Answer to Question #174745 in Quantum Mechanics for ...

Question

 The uncertainty in the momentum Δp of a golf ball stroked by a player which is traveling at 50m/s is  1×10−6 of its momentum. What is its uncertainty in position Δx? Mass of the golf ball was measured as 250 g 

Answer 1

According to the Heisenberg's uncertainty principle

$$\Delta p\Delta x \ge\dfrac{\hbar}{2}$$

where $\hbar \approx 1.05\times 10^{-34}J\cdot s$ is the reduced Planck constant. According to the task

$$\Delta p = 10^{-6}\cdot mv$$

where $m = 250g = 0.25kg$ is the mass of the ball, $v = 50m/s$ is its speed.

Thus, the smalles possible uncertainty in position is:

$$\Delta x = \dfrac{\hbar}{2\Delta p} = \dfrac{\hbar}{2\cdot 10^{-6}\cdot mv}\\ \Delta x = \dfrac{ 1.05\times 10^{-34}}{2\cdot 10^{-6}\cdot 0.25\cdot 50} \approx 4.2\times 10^{-30}m$$

Answer. $4.2\times 10^{-30}m$,