","comment":{"@type":"Comment","text":" Coordinates of center of mass of a system of particles can be calculated as summation of mass and distance from the individual masses divided by summation of masses of objects. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\left( 0.5,0.3 \\right)$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$\\left( 0.2,0.6 \\right)$$","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" $$\\left( 0.6,0.2 \\right)$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\left( 0.3,0.5 \\right)$$","position":0,"answerExplanation":{"@type":"Comment","text":" The centre of mass will be, $$ \\\\ x=\\dfrac{{{m}_{1}}{{x}_{1}}+{{m}_{2}}{{x}_{2}}+{{m}_{3}}{{x}_{3}}+{{m}_{4}}{{x}_{4}}}{{{m}_{1}}+{{m}_{2}}+{{m}_{3}}+{{m}_{4}}}\\\\ \\Rightarrow x=\\dfrac{6\\times 0+4\\times 1+2\\times 1+8\\times 0}{6+4+2+8}=0.3\\\\ \\Rightarrow y=\\dfrac{{{m}_{1}}{{y}_{1}}+{{m}_{2}}{{y}_{2}}+{{m}_{3}}{{y}_{3}}+{{m}_{4}}{{y}_{4}}}{{{m}_{1}}+{{m}_{2}}+{{m}_{3}}+{{m}_{4}}}\\\\ \\therefore y=\\dfrac{6\\times 0+4\\times 0+2\\times 1+8\\times 1}{6+4+2+8}=0.5$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Center of mass Quiz 1","text":"A bomb disposal team was sent to a station where a suspicious bag had been found in a corner. As they opened the bag, they discovered a ball that split into two pieces weighing $$2\\,kg$$ and $$4\\,kg$$, with a velocity of $$5\\,m/s$$ for the $$2\\,kg$$ component. The squad is aware that it was a prank played by a child, so they want to do something productive, so they want to measure the kinetic energy of the greater part of the ball, so that the kinetic energy can be-","comment":{"@type":"Comment","text":" Initial momentum of the system is zero as particles as at rest. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$8.5\\text{ }J$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$4.5\\text{ }J$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$16.5\\text{ }J$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$12.5\\text{ }J$$","position":2,"answerExplanation":{"@type":"Comment","text":" Applying momentum conservation we have $$ \\\\ {{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}={{m}_{1}}{{v}_{3}}+m{{v}_{4}}\\\\ $$ Initial velocities of particles is zero so $$ \\\\ 0=2\\times 5-4\\times v\\\\ \\Rightarrow v=2.5\\,m/s\\\\ \\text{Kinetic energy} = \\dfrac{1}{2}m{{v}^{2}}=\\dfrac{1}{2}\\times 4\\times {{(2.5)}^{2}}\\\\ \\therefore\\text{Kinetic energy} =12.5\\,J$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Center of mass Quiz 1","text":"As seen in the diagram, a plate with uniform mass density is mounted on a plane surface with an x-y plane. The system's centre of mass will be placed. $$ \\\\ $$ ","comment":{"@type":"Comment","text":"Coordinates of center of mass are calculated by taking points of center of mass of individual plates multiply by their areas. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\left( 1.5,0.5 \\right)$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$\\left( 1.5,1.5 \\right)$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$\\left( 0.5,1.5 \\right)$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\left( 1.5,0.75 \\right)$$","position":2,"answerExplanation":{"@type":"Comment","text":" The $$x$$ coordinate of center of mass will be given by, $$ \\\\ x=\\dfrac{{{A}_{1}}{{x}_{1}}+{{A}_{2}}{{x}_{2}}+{{A}_{3}}{{x}_{3}}+{{A}_{4}}{{x}_{4}}}{{{A}_{1}}+{{A}_{2}}+{{A}_{3}}+{{A}_{4}}}\\\\ \\Rightarrow x=\\dfrac{A\\times 0.5+A\\times 1.5+A\\times 2.5+A\\times 1.5}{A+A+A+A}\\\\ \\Rightarrow x =\\dfrac{6}{4}\\\\ \\Rightarrow x =1.5\\\\ $$ Similarly y coordinate $$ \\\\ y=\\dfrac{{{A}_{1}}{{y}_{1}}+{{A}_{2}}{{y}_{2}}+{{A}_{3}}{{y}_{3}}+{{A}_{4}}{{y}_{4}}}{{{A}_{1}}+{{A}_{2}}+{{A}_{3}}+{{A}_{4}}}\\\\ \\Rightarrow y=\\dfrac{A\\times 0.5+A\\times 0.5+A\\times 0.5+A\\times 1.5}{A+A+A+A}\\\\ \\Rightarrow y =\\dfrac{3}{4}\\\\ \\therefore y=0.75$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Center of mass Quiz 1","text":" Ram throws a rock at a velocity of $$5\\,ms^{-1}$$ at an angle of $$60^\\circ $$ with the horizontal to test his strength, but the stone splits in two parts of $$\\dfrac{1}{3}^{rd}$$ and $$\\dfrac{2}{3}^{rd}$$ of its original mass at some point during its trajectory.The smaller mass moves along the negative x direction with velocity of $$2\\text{ }m{{s}^{-1}}$$ the velocity of larger part will be- ","comment":{"@type":"Comment","text":"The total momentum along the x direction will remain conserved so a velocity component in horizontal direction is taken. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$3.75\\text{ }m{{s}^{-1}}$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$1.25\\text{ }m{{s}^{-1}}$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$5.75\\text{ }m{{s}^{-1}}$$","position":2}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$4.75\\text{ }m{{s}^{-1}}$$","position":3,"answerExplanation":{"@type":"Comment","text":" Applying momentum conservation in horizontal direction we get $$ \\\\ mv\\cos ({{60}^{\\circ }})=\\dfrac{2}{3}m{{v}_{1}}+\\dfrac{1}{3}m(-{{v}_{2}})\\\\ \\Rightarrow 5\\times \\dfrac{1}{2}=\\dfrac{2}{3}(v)-\\dfrac{1}{3}\\times 2\\\\ \\therefore v=4.75\\text{ }m{{s}^{-1}}$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Center of mass Quiz 1","text":" A pulley-ball system is set up by a man who wants to experiment the movement of the centre of mass of the system as shown in figure. If this system is released from rest the acceleration of COM of the system will be $$ \\\\ $$
","comment":{"@type":"Comment","text":" Acceleration of center of mass of moving system is calculated with summation of masses and acceleration of masses divided by summation of total mass. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{9}{10}\\,m/{{s}^{2}}$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{5}{7}\\,m/{{s}^{2}}$$","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{7}{5}\\,m/{{s}^{2}}$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{10}{9}\\,m/{{s}^{2}}$$","position":0,"answerExplanation":{"@type":"Comment","text":" Let T be tension in threads so, $$ \\\\ 10-T=1\\times a\\\\ \\Rightarrow a=\\dfrac{10}{3}\\\\ $$ The acceleration of both balls will be equal but in opposite direction so, $$ \\\\ \\Rightarrow {{a}_{c}}=\\dfrac{{{m}_{1}}{{a}_{1}}+{{m}_{2}}{{a}_{2}}}{{{m}_{1}}+{{m}_{2}}}\\\\ \\Rightarrow {{a}_{c}}=\\dfrac{2\\times \\dfrac{10}{3}-\\dfrac{10}{3}}{1+2}\\\\ \\therefore{{a}_{c}} =\\dfrac{10}{9}$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]}]}