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| 1. |
If kinetic energy of a body increases by 300% by what will be linear momentum of the body increase |
| Answer» Consider a particle\xa0having\xa0mass m moving with a velocity v so, that its kinetic energy, K.E\xa0=\xa0{tex}\\frac{1}{2}{/tex}mv2 and momentum, p = mv.Thus,\xa0{tex}\\mathrm{K.E}=\\frac{p^{2}}{2 m} \\text { or } p=\\sqrt{2 m K .E}{/tex}when, kinetic energy\xa0is increased by 300%, then new kinetic energy is given by:K.E\' = K.E + 300% of E =KE + 3KE = 4K.ENew momentum p\' =\xa0{tex}\\sqrt{2 m \\mathrm{K.E}^{\\prime}}=\\sqrt{2 m \\times 4 \\mathrm{K.E}}{/tex}{tex}=2 \\sqrt{2 m \\mathrm{K.E}}=2 p{/tex} Therefore,\xa0the Percentage increase in momentum\xa0{tex}=\\frac{p^{\\prime}-p}{p} \\times 100=\\frac{2 p-p}{p} \\times 100=100 \\%{/tex} | |