Question

If the error in the measurement of momentum of a particle is (+ 100%), then the error in the measurement of kinetic energy is
A) $ 100\% $
B) $ 200\% $
C) $ 300\% $
D) $ 400\% $

Answer 1

Hint : We will use the relation between momentum and energy of an object to find the dependence of energy on momentum. Then we will use the laws of error propagation to determine the percentage error in the measurement of kinetic energy knowing the error in measurement of momentum of the particle.

Formula Used: In this solution we will be using the following formula,
 $ K = \dfrac{1}{2}m{v^2} $ where $ K $ is the $ m $ is the mass and $ v $ is the velocity.

Complete step by step answer
We know that the kinetic energy of an object of mass $ m $ and velocity $ v $ is calculated as
 $ K = \dfrac{1}{2}m{v^2} $
Multiplying the numerator and denominator of the right side by $ m $ , we get
 $ K = \dfrac{1}{2}\dfrac{{{m^2}{v^2}}}{m} $
Since the momentum of the particle will be $ P = mv $ , we can write
 $ K = \dfrac{{{p^2}}}{{2m}} $
Taking the logarithm on both sides, we get
 $ \log K = 2\log p - \log (2m) $
On differentiating both sides, we get
 $ \dfrac{{dK}}{K} = 2\dfrac{{dP}}{P} $
Since the percentage error in measuring kinetic energy is $ 100\% $ , we can say
 $ \dfrac{{dK}}{K} \times 100 = 100\% $
And we can multiply both sides by 100 and write
 $ \dfrac{{\Delta K}}{K} \times 100 = 2 \times \dfrac{{\Delta P}}{P} \times 100 $
Which can be interpreted as
 $ \% \,{\text{error}}\,{\text{in}}\,{\text{kinetic energy = 2}} \times \% \,{\text{error}}\,{\text{in}}\,{\text{momentum}} $
So,
 $ \dfrac{{\Delta K}}{K} \times 100 = 2 \times \dfrac{{\Delta P}}{P} \times 100 $
 $ \Rightarrow \dfrac{{\Delta K}}{K} \times 100 = 2 \times 100 = 200\% $
Hence the percentage error in the measurement of kinetic energy is $ 200\% $ when the percentage error in measurement of momentum is $ 100\% $ which corresponds to option (B).

Note
Here we have assumed that the measurement error in the momentum of the body takes into account the error only in the velocity of the body and that the mass of the body remains constant and that there is no error in measuring it. If there was uncertainty in the measurement of the mass of the body, we would also have to take into account the error in measuring the mass of the body. We should avoid the mistake of selecting the answer $ 300\% $ by adding $ 200\% $ error to the $ 100\% $ error of kinetic energy since both the errors are independent of each other and each individually depends on the error in velocity only.