Question

If the linear momentum is increased by 5% the kinetic energy will increase 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%