{"id":13693,"date":"2022-01-30T14:04:37","date_gmt":"2022-01-30T08:34:37","guid":{"rendered":"https:\/\/infinitylearn.com\/surge\/?p=13693"},"modified":"2025-05-15T17:12:51","modified_gmt":"2025-05-15T11:42:51","slug":"jee-main-physics-kinematics-previous-year-questions-with-solutions","status":"publish","type":"post","link":"https:\/\/infinitylearn.com\/surge\/iit-jee\/study-materials\/physics\/jee-main-physics-kinematics-previous-year-questions-with-solutions\/","title":{"rendered":"JEE Main Physics Kinematics Previous Year Questions"},"content":{"rendered":"<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_37 counter-hierarchy ez-toc-counter ez-toc-grey ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title\">Table of Contents<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" style=\"display: none;\"><label for=\"item\" aria-label=\"Table of Content\"><span style=\"display: flex;align-items: center;width: 35px;height: 30px;justify-content: center;\"><svg style=\"fill: #999;color:#999\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" class=\"list-377408\" width=\"20px\" height=\"20px\" viewBox=\"0 0 24 24\" fill=\"none\"><path d=\"M6 6H4v2h2V6zm14 0H8v2h12V6zM4 11h2v2H4v-2zm16 0H8v2h12v-2zM4 16h2v2H4v-2zm16 0H8v2h12v-2z\" fill=\"currentColor\"><\/path><\/svg><svg style=\"fill: #999;color:#999\" class=\"arrow-unsorted-368013\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" width=\"10px\" height=\"10px\" viewBox=\"0 0 24 24\" version=\"1.2\" baseProfile=\"tiny\"><path d=\"M18.2 9.3l-6.2-6.3-6.2 6.3c-.2.2-.3.4-.3.7s.1.5.3.7c.2.2.4.3.7.3h11c.3 0 .5-.1.7-.3.2-.2.3-.5.3-.7s-.1-.5-.3-.7zM5.8 14.7l6.2 6.3 6.2-6.3c.2-.2.3-.5.3-.7s-.1-.5-.3-.7c-.2-.2-.4-.3-.7-.3h-11c-.3 0-.5.1-.7.3-.2.2-.3.5-.3.7s.1.5.3.7z\"\/><\/svg><\/span><\/label><input type=\"checkbox\" id=\"item\"><\/a><\/span><\/div>\n<nav><ul class='ez-toc-list ez-toc-list-level-1' style='display:block'><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-1\" href=\"https:\/\/infinitylearn.com\/surge\/iit-jee\/study-materials\/physics\/jee-main-physics-kinematics-previous-year-questions-with-solutions\/#JEE_Main_2024_Physics_Kinematics_Previous_Year_Questions_with_Solutions\" title=\"JEE Main 2024 Physics Kinematics Previous Year Questions with Solutions\">JEE Main 2024 Physics Kinematics Previous Year Questions with Solutions<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/infinitylearn.com\/surge\/iit-jee\/study-materials\/physics\/jee-main-physics-kinematics-previous-year-questions-with-solutions\/#Assertion_Reasoning_type\" title=\"Assertion &amp; Reasoning type\">Assertion &amp; Reasoning type<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/infinitylearn.com\/surge\/iit-jee\/study-materials\/physics\/jee-main-physics-kinematics-previous-year-questions-with-solutions\/#True_False_Type\" title=\"True \/ False Type\">True \/ False Type<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/infinitylearn.com\/surge\/iit-jee\/study-materials\/physics\/jee-main-physics-kinematics-previous-year-questions-with-solutions\/#Fill_in_the_blanks\" title=\"Fill in the blanks\">Fill in the blanks<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"JEE_Main_2024_Physics_Kinematics_Previous_Year_Questions_with_Solutions\"><\/span>JEE Main 2024 Physics Kinematics Previous Year Questions with Solutions<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>For JEE Main other Engineering Entrance Exam Preparation,  JEE Main Physics Kinematics Previous Year Questions with Solutions is given below.<\/p>\n<p><strong>New: <a href=\"https:\/\/infinitylearn.com\/surge\/iit-jee\/jee-main-city-intimation-slip-2024\/\">JEE Main Session 2 city intimation slip download link<\/a><\/strong> | <a href=\"https:\/\/infinitylearn.com\/surge\/iit-jee\/jee-mains-admit-card-2024-session-2\/\"><strong>JEE Main Session 2 Admit Card 2024 Live<\/strong><\/a><\/p>\n<p><strong>Don\u2019t Miss: JEE Main: <a href=\"https:\/\/infinitylearn.com\/surge\/iit-jee\/jee-main-2024-session-2-4-april-shift-1-2-question-paper-analysis-difficulty-level-and-pdf-solutions\/\">4 April Question Paper Analysis<\/a> | <a