Kinetic energy of a body is incresed by 44%.What is the percent increse in momentum?
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We know
E = p^2/2m
as the increase in percentage is 44 %
then ,
(E2/E1) x 100 = 44
or
E2 = (44/100)E1
now
here mass remians constant
thus ,
p2/p1 = (2mE2)^1/2 / (2mE1)1^1/2
p2/p1 = (E2/E1)^1/2
p2/p1 = (44/100)^1/2
p2/p1 = 6.6/10 = 0.66
percent increase will be 0.66 x 100 = 66%
E = p^2/2m
as the increase in percentage is 44 %
then ,
(E2/E1) x 100 = 44
or
E2 = (44/100)E1
now
here mass remians constant
thus ,
p2/p1 = (2mE2)^1/2 / (2mE1)1^1/2
p2/p1 = (E2/E1)^1/2
p2/p1 = (44/100)^1/2
p2/p1 = 6.6/10 = 0.66
percent increase will be 0.66 x 100 = 66%
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