1KE=p2
K1=x2/2m
K2=121x2/200m
k2-k1=21x2/200m
K2-K1K1*100=21% (ANS)
When the momentum of a body increased by 10%, its kinetic energy (K.E.) by
a)20%
b)40%
c)44%
d)none of these
PLEASE GIVE FULL DETAILS ABOUT THE PROBLEM .
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6 Answers
1057first of all we have to see that the inc in momentum is only due inc in velocity(mass is constant)
let mass be m nd original vel be v
v' be the changed velocity
mv=x
mv'=11x10
so vv'=1110
% inc in KE=12mv'2 -12mv212mv2 into 100
12 and m get cancelled
we are left with v'2-v2v2 into 100
so the answer turns out be 112 -102102 into 100
which is 21%
1057@vivek hum aur kya kiya?????? :P :P
i jus explained it a bit more.....