href=\"https:\/\/infinitylearn.com\/surge\/iit-jee\/jee-main-2024-session-2-5-april-shift-1-2-question-paper-analysis-difficulty-level-and-pdf-solutions\/\">5 April Question Paper Analysis<\/a> | <a href=\"https:\/\/infinitylearn.com\/surge\/iit-jee\/jee-main-2024-analysis-for-6-april-shift-1-2-exams\/\">6 April Question Paper Analysis<\/a><\/strong><\/p>\n<p><strong>Latest: JEE Main 2024: <a href=\"https:\/\/infinitylearn.com\/jee-main-rank-college-predictor\">Rank predictor<\/a> |<a href=\"https:\/\/infinitylearn.com\/surge\/iit-jee\/jee-main-college-predictor\/\"> College predictor<\/a><\/strong> | <a href=\"https:\/\/infinitylearn.com\/surge\/blog\/iit-jee\/what-is-the-fee-structure-in-iit\/\"><strong>IIT JEE Fee Structure 2024<\/strong><\/a><\/p>\n<p><strong>Q1. A piece of wood of mass 0.03 kg is dropped from the top of a 100 m height building. At the same time, a bullet of mass 0.02 Kg is fired vertically upward, with a velocity 100 m\/s, from the ground. The bullet gets embedded in the wood. Then the maximum height to which the combined system reaches above the top of the building before falling below is (g = 10 m\/s)<\/strong><\/p>\n<ol>\n<li>10 m<\/li>\n<li>30 m<\/li>\n<li>20 m<\/li>\n<li>40 m<\/li>\n<\/ol>\n<p><strong>Solution: <\/strong>Suppose both collide at the point P after time t. Time taken for the particles to collide,<\/p>\n<p>t = d\/v<sub>rel <\/sub>= 100\/100 = 1s<\/p>\n<p>Speed of wood just before collision =gt = 10m\/s<\/p>\n<p>Speed of bullet just before collision<\/p>\n<p>v -gt = 100 -10 = 90 m\/s<\/p>\n<p>Before<\/p>\n<p>0.03 kg \u2193 10 m\/s<\/p>\n<p>0.02 kg \u2191 90 m\/s<\/p>\n<p>After<\/p>\n<p>\u2191 v<\/p>\n<p>0.05 kg<\/p>\n<p>Now, the conservation of linear momentum just before and after the collision<\/p>\n<p>-(0.03)(10) + (0.02)(90) = (0.05)v \u21d2 v = 30 m\/s<\/p>\n<p>The maximum height reached by the body a = v<sup>2<\/sup>\/2g<\/p>\n<p>= (30)<sup>2<\/sup>\/2(10)<\/p>\n<p>= 45 m<\/p>\n<p>(100 -h) = \u00bd gt<sup>2<\/sup> = \u00bd x 10 x1 \u21d2h = 95 m<\/p>\n<p>Height above tower = 40 m<\/p>\n<p><strong>Answer: (d) <\/strong><\/p>\n<div class=\"card\" style=\"text-align: center;\"><a class=\"btn btn-primary\" href=\"https:\/\/infinitylearn.com\/test-series\/jee-12-aits?utm_source=surge&amp;utm_medium=Org&amp;utm_campaign=AITS&amp;mrk_coupon_code=AITS30&amp;tracking_source=il_surge\"><br \/>\n<img loading=\"lazy\" class=\" lazyloaded\" style=\"width: 100%; border-radius: 10px;\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2024\/12\/aitsclass12.png\" alt=\"class 12 jee test series\" width=\"1201\" height=\"636\" data-src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2024\/12\/aitsclass12.png\" \/><\/a><a class=\"btn btn-primary\" href=\"https:\/\/infinitylearn.com\/test-series\/jee-12-aits?utm_source=surge&amp;utm_medium=Org&amp;utm_campaign=AITS&amp;mrk_coupon_code=AITS30&amp;tracking_source=il_surge\"><button><strong>Get 30 Off! Use Code: AITS30<\/strong><\/button><\/a><\/p>\n<div style=\"text-align: center;\">Boost Your Preparation With Our All India Test Series for Class 12 JEE Main &amp; Advanced 2025<\/div>\n<\/div>\n<p><strong>Q2. A passenger train of length 60 m travels at a speed of 80 km\/hr. Another freight train of length 120 m travels at a speed of 30 km\/hr. The ratio of times taken by the passenger train to completely cross the freight train when: (i) they are moving in the same direction, and (ii) in the opposite direction, is<\/strong><\/p>\n<p>(a) 25\/11<\/p>\n<p>(b) 3\/2<\/p>\n<p>(c) 5\/2<\/p>\n<p>(d) 11\/5<\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>The total distance to be travelled by train is 60 + 120 = 180 m.<\/p>\n<p>When the trains are moving in the same direction, the relative velocity is v<sub>1<\/sub> \u2013 v<sub>2<\/sub> = 80 \u2013 30 = 50 km hr<sup>\u20131<\/sup>.