Question

If the error in the measurement of momentum is $20\%$ , then the error in the calculation of kinetic energy is (assume the error in measurement of $m$ as zero)

• A. $20\%$
• B. $44\%$
• C. $40\%$
• D. $200\%$

Answer 1

Answer: (B)

$KE=\frac{1}{2}\frac{p^2}{m}$
$K{E}_1=\frac{1}{2}\frac{p^2}{m}$
$p_1=\frac{6}{5}p$
$K{E}_2=\frac{1}{2}\left(\frac{36}{25}\right)\frac{p^2}{m}$
$\Delta KE=K{E}_2-K{E}_1$
$=\frac{1}{2}\left[\frac{{\left(\frac{6}{5}p\right)}^2-p^2}{m}\right]$
$=\frac{11}{25}\frac{p^2}{m}$
Percentage change in kinetic energy $=\frac{\Delta KE}{K{E}_1}\times 100$
$=\frac{11}{25}\times 100=44\%$