What will be the uncertainty in velocity of a bullet with a mass of 10 g whose position is known with `pm0.01mm`?

Options

A. `5.275xx10^(-33)ms^(-1)`
B. `5.275xx10^(-25)ms^(-1)`
C. `5.275xx10^(-5)ms^(-1)`
D. `5.275xx10^(-28)ms^(-1)`

Explanation

To solve the problem of finding the uncertainty in the velocity of a bullet with a known mass and position uncertainty, we will use Heisenberg's uncertainty principle. Here’s a step-by-step solution:

### Step 1: Convert the mass of the bullet to kilograms
The mass of the bullet is given as 10 grams. We need to convert this to kilograms for standard SI units.

\[ \text{Mass (m)} = 10 \, \text{g} = 10 \times 10^{-3} \, \text{kg} = 0.01 \, \text{kg} \]

**Hint:** Remember that 1 gram = \(10^{-3}\) kilograms.

### Step 2: Convert the position uncertainty to meters
The position uncertainty is given as ±0.01 mm. We need to convert this to meters.

\[ \Delta x = 0.01 \, \text{mm} = 0.01 \times 10^{-3} \, \text{m} = 1 \times 10^{-5} \, \text{m} \]

**Hint:** Recall that 1 millimeter = \(10^{-3}\) meters.

### Step 3: Write down Heisenberg's uncertainty principle
According to Heisenberg's uncertainty principle, the relationship between the uncertainty in position (\(\Delta x\)) and the uncertainty in momentum (\(\Delta p\)) is given by:

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

where \(h\) is Planck's constant.

**Hint:** Planck's constant \(h\) is approximately \(6.626 \times 10^{-34} \, \text{Js}\).

### Step 4: Express momentum uncertainty in terms of velocity uncertainty
The momentum \(p\) is given by the product of mass and velocity:

\[ \Delta p = m \Delta v \]

Substituting this into the uncertainty principle gives:

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

### Step 5: Rearrange to find the uncertainty in velocity (\(\Delta v\))
Rearranging the equation to solve for \(\Delta v\):

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

### Step 6: Substitute the known values
Now, we can substitute the values into the equation:

- \(h = 6.626 \times 10^{-34} \, \text{Js}\)
- \(m = 10 \times 10^{-3} \, \text{kg} = 0.01 \, \text{kg}\)
- \(\Delta x = 1 \times 10^{-5} \, \text{m}\)
- \(\pi \approx 3.14\)

Now substituting these values:

\[ \Delta v \geq \frac{6.626 \times 10^{-34}}{4 \times 3.14 \times 0.01 \times 1 \times 10^{-5}} \]

### Step 7: Calculate the uncertainty in velocity
Calculating the denominator:

\[ 4 \times 3.14 \times 0.01 \times 1 \times 10^{-5} = 1.256 \times 10^{-6} \]

Now substituting back into the equation:

\[ \Delta v \geq \frac{6.626 \times 10^{-34}}{1.256 \times 10^{-6}} \approx 5.275 \times 10^{-28} \, \text{m/s} \]

### Conclusion
Thus, the uncertainty in the velocity of the bullet is approximately \(5.275 \times 10^{-28} \, \text{m/s}\).

**Final Answer:** The uncertainty in velocity is \(5.275 \times 10^{-28} \, \text{m/s}\).

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