<\/p>\n<p>So time taken to cross each other, t<sub>1<\/sub> = 180\/(50 x 10<sup>3<\/sup>\/3600) = [(18 x 18)\/25] s<\/p>\n<p>When the trains are moving in opposite direction, relative velocity is |v1 \u2013 (\u2013v2 )| = 80 + 30 = 110 km hr<sup>\u20131<\/sup><\/p>\n<p>So time taken to cross each other<\/p>\n<p>t<sub>2<\/sub>= 180\/(110 x 10<sup>3<\/sup>\/3600) = [(18 x 36)\/110] s<\/p>\n<p>t<sub>1<\/sub>\/t<sub>2<\/sub>= [(18 x 18)\/25] \/ [(18 x 36)\/110] = 11\/5<\/p>\n<p><strong>Answers: (d) <\/strong><\/p>\n<div class=\"card\" style=\"text-align: center;\"><a class=\"btn btn-primary\" href=\"https:\/\/infinitylearn.com\/test-series\/jee-13-aits?utm_source=surge&amp;utm_medium=Org&amp;utm_campaign=AITS&amp;mrk_coupon_code=AITS30&amp;tracking_source=il_surge\"><br \/>\n<img loading=\"lazy\" class=\" lazyloaded\" style=\"width: 100%; border-radius: 10px;\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2024\/12\/aitsclass12.png\" alt=\"class 12 jee test series\" width=\"1201\" height=\"636\" data-src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2024\/12\/aitsclass12.png\" \/><\/a><a class=\"btn btn-primary\" href=\"https:\/\/infinitylearn.com\/test-series\/jee-12-aits?utm_source=surge&amp;utm_medium=Org&amp;utm_campaign=AITS&amp;mrk_coupon_code=AITS30&amp;tracking_source=il_surge\"><button><strong>Get 30 Off! Use Code: AITS30<\/strong><\/button><\/a><\/p>\n<div style=\"text-align: center;\">Boost Your Preparation With Our All India Test Series for Class 13 JEE Main &amp; Advanced 2025<\/div>\n<\/div>\n<p><strong>Q3. An automobile travelling at 40 km\/h can be stopped at a distance of 40 m by applying brakes. If the same automobile is travelling at 80 km\/h, the minimum stopping distance, in metres, is (assume no skidding)<\/strong><\/p>\n<p>(a) 100 m<\/p>\n<p>(b) 75 m<\/p>\n<p>(c) 160 m<\/p>\n<p>(d) 150 m<\/p>\n<p><strong>Solution :<\/strong><\/p>\n<p>Using v<sup>2<\/sup> = u<sup>2<\/sup> \u2013 2as<\/p>\n<p>0 = u<sup>2<\/sup> \u2013 2as<\/p>\n<p>S = u<sup>2 <\/sup>\/2a<\/p>\n<p>S<sub>1<\/sub>\/S<sub>2<\/sub> = u<sub>1<\/sub><sup>2<\/sup>\/u<sub>2<\/sub><sup>2<\/sup><\/p>\n<p>S<sub>2 <\/sub>= (u<sub>1<\/sub><sup>2<\/sup>\/u<sub>2<\/sub><sup>2<\/sup>)S<sub>1 <\/sub>= (2)<sup>2<\/sup>(40) = 160 m<\/p>\n<p><strong>Answer: (c)<\/strong><\/p>\n<p style=\"text-align: center;\"><a title=\"JEE Main Exam Centres 2024 (Released)\" href=\"https:\/\/infinitylearn.com\/surge\/iit-jee\/jee-main-exam-centres-2024\/\" target=\"_blank\" rel=\"noopener\"><strong><button>JEE Main Exam Centres 2024 (Released): Click to Check<\/button><\/strong><\/a><\/p>\n<p><strong>Q4. A parachutist after bailing out falls 50 m without friction. When the parachute opens, it decelerates at 2 m\/s<sup>2<\/sup>. He reaches the ground with a speed of 3 m\/s. At what height, did he bailout?<\/strong><\/p>\n<p>(a) 293 m<\/p>\n<p>(b) 111 m<\/p>\n<p>(c) 91 m<\/p>\n<p>(d) 182 m<\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Initially, the parachutist falls under gravity u <sup>2<\/sup> = 2ah = 2 \u00d7 9.8 \u00d7 50 = 980 m<sup>2<\/sup>s <sup>\u20132<\/sup><\/p>\n<p>He reaches the ground with speed = 3 m\/s,<\/p>\n<p>a = \u20132 m s<sup>\u20132<\/sup> \u21d2 (3)<sup>2<\/sup> = u <sup>2 <\/sup>\u2013 2 \u00d7 2 \u00d7 h<sub>1<\/sub><\/p>\n<p>9 = 980 \u2013 4 h<sub>1<\/sub><\/p>\n<p>h<sub>1<\/sub> = 971\/4<\/p>\n<p>h<sub>1<\/sub> = 242.75 m<\/p>\n<p>Total height = 50 + 242.75 = 292.75 = 293 m.<\/p>\n<div class=\"d-grid gap-2\"><a href=\"https:\/\/infinitylearn.com\/surge\/iit-jee\"> <button class=\"btn btn-primary\" style=\"width: 100%; cursor: pointer;\" type=\"button\"><strong>Also Read: JEE 2024<\/strong> <\/button> <\/a><\/div>\n<p><strong>Q5. A car, starting from rest, accelerates at the rate f through a distance s, then continues at a constant speed for time t and then decelerates at the rate f\/2 to come to rest. If the total distance traversed in 15 s, then<\/strong><\/p>\n<p>(a) s= \u00bdft<sup>2<\/sup><\/p>\n<p>(b) s= (\u00bc)ft<sup>2<\/sup><\/p>\n<p>(c) s = ft<\/p>\n<p>(d) s= (1\/72)ft<sup>2<\/sup><\/p>\n<p><strong>Solution: <\/strong>For the first part of the journey, s = s<sub>1<\/sub>,<\/p>\n<p>s<sub>1<\/sub> = \u00bd ft<sub>1<\/sub><sup>2<\/sup>\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026(1)<\/p>\n<p>v = f t<sub>1<\/sub> \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026(2)<\/p>\n<p>For second part of journey,<\/p>\n<p>s<sub>2<\/sub> = vt or s<sub>2<\/sub> = f t<sub>1<\/sub> t \u2026\u2026\u2026\u2026\u2026(3)<\/p>\n<p>For the third part of the journey,<\/p>\n<p>s<sub>3<\/sub> = \u00bd(f\/2)(2t<sub>1<\/sub>)<sup>2<\/sup>= \u00bd x ft<sub>1<\/sub><sup>2<\/sup><\/p>\n<p>s<sub>3<\/sub> = 2s<sub>1 <\/sub>= 2s \u2026\u2026\u2026\u2026\u2026\u2026\u2026.