Question

The uncertainty in the position of a baseball of mass$\mathbf{0}\mathbf{.}\mathbf{145}\mathbf{}\mathit{k}\mathit{g}\mathbf{}\mathit{i}\mathit{s}\mathbf{}\mathit{\mu }\mathit{m}$.What is the minimum uncertainty in its speed?

Short answer

Estimate the uncertainty in the speed of the baseball is $∆v\ge 3.6\times {10}^{-28}m/s$.

Step-by-step solution

Step 1:Given.

$$\begin{gathered} massoftheparticleism=0.145kg \\ uncertainityinthepositionis∆x=1.0\times {10}^{-6} \\ Planck’scons\tan tis\sout{h}=1.0545\times {10}^{-34}J·s \end{gathered}$$

Concept Introduction

The following equation describes the Heisenberg uncertainty principle for momentum:

$\mathbf{∆}\mathit{p}\mathbf{∆}\mathit{x}\mathbf{\ge }\frac{\sout{\mathbf{h}}}{\mathbf{2}}$………………..(1)

The following equation describes the Heisenberg uncertainty principle for momentum:

$$\begin{gathered} \mathit{m}\mathbf{∆}\mathit{v}\mathbf{∆}\mathit{x}\mathbf{\ge }\frac{\sout{\mathbf{h}}}{\mathbf{2}} \\ \mathbf{∆}\mathit{v}\mathbf{\ge }\frac{\sout{\mathbf{h}}}{\mathbf{2}\mathbf{m}\mathbf{∆}\mathbf{x}} \end{gathered}$$……………….(2)

Solving for ∆v.

$\begin{array}{rcl}∆v& \ge & \frac{\sout{h}}{2m∆x}\\ ∆v& \ge & \frac{1.0545\times {10}^{-34}J·s}{2\times \left(0.145kg\right)\times \left(1.0\times {10}^{-6}m\right)}\\ ∆v& \ge & 3.6\times {10}^{-28}m/s\end{array}$

Estimate the uncertainty in the speed of the baseball is $\begin{array}{rcl}∆v& \ge & 3.6\times {10}^{-28}m/s\end{array}$.