Question

At a baseball game, a radar gun measures the speed of a 144-g baseball to be \(137.32 \pm 0.10 \mathrm{km} / \mathrm{h} .\) (a) What is the minimum uncertainty of the position of the baseball? (b) If the speed of a proton is measured to the same precision, what is the minimum uncertainty in its position?

Short answer

Based on the given information and the steps provided, we can calculate the minimum uncertainty in position for the baseball and proton as follows: Step 1: Calculate the gravitational force acting on the baseball. The mass of the baseball (m_b) is given as 144 g, which we can convert to kg: \(m_b = \frac{144}{1000} \mathrm{kg} = 0.144 \mathrm{kg}\) Step 2: Convert the speed and uncertainty of speed to m/s. \(v_b = 137.32 \frac{1000}{3600} \mathrm{m} / \mathrm{s} \approx 38.144 \mathrm{m/s}\) \(\Delta v_b = 0.10 \frac{1000}{3600} \mathrm{m} / \mathrm{s} \approx 0.0278 \mathrm{m/s}\) Step 3: Calculate the uncertainty in momentum for both the baseball and the proton. For the baseball: \(\Delta p_b = m_b \Delta v_b \approx 0.144 \mathrm{kg} * 0.0278 \mathrm{m/s} \approx 0.0040 \mathrm{kg\cdot m/s}\) For the proton: \(\Delta p_p = m_p \Delta v_b \approx 1.6726219 \times 10^{-27} \mathrm{kg} * 0.0278 \mathrm{m/s} \approx 4.65 \times 10^{-29} \mathrm{kg\cdot m/s}\) Step 4: Calculate the minimum uncertainty in position for both the baseball and the proton. Using the reduced Planck constant, \(\hbar \approx 1.0545718 \times 10^{-34} \mathrm{kg\cdot m^2 / s}\), (a) For the baseball: \(\Delta x_b \geq \frac{1.0545718 \times 10^{-34} \mathrm{kg\cdot m^2 / s}}{2 * 0.0040 \mathrm{kg\cdot m/s}} \approx 1.31 \times 10^{-32} \mathrm{m}\) (b) For the proton: \(\Delta x_p \geq \frac{1.0545718 \times 10^{-34} \mathrm{kg\cdot m^2 / s}}{2 * 4.65 \times 10^{-29} \mathrm{kg\cdot m/s}} \approx 1.13 \times 10^{-6} \mathrm{m}\) So, the minimum uncertainty in position for the baseball is approximately \(1.31 \times 10^{-32} \mathrm{m}\), and the minimum uncertainty in position for the proton is approximately \(1.13 \times 10^{-6} \mathrm{m}\).

Step-by-step solution

Calculate the gravitational force acting on the baseball

Before calculating the uncertainty in position using Heisenberg's Uncertainty Principle, we eliminate the gravitational force acting on the baseball. The mass of the baseball (m_b) is given as 144 g, which we need to convert into kg. \(m_b = \frac{144}{1000} \mathrm{kg}\)

Convert the speed and uncertainty of speed to m/s

The speed of the baseball (v_b) is given as \(137.32 \mathrm{km} / \mathrm{h}\). We will convert it to meters per second (m/s) using the conversion factor: \(v_b = 137.32 \frac{1000}{3600} \mathrm{m} / \mathrm{s}\) Similarly, we convert the uncertainty in the speed of the baseball (\(\Delta v_b\)) to m/s: \(\Delta v_b = 0.10 \frac{1000}{3600} \mathrm{m} / \mathrm{s}\)

Calculate the uncertainty in momentum for both the baseball and the proton

Now, we will calculate the uncertainty in the momentum for both the baseball and the proton using their respective masses and uncertainty in speed. For the baseball: \(\Delta p_b = m_b \Delta v_b\) For the proton, we have m_p = \(1.6726219 \times 10^{-27} \mathrm{kg}\) (mass of a proton). The uncertainty in the speed of the proton is the same as that of the baseball, so: \(\Delta p_p = m_p \Delta v_b\)

Calculate the minimum uncertainty in position for both the baseball and the proton

Using the Heisenberg Uncertainty Principle, we can now calculate the minimum uncertainty in position for both the baseball and the proton: (a) For the baseball: \(\Delta x_b \geq \frac{\hbar}{2 \Delta p_b}\) (b) For the proton: \(\Delta x_p \geq \frac{\hbar}{2 \Delta p_p}\) Plug in the values of \(\hbar\), \(\Delta p_b\), and \(\Delta p_p\) to find the minimum uncertainties in position for the baseball and the proton.