(4)<\/p>\n<p>s<sub>1<\/sub> + s<sub>2<\/sub> + s<sub>3<\/sub> =15s<\/p>\n<p>Or s + ft<sub>1<\/sub>t + 2s = 15s<\/p>\n<p>ft<sub>1<\/sub>t = 12 s<\/p>\n<p>From (1) and (5) we get<\/p>\n<p>(s\/12 s )= ft<sub>1<\/sub><sup>2<\/sup>\/(2 x ft<sub>1<\/sub>t)<\/p>\n<p>Or t<sub>1 <\/sub>= t\/6<\/p>\n<p>Or s= \u00bd ft<sub>1<\/sub><sup>2<\/sup><\/p>\n<p>s= \u00bd f(t\/6)<sup>2<\/sup><\/p>\n<p>s= ft<sup>2<\/sup>\/72<\/p>\n<p><strong>Answer: (d) s= (1\/72)ft<sup>2<\/sup><\/strong><\/p>\n<p style=\"text-align: center;\"><a title=\"JEE Main City Intimation Slip\" href=\"https:\/\/infinitylearn.com\/surge\/iit-jee\/jee-main-city-intimation-slip-2024\/\" target=\"_blank\" rel=\"noopener\"><strong><button>JEE Main City Intimation Slip 2024 Session 2 (Out): Click to Check<\/button><\/strong><\/a><\/p>\n<p><strong>Q6. An automobile travelling with a speed of 60 km\/h, can brake to stop within a distance of 20 m. If the car is going twice as fast, i.e., 120 km\/h, the stopping distance will be<\/strong><\/p>\n<p>(a) 20 m<\/p>\n<p>(b) 40 m<\/p>\n<p>(c) 60 m<\/p>\n<p>(d) 80 m<\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let a be the retardation for both the vehicles.<\/p>\n<p>For automobile, v <sup>2<\/sup> = u <sup>2<\/sup> \u2013 2as<\/p>\n<p>u<sub>1<\/sub> <sup>2<\/sup> \u2013 2as<sub>1<\/sub> = 0<\/p>\n<p>u<sub>1<\/sub> <sup>2<\/sup> = 2as<sub>1<\/sub><\/p>\n<p>Similarly for car, u<sub>2<\/sub> <sup>2<\/sup> = 2as<sub>2<\/sub><\/p>\n<p>(u<sub>2<\/sub>\/u<sub>1<\/sub>)<sup>2<\/sup> = s<sub>2<\/sub>\/s<sub>1<\/sub> = (120\/60)<sup>2<\/sup> = s<sub>2<\/sub>\/20<\/p>\n<p>S<sub>2<\/sub> = 80 m<\/p>\n<p><strong>Answer: (d)<\/strong><\/p>\n<div class=\"table-responsive\">\n<table class=\"table table-bordered table-striped\" cellspacing=\"0\" cellpadding=\"5\">\n<tbody>\n<tr style=\"background-color: #89cff0; color: black;\">\n<td style=\"text-align: center;\" colspan=\"2\"><strong>Also Check<\/strong><\/td>\n<\/tr>\n<tr>\n<td><a href=\"https:\/\/infinitylearn.com\/surge\/iit-jee\/jee-mains\"><strong>JEE Main 2024<\/strong><\/a><\/td>\n<td><a href=\"https:\/\/infinitylearn.com\/surge\/iit-jee\/jee-syllabus\"><strong>JEE Main Syllabus 2024<\/strong><\/a><\/td>\n<\/tr>\n<tr>\n<td><a href=\"https:\/\/infinitylearn.com\/surge\/iit-jee\/jee-mains-exam-pattern\"><strong>JEE Main Exam Pattern 2024<\/strong><\/a><\/td>\n<td><a href=\"https:\/\/infinitylearn.com\/surge\/iit-jee\/jee-advanced\"><strong>JEE Advanced 2024<\/strong><\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<p><strong>Q7. A ball is released from the top of a tower of height h metre. It takes T second to reach the ground. What is the position of the ball in T\/3 second?<\/strong><\/p>\n<p>(a) (h\/9)metre from the ground<\/p>\n<p>(b) (7h\/9) metre from the ground<\/p>\n<p>(c) (8h\/9) metre from the ground<\/p>\n<p>(d) (17h\/18) metre from the ground<\/p>\n<p><strong>Solution: <\/strong>Equation of motion<\/p>\n<p>s= ut + gt<sup>2<\/sup><\/p>\n<p>h = 0 + \u00bd gT<sup>2<\/sup><\/p>\n<p>Or 2h = gT<sup>2<\/sup>\u2026\u2026\u2026(1)<\/p>\n<p>After T\/3 sec, s = 0 +\u00bd x g(T\/3)<sup>2<\/sup>= gT<sup>2<\/sup>\/18<\/p>\n<p>18 s = gT<sup>2<\/sup> \u2026\u2026\u2026\u2026(2)<\/p>\n<p>From (1) and (2), 18 s = 2h<\/p>\n<p>S = (h\/9) m from top.<\/p>\n<p>Height from ground = h \u2013 h\/9 = (8h\/9) m<\/p>\n<p><strong>Answer: (c)<\/strong><\/p>\n<p><strong>Q8. Which of the following statements is false for a particle moving in a circle with a constant angular speed?<\/strong><\/p>\n<p>(a) The velocity vector is tangent to the circle<\/p>\n<p>(b) The acceleration vector is tangent to the circle<\/p>\n<p>(c) The acceleration vector points to the centre of the circle<\/p>\n<p>(d) The velocity and acceleration vectors are perpendicular to each other<\/p>\n<p><strong>Answer: (b) The acceleration vector acts along the radius of the circle. The given statement is false.