{"id":4141,"date":"2016-05-06T11:49:00","date_gmt":"2016-05-06T11:49:00","guid":{"rendered":"http:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-science-forces-and-laws-of-motion-part-2\/"},"modified":"2018-06-07T12:42:40","modified_gmt":"2018-06-07T07:12:40","slug":"ncert-solutions-for-class-9-science-forces-and-laws-of-motion-part-2","status":"publish","type":"post","link":"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-science-forces-and-laws-of-motion-part-2\/","title":{"rendered":"NCERT Solutions for Class 9 Science Forces and Laws of Motion part 2"},"content":{"rendered":"<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_76 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\" style=\"cursor:inherit\">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\" aria-label=\"Toggle Table of Content\"><span class=\"ez-toc-js-icon-con\"><span class=\"\"><span class=\"eztoc-hide\" style=\"display:none;\">Toggle<\/span><span class=\"ez-toc-icon-toggle-span\"><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><\/span><\/span><\/a><\/span><\/div>\n<nav><ul class='ez-toc-list ez-toc-list-level-1 eztoc-toggle-hide-by-default' ><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-1\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-science-forces-and-laws-of-motion-part-2\/#NCERT_Class_9_Science_Chapter_wise_Solutions\" >NCERT Class 9 Science Chapter wise Solutions<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-science-forces-and-laws-of-motion-part-2\/#NCERT_Solutions_for_Class_9_Science_Forces_and_Laws_of_Motion_part_2\" >NCERT Solutions for Class 9 Science Forces and Laws of Motion part 2<\/a><ul class='ez-toc-list-level-6' ><li class='ez-toc-heading-level-6'><ul class='ez-toc-list-level-6' ><li class='ez-toc-heading-level-6'><ul class='ez-toc-list-level-6' ><li class='ez-toc-heading-level-6'><ul class='ez-toc-list-level-6' ><li class='ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-science-forces-and-laws-of-motion-part-2\/#1_An_object_experiences_a_net_zero_external_unbalanced_force_Is_it_possible_for_the_object_to_be_travelling_with_a_non-zero_velocity_If_yes_state_the_conditions_that_must_be_placed_on_the_magnitude_and_direction_of_the_velocity_If_no_provide_a_reason\" >1. An object experiences a net zero external unbalanced force. Is it possible for the object to be travelling with a non-zero velocity? If yes, state the conditions that must be placed on the magnitude and direction of the velocity. If no, provide a reason.<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-science-forces-and-laws-of-motion-part-2\/#2_When_a_carpet_is_beaten_with_a_stick_dust_comes_out_of_it_Explain\" >2. When a carpet is beaten with a stick, dust comes out of it. Explain.<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-5\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-science-forces-and-laws-of-motion-part-2\/#3_Why_is_it_advised_to_tie_any_luggage_kept_on_the_roof_of_a_bus_with_a_rope\" >3. Why is it advised to tie any luggage kept on the roof of a bus with a rope?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-6\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-science-forces-and-laws-of-motion-part-2\/#4_A_batsman_hits_a_cricket_ball_which_then_rolls_on_a_level_ground_After_covering_a_short_distance_the_ball_comes_to_rest_The_ball_slows_to_a_stop_because\" >4. A batsman hits a cricket ball which then rolls on a level ground. After covering a short distance, the ball comes to rest. The ball slows to a stop because<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-7\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-science-forces-and-laws-of-motion-part-2\/#5_A_truck_starts_from_rest_and_rolls_down_a_hill_with_a_constant_acceleration_It_travels_a_distance_of_400_m_in_20_s_Find_its_acceleration_Find_the_force_acting_on_it_if_its_mass_is_7_metric_tonnes_Hint_1_metric_tonne_1000_kg\" >5. A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of 400 m in 20 s. Find its acceleration. Find the force acting on it if its mass is 7 metric tonnes (Hint: 1 metric tonne = 1000 kg.)<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-8\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-science-forces-and-laws-of-motion-part-2\/#6_A_stone_of_1_kg_is_thrown_with_a_velocity_of_across_the_frozen_surface_of_a_lake_and_comes_to_rest_after_travelling_a_distance_of_50_m_What_is_the_force_of_friction_between_the_stone_and_the_ice\" >6. A stone of 1 kg is thrown with a velocity of across the frozen surface of a lake and comes to rest after travelling a distance of 50 m. What is the force of friction between the stone and the ice?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-9\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-science-forces-and-laws-of-motion-part-2\/#7_A_8000_kg_engine_pulls_a_train_of_5_wagons_each_of_2000_kg_along_a_horizontal_track_If_the_engine_exerts_a_force_of_40000_N_and_the_track_offers_a_friction_force_of_5000_N_then_calculate\" >7. A 8000 kg engine pulls a train of 5 wagons, each of 2000 kg, along a horizontal track. If the engine exerts a force of 40000 N and the track offers a friction force of 5000 N, then calculate:<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-10\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-science-forces-and-laws-of-motion-part-2\/#8_An_automobile_vehicle_has_a_mass_of_1500_kg_What_must_be_the_force_between_the_vehicle_and_road_if_the_vehicle_is_to_be_stopped_with_a_negative_acceleration_of\" >8. An automobile vehicle has a mass of 1500 kg. What must be the force between the vehicle and road if the vehicle is to be stopped with a negative acceleration of?