero momentum. - After the boy jumps onto the cart, they move together as a single system with a common velocity. - According to the law of conservation of momentum, the total momentum of the system before the collision is equal to the total momentum after the collision, provided no external forces act on the system (frictionless wheels are mentioned). - Initial momentum = (mass of boy \u00d7 velocity of boy) + (mass of cart \u00d7 velocity of cart) = (52 kg \u00d7 2 m\/s) + (3 kg \u00d7 0 m\/s) = 104 kg m\/s. - Let the final velocity of the combined system (boy + cart) be 'v'. The combined mass is 52 kg + 3 kg = 55 kg. - Final momentum = (combined mass \u00d7 final velocity) = 55 kg \u00d7 v. - By conservation of momentum: 104 kg m\/s = 55 kg \u00d7 v. - v = 104 \/ 55 \u2248 1.89 m\/s.","og_url":"https:\/\/exam.pscnotes.com\/mcq\/a-boy-of-mass-52-kg-jumps-with-a-horizontal-velocity-of-2-m-s-onto-a-s\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T06:55:43+00:00","author":"rawan239","twitter_card":"summary_large_image","twitter_misc":{"Written by":"rawan239","Est. reading time":"1 minute"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"WebPage","@id":"https:\/\/exam.pscnotes.com\/mcq\/a-boy-of-mass-52-kg-jumps-with-a-horizontal-velocity-of-2-m-s-onto-a-s\/","url":"https:\/\/exam.pscnotes.com\/mcq\/a-boy-of-mass-52-kg-jumps-with-a-horizontal-velocity-of-2-m-s-onto-a-s\/","name":"A boy of mass 52 kg jumps with a horizontal velocity of 2 m\/s onto a s","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T06:55:43+00:00","dateModified":"2025-06-01T06:55:43+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"The correct answer is 1.89 m\/s. This is calculated using the principle of conservation of linear momentum for the inelastic collision between the boy and the cart. - The system consists of the boy and the stationary cart. - Before the jump, the boy has momentum (mass \u00d7 velocity) and the cart has zero momentum. - After the boy jumps onto the cart, they move together as a single system with a common velocity. - According to the law of conservation of momentum, the total momentum of the system before the collision is equal to the total momentum after the collision, provided no external forces act on the system (frictionless wheels are mentioned). - Initial momentum = (mass of boy \u00d7 velocity of boy) + (mass of cart \u00d7 velocity of cart) = (52 kg \u00d7 2 m\/s) + (3 kg \u00d7 0 m\/s) = 104 kg m\/s. - Let the final velocity of the combined system (boy + cart) be 'v'. The combined mass is 52 kg + 3 kg = 55 kg. - Final momentum = (combined mass \u00d7 final velocity) = 55 kg \u00d7 v. - By conservation of momentum: 104 kg m\/s = 55 kg \u00d7 v. - v = 104 \/ 55 \u2248 1.89 m\/s.","breadcrumb":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/a-boy-of-mass-52-kg-jumps-with-a-horizontal-velocity-of-2-m-s-onto-a-s\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/exam.pscnotes.com\/mcq\/a-boy-of-mass-52-kg-jumps-with-a-horizontal-velocity-of-2-m-s-onto-a-s\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/exam.pscnotes.com\/mcq\/a-boy-of-mass-52-kg-jumps-with-a-horizontal-velocity-of-2-m-s-onto-a-s\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/exam.pscnotes.com\/mcq\/"},{"@type":"ListItem","position":2,"name":"UPSC NDA-1","item":"https:\/\/exam.pscnotes.com\/mcq\/category\/upsc-nda-1\/"},{"@type":"ListItem","position":3,"name":"A boy of mass 52 kg jumps with a horizontal velocity of 2 m\/s onto a s"}]},{"@type":"WebSite","@id":"https:\/\/exam.pscnotes.com\/mcq\/#website","url":"https:\/\/exam.pscnotes.com\/mcq\/","name":"MCQ and Quiz for Exams","description":"","potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/exam.pscnotes.com\/mcq\/?s={search_term_string}"},"query-input":"required name=search_term_string"}],"inLanguage":"en-US"},{"@type":"Person","@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209","name":"rawan239","image":{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/image\/","url":"https:\/\/secure.gravatar.com\/avatar\/761a7274f9cce048fa5b921221e7934820d74514df93ef195a9d22af0c1c9001?s=96&d=mm&r=g","contentUrl":"https:\/\/secure.gravatar.com\/avatar\/761a7274f9cce048fa5b921221e7934820d74514df93ef195a9d22af0c1c9001?s=96&d=mm&r=g","caption":"rawan239"},"sameAs":["https:\/\/exam.pscnotes.com"],"url":"https:\/\/exam.pscnotes.com\/mcq\/author\/rawan239\/"}]}},"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/87860","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/comments?post=87860"}],"version-history":[{"count":0,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/87860\/revisions"}],"wp:attachment":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/media?parent=87860"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/categories?post=87860"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/tags?post=87860"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}