Collection of Solved Problems in Physics

Baseball ball

Task number: 4401

• Hint 1

  We know that, using the principle of uncertainty, with the specification of a particle’s position comes the price of lost accuracy in measuring another quantity. What is the quantity (or quantities) in question?

• Solution of hint 1

  An increased precision in the measurement of the particle’s position comes at the cost of accuracy of measuring its motion. Therefore, the uncertainty of velocity, as well as momentum and kinetic energy, rises (these three variables can be derived from each other).

  Do you know any particular version of some principle of uncertainty?

• Solution of hint 1

  The principle of uncertainty for the location and momentum has the form (in 1D) \[\Delta x \,\cdot\, \Delta p \ge \frac{\hbar}{2} .\]

  As uncertainty of the momentum, consider in this case the absolute uncertainty of determination of the momentum.

• Hint 2

  What is the relative uncertainty of a measurement? How does the relative uncertainty propagate from one quantity to another? What if the one quantity is the product of two other which already carry some uncertainty?

• Solution of hint 2

  The relative uncertainty is a dimensionless quantity (often in percentage). We can calculate it from absolute uncertainty of the measurement divided by the measured value. We can denote it with the letter η.

  If the variable C is the product of the variables A and B we can use the formula

  η(C) = η(A) + η(B)

  .

• Notation

  m = 150 g	Ball’s mass
  Δm = 1 g	Absolute uncertainty of the mass
  v = 40 m·s−1	Ball’s velocity
  Δv = 1 m·s−1	Absolute uncertainty of the velocity
  Δx = ? (m)	Uncertainty of the location

• Solution

  We need to know the uncertainty Δp to estimate Δx from the principle of uncertainty \[\Delta x \,\cdot\, \Delta p \ge \frac{\hbar}{2}\] . As uncertainty of the momentum, we consider the absolute uncertainty of the determination of the momentum. Since the ball’s momentum is the product of its mass and velocity  p = m·v we can use this formula to calculate its relative uncertainty of these variables η(p) = η(m) + η(v).
  We have also formula from the definition of relation between absolute and relative uncertainty Δp = η(p)·p. With the use of the principle of uncertainty we can calculate Δx \[\Delta x \,\cdot\, \Delta p \ge \frac{\hbar}{2}\] and we obtain

  \[\Delta x \ge \frac{\hbar}{2} \,\frac{1}{\Delta p} ,\] \[\Delta x \ge \frac{\hbar}{2} \,\frac{1}{\eta(p)\cdot p} ,\] \[\Delta x \ge \frac{\hbar}{2} \,\frac{1}{(\eta(m)+\eta(v))\cdot mv} ,\] \[\Delta x \ge \frac{\hbar}{2} \,\frac{1}{\left(\frac{\Delta m}{m}+\frac{\Delta v}{v}\right)\cdot mv} ,\] \[\Delta x \ge \frac{1.055\,\cdot\,10^{-34}}{2\,\cdot\,\left(\frac{1}{150}+\frac{1}{40}\right)\,\cdot\, 0.15\,\cdot\, 40}\ \mbox{m} ,\] \[\Delta x \ge 2.8\,\cdot\,10^{-34}\ \mbox{m} .\]

  Not only does this uncertainty fall below the dimensions of a nucleus of the atom, it is also approaching the Planck-Wheeler length*. The space and time apparantely stop to exist in the dimensions smaller than this length, or rather they start to have completely different properties, which no one has been so far able to estimate, much less reasonably describe.

  We can conclude that the restriction from the quantum mechanics principles of uncertainty doesn’t matter in this macroscopic situation.

  * See http://en.wikipedia.org/wiki/Planck_length.

• Answer

  In principle, the minimum uncertainty of the location of the ball is equal to approximately 10⁻³⁴ m, which is much smaller than the devices can detect. The real uncertainty in the measurement caused by the measuring devices and the human’s eye would be much bigger. Even in very precise measurements, it could be in order of millimetres or the tens of millimetres i.e. 10⁻³ m to 10⁻⁴ m.