Question

A 12 gram bullet leaves the barrel of a rifle at 180 m/s. If the vertical position of the bullet is known to a precision of 0.65 cm (radius of the rifle's barrel). What is the minimum uncertainty in the bullet's vertical momentum?

Answer 1

\( Mass,\ m=12g \)

\( Velocity,\ v=180m/s \)

\( Uncertainty\ in\ position,\ Δx=0.65cm=0.65×10^{-2} m \)

\( Uncertainty\ in\ momentum = Δp \)

\( According\ to\ Heisenberg’s\ uncertainty\ principle, \)

 \( Δx.Δp≥h/4π \)

         \( Δp≥h/(4π.Δx) \)

\( Minimum\ uncertainty\ in\ momentum,\ Δp=h/(4π.Δx) \)

\( =(6.63 ×10^{-34} kg m^2⁄s)/(4π×0.65×10^{-2} m)=0.81×10^{-32} kg m/s \)

 

 

We use the Heisenberg's uncertainty principle to determine the uncertainty in momentum of bullet when uncertainty in its position has been given.