{"id":88066,"date":"2025-06-01T07:00:47","date_gmt":"2025-06-01T07:00:47","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=88066"},"modified":"2025-06-01T07:00:47","modified_gmt":"2025-06-01T07:00:47","slug":"a-ball-of-0-1-kg-mass-is-dropped-on-a-hard-floor-from-a-height-of-0-45","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/a-ball-of-0-1-kg-mass-is-dropped-on-a-hard-floor-from-a-height-of-0-45\/","title":{"rendered":"A ball of 0.1 kg mass is dropped on a hard floor from a height of 0.45"},"content":{"rendered":"

A ball of 0.1 kg mass is dropped on a hard floor from a height of 0.45 m and rises to a height of 0.20 m. If it was in touch with the floor for 0.1 s, the net force it applied on the floor while bouncing is : (take the gravitational acceleration g = 10 m s\u207b\u00b2)<\/p>\n

[amp_mcq option1=”1.0 N” option2=”6.0 N” option3=”3.0 N” option4=”5.0 N” correct=”option2″]<\/p>\n

\n
\n
This question was previously asked in<\/div>\n
UPSC NDA-1 – 2024<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nThe correct option is B. The net force the ball applied on the floor while bouncing is 6.0 N.
\n<\/section>\n
\nFirst, calculate the velocity of the ball just before impact (v\u2081) and just after impact (v\u2082). Using energy conservation (v\u00b2 = 2gh), v\u2081 = sqrt(2 * 10 * 0.45) = 3 m\/s (downward) and v\u2082 = sqrt(2 * 10 * 0.20) = 2 m\/s (upward). Let’s take upward as positive. v\u2081 = -3 m\/s, v\u2082 = +2 m\/s. The change in momentum of the ball is \u0394p = m(v\u2082 – v\u2081) = 0.1 kg * (2 – (-3)) m\/s = 0.1 * 5 = 0.5 kg m\/s (upward). The average net force on the ball during contact is F_net_on_ball = \u0394p \/ \u0394t = 0.5 Ns \/ 0.1 s = 5 N (upward). This net force is the sum of the upward normal force (N) from the floor and the downward gravitational force (mg): F_net_on_ball = N – mg. So, 5 N = N – (0.1 kg * 10 m\/s\u00b2) = N – 1 N. This gives the average normal force from the floor on the ball, N = 6 N (upward). By Newton’s third law, the force applied by the ball on the floor is equal in magnitude and opposite in direction to the normal force from the floor on the ball. Thus, the force applied on the floor is 6 N (downward). The question asks for the magnitude of this force.
\n<\/section>\n
\nThe net force applied by the ball on the floor is essentially the reaction force to the average normal force exerted by the floor on the ball during the collision. Gravity also acts on the ball during the contact time, but the primary force during bouncing is the large normal force from the floor. The calculation involving the change in momentum accounts for the effect of all forces during the contact period, including gravity.
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

A ball of 0.1 kg mass is dropped on a hard floor from a height of 0.45 m and rises to a height of 0.20 m. If it was in touch with the floor for 0.1 s, the net force it applied on the floor while bouncing is : (take the gravitational acceleration g = … <\/p>\n

