Question

if linear momentum is increased by 5%then kinetic energy will be increased by

Answer 1

 

We know the relation between linear momentum, P and kinetic energy, K is given by,
K = P²/2m , where m is mass of object. 

now, taking log base e both sides, 

logK = log(P²/2m) = logP² - log2m 

or, logK = 2logP - logm - log2 

differentiate both sides, 

dK/K = 2dP/P - dm/m - 0 

or, ± ∆K/K = ± 2 × ∆P/P ± ∆m/m

[ we took here ± because it can be possible increasing or decreasing so, in general expression we use both sign e.g., ± ] 

but for getting maximum error in Kinetic energy 
use, ∆K/K = 2 × ∆P/P + ∆m/m 

e.g., % error in K = 2 × % error in P + % error in m 

given, % error in P = 5 % 
% error in m = 0 

so, % error in K = 2 × 5 % = 10% 

hence , kinetic energy will increase by 10%