{"id":87060,"date":"2025-06-01T04:27:17","date_gmt":"2025-06-01T04:27:17","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=87060"},"modified":"2025-06-01T04:27:17","modified_gmt":"2025-06-01T04:27:17","slug":"which-one-of-the-following-is-the-quantity-of-transfer-of-linear-momen","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/which-one-of-the-following-is-the-quantity-of-transfer-of-linear-momen\/","title":{"rendered":"Which one of the following is the quantity of transfer of linear momen"},"content":{"rendered":"

Which one of the following is the quantity of transfer of linear momentum to the floor, when a dumbbell of mass 500 g falls from a height of 5 m and stops after hitting the floor (take $g=10 \\text{ m\/s}^2$)?<\/p>\n

[amp_mcq option1=”0\u00b75 kg-m\/s” option2=”5\u00b70 kg-m\/s” option3=”10\u00b70 kg-m\/s” option4=”1\u00b70 kg-m\/s” correct=”option2″]<\/p>\n

\n
\n
This question was previously asked in<\/div>\n
UPSC Geoscientist – 2023<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nThe correct answer is B) 5.0 kg-m\/s.
\n<\/section>\n
\nThe transfer of linear momentum to the floor is equal to the magnitude of the change in linear momentum of the dumbbell during the impact, but in the opposite direction.
\nFirst, calculate the velocity of the dumbbell just before hitting the floor. Using the equation $v^2 = u^2 + 2gh$, where $u=0$ (starts from rest), $g=10 \\text{ m\/s}^2$, and $h=5 \\text{ m}$.
\n$v^2 = 0^2 + 2 \\times 10 \\times 5 = 100 \\text{ m}^2\/\\text{s}^2$.
\n$v = \\sqrt{100} = 10 \\text{ m\/s}$. The direction is downwards.
\nThe mass of the dumbbell is $m = 500 \\text{ g} = 0.5 \\text{ kg}$.
\nThe momentum of the dumbbell just before impact is $p_{initial} = m \\times v = 0.5 \\text{ kg} \\times 10 \\text{ m\/s} = 5 \\text{ kg-m\/s}$ (downwards).
\nAfter hitting the floor, the dumbbell stops, so its final velocity is 0.
\nThe momentum of the dumbbell after impact is $p_{final} = m \\times 0 = 0$.
\nThe change in momentum of the dumbbell is $\\Delta p_{dumbbell} = p_{final} – p_{initial} = 0 – (5 \\text{ kg-m\/s downwards}) = -5 \\text{ kg-m\/s downwards} = 5 \\text{ kg-m\/s upwards}$.
\nBy the impulse-momentum theorem and Newton’s third law, the momentum transferred to the floor is equal in magnitude and opposite in direction to the change in momentum of the dumbbell.
\nMomentum transfer to floor = $-\\Delta p_{dumbbell} = -(5 \\text{ kg-m\/s upwards}) = 5 \\text{ kg-m\/s downwards}$.
\nThe question asks for the *quantity* of transfer, which is the magnitude. The magnitude is 5 kg-m\/s.
\n<\/section>\n
\nThe impulse delivered to the floor is equal to the force exerted by the dumbbell on the floor integrated over the time of impact. This impulse causes the change in momentum of the floor. For the system of dumbbell + Earth, the total momentum is conserved during the impact (assuming external forces like gravity are negligible during the short impact time). The momentum lost by the dumbbell is gained by the Earth (via the floor). The Earth’s mass is so large that its velocity change is negligible, but it does gain momentum.
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

Which one of the following is the quantity of transfer of linear momentum to the floor, when a dumbbell of mass 500 g falls from a height of 5 m and stops after hitting the floor (take $g=10 \\text{ m\/s}^2$)? [amp_mcq option1=”0\u00b75 kg-m\/s” option2=”5\u00b70 kg-m\/s” option3=”10\u00b70 kg-m\/s” option4=”1\u00b70 kg-m\/s” correct=”option2″] This question was previously asked … <\/p>\n

