Work Energy Power

JEE Main level, single correct, medium

79 of 183

Question 79 of 183

If momentum of a body is increased by 2020%, then its kinetic energy increases by :

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Answer: (C) 4444%

A body of mass mm, moving with momentum pp and kinetic energy KK are related as follows,
K=p22mK=\dfrac{{p}^{2}}{2m}.
As the momentum of the body is increased by 2020%, then its momentum becomes,
(p+20100p)=1.2p\left(p+\dfrac{20}{100}p\right)=1.2p
and the mass of the body remains unchanged. So, kinetic energy is directly related to the square of momentum.
Hence, new kinetic energy (K′)\left({K}^{'}\right) will be,
K′=(1.2p)22m{K}^{'}=\dfrac{{\left(1.2p\right)}^{2}}{2m}
K′=1.44K{K}^{'}=1.44K.
So, the percentage change in kinetic energy
=K′−KK×100=\dfrac{{K}^{'}-K}{K}\times 100
=1.44K−KK×100=\dfrac{1.44K-K}{K}\times 100
=(1.44−1)×100=\left(1.44-1\right)\times 100
=44=44%.

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