","comment":{"@type":"Comment","text":"We know that ratio of final velocities of both ball will be $\\dfrac{{{v}_{1}}}{{{v}_{2}}}=\\dfrac{1-e}{1+e}$. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{1}{2}$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{1}{\\sqrt{3}}$$","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" Both (a) or (b)","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{1}{\\sqrt{2}}$$","position":1,"answerExplanation":{"@type":"Comment","text":" $$K.{{E}_{f}}=0.75K.{{E}_{i}}\\\\\\Rightarrow \\dfrac{1}{2}m{{v}_{1}}^{2}+\\dfrac{1}{2}m{{v}_{2}}^{2}=0.75\\times \\dfrac{1}{2}m{{v}^{2}}\\\\\\Rightarrow mv=m{{v}_{1}}+m{{v}_{2}}\\\\\\Rightarrow {{v}_{1}}=(\\dfrac{1+e}{2})v\\\\\\Rightarrow {{v}_{2}}=(\\dfrac{1-e}{2})v\\\\ $$ Putting these values in kinetic energy equation, we have $$ \\\\ {{\\left( 1+e \\right)}^{2}}+{{\\left( 1-e \\right)}^{2}}=3\\\\\\Rightarrow 2+2{{e}^{2}}=3\\\\\\therefore {{e}^{2}}=\\dfrac{1}{2}$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Inelastic collisions Quiz 1","text":"A man throws a ball towards a wall with a velocity of $$4\\hat{i}+6\\hat{j}$$. The coefficient of restitution between wall and ball is 1/4 . The final velocity vector calculated by the man after the impact will be-","comment":{"@type":"Comment","text":"Velocity along the wall will remain the same as there is no collision happening vertically to the wall but there is a head-on horizontal collision in which velocity is changing due to the coefficient of restitution. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\hat{i}+6 \\hat{j}$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$\\hat{i}+4 \\hat{j}$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$-\\hat{i}-4 \\hat{j}$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$-\\hat{i}+6 \\hat{j}$$","position":2,"answerExplanation":{"@type":"Comment","text":" Coefficient of restitution $$(e) =\\dfrac{\\text{velocity of separation}}{\\text{velocity of approach}} $$ $$ \\\\ $$ Vertical component along the wall will remain same but for horizontal component, $$ \\\\e=\\dfrac{\\text{velocity of separation}}{\\text{velocity of approach}}\\\\\\Rightarrow e =\\dfrac{v-0}{0-4}\\\\\\Rightarrow e =-\\dfrac{v}{4} \\\\\\Rightarrow \\dfrac{1}{4}=-\\dfrac{v}{4} \\\\\\therefore v=-1\\\\$$ So, final velocity component will be $$ -\\hat{i}+6\\hat{j}$$.","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Inelastic collisions Quiz 1","text":"A box of mass $$2\\,kg$$ is attached to a ceiling with weightless string as shown in figure. A boy at the ground throws a small stone of mass $$1\\,kg$$ so that it hits a block with a speed of $$5\\,m/s$$ and stick to it, the final speed of combined system of block + stone will be- ","comment":{"@type":"Comment","text":"Stone is sticking to the box after colliding with it, so the total momentum of the system will be conserved."},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{1}{2}m/s$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{2}{3}m/s$$","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{3}{2}m/s$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{5}{3}m/s$$","position":1,"answerExplanation":{"@type":"Comment","text":" $$\\text{Applying momentum conservation-}\\\\{{m}_{1}}{{v}_{1}}=({{m}_{2}}+{{m}_{1}})v\\\\\\Rightarrow 1\\times 5=(2+1)v\\\\\\therefore v=\\dfrac{5}{3}\\,m/s$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Inelastic collisions Quiz 1","text":"A basketball player drops the ball from a height of $$5\\,m$$ which collides with ground and rebounds and again reaches some height $$H$$, the coefficient of restitution is $$0.7$$. The value of $$H$$ calculated by the player will be- ","comment":{"@type":"Comment","text":"The above formula can easily be calculated by the momentum and energy conservation of the system. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" 3.45 m","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" 1.55 m","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" 6.5 m","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" 2.45 m","position":0,"answerExplanation":{"@type":"Comment","text":" We know that the next height after the rebound will be $$\\\\ {{H}_{2}}={{e}^{2}}{{H}_{1}}\\\\\\Rightarrow {{H}_{2}}={{0.7}^{2}}\\times 5\\\\\\therefore {{H}_{2}}=2.45\\,m$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]}]}