$$ \\\\ \\text{Initial velocity of ball at point A(u)=0} \\\\\\text{distance(s)=0}\\text{0.8m} \\\\\\text{so final velocity at point B, v can be calculated as} \\\\{{\\text{v}}^{2}}-{{u}^{2}}=2as (\\text{take }a=10m/{{s}^{2}}) \\\\\\therefore v=\\sqrt{2\\times 10\\times 0.8}=4m/s \\\\\\text{Let u } ' \\text{ be the velocity after rebounding the ball} \\\\\\text{Also final velocity at point C} \\text{(v } ' \\text{ )=0} \\\\\\text{using (v } ' {{\\text{)}}^{2}}-{{(u')}^{2}}=2as (\\text{here }s=0.2m) \\\\\\therefore u=\\sqrt{2\\times 10\\times 0.2}=2m/s \\\\\\text{Hence } \\text{Impulse } \\text{=change } \\text{in } \\text{momentum} \\\\\\text{Impulse=mv-(-mu } ' \\text{ )=mv+mu } ' \\text{ =m(v+u } ' \\text{ )} \\\\ \\text{take m =0}\\text{.08kg}, \\\\\\therefore\\text{Impulse=0}\\text{0.08(4+2)=0}\\text{.48 Ns}$$ ","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Newton's second law for rotation Quiz 1","text":"A customer came in the rope shop, where he told to the shopkeeper that he wants a very thick rope so that the rope can easily support the lift which is present in his building and the weight of the lift is 4500Kg and he told that the maximum upward acceleration of the lift is $$\\text{2 m/}{{\\text{s}}^{\\text{2}}}$$and he also told that the rope must have breaking stress of $$5.4 \\times {10^{8}} N{{\\text{m}}^{\\text{-2}}}$$ so that it can easily support the lift. Now help the shopkeeper to find the rope by calculating the diameter of the rope.$$\\text{(Take g=10m/}{{\\text{s}}^{2}})$$ ","comment":{"@type":"Comment","text":"Apply the formula Stress=$\\dfrac{Force}{Area}$ and then calculate the diameter."},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":"$$ 1.20\\times {{10}^{-2}}m$$ ","position":0},{"@type":"Answer","encodingFormat":"text/html","text":"$$ 1.2\\times {{10}^{2}}m$$ ","position":1},{"@type":"Answer","encodingFormat":"text/html","text":"$$ 1.02\\times {{10}^{-2}}m$$ ","position":2}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":"$$ 1.12\\times {{10}^{-2}}m$$","position":3,"answerExplanation":{"@type":"Comment","text":" $$\\text{In this mass of the lift(m)=4500Kg} \\\\\\text{upward acceleration(a)=2m/}{{\\text{s}}^{2}} \\\\\\text{breaking stress=5}\\text{.4}\\times \\text{1}{{\\text{0}}^{8}} \\\\\\text{Since the lift moves upwards, the tension in the rope is} \\\\\\text{T=m(g+a)=4500(10+2)=54000N} \\\\\\text{Also we know, stress=}\\dfrac{Force}{Area}=\\dfrac{T}{\\pi \\dfrac{{{D}^{2}}}{4}} \\\\5.4\\times {{10}^{8}}=\\dfrac{4\\times 54000\\times 7}{22\\times {{D}^{2}}} \\\\{{D}^{2}}=\\dfrac{4\\times 54000\\times 7}{22\\times 5.4\\times {{10}^{8}}}\\\\\\Rightarrow {{D}^{2}} =\\dfrac{\\text{15,12,000}}{118.8\\times {{10}^{8}}}\\\\\\Rightarrow {{D}^{2}}=\\dfrac{28}{22}\\times {{10}^{-4}}\\\\\\therefore {{D}^{2}}=1.27\\times \\text{1}{{\\text{0}}^{-4}} \\\\\\text{So we get value of Diameter(D)=1}\\text{.12}\\times \\text{1}{{\\text{0}}^{-2}} m$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Newton's second law for rotation Quiz 1","text":"Ashu rides a roller coaster ride in the circus. When he sits on the seat of the ride after some time the ride starts and it moves in a fixed circular path. To check the velocity of the ride Ashu invented a device which measures the velocity and when he checks the velocity, he finds that the ride moves with a velocity of $$\\text{23} \\text{m/s}$$. If the mass of the ride is 500g then what will be the angular momentum of the ride. $$\\text{(Assume that the perpendicular distance from the axis of rotation of the ride is 20m)}$$","comment":{"@type":"Comment","text":"First calculate linear momentum then calculate the value of angular momentum. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$340 kg{{\\text{m}}^{\\text{2}}}\\text{/s}$$ ","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$420 kg{{\\text{m}}^{\\text{2}}}\\text{/s}$$ ","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" $$190 kg{{\\text{m}}^{\\text{2}}}\\text{/s}$$ ","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$230 \\text{kg}{{\\text{m}}^{\\text{2}}}\\text{/s}$$","position":0,"answerExplanation":{"@type":"Comment","text":" $$\\text{Angular momentum=linear momentum } \\times \\text{ perpendicular distance from the axis of rotation} \\\\\\text{Linear momentum= Mass } \\times \\text{ Velocity} \\\\\\text{Mass=500g=0}\\text{.5Kg, velocity=23m/s} \\\\\\text{perpendicular distance from the axis of rotation=20m} \\\\\\text{So put all these values in the equation then } \\\\\\text{Angular momentum will be: 0}\\text{.5}\\times \\text{23}\\times \\text{20=230 kg}{{\\text{m}}^{\\text{2}}}\\text{/s}$$ ","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]}]}