{"id":88919,"date":"2025-06-01T07:24:50","date_gmt":"2025-06-01T07:24:50","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=88919"},"modified":"2025-06-01T07:24:50","modified_gmt":"2025-06-01T07:24:50","slug":"a-bullet-of-mass-10-g-is-horizontally-fired-with-velocity-300-m-s-1","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/a-bullet-of-mass-10-g-is-horizontally-fired-with-velocity-300-m-s-1\/","title":{"rendered":"A bullet of mass 10 g is horizontally fired with velocity 300 m s$^{-1"},"content":{"rendered":"

A bullet of mass 10 g is horizontally fired with velocity 300 m s$^{-1}$ from a pistol of mass 1 kg. What is the recoil velocity of the pistol?<\/p>\n

[amp_mcq option1=”0\u00b73 m s$^{-1}$” option2=”3 m s$^{-1}$” option3=”\u22123 m s$^{-1}$” option4=”\u22120\u00b73 m s$^{-1}$” correct=”option3″]<\/p>\n

\n
\n
This question was previously asked in<\/div>\n
UPSC NDA-2 – 2022<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nThe recoil velocity of the pistol is -3 m\/s.
\n<\/section>\n
\n– This problem can be solved using the principle of conservation of linear momentum.
\n– The total momentum of the system (pistol + bullet) before firing is zero, as both are at rest.
\n– According to the conservation of momentum, the total momentum of the system after firing must also be zero.
\n– Let $m_b$ and $v_b$ be the mass and velocity of the bullet, and $m_p$ and $v_p$ be the mass and recoil velocity of the pistol.
\n– Momentum before firing = 0
\n– Momentum after firing = $m_b v_b + m_p v_p$
\n– By conservation of momentum: $m_b v_b + m_p v_p = 0$
\n– Given: $m_b = 10 \\text{ g} = 0.01 \\text{ kg}$, $v_b = 300 \\text{ m\/s}$, $m_p = 1 \\text{ kg}$.
\n– Substituting the values: $(0.01 \\text{ kg})(300 \\text{ m\/s}) + (1 \\text{ kg}) v_p = 0$
\n– $3 \\text{ kg} \\cdot \\text{m\/s} + v_p \\text{ kg} = 0$
\n– $v_p = -3 \\text{ m\/s}$
\n<\/section>\n
\nThe negative sign for the recoil velocity indicates that the pistol moves in the opposite direction to the bullet. The magnitude of the recoil velocity is 3 m\/s. This demonstrates Newton’s third law of motion (action-reaction) applied to momentum conservation.
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

A bullet of mass 10 g is horizontally fired with velocity 300 m s$^{-1}$ from a pistol of mass 1 kg. What is the recoil velocity of the pistol? [amp_mcq option1=”0\u00b73 m s$^{-1}$” option2=”3 m s$^{-1}$” option3=”\u22123 m s$^{-1}$” option4=”\u22120\u00b73 m s$^{-1}$” correct=”option3″] This question was previously asked in UPSC NDA-2 – 2022 Download PDFAttempt … <\/p>\n