Detailed SolutionWhich one of the following is the quantity of transfer of linear momen<\/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":[1091],"tags":[1105,1129,1128],"class_list":["post-87060","post","type-post","status-publish","format-standard","hentry","category-upsc-geoscientist","tag-1105","tag-mechanics","tag-physics","no-featured-image-padding"],"yoast_head":"\nWhich one of the following is the quantity of transfer of linear momen<\/title>\n<meta name=\"description\" content=\"The correct answer is B) 5.0 kg-m\/s. The transfer of linear momentum to the floor is equal to the magnitude of the change in linear momentum of the dumbbell during the impact, but in the opposite direction. First, calculate the velocity of the dumbbell just before hitting the floor. Using the equation $v^2 = u^2 + 2gh$, where $u=0$ (starts from rest), $g=10 text{ m\/s}^2$, and $h=5 text{ m}$. $v^2 = 0^2 + 2 times 10 times 5 = 100 text{ m}^2\/text{s}^2$. $v = sqrt{100} = 10 text{ m\/s}$. The direction is downwards. The mass of the dumbbell is $m = 500 text{ g} = 0.5 text{ kg}$. The momentum of the dumbbell just before impact is $p_{initial} = m times v = 0.5 text{ kg} times 10 text{ m\/s} = 5 text{ kg-m\/s}$ (downwards). After hitting the floor, the dumbbell stops, so its final velocity is 0. The momentum of the dumbbell after impact is $p_{final} = m times 0 = 0$. The change in momentum of the dumbbell is $Delta p_{dumbbell} = p_{final} - p_{initial} = 0 - (5 text{ kg-m\/s downwards}) = -5 text{ kg-m\/s downwards} = 5 text{ kg-m\/s upwards}$. By the impulse-momentum theorem and Newton's third law, the momentum transferred to the floor is equal in magnitude and opposite in direction to the change in momentum of the dumbbell. Momentum transfer to floor = $-Delta p_{dumbbell} = -(5 text{ kg-m\/s upwards}) = 5 text{ kg-m\/s downwards}$. The question asks for the *quantity* of transfer, which is the magnitude. The magnitude is 5 kg-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\/which-one-of-the-following-is-the-quantity-of-transfer-of-linear-momen\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Which one of the following is the quantity of transfer of linear momen\" \/>\n<meta property=\"og:description\" content=\"The correct answer is B) 5.0 kg-m\/s. The transfer of linear momentum to the floor is equal to the magnitude of the change in linear momentum of the dumbbell during the impact, but in the opposite direction. First, calculate the velocity of the dumbbell just before hitting the floor. Using the equation $v^2 = u^2 + 2gh$, where $u=0$ (starts from rest), $g=10 text{ m\/s}^2$, and $h=5 text{ m}$. $v^2 = 0^2 + 2 times 10 times 5 = 100 text{ m}^2\/text{s}^2$. $v = sqrt{100} = 10 text{ m\/s}$. The direction is downwards. The mass of the dumbbell is $m = 500 text{ g} = 0.5 text{ kg}$. The momentum of the dumbbell just before impact is $p_{initial} = m times v = 0.5 text{ kg} times 10 text{ m\/s} = 5 text{ kg-m\/s}$ (downwards). After hitting the floor, the dumbbell stops, so its final velocity is 0. The momentum of the dumbbell after impact is $p_{final} = m times 0 = 0$. The change in momentum of the dumbbell is $Delta p_{dumbbell} = p_{final} - p_{initial} = 0 - (5 text{ kg-m\/s downwards}) = -5 text{ kg-m\/s downwards} = 5 text{ kg-m\/s upwards}$. By the impulse-momentum theorem and Newton's third law, the momentum transferred to the floor is equal in magnitude and opposite in direction to the change in momentum of the dumbbell. Momentum transfer to floor = $-Delta p_{dumbbell} = -(5 text{ kg-m\/s upwards}) = 5 text{ kg-m\/s downwards}$. The question asks for the *quantity* of transfer, which is the magnitude. The magnitude is 5 kg-m\/s.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/which-one-of-the-following-is-the-quantity-of-transfer-of-linear-momen\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T04:27:17+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=\"2 minutes\" \/>\n<!-- \/ Yoast SEO Premium plugin. -->","yoast_head_json":{"title":"Which one of the following is the quantity of transfer of linear momen","description":"The correct answer is B) 5.0 kg-m\/s. The transfer of linear momentum to the floor is equal to the magnitude of the change in linear momentum of the dumbbell during the impact, but in the opposite direction. First, calculate the velocity of the dumbbell just before hitting the floor. Using the equation $v^2 = u^2 + 2gh$, where $u=0$ (starts from rest), $g=10 text{ m\/s}^2$, and $h=5 text{ m}$. $v^2 = 0^2 + 2 times 10 times 5 = 100 text{ m}^2\/text{s}^2$. $v = sqrt{100} = 10 text{ m\/s}$. The direction is downwards. The mass of the dumbbell is $m = 500 text{ g} = 0.5 text{ kg}$. The momentum of the dumbbell just before impact is $p_{initial} = m times v = 0.5 text{ kg} times 10 text{ m\/s} = 5 text{ kg-m\/s}$ (downwards). After hitting the floor, the dumbbell stops, so its final velocity is 0. The momentum of the dumbbell after impact is $p_{final} = m times 0 = 0$. The change in momentum of the dumbbell is $Delta p_{dumbbell} = p_{final} - p_{initial} = 0 - (5 text{ kg-m\/s downwards}) = -5 text{ kg-m\/s downwards} = 5 text{ kg-m\/s upwards}$. By the impulse-momentum theorem and Newton's third law, the momentum transferred to the floor is equal in magnitude and opposite in direction to the change in momentum of the dumbbell. Momentum transfer to floor = $-Delta p_{dumbbell} = -(5 text{ kg-m\/s upwards}) = 5 text{ kg-m\/s