If the error in the measurement of the momentum of a particle is 100 - than- what will be the error in measurement of kinetic energy of that particle

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 sid ...

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Standard XII

Physics

Bernoulli's Principle

If the error ...

Question

If the error in the measurement of the momentum of a particle is 100℅ then, what will be the error in measurement of kinetic energy of that particle?

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Solution

As we know ;
$m o m e n t u m, P = m v ⟶ (1)$
$k i n e t i c e n e r g y, K E = \frac{1}{2} m v^{2} ⟶ (2)$
$f r o m (1) & (2) w e c a n s a y$
$p^{2} = 2 m K E$
$e r r o r i n m e a s u r e m e n t c a n b e g i v e n a s$
$2 \frac{Δ p}{p} = \frac{Δ K E}{K E}; t h e r e i s n o e r r o r c o n s i d e r i n m a s s$
$\frac{Δ K E}{K E} = 2 \frac{Δ P}{P}$
$\frac{Δ K E}{K E} = 2 (100 %) = 200 %$ .

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If the error in the measurement of the momentum of a particle is 100℅ then, what will be the error in measurement of kinetic energy of that particle?

As we know ;momentum,P=mv⟶(1)kineticenergy,KE=12mv2⟶(2)from(1)&(2)wecansayp2=2mKEerrorinmeasurementcanbegivenas2Δpp=ΔKEKE;thereisnoerrorconsiderinmass ΔKEKE=2ΔPP ΔKEKE=2(100%)=200%.