<\/strong><\/p>\n<p style=\"text-align: center;\"><button style=\"width: 50%;\"><a class=\"btn btn-primary\" href=\"https:\/\/infinitylearn.com\/jee-class-11-online-course\"><strong>Enroll Now to JEE Class 11 Course<\/strong><\/a><\/button><\/p>\n<p><strong>Q9. From a building, two balls A and B are thrown such that A is thrown upwards and B downwards (both vertically). If v<sub>A<\/sub> and v<sub>B<\/sub> are their respective velocities on reaching the ground, then<\/strong><\/p>\n<p>(a) v<sub>B<\/sub> &gt; v<sub>A<\/sub><\/p>\n<p>(b) v<sub>A<\/sub> = v<sub>B<\/sub><\/p>\n<p>(c) v<sub>A<\/sub> &gt; v<sub>B<\/sub><\/p>\n<p>(d) their velocities depend on their masses<\/p>\n<p><strong>Solution<\/strong><\/p>\n<p>Ball A projected upwards with velocity u, falls back with velocity u downwards. It completes its journey to the ground under gravity.<\/p>\n<p>v<sub>A<\/sub> <sup>2<\/sup> = u<sup>2<\/sup> + 2gh \u2026(1)<\/p>\n<p>Ball B starts with downwards velocity u and reaches ground after travelling a vertical distance h v<sub>B<\/sub> 2 = u<sup>2<\/sup> + 2gh \u2026(2)<\/p>\n<p>From (1) and (2)<\/p>\n<p>v<sub>A <\/sub>= v<sub>B<\/sub><\/p>\n<p><strong>Answer: (b)<\/strong><\/p>\n<p style=\"text-align: center;\"><strong><span class=\"TextRun SCXW209897747 BCX0\" lang=\"EN-US\" xml:lang=\"EN-US\" data-contrast=\"none\"><span class=\"NormalTextRun SCXW209897747 BCX0\">Click Here \u2013<\/span><span class=\"NormalTextRun SCXW209897747 BCX0\">If you are looking for the <\/span><\/span><a class=\"Hyperlink SCXW209897747 BCX0\" href=\"https:\/\/infinitylearn.com\/surge\/iit-jee-neet-coaching-lucknow-up\/\" target=\"_blank\" rel=\"noreferrer noopener\"><span class=\"TextRun Underlined SCXW209897747 BCX0\" lang=\"EN-US\" xml:lang=\"EN-US\" data-contrast=\"none\"><span class=\"NormalTextRun SCXW209897747 BCX0\" data-ccp-charstyle=\"Hyperlink\">best IIT Coaching in Lucknow<\/span><\/span><\/a><\/strong><\/p>\n<p><strong>Q10: If a body loses half of its velocity on penetrating 3 cm in a wooden block, then how much will it penetrate more before coming to rest?<\/strong><\/p>\n<p>(a) 1 cm<\/p>\n<p>(b) 2 cm<\/p>\n<p>(c) 3 cm<\/p>\n<p>(d) 4 cm<\/p>\n<p><strong>Solution<\/strong><\/p>\n<p>For first part of penetration, by equation of motion,<\/p>\n<p>(u\/2)<sup>2<\/sup> = u<sup>2<\/sup> -2a(3)<\/p>\n<p>3u<sup>2<\/sup> = 24a \u21d2 u<sup>2<\/sup> = 8a \u2026(1)<\/p>\n<p>For latter part of penetration,<\/p>\n<p>0= (u\/2)<sup>2<\/sup> -2ax<\/p>\n<p>or u<sup>2<\/sup> = 8ax\u2026\u2026\u2026\u2026\u2026(2)<\/p>\n<p>From (1) and (2)<\/p>\n<p>8ax = 8a x = 1 cm<\/p>\n<p><strong>Answer: (a)<\/strong><\/p>\n<p style=\"text-align: center;\"><strong>Click Here to Read \u2013 <a href=\"https:\/\/infinitylearn.com\/surge\/blog\/iit-jee\/vernier-caliper-least-count\/\">Vernier Caliper Least Count<\/a><\/strong><\/p>\n<p><strong>Q11. From a tower of height H, a particle is thrown vertically upwards with a speed u. The time taken by the particle, to hit the ground is n times that taken by it to reach the highest point of its path. The relation between H, u and n is<\/strong><\/p>\n<p>(a) gH = (n \u2013 2)u<sup>2<\/sup><\/p>\n<p>(b) 2gH = n 2u<sup>2<\/sup><\/p>\n<p>(c) gH = (n \u2013 2)2u<sup>2<\/sup><\/p>\n<p>(d) 2gH = nu<sup>2<\/sup>(n \u2013 2)<\/p>\n<p><strong>Solution<\/strong><\/p>\n<p>Time taken by the particle to reach the topmost point is,<\/p>\n<p>t =u\/g \u2026 (1)<\/p>\n<p>Time taken by the particle to reach the ground = nt<\/p>\n<p>Using, s = ut + \u00bd at<sup>2<\/sup><\/p>\n<p>-H = u(nt) \u2013 \u00bd g(nt)<sup>2<\/sup><\/p>\n<p>-H = u x n (u\/g) = \u00bd g<sup>2<\/sup>(u\/g)<sup>2 <\/sup>[using (1)]\n<p>\u21d2-2gH = 2nu<sup>2<\/sup> \u2013 n<sup>2<\/sup>u<sup>2<\/sup><\/p>\n<p>\u21d2 2gH = nu<sup>2<\/sup>(n -2)<\/p>\n<p><strong>Answer: (d) 2gH = nu<sup>2<\/sup>(n \u2013 2)<\/strong><\/p>\n<p><strong>Q12: For a tower of height H, a particle is thrown vertically upwards with a speed u. The time taken by the particle to hit the ground is n times that taken by it to reach the highest point of its path. The relation between H,u and n is<\/strong><\/p>\n<ol>\n<li>2gH = nu<sup>2 <\/sup>(n-2)<\/li>\n<li>gH = (n-2)u<sup>2<\/sup><\/li>\n<li>2gH = n<sup>2<\/sup>u<sup>2<\/sup><\/li>\n<li>gH = (n \u2013 2)<sup>2<\/sup>u<sup>2<\/sup><\/li>\n<\/ol>\n<p><strong>Solution:<\/strong><\/p>\n<p>Time to reach highest point = t = u\/g<\/p>\n<p>Time to reach ground = nt<\/p>\n<p>S = ut + \u00bd at<sup>2<\/sup><\/p>\n<p>-H = u(nt) \u2013 \u00bd g (nt)<sup>2<\/sup><\/p>\n<p>2gH = nu<sup>2 <\/sup>(n-2)<\/p>\n<p><strong>Answer: (a)<\/strong><\/p>\n<p>Join our <a href=\"https:\/\/infinitylearn.com\/jee\" target=\"_blank\" rel=\"noopener\"><strong>JEE course<\/strong><\/a> to make your IIT dream come true. If you live in Patna then prepare from the <strong><a href=\"https:\/\/infinitylearn.com\/surge\/patna\/\">best IIT JEE coaching center in Patna<\/a><\/strong>.