Detailed SolutionA bullet of mass 10 g is horizontally fired with velocity 300 m s$^{-1<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1094],"tags":[1108,1129,1128],"class_list":["post-88919","post","type-post","status-publish","format-standard","hentry","category-upsc-nda-2","tag-1108","tag-mechanics","tag-physics","no-featured-image-padding"],"yoast_head":"\nA bullet of mass 10 g is horizontally fired with velocity 300 m s$^{-1<\/title>\n<meta name=\"description\" content=\"The recoil velocity of the pistol is -3 m\/s. - This problem can be solved using the principle of conservation of linear momentum. - The total momentum of the system (pistol + bullet) before firing is zero, as both are at rest. - According to the conservation of momentum, the total momentum of the system after firing must also be zero. - Let $m_b$ and $v_b$ be the mass and velocity of the bullet, and $m_p$ and $v_p$ be the mass and recoil velocity of the pistol. - Momentum before firing = 0 - Momentum after firing = $m_b v_b + m_p v_p$ - By conservation of momentum: $m_b v_b + m_p v_p = 0$ - Given: $m_b = 10 text{ g} = 0.01 text{ kg}$, $v_b = 300 text{ m\/s}$, $m_p = 1 text{ kg}$. - Substituting the values: $(0.01 text{ kg})(300 text{ m\/s}) + (1 text{ kg}) v_p = 0$ - $3 text{ kg} cdot text{m\/s} + v_p text{ kg} = 0$ - $v_p = -3 text{ m\/s}$\" \/>\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-bullet-of-mass-10-g-is-horizontally-fired-with-velocity-300-m-s-1\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"A bullet of mass 10 g is horizontally fired with velocity 300 m s$^{-1\" \/>\n<meta property=\"og:description\" content=\"The recoil velocity of the pistol is -3 m\/s. - This problem can be solved using the principle of conservation of linear momentum. - The total momentum of the system (pistol + bullet) before firing is zero, as both are at rest. - According to the conservation of momentum, the total momentum of the system after firing must also be zero. - Let $m_b$ and $v_b$ be the mass and velocity of the bullet, and $m_p$ and $v_p$ be the mass and recoil velocity of the pistol. - Momentum before firing = 0 - Momentum after firing = $m_b v_b + m_p v_p$ - By conservation of momentum: $m_b v_b + m_p v_p = 0$ - Given: $m_b = 10 text{ g} = 0.01 text{ kg}$, $v_b = 300 text{ m\/s}$, $m_p = 1 text{ kg}$. - Substituting the values: $(0.01 text{ kg})(300 text{ m\/s}) + (1 text{ kg}) v_p = 0$ - $3 text{ kg} cdot text{m\/s} + v_p text{ kg} = 0$ - $v_p = -3 text{ m\/s}$\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/a-bullet-of-mass-10-g-is-horizontally-fired-with-velocity-300-m-s-1\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T07:24:50+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 bullet of mass 10 g is horizontally fired with velocity 300 m s$^{-1","description":"The recoil velocity of the pistol is -3 m\/s. - This problem can be solved using the principle of conservation of linear momentum. - The total momentum of the system (pistol + bullet) before firing is zero, as both are at rest. - According to the conservation of momentum, the total momentum of the system after firing must also be zero. - Let $m_b$ and $v_b$ be the mass and velocity of the bullet, and $m_p$ and $v_p$ be the mass and recoil velocity of the pistol. - Momentum before firing = 0 - Momentum after firing = $m_b v_b + m_p v_p$ - By conservation of momentum: $m_b v_b + m_p v_p = 0$ - Given: $m_b = 10 text{ g} = 0.01 text{ kg}$, $v_b = 300 text{ m\/s}$, $m_p = 1 text{ kg}$. - Substituting the values: $(0.01 text{ kg})(300 text{ m\/s}) + (1 text{ kg}) v_p = 0$ - $3 text{ kg} cdot text{m\/s} + v_p text{ kg} = 0$ - $v_p = -3 text{ m\/s}$","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-bullet-of-mass-10-g-is-horizontally-fired-with-velocity-300-m-s-1\/","og_locale":"en_US","og_type":"article","og_title":"A bullet of mass 10 g is horizontally fired with velocity 300 m s$^{-1","og_description":"The recoil velocity of the pistol is -3 m\/s. - This problem can be solved using the principle of conservation of linear momentum. - The total momentum of the system (pistol + bullet) before firing is zero, as both are at rest. - According to the conservation of momentum, the total momentum of the system after firing must also be zero. - Let $m_b$ and $v_b$ be the mass and velocity of the bullet, and $m_p$ and $v_p$ be the mass and recoil velocity of the pistol. - Momentum before firing = 0 - Momentum after firing = $m_b v_b + m_p v_p$ - By conservation of momentum: $m_b v_b + m_p v_p = 0$ - Given: $m_b = 10 text{ g} = 0.01 text{ kg}$, $v_b = 300 text{ m\/s}$, $m_p = 1 text{ kg}$. - Substituting the values: $(0.01 text{ kg})(300 text{ m\/s}) + (1 text{ kg}) v_p = 0$ - $3 text{ kg} cdot text{m\/s} + v_p text{ kg} = 0$ - $v_p = -3 text{ m\/s}$","og_url":"https:\/\/exam.pscnotes.com\/mcq\/a-bullet-of-mass-10-g-is-horizontally-fired-with-velocity-300-m-s-1\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T07:24:50+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-bullet-of-mass-10-g-is-horizontally-fired-with-velocity-300-m-s-1\/","url":"https:\/\/exam.pscnotes.com\/mcq\/a-bullet-of-mass-10-g-is-horizontally-fired-with-velocity-300-m-s-1\/","name":"A bullet of mass 10 g is horizontally fired with velocity 300 m s$^{-1","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T07:24:50+00:00","dateModified":"2025-06-01T07:24:50+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"The recoil velocity of the pistol is -3 m\/s. - This problem can be solved using the principle of conservation of linear momentum. - The total momentum of the system (pistol + bullet) before firing is zero, as both are at rest. - According to the conservation of momentum, the total momentum of the system after firing must also be zero. - Let $m_b$ and $v_b$ be the mass and velocity of the bullet, and $m_p$ and $v_p$ be the mass and recoil velocity of the pistol. - Momentum before firing = 0 - Momentum after firing = $m_b v_b + m_p v_p$ - By conservation of momentum: $m_b v_b + m_p v_p = 0$ - Given: $m_b = 10 \\text{ g} = 0.01 \\text{ kg}$, $v_b = 300 \\text{ m\/s}$, $m_p = 1 \\text{ kg}$. - Substituting the values: $(0.01 \\text{ kg})(300 \\text{ m\/s}) + (1 \\text{ kg}) v_p = 0$ - $3 \\text{ kg} \\cdot \\text{m\/s} + v_p \\text{ kg} = 0$ - $v_p = -3 \\text{ m\/s}$","breadcrumb":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/a-bullet-of-mass-10-g-is-horizontally-fired-with-velocity-300-m-s-1\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/exam.pscnotes.com\/mcq\/a-bullet-of-mass-10-g-is-horizontally-fired-with-velocity-300-m-s-1\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/exam.pscnotes.com\/mcq\/a-bullet-of-mass-10-g-is-horizontally-fired-with-velocity-300-m-s-1\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/exam.pscnotes.com\/mcq\/"},{"@type":"ListItem","position":2,"name":"UPSC NDA-2","item":"https:\/\/exam.pscnotes.com\/mcq\/category\/upsc-nda-2\/"},{"@type":"ListItem","position":3,"name":"A bullet of mass 10 g is horizontally fired with velocity 300 m s$^{-1"}]},{"@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\/88919","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=88919"}],"version-history":[{"count":0,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/88919\/revisions"}],"wp:attachment":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/media?parent=88919"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/categories?post=88919"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/tags?post=88919"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}