If kinetic energy of a body …
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Posted by Pawan Sidhu 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
Consider a particle having mass m moving with a velocity v so, that its kinetic energy, K.E = {tex}\frac{1}{2}{/tex}mv2 and momentum, p = mv.
Thus, {tex}\mathrm{K.E}=\frac{p^{2}}{2 m} \text { or } p=\sqrt{2 m K .E}{/tex}
when, kinetic energy is increased by 300%, then new kinetic energy is given by:
K.E' = K.E + 300% of E =KE + 3KE = 4K.E
New momentum p' = {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, the Percentage increase in momentum {tex}=\frac{p^{\prime}-p}{p} \times 100=\frac{2 p-p}{p} \times 100=100 \%{/tex}
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