How do you propagate uncertainty for kinetic energy when both mass and velocity are measured?

Conditions

• The kinetic energy formula is $K = \frac{1}{2}mv^2$.
• Both mass $m$ and velocity $v$ carry independent absolute uncertainties $\delta m$ and $\delta v$.
• The uncertainties are small enough for a first-order linear approximation.

Reasoning, step by step

1. Start with the kinetic energy formula $K = \frac{1}{2}mv^2$.
2. Write the two-variable quadrature propagation formula: $\delta K = \sqrt{\left(\frac{\partial K}{\partial m}\delta m\right)^2 + \left(\frac{\partial K}{\partial v}\delta v\right)^2}$.
3. Compute the partial derivative with respect to mass: $\frac{\partial K}{\partial m} = \frac{1}{2}v^2$.
4. Compute the partial derivative with respect to velocity: $\frac{\partial K}{\partial v} = mv$.
5. Substitute the derivatives and the given numerical values into the formula.
6. Evaluate the expression to find the final propagated uncertainty $\delta K$.

Example

For a mass of $10.12 \pm 0.11\,\text{kg}$ and a velocity of $2.34 \pm 0.04\,\text{m/s}$, the propagated uncertainty is calculated as $\delta K = \sqrt{\left(\frac{1}{2}(2.34)^2(0.11)\right)^2 + \left((10.12)(2.34)(0.04)\right)^2}$, which evaluates to $0.99\,\text{J}$.

Common misconceptions

• Assuming the uncertainties simply add together ($\delta K = \frac{\partial K}{\partial m}\delta m + \frac{\partial K}{\partial v}\delta v$), which represents the worst-case scenario rather than independent random errors.
• Forgetting to square the individual terms before summing them under the square root.
• Confusing partial derivatives with ordinary derivatives by failing to hold the other variable constant during differentiation.

Watch the explanation

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