Question

Kinetic energy \(\mathrm{K}\) and linear momentum \(\mathrm{P}\) are related as \(\mathrm{K}=\left(\mathrm{P}^{2} / 2 \mathrm{~m}\right) .\) What is the equation of the relative error \(\Delta \mathrm{k} / \mathrm{k}\) in measurement of the \(\mathrm{K} ?\) (mass in constant) (a) \((\mathrm{P} / \Delta \mathrm{P})\) (b) \(2(\Delta \mathrm{P} / \mathrm{P})\) (c) \((\mathrm{P} / 2 \Delta \mathrm{P})\) (d) \(4(\Delta \mathrm{P} / \mathrm{P})\)

Short answer

The equation for the relative error in the measurement of K is \(2(\Delta P / P)\), which corresponds to option (b).

Step-by-step solution

We are given the equation: \(K = \frac{P^2}{2m}\) #Step 2: Differentiate K with respect to P#

To differentiate K with respect to P, we can use the power rule: \(\frac{d}{dx} x^n = nx^{n-1}\) \(\frac{dK}{dP} = \frac{d}{dP} \frac{P^2}{2m} = \frac{2P}{2m} = \frac{P}{m}\) #Step 3: Find the equation for ΔK#

Now we need to find the equation for ΔK, the change in the kinetic energy. At P = P₀, K = K₀. If P changes by ΔP, K changes by ΔK. Knowing that, we can write: \(\Delta K = \frac{dK}{dP} \Delta P = \frac{P}{m} \Delta P\) #Step 4: Divide ΔK by K to find the relative error#

The relative error is given by \(\frac{\Delta K}{K}\). We can now plug in the equations we derived in the previous steps: \(\frac{\Delta K}{K} = \frac{\frac{P}{m}\Delta P}{\frac{P^2}{2m}} = \frac{2 \Delta P}{P}\) Answer: The equation for the relative error in the measurement of K is \(\boxed{2(\Delta P / P)}\), which corresponds to option (b).

Key concepts

Error Analysis

Error analysis helps us determine the uncertainty or 'error' in scientific measurements. When you're measuring physical quantities like kinetic energy, it's crucial to understand how errors in your measurements affect the final result. In this exercise, we are focused on kinetic energy (\(K\)) and how an error in measuring momentum (\(P\)) affects \(K\). The relative error is a way to express this impact as a fraction between the error in a measurement and the true value.

The relative error formula we derived is \( \frac{\Delta K}{K} = \frac{2 \Delta P}{P} \). This tells us that the relative error in kinetic energy depends directly on the relative error in the measurement of momentum. The \(2\) in the formula indicates that any error in \(P\) will affect \(K\) twice as much. This kind of insight is critical for experiments where precision is essential, allowing us to predict how errors propagate through equations. Understanding and calculating these errors enables better accuracy in scientific research.

Differentiation

Differentiation is a mathematical process used to determine how a function changes when its inputs change. In physics, it's often used to find rates of change, like acceleration or velocity. In this problem, we differentiate kinetic energy \(K\) with respect to momentum \(P\), to understand how changes in \(P\) influence \(K\).

By using the power rule, an essential technique in differentiation, we find \( \frac{dK}{dP} = \frac{P}{m} \). This expression tells us the rate of change of kinetic energy with respect to momentum. Essentially, it's a way to quantify how small changes in momentum will affect the kinetic energy.

• The power rule employed here is a fundamental technique in calculus: \(\frac{d}{dx} x^n = nx^{n-1}\).
• This process is crucial for deriving formulas that describe physical behaviors and alterations due to external changes.

Differentiation thus enables predictions about how variables like kinetic energy will respond to small perturbations in conditions like momentum.

Physics Formulas

Physics formulas are the backbone of solving problems in physics. They offer a structured way to relate different physical quantities, such as kinetic energy and momentum in this exercise.

The initial formula \( K = \frac{P^2}{2m} \) is fundamental because it expresses kinetic energy in terms of momentum and mass. This relationship is derived from principles of mechanics and allows us to understand how different variables interact.

Rewriting or differentiating these formulas gives us insight into changes. For example, knowing that \( \Delta K = \frac{P}{m} \Delta P \) gives us a direct method to assess how a change in momentum (\(\Delta P\)) will influence kinetic energy (\(\Delta K\)).

• Formulas are like tools, and learning to manipulate them can unveil new information.
• They provide a language to succinctly express natural phenomena.

With a solid grasp of these equations, you can predict outcomes, calculate necessary quantities, and truly appreciate the interconnectedness of physical properties.