<\/p>\n<h3><span class=\"ez-toc-section\" id=\"Assertion_Reasoning_type\"><\/span>Assertion &amp; Reasoning type<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>Instructions<\/strong> : The following question contains statement-I (assertion) and statement -2 (reason) of these statements, mark correct choice if<\/p>\n<p>a) Statement-1 and 2 are true and statement-2 is a correct explanation for statement-1<\/p>\n<p>b) Statement-l and 2 are true and statement-2 is not a correct explanation for statement-1<\/p>\n<p>c) Statement-1 is true, statement-2 is false<\/p>\n<p>d) Statement-1 is false, statement-2 is true.<\/p>\n<p style=\"text-align: center;\"><button style=\"width: 50%;\"><a class=\"btn btn-primary\" href=\"https:\/\/infinitylearn.com\/jee-class-12-online-course\"><strong>Enroll Now to JEE Class 12 Course<\/strong><\/a><\/button><\/p>\n<p><strong>16. Statement -1 : For an observer looking out through the window of a fast moving train the nearby objects appear to move in the opposite direction to the train, while the distant objects appear to be stationary.<\/strong><\/p>\n<p><strong>18. Is the time variation of position shown in the figure observed in nature ?     [1979]<\/strong><\/p>\n<p><strong>Ans. <\/strong>If we draw a straight line parallel to x-axis, it cuts the given figure at 2 points indicating that at one time, the,particle is at two different positions which is absurd. Thus such a motion as shown in the figures is never observed in nature.<\/p>\n<p><strong>19. Particles P and Q of mass 20 g and 40g respectively are simultaneously projected from points A and B on the ground.The initial velocities of P and Q make 45\u00b0 and 135\u00b0 angles respectively with the horizontal AB as shown in the figure. Each particle has an initial speed of 49m\/s. The separation AB is 245m. Both particle travel in the same vertical plane and undergo a collision. After the collision, P retraces its path. Determine the position of Q when it hits the ground. How much time after the collision does the particle Q take to reach the ground ? Take g = 9.8m \/ s<sup>2<\/sup>  [1982-8 marks]<\/strong><\/p>\n<p><strong>Ans. <\/strong><\/p>\n<p>Given:<\/p>\n<ul>\n<li>Mass of particle P, mP = 20 g = 0.02 kg<\/li>\n<li>Mass of particle Q, mQ = 40 g = 0.04 kg<\/li>\n<li>Initial speed of both particles: 49 m\/s<\/li>\n<li>Angle of projection of P: 45\u00b0<\/li>\n<li>Angle of projection of Q: 135\u00b0<\/li>\n<li>Horizontal separation between points A and B: 245 m<\/li>\n<li>Acceleration due to gravity, g = 9.8 m\/s\u00b2<\/li>\n<\/ul>\n<p>Step 1: Initial velocity components<\/p>\n<p><strong>Particle P:<\/strong><\/p>\n<ul>\n<li>uPx = 49 \u00d7 cos 45\u00b0 = 34.65 m\/s<\/li>\n<li>uPy = 49 \u00d7 sin 45\u00b0 = 34.65 m\/s<\/li>\n<\/ul>\n<p><strong>Particle Q:<\/strong><\/p>\n<ul>\n<li>uQx = 49 \u00d7 cos 135\u00b0 = -34.65 m\/s<\/li>\n<li>uQy = 49 \u00d7 sin 135\u00b0 = 34.65 m\/s<\/li>\n<\/ul>\n<p>Step 2: Time of collision (tc)<\/p>\n<p>At collision, horizontal positions of P and Q are equal:<\/p>\n<p>34.65 \u00d7 tc = 245 &#8211; 34.65 \u00d7 tc<\/p>\n<p>69.3 \u00d7 tc = 245 \u21d2 tc = 3.535 seconds<\/p>\n<p>Step 3: Vertical position at collision<\/p>\n<p>For both particles, vertical position at tc:<\/p>\n<p>y = 34.65 \u00d7 3.535 &#8211; 0.5 \u00d7 9.8 \u00d7 (3.535)\u00b2 = 61.2 meters<\/p>\n<p>Step 4: Velocities just before collision<\/p>\n<ul>\n<li>Velocity of P: vPx = 34.65 m\/s, vPy = 0 m\/s<\/li>\n<li>Velocity of Q: vQx = -34.65 m\/s, vQy = 0 m\/s<\/li>\n<\/ul>\n<p>Step 5: Velocity of P after collision<\/p>\n<p>P retraces its path, so:<\/p>\n<p>vPx&#8217; = -34.65 m\/s, vPy&#8217; = 0 m\/s<\/p>\n<p>Step 6: Velocity of Q after collision (using conservation of momentum)<\/p>\n<p>0.02 \u00d7 34.65 + 0.04 \u00d7 (-34.65) = 0.02 \u00d7 (-34.65) + 0.04 \u00d7 vQx&#8217;<\/p>\n<p>-0.693 = -0.693 + 0.04 \u00d7 vQx&#8217; \u21d2 vQx&#8217; = 0 m\/s<\/p>\n<p>Vertical velocity of Q remains vQy&#8217; = 0 m\/s<\/p>\n<p>Step 7: Position of Q when it hits the ground<\/p>\n<p>Q falls vertically from height 61.2 m with zero horizontal velocity.