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Grade 12th passMechanics

if the increase in the kinetic energy of a body is 22% then the increase in the momentum is nearly what %?

Profile image of akshay kumar patra
9 Years agoGrade 12th pass
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2 Answers

Profile image of NITAI PATRA
9 Years ago
E=1/2(mv2)
or,E=1/2(m2v2)/m
or,E=P2/m as P=mv
or,dE=PdP/m
so,
dE/E=2(dP/P)
If dE/E is 22%. then dP/P is 11% as 
Profile image of Debayan Ghosh
5 Years ago
P=mv
E=1/2mv²
P=√2mE, if m is constant
So, P2/P1 = √E2/E1=√1.22E/E
 ›P2=1.1P1
›P2 = P1+0.1 P1 =P1+ 10% of P1