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-11\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-science-forces-and-laws-of-motion-part-2\/#9_What_is_the_momentum_of_an_object_of_mass_m_moving_with_a_velocity_v\" >9. What is the momentum of an object of mass m, moving with a velocity v?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-12\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-science-forces-and-laws-of-motion-part-2\/#10_Using_a_horizontal_force_of_200_N_we_intend_to_move_a_wooden_cabinet_across_a_floor_at_a_constant_velocity_What_is_the_friction_force_that_will_be_exerted_on_the_cabinet\" >10. Using a horizontal force of 200 N, we intend to move a wooden cabinet across a floor at a constant velocity. What is the friction force that will be exerted on the cabinet?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-13\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-science-forces-and-laws-of-motion-part-2\/#11_Two_objects_each_of_mass_15_kg_are_moving_in_the_same_straight_line_but_in_opposite_directions_The_velocity_of_each_object_is_before_the_collision_during_which_they_stick_together_What_will_be_the_velocity_of_the_combined_object_after_collision\" >11. Two objects, each of mass 1.5 kg, are moving in the same straight line but in opposite directions. The velocity of each object is before the collision during which they stick together. What will be the velocity of the combined object after collision?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-14\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-science-forces-and-laws-of-motion-part-2\/#12_According_to_the_third_law_of_motion_when_we_push_on_an_object_the_object_pushes_back_on_us_with_an_equal_and_opposite_force_If_the_object_is_a_massive_truck_parked_along_the_roadside_it_will_probably_not_move_A_student_justifies_this_by_answering_that_the_two_opposite_and_equal_forces_cancel_each_other_Comment_on_this_logic_and_explain_why_the_truck_does_not_move\" >12. According to the third law of motion when we push on an object, the object pushes back on us with an equal and opposite force. If the object is a massive truck parked along the roadside, it will probably not move. A student justifies this by answering that the two opposite and equal forces cancel each other. Comment on this logic and explain why the truck does not move.<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-15\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-science-forces-and-laws-of-motion-part-2\/#13_A_hockey_ball_of_mass_200_g_travelling_at_is_struck_by_a_hockey_stick_so_as_to_return_it_along_its_original_path_with_a_velocity_at_Calculate_the_change_of_momentum_occurred_in_the_motion_of_the_hockey_ball_by_the_force_applied_by_the_hockey_stick\" >13. A hockey ball of mass 200 g travelling at is struck by a hockey stick so as to return it along its original path with a velocity at. Calculate the change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-16\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-science-forces-and-laws-of-motion-part-2\/#NCERT_Solutions_for_Class_9_Science_Forces_and_Laws_of_Motion_part_2-2\" >NCERT Solutions for Class 9 Science Forces and Laws of Motion part 2<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-17\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-science-forces-and-laws-of-motion-part-2\/#15_An_object_of_mass_1_kg_travelling_in_a_straight_line_with_a_velocity_of_collides_with_and_sticks_to_a_stationary_wooden_block_of_mass_5_kg_Then_they_both_move_off_together_in_the_same_straight_line_Calculate_the_total_momentum_just_before_the_impact_and_just_after_the_impact_Also_calculate_the_velocity_of_the_combined_object\" >15. An object of mass 1 kg travelling in a straight line with a velocity of collides with, and sticks to, a stationary wooden block of mass 5 kg. Then they both move off together in the same straight line. Calculate the total momentum just before the impact and just after the impact. Also, calculate the velocity of the combined object.<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-18\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-science-forces-and-laws-of-motion-part-2\/#16_An_object_of_mass_100_kg_is_accelerated_uniformly_from_a_velocity_of_to_in_6_s_Calculate_the_initial_and_final_momentum_of_the_object_Also_find_the_magnitude_of_the_force_exerted_on_the_object\" >16. An object of mass 100 kg is accelerated uniformly from a velocity of to  in 6 s. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object.