Work Energy Power

JEE Main level, numerical answer, medium

71 of 183

Question 71 of 183

The momentum of a body is increased by 5050%. The percentage increase in the kinetic energy of the body is________%.

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Answer: 125

The relation of kinetic energy and momentum is given by
K=p22mK=\dfrac{{p}^{2}}{2m}
The new value of momentum is p′=1.5pp'=1.5p
The new value of kinetic energy is
⇒K′=(1.5p)22m=2.25K\Rightarrow K'=\dfrac{{\left(1.5p\right)}^{2}}{2m}=2.25K
Hence, the percentage increase in kinetic energy is
K′−KK×100=2.25K−KK=125\dfrac{K'-K}{K}\times 100=\dfrac{2.25K-K}{K}=125%

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