the linear momentum of a body is increased by 10%what is the percentage change in its KE?
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p=mv
p=mv+10%
k.e=1/2mv^2
v=p/m
k.e=1/2m(p/m)^2
k.e=p^2/2m
k.e=mv+10%/2m
K = p² / (2m)
When p is increased by 10%, the new kinetic energy will be,
K’=(1.1p)^2/2m=1.21(p^2/2m)
% increase=(change/original value)x100.=[(1.21–1)(p^2/2m/(p^2/2m)]x100=21%.
p=mv+10%
k.e=1/2mv^2
v=p/m
k.e=1/2m(p/m)^2
k.e=p^2/2m
k.e=mv+10%/2m
K = p² / (2m)
When p is increased by 10%, the new kinetic energy will be,
K’=(1.1p)^2/2m=1.21(p^2/2m)
% increase=(change/original value)x100.=[(1.21–1)(p^2/2m/(p^2/2m)]x100=21%.
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