KE of a body is increased by 44 % what is the percentage increase in momentum
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have earlier shown the relation between kinetic energy and momentum in my ans of question of the same type which can be seen from my answers.
The relation is
Ke=p^2÷2m
Now,
Ke'=ke+44%ofke
=Ke+0.44ke
Ke' =1.44ke
P'^2÷2m=1.44p^2÷2m
P'2=1.44p^2
P'=1.2p
Now %increase in momentum,
={(P'-p)÷p}×100
={(1.2p-p)÷p}×100
={0.2p÷p}×100
=0.2×100
=20%
Hence momentum will increase by 20%
Thanks for reading and upvoting…
The relation is
Ke=p^2÷2m
Now,
Ke'=ke+44%ofke
=Ke+0.44ke
Ke' =1.44ke
P'^2÷2m=1.44p^2÷2m
P'2=1.44p^2
P'=1.2p
Now %increase in momentum,
={(P'-p)÷p}×100
={(1.2p-p)÷p}×100
={0.2p÷p}×100
=0.2×100
=20%
Hence momentum will increase by 20%
Thanks for reading and upvoting…
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