<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-19\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-science-forces-and-laws-of-motion-part-2\/#17_Akhtar_Kiran_and_Rahul_were_riding_in_a_motorcar_that_was_moving_with_a_high_velocity_on_an_expressway_when_an_insect_hit_the_windshield_and_got_stuck_on_the_windscreen_Akhtar_and_Kiran_started_pondering_over_the_situation_Kiran_suggested_that_the_insect_suffered_a_greater_change_in_momentum_as_compared_to_the_change_in_momentum_of_the_motorcar_because_the_change_in_the_velocity_of_the_insect_was_much_more_than_that_of_the_motorcar_Akhtar_said_that_since_the_motorcar_was_moving_with_a_larger_velocity_it_exerted_a_larger_force_on_the_insect_And_as_a_result_the_insect_died_Rahul_while_putting_an_entirely_new_explanation_said_that_both_the_motorcar_and_the_insect_experienced_the_same_force_and_a_change_in_their_momentum_Comment_on_these_suggestions\" >17. Akhtar, Kiran and Rahul were riding in a motorcar that was moving with a high velocity on an expressway when an insect hit the windshield and got stuck on the windscreen. Akhtar and Kiran started pondering over the situation. Kiran suggested that the insect suffered a greater change in momentum as compared to the change in momentum of the motorcar (because the change in the velocity of the insect was much more than that of the motorcar). Akhtar said that since the motorcar was moving with a larger velocity, it exerted a larger force on the insect. And as a result the insect died. Rahul while putting an entirely new explanation said that both the motorcar and the insect experienced the same force and a change in their momentum. Comment on these suggestions.<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-20\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-science-forces-and-laws-of-motion-part-2\/#18_How_much_momentum_will_a_dumb-bell_of_mass_10_kg_transfer_to_the_floor_if_it_falls_from_a_height_of_80_cm_Take_its_downward_acceleration_to_be\" >18. How much momentum will a dumb-bell of mass 10 kg transfer to the floor if it falls from a height of 80 cm? Take its downward acceleration to be.<\/a><\/li><\/ul><\/li><\/ul><\/li><\/ul><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-21\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-science-forces-and-laws-of-motion-part-2\/#NCERT_Solutions_for_Class_9_Science\" >NCERT Solutions for Class 9 Science<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-22\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-science-forces-and-laws-of-motion-part-2\/#CBSE_app_for_Class_9\" >CBSE app for Class 9<\/a><\/li><\/ul><\/nav><\/div>\n<p>NCERT Solutions for Class 9 Science Forces and Laws of Motion part 2 Class 9 Science book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. Class 9 Science chapter wise NCERT solution for Science Book for all the chapters can be downloaded from our website and myCBSEguide mobile app for free.<\/p>\n<p style=\"text-align: center;\"><strong>NCERT solutions for Science\u00a0Forces and Laws of Motion <a class=\"button\" href=\"https:\/\/mycbseguide.com\/downloads\/cbse-class-09-science\/1218\/ncert-solutions\/5\/\">Download as PDF<\/a><\/strong><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/media-mycbseguide.s3.ap-south-1.amazonaws.com\/images\/blog\/09_Science_Book.jpg\" alt=\"NCERT Solutions for Class 9 Science Forces and Laws of Motion part 2\" width=\"169\" height=\"227\" \/><\/p>\n<h2><span class=\"ez-toc-section\" id=\"NCERT_Class_9_Science_Chapter_wise_Solutions\"><\/span><span class=\"underline-text\">NCERT Class 9 Science Chapter wise Solutions<\/span><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<ul>\n<li>01 \u2013 \u00a0Matter in Our Surroundings<\/li>\n<li>02 \u2013 Is Mattter Around us Pure<\/li>\n<li>03 \u2013 Atoms and Molecules<\/li>\n<li>04 \u2013 Structure of the Atom<\/li>\n<li>05 \u2013 The Fundamental Unit of Life<\/li>\n<li>06 \u2013 Tissues<\/li>\n<li>07 \u2013 Diversity in Living Organisms<\/li>\n<li>08 \u2013 Motion<\/li>\n<li>09 \u2013 Force and Laws of Motion<\/li>\n<li>10 \u2013 Gravitation<\/li>\n<li>11 \u2013 Word and Energy<\/li>\n<li>12 \u2013 Sound<\/li>\n<li>13 \u2013 Why Do We Fall Ill<\/li>\n<li>14 \u2013 Natural Resources<\/li>\n<li>15 \u2013 Improvement in Food Resources<\/li>\n<\/ul>\n<h2><span class=\"ez-toc-section\" id=\"NCERT_Solutions_for_Class_9_Science_Forces_and_Laws_of_Motion_part_2\"><\/span>NCERT Solutions for Class 9 Science Forces and Laws of Motion part 2<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"1_An_object_experiences_a_net_zero_external_unbalanced_force_Is_it_possible_for_the_object_to_be_travelling_with_a_non-zero_velocity_If_yes_state_the_conditions_that_must_be_placed_on_the_magnitude_and_direction_of_the_velocity_If_no_provide_a_reason\"><\/span><strong>1. An object experiences a net zero external unbalanced force. Is it possible for the object to be travelling with a non-zero velocity? If yes, state the conditions that must be placed on the magnitude and direction of the velocity. If no, provide a reason.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>No, it is not possible for the object to be travelling with a non-zero velocity if an object experiences a net zero external unbalanced force since unbalanced forces cannot be equal to zero.<\/p>\n<hr \/>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"2_When_a_carpet_is_beaten_with_a_stick_dust_comes_out_of_it_Explain\"><\/span><strong>2. When a carpet is beaten with a stick, dust comes out of it. Explain.