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)

$KE=\frac{1}{2}\frac{p^{2}}{m}$
$KE_{1}=\frac{1}{2}\frac{p^{2}}{m}$
$p_{1}=\frac{6}{5} \, p$
$KE _2=\frac{1}{2} \quad\left(\frac{36}{25}\right) \frac{ p ^2}{ m }$
$ΔKE= \, KE_{2}-KE_{1}$
$=\frac{1}{2} \, \left[\frac{\left(\frac{6}{5} p\right)^{2} - p^{2}}{m}\right]$
$=\frac{11}{25} \, \frac{p^{2}}{m}$
Percentage change in kinetic energy $=\frac{ΔKE}{KE_{1}}\times 100$
$=\frac{11}{25}\times 100=44\%$