Detailed SolutionA ball of 0.1 kg mass is dropped on a hard floor from a height of 0.45<\/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":[1103,1129,1128],"class_list":["post-88066","post","type-post","status-publish","format-standard","hentry","category-upsc-nda-1","tag-1103","tag-mechanics","tag-physics","no-featured-image-padding"],"yoast_head":"\nA ball of 0.1 kg mass is dropped on a hard floor from a height of 0.45<\/title>\n<meta name=\"description\" content=\"The correct option is B. The net force the ball applied on the floor while bouncing is 6.0 N. First, calculate the velocity of the ball just before impact (v\u2081) and just after impact (v\u2082). Using energy conservation (v\u00b2 = 2gh), v\u2081 = sqrt(2 * 10 * 0.45) = 3 m\/s (downward) and v\u2082 = sqrt(2 * 10 * 0.20) = 2 m\/s (upward). Let's take upward as positive. v\u2081 = -3 m\/s, v\u2082 = +2 m\/s. The change in momentum of the ball is \u0394p = m(v\u2082 - v\u2081) = 0.1 kg * (2 - (-3)) m\/s = 0.1 * 5 = 0.5 kg m\/s (upward). The average net force on the ball during contact is F_net_on_ball = \u0394p \/ \u0394t = 0.5 Ns \/ 0.1 s = 5 N (upward). This net force is the sum of the upward normal force (N) from the floor and the downward gravitational force (mg): F_net_on_ball = N - mg. So, 5 N = N - (0.1 kg * 10 m\/s\u00b2) = N - 1 N. This gives the average normal force from the floor on the ball, N = 6 N (upward). By Newton's third law, the force applied by the ball on the floor is equal in magnitude and opposite in direction to the normal force from the floor on the ball. Thus, the force applied on the floor is 6 N (downward). The question asks for the magnitude of this force.\" \/>\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-ball-of-0-1-kg-mass-is-dropped-on-a-hard-floor-from-a-height-of-0-45\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"A ball of 0.1 kg mass is dropped on a hard floor from a height of 0.45\" \/>\n<meta property=\"og:description\" content=\"The correct option is B. The net force the ball applied on the floor while bouncing is 6.0 N. First, calculate the velocity of the ball just before impact (v\u2081) and just after impact (v\u2082). Using energy conservation (v\u00b2 = 2gh), v\u2081 = sqrt(2 * 10 * 0.45) = 3 m\/s (downward) and v\u2082 = sqrt(2 * 10 * 0.20) = 2 m\/s (upward). Let's take upward as positive. v\u2081 = -3 m\/s, v\u2082 = +2 m\/s. The change in momentum of the ball is \u0394p = m(v\u2082 - v\u2081) = 0.1 kg * (2 - (-3)) m\/s = 0.1 * 5 = 0.5 kg m\/s (upward). The average net force on the ball during contact is F_net_on_ball = \u0394p \/ \u0394t = 0.5 Ns \/ 0.1 s = 5 N (upward). This net force is the sum of the upward normal force (N) from the floor and the downward gravitational force (mg): F_net_on_ball = N - mg. So, 5 N = N - (0.1 kg * 10 m\/s\u00b2) = N - 1 N. This gives the average normal force from the floor on the ball, N = 6 N (upward). By Newton's third law, the force applied by the ball on the floor is equal in magnitude and opposite in direction to the normal force from the floor on the ball. Thus, the force applied on the floor is 6 N (downward). The question asks for the magnitude of this force.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/a-ball-of-0-1-kg-mass-is-dropped-on-a-hard-floor-from-a-height-of-0-45\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T07:00:47+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":"A ball of 0.1 kg mass is dropped on a hard floor from a height of 0.45","description":"The correct option is B. The net force the ball applied on the floor while bouncing is 6.0 N. First, calculate the velocity of the ball just before impact (v\u2081) and just after impact (v\u2082). Using energy conservation (v\u00b2 = 2gh), v\u2081 = sqrt(2 * 10 * 0.45) = 3 m\/s (downward) and v\u2082 = sqrt(2 * 10 * 0.20) = 2 m\/s (upward). Let's take upward as positive. v\u2081 = -3 m\/s, v\u2082 = +2 m\/s. The change in momentum of the ball is \u0394p = m(v\u2082 - v\u2081) = 0.1 kg * (2 - (-3)) m\/s = 0.1 * 5 = 0.5 kg m\/s (upward). The average net force on the ball during contact is F_net_on_ball = \u0394p \/ \u0394t = 0.5 Ns \/ 0.1 s = 5 N (upward). This net force is the sum of the upward normal force (N) from the floor and the downward gravitational force (mg): F_net_on_ball = N - mg. So, 5 N = N - (0.1 kg * 10 m\/s\u00b2) = N - 1 N. This gives the average normal force from the floor on the ball, N = 6 N (upward). By Newton's third law, the force applied by the ball on the floor is equal in magnitude and opposite in direction to the