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>When a carpet is beaten with a stick, dust comes out of it because carpet fibres vibrate in forward and backward direction as carpet is beaten but the loosely bound dust particles due to inertia remain at rest and so they come out.<\/p>\n<hr \/>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"3_Why_is_it_advised_to_tie_any_luggage_kept_on_the_roof_of_a_bus_with_a_rope\"><\/span><strong>3. Why is it advised to tie any luggage kept on the roof of a bus with a rope?<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>It is advised to tie any luggage kept on the roof of a bus with a rope because when bus moves the luggage also gets moving with the velocity same as that of the bus and in the same direction but when bus changes direction or deaccelerates, due to inertia of motion luggage moves in the same direction and may get thrown away from roof of buses.<\/p>\n<hr \/>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"4_A_batsman_hits_a_cricket_ball_which_then_rolls_on_a_level_ground_After_covering_a_short_distance_the_ball_comes_to_rest_The_ball_slows_to_a_stop_because\"><\/span><strong>4. A batsman hits a cricket ball which then rolls on a level ground. After covering a short distance, the ball comes to rest. The ball slows to a stop because<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>(a) the batsman did not hit the ball hard enough.<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(b) velocity is proportional to the force exerted on the ball.<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(c) there is a force on the ball opposing the motion.<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(d) there is no unbalanced force on the ball, so the ball would want to come to rest.<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>Ans. (c)<\/strong> there is a force on the ball opposing the motion.<\/p>\n<hr \/>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"5_A_truck_starts_from_rest_and_rolls_down_a_hill_with_a_constant_acceleration_It_travels_a_distance_of_400_m_in_20_s_Find_its_acceleration_Find_the_force_acting_on_it_if_its_mass_is_7_metric_tonnes_Hint_1_metric_tonne_1000_kg\"><\/span><strong>5. A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of 400 m in 20 s. Find its acceleration. Find the force acting on it if its mass is 7 metric tonnes (<em>Hint<\/em>: 1 metric tonne = 1000 kg.)<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>According to the question,<\/p>\n<p style=\"text-align: justify;\">initial velocity of truck (u) = 0<\/p>\n<p style=\"text-align: justify;\">distance = s= 400 m and time = 20 s<\/p>\n<p style=\"text-align: justify;\">mass of truck = 7metric tones = 7000kg<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 22px; width: 113px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch09\/image017.png\" \/><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 22px; width: 167px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch09\/image018.png\" \/><\/p>\n<p style=\"text-align: justify;\">400 =0 + 200a<\/p>\n<p style=\"text-align: justify;\">400 = 200a<\/p>\n<p style=\"text-align: justify;\">a = 400\/200 <img decoding=\"async\" style=\"height: 22px; width: 70px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch09\/image019.png\" \/><\/p>\n<p style=\"text-align: justify;\">therefore, <img decoding=\"async\" style=\"height: 19px; width: 139px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch09\/image020.png\" \/>= 14000 N<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 9 Science Forces and Laws of Motion part 2<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"6_A_stone_of_1_kg_is_thrown_with_a_velocity_of_across_the_frozen_surface_of_a_lake_and_comes_to_rest_after_travelling_a_distance_of_50_m_What_is_the_force_of_friction_between_the_stone_and_the_ice\"><\/span><strong>6. A stone of 1 kg is thrown with a velocity of <img decoding=\"async\" style=\"height: 22px; width: 60px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch09\/image021.png\" \/>across the frozen surface of a lake and comes to rest after travelling a distance of 50 m. What is the force of friction between the stone and the ice?<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>since <img decoding=\"async\" style=\"height: 22px; width: 97px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch09\/image022.png\" \/><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 22px; width: 122px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch09\/image023.png\" \/>(object comes to rest so v=0)<\/p>\n<p style=\"text-align: justify;\">-100a = 400<\/p>\n<p style=\"text-align: justify;\">a = 400\/-100 = <img decoding=\"async\" style=\"height: 22px; width: 72px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch09\/image024.png\" \/><\/p>\n<p style=\"text-align: justify;\">therefore, the force of friction between the stone and the ice<\/p>\n<p style=\"text-align: justify;\">= -4 N<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 9 Science Forces and Laws of Motion part 2<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"7_A_8000_kg_engine_pulls_a_train_of_5_wagons_each_of_2000_kg_along_a_horizontal_track_If_the_engine_exerts_a_force_of_40000_N_and_the_track_offers_a_friction_force_of_5000_N_then_calculate\"><\/span><strong>7. A 8000 kg engine pulls a train of 5 wagons, each of 2000 kg, along a horizontal track. If the engine exerts a force of 40000 N and the track offers a friction force of 5000 N, then calculate:<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>(a) the net accelerating force;<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(b) the acceleration of the train; and<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(c) the force of wagon 1 on wagon 2.