<\/p>\n<p>Time to fall:<\/p>\n<p>tf = \u221a(2 \u00d7 61.2 \/ 9.8) = 3.535 seconds<\/p>\n<p>Step 8: Horizontal position of Q when it lands<\/p>\n<p>Position remains constant during fall:<\/p>\n<p>xQ = 245 &#8211; 34.65 \u00d7 3.535 = 122.6 meters<\/p>\n<p>Final Answers:<\/p>\n<ul>\n<li><strong>Position of Q on ground:<\/strong> 122.6 meters from point A<\/li>\n<li><strong>Time taken by Q to reach ground after collision:<\/strong> 3.535 seconds<\/li>\n<\/ul>\n<p><strong>20. A body falling freely from a given height H hits an inclined plane in its path at a height h. As a result of this impact the direction of the velocity of the body becomes horizontal. For what value of (h\/H) the body will take maximum time to reach the ground?<\/strong><\/p>\n<p><strong>Ans. <\/strong>PQR is an inclined plane. The body falls under gravity vertically from A to B till it strikes the inclined plane. At B, its velocity is horizontal. From B to C, the body follows a parabolic path. At C, it reaches the ground.<\/p>\n<p><strong>23. A cart is moving along x-direction with a velocity of 4m\/s. A person on the cart, throws a stone with a velocity of 6m\/s, relative to himself. In the frame of reference of the cart, the stone is thrown in y-z plane making an angle of 30\u00b0 with vertical z-axis. At the highest point of its trajectory, the stone hits an object of equal mass, hung vertically from branch of a tree, by means of a string of length L. A completely inelastic collision occurs in which the stone gets embedded in the object. Determine:[1997-5 marks]<\/strong><\/p>\n<p><strong>i) the speed of the combined mass immediately after the collision with respect to an observer on the ground<\/strong><\/p>\n<p><strong>ii)the length L of the string such that tension in the string becomes zero when the string becomes horizontal during the subsequent motion of the combined mass.<\/strong><\/p>\n<p><strong>Ans.<\/strong>i) The cart is moving in x-y plane<\/p>\n<p>The stone, thrown from cart, travels in y-z plane while z-axis is vertical axis. The stone makes an angle of 30\u00b0 with z-axis. Its path is parabolic. At the highest point of its trajectory, the vertical velocity of the stone will be zero. The velocity of the stone is thus confined to (x, y) plane at the highest point. Velocity of cart is along x-axis<\/p>\n<p>ii) It is given that the tension in the string becomes zero at horizontal position. The combined mass therefore is at rest in this position. During the subsequent motion of the combined mass, the energy is conserved.<\/p>\n<p style=\"text-align: center;\"><button style=\"width: 50%;\"><a class=\"btn btn-primary\" href=\"https:\/\/infinitylearn.com\/jee-class-13-online-course\"><strong>Enroll Now to JEE Dropper Course<\/strong><\/a><\/button><\/p>\n<p><strong>26. A particle is suspended vertically from a point O by an inextensible  massless string of length L. A vertical line AB is at a distance L\/8 from O as shown in figure. The object is given a horizontal velocity u. At some point, its motion ceases to be circular and eventually the object passes through the line AB. At the instant of crossing AB, its velocity is horizontal. Find u. [1999-10 marks]<\/strong><\/p>\n<p><strong>Ans. <\/strong>OP denotes string of length L. The particle starts from P with a horizontal velocity u. It travels along a circular path till Q. Then it passes the line AB. At Q, motion ceases to be circular. At C, its velocity is horizontal, where the particle crosses AB. After Q, the particle performs a projectile motion. At C, the velocity becomes horizontal. Thus C is at the highest point of projectile motion<\/p>\n<h3><span class=\"ez-toc-section\" id=\"True_False_Type\"><\/span>True \/ False Type<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>30. Two balls of different masses are thrown vertically upwards with the same speed. They pass through the point of projection in their downward motion with the same speed (Neglect air resistance).