What is the minimum uncertainty in theh position of a bullet of mass 5g that is known to have a speed somewhere between 550,00000 and 550,00001 `m s^(-1)`?

Explanation

To find the minimum uncertainty in the position of a bullet of mass 5g, given its speed is between 55000000 m/s and 55000001 m/s, we will use the Heisenberg Uncertainty Principle. Here’s how to solve the problem step by step:

### Step 1: Convert the mass of the bullet to kilograms
The mass of the bullet is given as 5 grams. We need to convert this to kilograms since the standard unit of mass in physics is kilograms.

\[ \text{Mass} = 5 \, \text{g} = 5 \times 10^{-3} \, \text{kg} \]

### Step 2: Calculate the uncertainty in velocity (ΔV)
The speed of the bullet is given as a range between 55000000 m/s and 55000001 m/s. The uncertainty in velocity (ΔV) can be calculated as:

\[ \Delta V = V_2 - V_1 = 55000001 \, \text{m/s} - 55000000 \, \text{m/s} = 1 \, \text{m/s} \]

### Step 3: Use the Heisenberg Uncertainty Principle
According to the Heisenberg Uncertainty Principle, the relationship between the uncertainty in position (Δx) and the uncertainty in momentum (Δp) is given by:

\[ \Delta x \cdot \Delta p \geq \frac{h}{4\pi} \]

Where:
- \( \Delta p = m \cdot \Delta V \) (momentum uncertainty)
- \( h \) is Planck's constant, which is approximately \( 6.626 \times 10^{-34} \, \text{Js} \)

### Step 4: Calculate the uncertainty in momentum (Δp)
Now, we can calculate the uncertainty in momentum:

\[ \Delta p = m \cdot \Delta V = (5 \times 10^{-3} \, \text{kg}) \cdot (1 \, \text{m/s}) = 5 \times 10^{-3} \, \text{kg m/s} \]

### Step 5: Rearrange the uncertainty principle to find Δx
We can rearrange the Heisenberg Uncertainty Principle to solve for Δx:

\[ \Delta x \geq \frac{h}{4\pi \Delta p} \]

### Step 6: Substitute the values into the equation
Substituting the known values into the equation:

\[ \Delta x \geq \frac{6.626 \times 10^{-34} \, \text{Js}}{4 \cdot 3.14 \cdot (5 \times 10^{-3} \, \text{kg m/s})} \]

Calculating the denominator:

\[ 4 \cdot 3.14 \cdot (5 \times 10^{-3}) = 0.0628 \, \text{kg m/s} \]

Now substituting back:

\[ \Delta x \geq \frac{6.626 \times 10^{-34}}{0.0628} \approx 1.055 \times 10^{-32} \, \text{m} \]

### Final Answer
The minimum uncertainty in the position of the bullet is approximately:

\[ \Delta x \approx 1.055 \times 10^{-32} \, \text{m} \]

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