The momentum of a body is increased by 25 the kinetic energy is increased by about
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The relation between kinetic energy, K and momentum p is the following.
K=p^2/2m…………(1)
Since, body is the same, mass remains constant.
If momentum increases by 25%, then new momentum will be
p’=1.25 p. Hence, new kinetic energy will be K’=(1.25)^2p^2/2m……….(2)
Then, (K’-K)/K=[p^2/2m(1.25)^2–1)]/(p^2/2m)=[(1.25)^2–1]=1.56–1.=0.56
Therefore % increase in kinetic energy=0.56x100=56%.
K=p^2/2m…………(1)
Since, body is the same, mass remains constant.
If momentum increases by 25%, then new momentum will be
p’=1.25 p. Hence, new kinetic energy will be K’=(1.25)^2p^2/2m……….(2)
Then, (K’-K)/K=[p^2/2m(1.25)^2–1)]/(p^2/2m)=[(1.25)^2–1]=1.56–1.=0.56
Therefore % increase in kinetic energy=0.56x100=56%.
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