<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>Ans. (a)<\/strong> The net accelerating force = Force exerted by engine &#8211; frictional force of track = 40000 &#8211; 5000 = 35000 N<\/p>\n<p style=\"text-align: justify;\"><strong>(b)<\/strong> the acceleration of the train = a = F\/m = <img decoding=\"async\" style=\"height: 26px; width: 119px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch09\/image026.png\" \/>= 35000\/10000 = <img decoding=\"async\" style=\"height: 22px; width: 62px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch09\/image027.png\" \/><\/p>\n<p style=\"text-align: justify;\"><strong>(c)<\/strong> the force of wagon 1 on wagon 2<\/p>\n<p style=\"text-align: justify;\">Wagon 1 will have to exert force on all 4 wagons next to it<\/p>\n<p style=\"text-align: justify;\">so mass of other 4 wagons = <img decoding=\"async\" style=\"height: 19px; width: 57px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch09\/image028.png\" \/> = 8000 kg<\/p>\n<p style=\"text-align: justify;\">F = ma = <img decoding=\"async\" style=\"height: 24px; width: 128px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch09\/image029.png\" \/>= 28000 N<\/p>\n<hr \/>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"8_An_automobile_vehicle_has_a_mass_of_1500_kg_What_must_be_the_force_between_the_vehicle_and_road_if_the_vehicle_is_to_be_stopped_with_a_negative_acceleration_of\"><\/span><strong>8. An automobile vehicle has a mass of 1500 kg. What must be the force between the vehicle and road if the vehicle is to be stopped with a negative acceleration of<img decoding=\"async\" style=\"height: 22px; width: 64px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch09\/image030.png\" \/>?<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>since <img decoding=\"async\" style=\"height: 19px; width: 153px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch09\/image031.png\" \/>= -2550 N (negative sign symbolises acceleration in opposite direction)<\/p>\n<hr \/>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"9_What_is_the_momentum_of_an_object_of_mass_m_moving_with_a_velocity_v\"><\/span><strong>9. What is the momentum of an object of mass <em>m, <\/em>moving with a velocity <em>v<\/em>?<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>(a)<img decoding=\"async\" style=\"height: 29px; width: 42px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch09\/image032.png\" \/><\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(b) <img decoding=\"async\" style=\"height: 22px; width: 29px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch09\/image033.png\" \/><\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(c)<img decoding=\"async\" style=\"height: 22px; width: 42px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch09\/image034.png\" \/><\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(d) <em>mv<\/em><\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>Ans. (d)<\/strong> mv<\/p>\n<hr \/>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"10_Using_a_horizontal_force_of_200_N_we_intend_to_move_a_wooden_cabinet_across_a_floor_at_a_constant_velocity_What_is_the_friction_force_that_will_be_exerted_on_the_cabinet\"><\/span><strong>10. Using a horizontal force of 200 N, we intend to move a wooden cabinet across a floor at a constant velocity. What is the friction force that will be exerted on the cabinet?<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>200 N<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 9 Science Forces and Laws of Motion part 2<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"11_Two_objects_each_of_mass_15_kg_are_moving_in_the_same_straight_line_but_in_opposite_directions_The_velocity_of_each_object_is_before_the_collision_during_which_they_stick_together_What_will_be_the_velocity_of_the_combined_object_after_collision\"><\/span><strong>11. Two objects, each of mass 1.5 kg, are moving in the same straight line but in opposite directions. The velocity of each object is <img decoding=\"async\" style=\"height: 22px; width: 64px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch09\/image035.png\" \/>before the collision during which they stick together. What will be the velocity of the combined object after collision?