Question

30. If the error in the measurement of momentum of a particle is 100%, then % error in
themeasurement of kinetic energy is:
(a)100%
(b)200%
(c)300%
(d)400%​

Answer 1

Answer:

Kinetic energy = m*v*v/2 = p^2/2m

Here p = momentum = m*v

Now, let a slight change in a particular quantity 'X’ be dX

KE = p^2/(2m)

Taking logarithm on both sides,

log(KE) = 2log(p) - 2log(2m)

Differentiating on both sides,

d(KE)/KE = 2dp/p …(change in Mass is 0)

From this equation, we can observe that the error in measurement of Kinetic energy is twice that of momentum.

Hence the answer is 200%