The kinetic energy of a body is increased by 21%. What is the percentage increase in the linear momentum of the body?
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k.e=1/2×mv^2
=1/2×m^2v^2/m
k.e=p^2/m
initial patential energy
p=root2mk where k.e=k
increase in k.e=20% of k=0.2k
total k.e=k+0.2k
=1.2k
Final momentum,p'=root1.2(2m)
% increase in linear momentum=p'-p/p×100
=root1.2 (2m)-root2mk /root2mk*100
=root2mk (1.1-1)/root2mk*100
=10%
=1/2×m^2v^2/m
k.e=p^2/m
initial patential energy
p=root2mk where k.e=k
increase in k.e=20% of k=0.2k
total k.e=k+0.2k
=1.2k
Final momentum,p'=root1.2(2m)
% increase in linear momentum=p'-p/p×100
=root1.2 (2m)-root2mk /root2mk*100
=root2mk (1.1-1)/root2mk*100
=10%
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