Question

7. 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

Explanation :

Here, we are given the error% of momemtum = +100%

We know that,

Momentum (p) = Mass (m) × Velocity (v)

Now,

We are asked the error% of Kinetic Energy.

KE = ½mv² (where; m = mass, v = velocity, KE = Kinetic Energy)

$\implies\sf KE=\dfrac{1}{2}mv^2$

Multiplying both the numerator & denominator by m,

$\implies\sf KE=\dfrac{1}{2}\dfrac{mv^2\times m}{m}$

$\implies\sf KE=\dfrac{m^2v^2}{2m}$

• p = mv
• So, m²v² = p²

$\implies\sf KE=\dfrac{p^2}{2m}$

Taking logarithm on both the sides,

$\sf\log KE=2\log p-\log2m$

Differentiating both the sides,

$\sf\dfrac{\Delta KE}{KE}=2\dfrac{\Delta p}{p}$

Error% in KE,

$\sf\dfrac{\Delta KE}{KE}\times100=2\dfrac{\Delta p}{p}\times100$

here,

• $\sf\dfrac{\Delta p}{p}\times100$ = 100%

$\\ \sf\dfrac{\Delta KE}{KE}\times100=2\times100$

$\\ \sf=200\%$

$\mapsto\boxed{\bf Error\%\ in\ KE=200\%.}$

Error% in Kinetic Energy is 200%.

Answer is option (b).