","comment":{"@type":"Comment","text":" Before and after the collision, measure the retarding force and balance the energy equation. Because of friction, kinetic energy will change."},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{15}{16}\\dfrac{m{{v}^{2}}}{t}$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{3}{8}\\dfrac{m{{v}^{2}}}{T}$$","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{3}{8}\\dfrac{m{{v}^{2}}}{t}$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{15}{32}\\dfrac{m{{v}^{2}}}{t}$$","position":0,"answerExplanation":{"@type":"Comment","text":" $$\\text{Energy of bullet before collision} =\\dfrac{1}{2}m{{v}^{2}} \\\\ \\Rightarrow\\text{Energy of bullet after collision} =\\dfrac{1}{2}m{{\\left( \\dfrac{v}{4} \\right)}^{2}} \\\\ \\Rightarrow\\text{Energy of bullet after collision}=\\dfrac{1}{16}\\left( \\dfrac{1}{2}m{{v}^{2}} \\right) \\\\ \\text{Loss in kinetic energy=Energy consume by wood of thickness}\\,''t'' \\\\ \\Rightarrow\\dfrac{15}{32}m{{v}^{2}}=f\\times t \\\\ \\therefore f=\\dfrac{15}{32}\\dfrac{m{{v}^{2}}}{t}$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Kinetic energy Quiz 1","text":" A man-sized ball of mass $$\"m\"$$ is thrown vertically upward with a velocity of $$\\sqrt{gH}$$ from a building of height $$\"H\"$$. The provided kinetic energy just before impact with the ground is","comment":{"@type":"Comment","text":"To find gain in kinetic energy, find velocity just before striking the around. Then the difference in kinetic energy will be equal to gain in kinetic energy. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$3mgH$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$4mgH$$","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" $$8mgH$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$mgH$$","position":0,"answerExplanation":{"@type":"Comment","text":" $$\\text{Initial velocity }=\\sqrt{gH}\\text{, }K.\\sum i=\\dfrac{1}{2}mgH \\\\ \\Rightarrow\\text{Final velocity}={{v}^{2}}={{\\left( -\\sqrt{gH} \\right)}^{2}}+2gH \\\\ \\Rightarrow v=\\sqrt{3gH} \\\\ \\Rightarrow\\text{Final kinetic energy}=\\dfrac{1}{2}m\\left( 3gH \\right)\\\\ \\Rightarrow\\text{Gain in kinetic energy}=\\dfrac{3}{2}mgH-\\dfrac{1}{2}mgH \\\\ \\therefore\\text{Gain in kinetic energy}=mgH$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Kinetic energy Quiz 1","text":" A man-made block of mass $$m=1\\,kg$$ with an initial velocity of $$v=10\\,m/s$$, is allowed to slide over a rough surface with a coefficient of friction of $$\\mu =0.1$$. Because of friction, it comes to a halt after $$T\\sec $$. The distance travelled by block arriving at a rest is","comment":{"@type":"Comment","text":"Use energy balance equation $\\dfrac{1}{2}m{{v}^{2}}=\\mu mgS$ to predict the distance travelled. Whole kinetic energy is consumed by frictional energy. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$100\\,m$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$25\\,m$$","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" $$12.5\\,m$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$50\\,m$$","position":1,"answerExplanation":{"@type":"Comment","text":" $$\\text{In given problem,} \\\\ \\text{Kinetic energy of block is consumed by friction,} \\\\ \\dfrac{1}{2}m{{v}^{2}}=\\left( \\mu mg \\right)\\times S \\\\ \\Rightarrow \\dfrac{1}{2}\\times 1\\times 10\\times 10=0.1\\times 1\\times 10\\times S \\\\ \\therefore S=50\\,\\text{m}$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Kinetic energy Quiz 1","text":" A ball of man $$''m''$$ moving with velocity $${{V}}$$ in horizontal direction strikes another hall of man $$''2m''$$ resting over the surface elastically. The loss in kinetic energy is ","comment":{"@type":"Comment","text":"Use momentum cons equation to get final velocity of system, then use $\\dfrac{1}{2}m{{v}^{2}}=K.E. $ to get energy before and after collision. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$50\\%$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$33\\%$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$75\\%$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$67\\%$$","position":2,"answerExplanation":{"@type":"Comment","text":" $$\\text{Energy before collision} \\\\ {{K.E}_{i}}=\\dfrac{1}{2}m{{v}^{2}} \\\\ \\text{Momentum is conserve as no external force acting on system}\\text{.} \\\\ MV=\\left( 3m \\right){{V}_{common}} \\\\ \\Rightarrow {{V}_{common}}=\\dfrac{V}{3} \\\\ \\Rightarrow K.{{E }_{f}}=\\dfrac{1}{2}\\left( 3m \\right)\\dfrac{{{V}^{2}}}{9}=\\dfrac{1}{3}\\left( \\dfrac{1}{2}m{{V}^{2}} \\right) \\\\ \\therefore\\text{Energy loss}=\\dfrac{\\left( \\dfrac{2}{3} \\right)\\times {{K.E}_{i}}}{{{K.E}_{i}}}\\times 100=67\\%\\left( approx. \\right)$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Kinetic energy Quiz 1","text":" Velocity time graph shows the variation in velocity with respect to time. According to graph the kinetic energy of particle is ","comment":{"@type":"Comment","text":"Kinetic energy of particles depends on magnitude of velocity, check velocity at various instances. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" increasing with time ","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" constant throughout","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" none of above","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" decreasing with time","position":1,"answerExplanation":{"@type":"Comment","text":"
$$ \\\\ $$ In given $$v-t$$ graph magnitude of velocity decreasing with time. $$ \\\\ $$ So, kinetic energy $$=\\dfrac{1}{2}m{{v}^{2}} $$ $$\\\\ $$ Also decreases with time.","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]}]}