The percentage error in the measurement of ma | vedclass

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(B) Given: Percentage error in mass, $\frac{\Delta m}{m} 	imes 100 = 0.73\%$. Percentage error in speed, $\frac{\Delta v}{v} 	imes 100 = 1.84\%$. Mo

Vedclass · Accepted Answer

$2.57\%, 4.41\%$

Vedclass · Answer

(B) Given: Percentage error in mass, $\frac{\Delta m}{m} 	imes 100 = 0.73\%$. Percentage error in speed, $\frac{\Delta v}{v} 	imes 100 = 1.84\%$. Momentum $p = mv$. The relative error in momentum is $\frac{\Delta p}{p} = \frac{\Delta m}{m} + \frac{\Delta v}{v}$. Percentage error in momentum = $0.73\% + 1.84\% = 2.57\%$. Kinetic energy $K = \frac{1}{2}mv^2$. The relative error in kinetic energy is $\frac{\Delta K}{K} = \frac{\Delta m}{m} + 2\frac{\Delta v}{v}$. Percentage error in kinetic energy = $0.73\% + 2(1.84\%) = 0.73\% + 3.68\% = 4.41\%$. Thus, the percentage errors are $2.57\%$ and $4.41\%$.

The percentage error in the measurement of mass of a body and its speed are $0.73\%$ and $1.84\%$ respectively. The percentage error in the measurement of its momentum and kinetic energy are respectively

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