","comment":{"@type":"Comment","text":" Velocity of mass when it touches the ball at ground can be calculated by conservation of mechanical energy. Also, $e=1$ for elastic collision so bodies of equal mass exchange their velocities after elastic collision. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\sqrt{\\dfrac{gL}{2}}$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{\\sqrt{3}}{2}gL $$","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" Zero","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\sqrt{2gL}$$","position":0,"answerExplanation":{"@type":"Comment","text":" As the body lowers from a height $$L$$, it gains kinetic energy and loses gravitational potential energy. $$ \\\\ $$ Total energy should be conserved, so, $$ \\\\ \\text{K}\\text{.}{{\\text{E}}_{i}}+P.{{E}_{i}}=\\text{K}\\text{.}{{\\text{E}}_{f}}+P.{{E}_{f}} \\\\ 0+mgL=\\dfrac{1}{2}m{{v}^{2}}+0 \\\\ \\Rightarrow \\dfrac{1}{2}m{{v}^{2}}=mgL \\\\ \\Rightarrow v=\\sqrt{2gL} \\\\ $$ $$ \\therefore v=\\sqrt{2gL}$$ is the velocity of the body before striking identical mass kept at rest on ground. If two identical bodies collide elastically, they exchange their velocities. So, velocity of mass attached to rod becomes zero, while the mass kept on ground starts moving with $$\\,\\sqrt{2gL}$$ velocity","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Momentum and collisions Quiz 1","text":"A body of mass $$\\left( m \\right)$$ and velocity $$V$$ strikes another body of mass $$\\left( 4m \\right)$$ pure inelastically. The loss in kinetic energy is ","comment":{"@type":"Comment","text":" Conserve momentum before and after collision, then calculate their respective kinetic energies."},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$75\\%$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$90\\%$$","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" $$60\\%$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$80\\%$$","position":1,"answerExplanation":{"@type":"Comment","text":" As momentum is conserved because of zero externally applied force, $$ \\\\ $$ Initial momentum $${{\\text{P}}_{i}}=mV,\\text{ }K.{{E}_{initial}}=\\dfrac{1}{2}m{{V}^{2}}$$ $$ \\\\ $$ Final momentum $${{\\text{P}}_{f}}=\\left( m+4m \\right){{V}_{common}}\\text{=5m}{{\\text{V}}_{common}} $$ because collision is purely inelastic so both bodies will stick together $$ \\\\ {{\\text{P}}_{i}}={{\\text{P}}_{f}} \\\\ \\Rightarrow mV=\\text{5m}{{\\text{V}}_{common}} \\\\ \\Rightarrow {{V}_{common}}=\\left( \\dfrac{V}{5} \\right), \\\\ $$ Final kinetic energy, $$K.{{E}_{f}}=\\dfrac{1}{2}\\left( 5m \\right)\\times \\dfrac{{{V}^{2}}}{25}=\\dfrac{1}{5}\\left( \\dfrac{1}{2}m{{V}^{2}} \\right) \\\\ $$ Loss in K.E $$ =\\dfrac{K.{{E}_{i}}-K.{{E}_{f}}}{K.{{E}_{i}}}$$ $$ \\\\ =\\dfrac{\\dfrac{1}{2}m{{V}^{2}}-\\dfrac{1}{5}\\left( \\dfrac{1}{2}m{{V}^{2}} \\right)}{\\dfrac{1}{2}m{{V}^{2}}}\\times 100 \\\\ \\Rightarrow \\text{Loss in }K.E= \\dfrac{\\dfrac{4}{5}\\left( \\dfrac{1}{2}m{{V}^{2}} \\right)}{\\dfrac{1}{2}m{{V}^{2}}}\\times 100 \\\\ \\therefore \\text{Loss in }K.E=80\\%$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Momentum and collisions Quiz 1","text":" In a given arrangement, a block is placed on the top most position of inclined surface of wedge of mass $$2\\,m$$.The angle of inclination is $$\\theta =60{}^\\circ $$, if height of the wedge is $$H$$ then find its velocity when the block of mass $$\\left( m \\right)$$ reaches ground after slipping without any friction is $$ \\\\ $$ ","comment":{"@type":"Comment","text":" There is no external force on the system along the x-axis, so momentum is conserved along the x-axis of the system. Also, mechanical energy of the system is conserved too. Potential energy of the system is given as $mgh$, kinetic energy is $\\dfrac{1}{2}m{{v}^{2}}$ and momentum is $P=mv$. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\sqrt{\\dfrac{gH}{3}}$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{\\sqrt{3gH}}{2}$$","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" $$\\sqrt{gH}$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{4}{3}\\sqrt{gH}$$","position":0,"answerExplanation":{"@type":"Comment","text":"
$$ \\\\ $$ Initial velocities are zero for both objects. $$ \\\\ $$ Thus initial momentum $$=0$$ Let, final velocity of block when it is about to reach ground be $$V$$, and that of the wedge be $$V'$$. $$ \\\\ $$ Velocity $$V$$ along x-axis $${{V}_{x}}$$ for the block is $$ \\\\ {{\\text{V}}_{x}}=V\\cos 60{}^\\circ =\\dfrac{V}{2}\\text{, } \\\\ $$ Final momentum $${{P}_{f}}=m\\left( \\dfrac{V}{2} \\right)+2mV' $$ $$ \\\\ $$ As there are no external forces along x-axis, $$ \\\\ {{\\text{P}}_{i}}={{P}_{f}} \\\\ 0=2m\\times V'+m\\dfrac{V}{2} \\\\ \\Rightarrow 2V'=-\\dfrac{V}{2} \\\\ \\Rightarrow V'=-\\dfrac{V}{4} \\\\ $$ From conservation of mechanical energy, $$ \\\\ K.{{E}_{i}}+P.{{E}_{i}}=K.{{E}_{f}}+P.{{E}_{f}} \\\\ \\Rightarrow K.{{E}_{i}}=0 \\\\ \\Rightarrow P.{{E}_{i}}=mgH \\\\ \\Rightarrow K.{{E}_{f}}=\\dfrac{1}{2}m{{V}^{2}}+\\dfrac{1}{2}\\left( 2m \\right)V{{'}^{2}} \\\\ \\Rightarrow P.{{E}_{f}}=0 \\\\ \\Rightarrow 0+mgH=\\dfrac{1}{2}m{{V}^{2}}+\\dfrac{1}{2}\\left( 2m \\right)V{{'}^{2}}+0 \\\\ \\Rightarrow gH=\\dfrac{{{V}^{2}}}{2}+V{{'}^{2}} \\\\ \\Rightarrow gH=\\dfrac{{{V}^{2}}}{2}+\\dfrac{{{V}^{2}}}{16} \\\\ \\Rightarrow gH=\\dfrac{9{{V}^{2}}}{16} \\\\ \\Rightarrow V=\\sqrt{\\dfrac{16gH}{9}} \\\\ \\therefore V=\\dfrac{4}{3}\\sqrt{gH}$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]}]}