<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>Momentum before collision took place = <img decoding=\"async\" style=\"height: 24px; width: 79px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch09\/image013.png\" \/><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 26px; width: 153px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch09\/image036.png\" \/>= 0<\/p>\n<p style=\"text-align: justify;\">Since the objects stick together after collision hence<\/p>\n<p style=\"text-align: justify;\">momentum after collision <img decoding=\"async\" style=\"height: 26px; width: 97px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch09\/image037.png\" \/><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 26px; width: 107px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch09\/image038.png\" \/><\/p>\n<p style=\"text-align: justify;\">= 3v<\/p>\n<p style=\"text-align: justify;\">momentum before collision = momentum after collision<\/p>\n<p style=\"text-align: justify;\">0 = 3v, v= 0\/3 = 0<\/p>\n<p style=\"text-align: justify;\">the velocity of the combined object after collision (v)= 0<\/p>\n<hr \/>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"12_According_to_the_third_law_of_motion_when_we_push_on_an_object_the_object_pushes_back_on_us_with_an_equal_and_opposite_force_If_the_object_is_a_massive_truck_parked_along_the_roadside_it_will_probably_not_move_A_student_justifies_this_by_answering_that_the_two_opposite_and_equal_forces_cancel_each_other_Comment_on_this_logic_and_explain_why_the_truck_does_not_move\"><\/span><strong>12. According to the third law of motion when we push on an object, the object pushes back on us with an equal and opposite force. If the object is a massive truck parked along the roadside, it will probably not move. A student justifies this by answering that the two opposite and equal forces cancel each other. Comment on this logic and explain why the truck does not move.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>According to the third law of motion when we push on an object, the object pushes back on us with an equal and opposite force result is the two opposite and equal forces cancel each other but when one of these forces is bigger than inertia so the object moves in the direction of force applied. As this student explains the truck is massive so the force applied cannot overcome force caused by inertia. Therefore, the truck does not move.<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 9 Science Forces and Laws of Motion part 2<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"13_A_hockey_ball_of_mass_200_g_travelling_at_is_struck_by_a_hockey_stick_so_as_to_return_it_along_its_original_path_with_a_velocity_at_Calculate_the_change_of_momentum_occurred_in_the_motion_of_the_hockey_ball_by_the_force_applied_by_the_hockey_stick\"><\/span><strong>13. A hockey ball of mass 200 g travelling at <img decoding=\"async\" style=\"height: 22px; width: 59px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch09\/image039.png\" \/>is struck by a hockey stick so as to return it along its original path with a velocity at<img decoding=\"async\" style=\"height: 22px; width: 52px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch09\/image040.png\" \/>. Calculate the change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>mass of hockey ball = 200g = 0.2 kg<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 24px; width: 160px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch09\/image041.png\" \/> (return velocity)<\/p>\n<p style=\"text-align: justify;\">initial momentum of hockey ball <img decoding=\"async\" style=\"height: 22px; width: 124px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch09\/image042.png\" \/>= 2 kg m\/s<\/p>\n<p style=\"text-align: justify;\">final momentum of hockey ball <img decoding=\"async\" style=\"height: 22px; width: 121px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch09\/image043.png\" \/>= -1 kg m\/s<\/p>\n<p style=\"text-align: justify;\">change in momentum of hockey ball = 2 \u2013 (-1) = 2 + 1 = 3 kg m\/s<\/p>\n<hr \/>\n<h6 style=\"text-align: center;\"><span class=\"ez-toc-section\" id=\"NCERT_Solutions_for_Class_9_Science_Forces_and_Laws_of_Motion_part_2-2\"><\/span>NCERT Solutions for Class 9 Science Forces and Laws of Motion part 2<span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>14. A bullet of mass 10 g travelling horizontally with a velocity of <img decoding=\"async\" style=\"height: 22px; width: 67px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch09\/image044.png\" \/>strikes a stationary wooden block and comes to rest in 0.03 s. Calculate the distance of penetration of the bullet into the block. Also calculate the magnitude of the force exerted by the wooden block on the bullet.