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.