The mass m of an object is measured as (2.0 \pm 0.1)\text{ kg} and its speed v is measured… — JAMB UTME

The mass $m$ of an object is measured as $(2.0 \pm 0.1)\text{ kg}$ and its speed $v$ is measured as $(5.0 \pm 0.2)\text{ m s}^{-1}$ . What is the maximum percentage error in the calculated kinetic energy of the object?

$13\%$ Answer
B
$9\%$
C
$8\%$
D
$21\%$

Answer

The maximum percentage error in the calculated kinetic energy is

13\%

For a calculated quantity

E = \frac{1}{2}mv^2

, the constant factor

\frac{1}{2}

has no error. The fractional error propagation formula gives

\frac{\Delta E}{E} = \frac{\Delta m}{m} + 2\frac{\Delta v}{v}

. Substituting

\frac{\Delta m}{m} = 0.05

(

5\%

) and

\frac{\Delta v}{v} = 0.04

(

4\%

) yields

5\% + 2(4\%) = 13\%

. Thus, the option stating

13\%

is correct.

Step-by-Step Solution

Calculate the fractional error and percentage error in mass

m

\frac{\Delta m}{m} = \frac{0.1}{2.0} = 0.05 = 5\%

Percentage error in mass is the absolute error divided by the measured value multiplied by 100.

Calculate the fractional error and percentage error in speed

v

\frac{\Delta v}{v} = \frac{0.2}{5.0} = 0.04 = 4\%

Percentage error in speed is the absolute error divided by the measured value multiplied by 100.

Apply the error propagation formula for kinetic energy

E = \frac{1}{2}m v^2

\frac{\Delta E}{E} = \frac{\Delta m}{m} + 2\left(\frac{\Delta v}{v}\right)

When physical quantities are raised to a power and multiplied, their fractional errors are multiplied by the respective power index and added.

Compute the maximum percentage error in kinetic energy

\%\text{ error in } E = 5\% + 2(4\%) = 5\% + 8\% = 13\%

Combining the weighted percentage errors yields the total maximum percentage error.

Key Concept

Error Propagation in Power and Product Relationships

Estimated Time:2m 0s

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