[1983-2 marks]<\/strong><\/p>\n<p><strong>Ans.  <\/strong><strong>True<\/strong><\/p>\n<p><strong>Explanation:<\/strong><\/p>\n<p>Since air resistance is neglected, the only force acting on both balls is gravity, which accelerates them equally regardless of their masses. Therefore, both balls will return to the point of projection with the same speed they were thrown upwards, irrespective of their different masses.<\/p>\n<p><strong>31.\u2019A projectile fired from the ground follows a parabolic path. The speed of the projectile, is minimum at the top of its path. [1984-2 marks]<\/strong><\/p>\n<p><strong>Ans.  <\/strong><strong>True<\/strong><\/p>\n<p><strong>Explanation:<\/strong><br \/>\nSince air resistance is neglected, the only force acting on both balls is gravity, which accelerates them equally regardless of their masses. Therefore, both balls will return to the point of projection with the same speed they were thrown upwards, irrespective of their different masses.<\/p>\n<p><strong>32.Two identical trains are moving on rails along the equator on the earth in opposite directions with the same speed. They will exert the same pressure on the rails      [1985-2 marks]<\/strong><\/p>\n<p><strong>Ans. <\/strong>False<\/p>\n<p>Explanation:<\/p>\n<p>Because of the Earth&#8217;s rotation, trains moving in opposite directions along the equator experience different effective gravitational forces due to the Coriolis effect and centrifugal force. The train moving eastward (in the direction of Earth&#8217;s rotation) experiences a slightly reduced effective weight, while the train moving westward experiences a slightly increased effective weight. As pressure depends on the normal force (weight) exerted on the rails, the pressures will not be exactly the same.<\/p>\n<h3><span class=\"ez-toc-section\" id=\"Fill_in_the_blanks\"><\/span>Fill in the blanks<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>33. A particle moves in a circle of radius R. In half the period of revolution its displacement is\u2026\u2026\u2026. and distance covered is\u2026\u2026\u2026 \u2026\u2026\u2026.. [1983-2 marks]<\/strong><\/p>\n<p><strong>Ans. <\/strong>Displacement: The displacement is the straight-line distance between the starting point and the point after half a revolution, which is the diameter of the circle.<\/p>\n<p>Displacement=2R<\/p>\n<p>Distance covered: The distance covered is half the circumference of the circle.<\/p>\n<p>Distance=\u03c0R<\/p>\n<p><strong>34. Four persons K,L,M,N are initially at the four corners of a square of side d. Each person now moves with a uniform speed vin such a way that K always moves directly towards L, L directly towards M,M directly towards N, and N directly towards K. The four persons will meet at a time \u2026\u2026\u2026\u2026.. [1984-2 marks]<\/strong><\/p>\n<p><strong>Ans. <\/strong>The four persons will meet at the center of the square after a time<\/p>\n<p>t= v\/d<\/p>\n<p>Explanation:<\/p>\n<p>Each person moves towards the next with speed v, and due to symmetry, their paths curve inward, forming a logarithmic spiral. The distance between them decreases uniformly, and they meet exactly after time t= v\/d<\/p>\n","protected":false},"excerpt":{"rendered":"<p>JEE Main 2024 Physics Kinematics Previous Year Questions with Solutions For JEE Main other Engineering Entrance Exam Preparation, JEE Main [&hellip;]<\/p>\n","protected":false},"author":7,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_yoast_wpseo_focuskw":"JEE Main 2024","_yoast_wpseo_title":"JEE Main 2024 Physics Kinematics Previous Year Questions with Solutions","_yoast_wpseo_metadesc":"Get JEE Main Physics Kinematics Previous Year Questions with Solutions PDF at infinity learn for JEE Main 2024 exam","custom_permalink":"iit-jee\/study-materials\/physics\/jee-main-physics-kinematics-previous-year-questions-with-solutions\/"},"categories":[4,21],"tags":[],"table_tags":[],"acf":[],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v17.9 - 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