<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>v =u + at<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 19px; width: 125px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch09\/image045.png\" \/><\/p>\n<p style=\"text-align: justify;\">a = -150\/0.03 = <img decoding=\"async\" style=\"height: 22px; width: 80px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch09\/image046.png\" \/><\/p>\n<p style=\"text-align: justify;\">the distance of penetration of the bullet into the block <img decoding=\"async\" style=\"height: 26px; width: 120px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch09\/image047.png\" \/><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 42px; width: 203px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch09\/image048.png\" \/><\/p>\n<p style=\"text-align: justify;\">= 4.5 &#8211; 2.25<\/p>\n<p style=\"text-align: justify;\">= 2.25 m<\/p>\n<p style=\"text-align: justify;\">the magnitude of the force exerted by the wooden block on the bullet<\/p>\n<p style=\"text-align: justify;\">m = 10g = 0.01kg<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 19px; width: 65px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch09\/image049.png\" \/><img decoding=\"async\" style=\"height: 24px; width: 150px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch09\/image050.png\" \/>= -50 N<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 9 Science Forces and Laws of Motion part 2<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"15_An_object_of_mass_1_kg_travelling_in_a_straight_line_with_a_velocity_of_collides_with_and_sticks_to_a_stationary_wooden_block_of_mass_5_kg_Then_they_both_move_off_together_in_the_same_straight_line_Calculate_the_total_momentum_just_before_the_impact_and_just_after_the_impact_Also_calculate_the_velocity_of_the_combined_object\"><\/span><strong>15. An object of mass 1 kg travelling in a straight line with a velocity of <img decoding=\"async\" style=\"height: 22px; width: 59px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch09\/image039.png\" \/>collides with, and sticks to, a stationary wooden block of mass 5 kg. Then they both move off together in the same straight line. Calculate the total momentum just before the impact and just after the impact. Also, calculate the velocity of the combined object.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>Wooden block is stationery (at rest) so its velocity <img decoding=\"async\" style=\"height: 26px; width: 68px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch09\/image051.png\" \/><\/p>\n<p style=\"text-align: justify;\">mass of combined object is = 1 kg + 5 kg = 6 kg<\/p>\n<p style=\"text-align: justify;\">total momentum before the impact <img decoding=\"async\" style=\"height: 19px; width: 91px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch09\/image052.png\" \/>= 10 kg m\/s<\/p>\n<p style=\"text-align: justify;\">law of conservation of momentum:<\/p>\n<p style=\"text-align: justify;\">total momentum just before the impact = total momentum after the impact= 10 kg m\/s<\/p>\n<p style=\"text-align: justify;\">therefore, the velocity of the combined object: <img decoding=\"async\" style=\"height: 19px; width: 93px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch09\/image053.png\" \/><\/p>\n<p style=\"text-align: justify;\">v = 10\/6 = 1.67 m\/s<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 9 Science Forces and Laws of Motion part 2<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"16_An_object_of_mass_100_kg_is_accelerated_uniformly_from_a_velocity_of_to_in_6_s_Calculate_the_initial_and_final_momentum_of_the_object_Also_find_the_magnitude_of_the_force_exerted_on_the_object\"><\/span><strong>16. An object of mass 100 kg is accelerated uniformly from a velocity of <img decoding=\"async\" style=\"height: 22px; width: 52px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch09\/image040.png\" \/>to <img decoding=\"async\" style=\"height: 22px; width: 52px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch09\/image054.png\" \/> in 6 s. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>Initial momentum of the object <img decoding=\"async\" style=\"height: 19px; width: 65px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch09\/image055.png\" \/>= 500 kg m\/s<\/p>\n<p style=\"text-align: justify;\">Final momentum of the object = <img decoding=\"async\" style=\"height: 19px; width: 47px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch09\/image056.png\" \/>= 800 kg m\/s<\/p>\n<p style=\"text-align: justify;\">since v = u + at<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 19px; width: 79px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch09\/image057.png\" \/><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 22px; width: 150px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch09\/image058.png\" \/><\/p>\n<p style=\"text-align: justify;\">since <img decoding=\"async\" style=\"height: 19px; width: 136px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch09\/image059.png\" \/>= 50 N<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 9 Science Forces and Laws of Motion part 2<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"17_Akhtar_Kiran_and_Rahul_were_riding_in_a_motorcar_that_was_moving_with_a_high_velocity_on_an_expressway_when_an_insect_hit_the_windshield_and_got_stuck_on_the_windscreen_Akhtar_and_Kiran_started_pondering_over_the_situation_Kiran_suggested_that_the_insect_suffered_a_greater_change_in_momentum_as_compared_to_the_change_in_momentum_of_the_motorcar_because_the_change_in_the_velocity_of_the_insect_was_much_more_than_that_of_the_motorcar_Akhtar_said_that_since_the_motorcar_was_moving_with_a_larger_velocity_it_exerted_a_larger_force_on_the_insect_And_as_a_result_the_insect_died_Rahul_while_putting_an_entirely_new_explanation_said_that_both_the_motorcar_and_the_insect_experienced_the_same_force_and_a_change_in_their_momentum_Comment_on_these_suggestions\"><\/span><strong>17. Akhtar, Kiran