Consider a golf ball with a mass of 45.9 grams traveling at 200 km/hr. If an experiment is designed to measure the position of the golf ball at some instant of time with a precision of 1 mm, then what will be the uncertainty in the speed of the golf ball? What percentage of the speed of the golf ball does this uncertainty represent?

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DeniceSandidge DeniceSandidge · Answer 1 · 2019-09-15T08:11:35+02:00

Answer:

speed of golf ball is 1.15 × [tex]10^{-30}[/tex] m/s

and % of uncertainty in speed = 2.07 × [tex]10^{-30}[/tex] %

Explanation:

given data

mass = 45.9 gram = 0.0459 kg

speed = 200 km/hr = 55.5 m/s

uncertainty position Δx = 1 mm = [tex]10^{-3}[/tex] m

to find out

speed of the golf ball and % of speed of the golf ball

solution

we will apply here heisenberg uncertainty principle that is

uncertainty position ×uncertainty momentum ≥ [tex]\frac{h}{4\pi }[/tex] ......1

Δx × ΔPx ≥ [tex]\frac{h}{4\pi }[/tex]

here uncertainty momentum ΔPx = mΔVx

and uncertainty velocity = ΔVx

and h = 6.626 × [tex]10^{-34}[/tex] Js

so put here all these value in equation 1

[tex]10^{-3}[/tex] × 0.0459 × ΔVx = [tex]\frac{6.626*10^{-34}}{4\pi }[/tex]

ΔVx = 1.15 × [tex]10^{-30}[/tex] m/s

and

so % of uncertainty in speed = ΔV / m

% of uncertainty in speed = 1.15 × [tex]10^{-30}[/tex] / 55.5

% of uncertainty in speed = 2.07 × [tex]10^{-30}[/tex] %