Question

a 0.15 kg baseball thrown at 100 mph has a momentum of 6.8 kg·m/s. if the uncertainty in measuring the momentum is 1.0×10^-7 of the momentum, calculate the uncertainty in the baseballs position. round your answer to 2 significant digits.

Explanation:

Step1: Calculate the uncertainty in momentum ($\Delta p$)

Given the momentum $p = 6.8\,\text{kg}\cdot\frac{\text{m}}{\text{s}}$ and the uncertainty in momentum is $1.0\times 10^{-7}$ of the momentum.
So, $\Delta p=1.0\times 10^{-7}\times p$
Substitute $p = 6.8\,\text{kg}\cdot\frac{\text{m}}{\text{s}}$ into the formula:
$\Delta p=1.0\times 10^{-7}\times6.8\,\text{kg}\cdot\frac{\text{m}}{\text{s}} = 6.8\times 10^{-7}\,\text{kg}\cdot\frac{\text{m}}{\text{s}}$

Step2: Use the Heisenberg uncertainty principle ($\Delta x\Delta p\geq\frac{h}{4\pi}$)

We want to find $\Delta x$, and from $\Delta x\Delta p\geq\frac{h}{4\pi}$, we can solve for $\Delta x$ (taking the equality for the minimum uncertainty, $\Delta x=\frac{h}{4\pi\Delta p}$).
The Planck's constant $h = 6.626\times 10^{-34}\,\text{J}\cdot\text{s}$
Substitute $h = 6.626\times 10^{-34}\,\text{J}\cdot\text{s}$ and $\Delta p = 6.8\times 10^{-7}\,\text{kg}\cdot\frac{\text{m}}{\text{s}}$ into the formula:
$\Delta x=\frac{6.626\times 10^{-34}}{4\pi\times6.8\times 10^{-7}}$
First, calculate the denominator: $4\pi\times6.8\times 10^{-7}\approx4\times3.14\times6.8\times 10^{-7}=85.408\times 10^{-7}$
Then, $\Delta x=\frac{6.626\times 10^{-34}}{85.408\times 10^{-7}}=\frac{6.626}{85.408}\times10^{-27}\approx 0.0776\times10^{-27}=7.8\times 10^{-29}\,\text{m}$

Answer:

$7.8\times 10^{-29}\,\text{m}$