and Rahul were riding in a motorcar that was moving with a high velocity on an expressway when an insect hit the windshield and got stuck on the windscreen. Akhtar and Kiran started pondering over the situation. Kiran suggested that the insect suffered a greater change in momentum as compared to the change in momentum of the motorcar (because the change in the velocity of the insect was much more than that of the motorcar). Akhtar said that since the motorcar was moving with a larger velocity, it exerted a larger force on the insect. And as a result the insect died. Rahul while putting an entirely new explanation said that both the motorcar and the insect experienced the same force and a change in their momentum. Comment on these suggestions.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>Since the mass of insect is negligible in comparison to the mass of motorcar therefore there will be no any change in the momentum of motorcar.<\/p>\n<hr \/>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"18_How_much_momentum_will_a_dumb-bell_of_mass_10_kg_transfer_to_the_floor_if_it_falls_from_a_height_of_80_cm_Take_its_downward_acceleration_to_be\"><\/span><strong>18. How much momentum will a dumb-bell of mass 10 kg transfer to the floor if it falls from a height of 80 cm? Take its downward acceleration to be<img decoding=\"async\" style=\"height: 22px; width: 60px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch09\/image060.png\" \/>.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\">Ans. height from which dumb bell falls = 80 cm = 0.8 m<\/p>\n<p style=\"text-align: justify;\">since we know<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 24px; width: 100px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch09\/image061.png\" \/><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 22px; width: 122px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch09\/image062.png\" \/>= 16<\/p>\n<p style=\"text-align: justify;\">v = <img decoding=\"async\" style=\"height: 21px; width: 28px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch09\/image063.png\" \/><\/p>\n<p style=\"text-align: justify;\">v = 4 m\/s<\/p>\n<p style=\"text-align: justify;\">momentum = mv <img decoding=\"async\" style=\"height: 19px; width: 57px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch09\/image064.png\" \/>= 40 kg m\/s<\/p>\n<h2><span class=\"ez-toc-section\" id=\"NCERT_Solutions_for_Class_9_Science\"><\/span>NCERT Solutions for Class 9 Science<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>NCERT Solutions Class 9 Science PDF (Download) Free from myCBSEguide app and myCBSEguide website. Ncert solution class 9 Science includes text book solutions .NCERT Solutions for CBSE Class 9 Science have total 15 chapters. 9 Science NCERT Solutions in PDF for free Download on our website. Ncert Science class 9 solutions PDF and Science ncert class 9 PDF solutions with latest modifications and as per the latest CBSE syllabus are only available in myCBSEguide.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"CBSE_app_for_Class_9\"><\/span>CBSE app for Class 9<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>To download\u00a0NCERT Solutions for class 9 Science, Computer Science, Home Science,Hindi ,English, Maths Social Science do check myCBSEguide app or website. myCBSEguide provides sample papers with solution, test papers for chapter-wise practice, NCERT solutions, NCERT Exemplar solutions, quick revision notes for ready reference, CBSE guess papers and CBSE important question papers. Sample Paper all are made available through\u00a0<a href=\"https:\/\/play.google.com\/store\/apps\/details?id=in.techchefs.MyCBSEGuide&amp;referrer=utm_source%3Dmycbse_bottom%26utm_medium%3Dtext%26utm_campaign%3Dmycbseads\"><strong>the best app for CBSE students<\/strong><\/a>\u00a0and myCBSEguide website.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>NCERT Solutions for Class 9 Science Forces and Laws of Motion part 2 Class 9 Science book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from &#8230; <a title=\"NCERT Solutions for Class 9 Science Forces and Laws of Motion part 2\" class=\"read-more\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-science-forces-and-laws-of-motion-part-2\/\" aria-label=\"More on NCERT Solutions for Class 9 Science Forces and Laws of Motion part 2\">Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":-1,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[281,1377],"tags":[283,1344,216,349],"class_list":["post-4141","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-ncert-solutions","category-science-cbse-class-09","tag-cbse-study-material","tag-class-9","tag-ncert-solutions","tag-science"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v26.0 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>NCERT Solutions for Class 9 Science Forces and Laws of Motion part 2<\/title>\n<meta name=\"description\" content=\"NCERT Solutions for Class 9 Science Forces and Laws of Motion part 2 in PDF format for free download. 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