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Q.
If the momentum of the body is increased $ 100% $ . Then the percentage increase in the kinetic energy will be
Punjab PMETPunjab PMET 1999
Solution:
Initial momentum $ {{p}_{1}}=100\,\,p $ Final momentum $ {{p}_{2}}=200\,\,p $ Initial kinetic energy $ {{K}_{1}}=K $ The kinetic energy is given by $ K=\frac{{{p}^{2}}}{2m}\propto {{p}^{2}} $ or $ \frac{{{K}_{1}}}{{{K}_{2}}}={{\left( \frac{{{p}_{1}}}{{{p}_{2}}} \right)}^{2}} $ or $ \frac{{{K}_{1}}}{{{K}_{2}}}={{\left( \frac{100p}{200p} \right)}^{2}}=\frac{1}{4} $ $ {{K}_{2}}=4\,\,{{K}_{1}} $ Increase in kinetic energy is $ =4{{K}_{1}}-{{K}_{1}}=3{{K}_{1}}=300% $