If the linear momentum is increased by 5%,the | vedclass

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(C) The relationship between kinetic energy $E$ and linear momentum $p$ is given by $E = \frac{p^2}{2m}$.For small percentage changes,we can use the d

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$10.25$

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(C) The relationship between kinetic energy $E$ and linear momentum $p$ is given by $E = \frac{p^2}{2m}$.For small percentage changes,we can use the differential method: $\frac{\Delta E}{E} = 2 \left( \frac{\Delta p}{p} ight)$.Given $\frac{\Delta p}{p} = 5\% = 0.05$,the change in kinetic energy is $\frac{\Delta E}{E} = 2 	imes 5\% = 10\%$.Alternatively,using the exact formula: $E' = \frac{(1.05p)^2}{2m} = 1.1025 E$.The percentage increase is $(1.1025 - 1) 	imes 100 = 10.25\%$. Since $10.25\%$ is the precise value,we select the closest standard approximation or the exact value if provided.

If the linear momentum is increased by $5\%$,the kinetic energy will increase by.......$\%$

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