downwards}$. The question asks for the *quantity* of transfer, which is the magnitude. The magnitude is 5 kg-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\/which-one-of-the-following-is-the-quantity-of-transfer-of-linear-momen\/","og_locale":"en_US","og_type":"article","og_title":"Which one of the following is the quantity of transfer of linear momen","og_description":"The correct answer is B) 5.0 kg-m\/s. The transfer of linear momentum to the floor is equal to the magnitude of the change in linear momentum of the dumbbell during the impact, but in the opposite direction. First, calculate the velocity of the dumbbell just before hitting the floor. Using the equation $v^2 = u^2 + 2gh$, where $u=0$ (starts from rest), $g=10 text{ m\/s}^2$, and $h=5 text{ m}$. $v^2 = 0^2 + 2 times 10 times 5 = 100 text{ m}^2\/text{s}^2$. $v = sqrt{100} = 10 text{ m\/s}$. The direction is downwards. The mass of the dumbbell is $m = 500 text{ g} = 0.5 text{ kg}$. The momentum of the dumbbell just before impact is $p_{initial} = m times v = 0.5 text{ kg} times 10 text{ m\/s} = 5 text{ kg-m\/s}$ (downwards). After hitting the floor, the dumbbell stops, so its final velocity is 0. The momentum of the dumbbell after impact is $p_{final} = m times 0 = 0$. The change in momentum of the dumbbell is $Delta p_{dumbbell} = p_{final} - p_{initial} = 0 - (5 text{ kg-m\/s downwards}) = -5 text{ kg-m\/s downwards} = 5 text{ kg-m\/s upwards}$. By the impulse-momentum theorem and Newton's third law, the momentum transferred to the floor is equal in magnitude and opposite in direction to the change in momentum of the dumbbell. Momentum transfer to floor = $-Delta p_{dumbbell} = -(5 text{ kg-m\/s upwards}) = 5 text{ kg-m\/s downwards}$. The question asks for the *quantity* of transfer, which is the magnitude. The magnitude is 5 kg-m\/s.","og_url":"https:\/\/exam.pscnotes.com\/mcq\/which-one-of-the-following-is-the-quantity-of-transfer-of-linear-momen\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T04:27:17+00:00","author":"rawan239","twitter_card":"summary_large_image","twitter_misc":{"Written by":"rawan239","Est. reading time":"2 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"WebPage","@id":"https:\/\/exam.pscnotes.com\/mcq\/which-one-of-the-following-is-the-quantity-of-transfer-of-linear-momen\/","url":"https:\/\/exam.pscnotes.com\/mcq\/which-one-of-the-following-is-the-quantity-of-transfer-of-linear-momen\/","name":"Which one of the following is the quantity of transfer of linear momen","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T04:27:17+00:00","dateModified":"2025-06-01T04:27:17+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"The correct answer is B) 5.0 kg-m\/s. The transfer of linear momentum to the floor is equal to the magnitude of the change in linear momentum of the dumbbell during the impact, but in the opposite direction. First, calculate the velocity of the dumbbell just before hitting the floor. Using the equation $v^2 = u^2 + 2gh$, where $u=0$ (starts from rest), $g=10 \\text{ m\/s}^2$, and $h=5 \\text{ m}$. $v^2 = 0^2 + 2 \\times 10 \\times 5 = 100 \\text{ m}^2\/\\text{s}^2$. $v = \\sqrt{100} = 10 \\text{ m\/s}$. The direction is downwards. The mass of the dumbbell is $m = 500 \\text{ g} = 0.5 \\text{ kg}$. The momentum of the dumbbell just before impact is $p_{initial} = m \\times v = 0.5 \\text{ kg} \\times 10 \\text{ m\/s} = 5 \\text{ kg-m\/s}$ (downwards). After hitting the floor, the dumbbell stops, so its final velocity is 0. The momentum of the dumbbell after impact is $p_{final} = m \\times 0 = 0$. The change in momentum of the dumbbell is $\\Delta p_{dumbbell} = p_{final} - p_{initial} = 0 - (5 \\text{ kg-m\/s downwards}) = -5 \\text{ kg-m\/s downwards} = 5 \\text{ kg-m\/s upwards}$. By the impulse-momentum theorem and Newton's third law, the momentum transferred to the floor is equal in magnitude and opposite in direction to the change in momentum of the dumbbell. Momentum transfer to floor = $-\\Delta p_{dumbbell} = -(5 \\text{ kg-m\/s upwards}) = 5 \\text{ kg-m\/s downwards}$. The question asks for the *quantity* of transfer, which is the magnitude. The magnitude is 5 kg-m\/s.","breadcrumb":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/which-one-of-the-following-is-the-quantity-of-transfer-of-linear-momen\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/exam.pscnotes.com\/mcq\/which-one-of-the-following-is-the-quantity-of-transfer-of-linear-momen\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/exam.pscnotes.com\/mcq\/which-one-of-the-following-is-the-quantity-of-transfer-of-linear-momen\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/exam.pscnotes.com\/mcq\/"},{"@type":"ListItem","position":2,"name":"UPSC Geoscientist","item":"https:\/\/exam.pscnotes.com\/mcq\/category\/upsc-geoscientist\/"},{"@type":"ListItem","position":3,"name":"Which one of the following is the quantity of transfer of linear momen"}]},{"@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\/87060","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=87060"}],"version-history":[{"count":0,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/87060\/revisions"}],"wp:attachment":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/media?parent=87060"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/categories?post=87060"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/tags?post=87060"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}