normal force from the floor on the ball. Thus, the force applied on the floor is 6 N (downward). The question asks for the magnitude of this force.","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-ball-of-0-1-kg-mass-is-dropped-on-a-hard-floor-from-a-height-of-0-45\/","og_locale":"en_US","og_type":"article","og_title":"A ball of 0.1 kg mass is dropped on a hard floor from a height of 0.45","og_description":"The correct option is B. The net force the ball applied on the floor while bouncing is 6.0 N. First, calculate the velocity of the ball just before impact (v\u2081) and just after impact (v\u2082). Using energy conservation (v\u00b2 = 2gh), v\u2081 = sqrt(2 * 10 * 0.45) = 3 m\/s (downward) and v\u2082 = sqrt(2 * 10 * 0.20) = 2 m\/s (upward). Let's take upward as positive. v\u2081 = -3 m\/s, v\u2082 = +2 m\/s. The change in momentum of the ball is \u0394p = m(v\u2082 - v\u2081) = 0.1 kg * (2 - (-3)) m\/s = 0.1 * 5 = 0.5 kg m\/s (upward). The average net force on the ball during contact is F_net_on_ball = \u0394p \/ \u0394t = 0.5 Ns \/ 0.1 s = 5 N (upward). This net force is the sum of the upward normal force (N) from the floor and the downward gravitational force (mg): F_net_on_ball = N - mg. So, 5 N = N - (0.1 kg * 10 m\/s\u00b2) = N - 1 N. This gives the average normal force from the floor on the ball, N = 6 N (upward). By Newton's third law, the force applied by the ball on the floor is equal in magnitude and opposite in direction to the normal force from the floor on the ball. Thus, the force applied on the floor is 6 N (downward). The question asks for the magnitude of this force.","og_url":"https:\/\/exam.pscnotes.com\/mcq\/a-ball-of-0-1-kg-mass-is-dropped-on-a-hard-floor-from-a-height-of-0-45\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T07:00:47+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\/a-ball-of-0-1-kg-mass-is-dropped-on-a-hard-floor-from-a-height-of-0-45\/","url":"https:\/\/exam.pscnotes.com\/mcq\/a-ball-of-0-1-kg-mass-is-dropped-on-a-hard-floor-from-a-height-of-0-45\/","name":"A ball of 0.1 kg mass is dropped on a hard floor from a height of 0.45","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T07:00:47+00:00","dateModified":"2025-06-01T07:00:47+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"The correct option is B. The net force the ball applied on the floor while bouncing is 6.0 N. First, calculate the velocity of the ball just before impact (v\u2081) and just after impact (v\u2082). Using energy conservation (v\u00b2 = 2gh), v\u2081 = sqrt(2 * 10 * 0.45) = 3 m\/s (downward) and v\u2082 = sqrt(2 * 10 * 0.20) = 2 m\/s (upward). Let's take upward as positive. v\u2081 = -3 m\/s, v\u2082 = +2 m\/s. The change in momentum of the ball is \u0394p = m(v\u2082 - v\u2081) = 0.1 kg * (2 - (-3)) m\/s = 0.1 * 5 = 0.5 kg m\/s (upward). The average net force on the ball during contact is F_net_on_ball = \u0394p \/ \u0394t = 0.5 Ns \/ 0.1 s = 5 N (upward). This net force is the sum of the upward normal force (N) from the floor and the downward gravitational force (mg): F_net_on_ball = N - mg. So, 5 N = N - (0.1 kg * 10 m\/s\u00b2) = N - 1 N. This gives the average normal force from the floor on the ball, N = 6 N (upward). By Newton's third law, the force applied by the ball on the floor is equal in magnitude and opposite in direction to the normal force from the floor on the ball. Thus, the force applied on the floor is 6 N (downward). The question asks for the magnitude of this force.","breadcrumb":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/a-ball-of-0-1-kg-mass-is-dropped-on-a-hard-floor-from-a-height-of-0-45\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/exam.pscnotes.com\/mcq\/a-ball-of-0-1-kg-mass-is-dropped-on-a-hard-floor-from-a-height-of-0-45\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/exam.pscnotes.com\/mcq\/a-ball-of-0-1-kg-mass-is-dropped-on-a-hard-floor-from-a-height-of-0-45\/#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 ball of 0.1 kg mass is dropped on a hard floor from a height of 0.45"}]},{"@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\/88066","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=88066"}],"version-history":[{"count":0,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/88066\/revisions"}],"wp:attachment":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/media?parent=88066"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/categories?post=88066"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/tags?post=88066"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}