Math Problem Statement

Momentum is defined as the product of the mass and the velocity. When m= (1.57±0.08)×10-12kg and v= (132±3) cm/s, calculate the magnitude of the momentum and its uncertainty. Use SI (mks) units.

Solution

The momentum pp of an object is given by the formula:

p=mvp = m \cdot v

where mm is the mass and vv is the velocity.

Step 1: Convert the velocity to SI units (m/s)

Given:

  • v=(132±3)cm/sv = (132 \pm 3) \, \text{cm/s}

Convert to meters per second: v=132cm/s=1.32m/sv = 132 \, \text{cm/s} = 1.32 \, \text{m/s} and the uncertainty: Δv=3cm/s=0.03m/s\Delta v = 3 \, \text{cm/s} = 0.03 \, \text{m/s}

Step 2: Apply the given mass

The mass is already in SI units: m=(1.57±0.08)×1012kgm = (1.57 \pm 0.08) \times 10^{-12} \, \text{kg}

Step 3: Calculate the momentum

Now calculate the momentum using the central values of mm and vv: p=mv=(1.57×1012)(1.32)=2.0724×1012kg m/sp = m \cdot v = (1.57 \times 10^{-12}) \cdot (1.32) = 2.0724 \times 10^{-12} \, \text{kg m/s}

Step 4: Calculate the uncertainty

To find the uncertainty in the momentum, use the formula for the propagation of uncertainties in multiplication:

Δpp=(Δmm)2+(Δvv)2\frac{\Delta p}{p} = \sqrt{\left(\frac{\Delta m}{m}\right)^2 + \left(\frac{\Delta v}{v}\right)^2}

Substitute the uncertainties: Δpp=(0.081.57)2+(0.031.32)2\frac{\Delta p}{p} = \sqrt{\left(\frac{0.08}{1.57}\right)^2 + \left(\frac{0.03}{1.32}\right)^2} Δpp=(0.05096)2+(0.02273)2=0.002597+0.000517=0.0031140.0558\frac{\Delta p}{p} = \sqrt{(0.05096)^2 + (0.02273)^2} = \sqrt{0.002597 + 0.000517} = \sqrt{0.003114} \approx 0.0558

So the relative uncertainty is approximately 0.05580.0558, or 5.58%5.58\%.

Step 5: Final uncertainty in momentum

Now calculate the absolute uncertainty in pp: Δp=0.0558×2.0724×1012=0.1156×1012=1.16×1013kg m/s\Delta p = 0.0558 \times 2.0724 \times 10^{-12} = 0.1156 \times 10^{-12} = 1.16 \times 10^{-13} \, \text{kg m/s}

Final Answer

The magnitude of the momentum is: p=(2.07±0.12)×1012kg m/sp = (2.07 \pm 0.12) \times 10^{-12} \, \text{kg m/s}


Would you like more details or have any questions?

Here are some related questions:

  1. How does momentum change if velocity doubles while mass remains constant?
  2. What is the significance of momentum conservation in collisions?
  3. How does uncertainty propagation apply to other physical quantities?
  4. How is momentum used in understanding Newton’s Second Law?
  5. What role does momentum play in rocket propulsion?

Tip: When working with uncertainties, always combine them using the appropriate formulas for the mathematical operations involved (e.g., addition, multiplication).

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Math Problem Analysis

Mathematical Concepts

Momentum
Uncertainty Propagation
Unit Conversion

Formulas

Momentum formula: p = m * v
Uncertainty propagation for multiplication: Δp/p = √[(Δm/m)^2 + (Δv/v)^2]

Theorems

Propagation of Uncertainty

Suitable Grade Level

Grades 11-12 or introductory college level