Question

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?

Answer 1

Answer:

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

and % of uncertainty in speed =  2.07 × $10^{-30}$ %

Explanation:

given data

mass = 45.9 gram = 0.0459 kg

speed = 200 km/hr = 55.5 m/s

uncertainty position Δx = 1 mm = $10^{-3}$ 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 ≥ $\frac{h}{4\pi }$    ......1

Δx × ΔPx  ≥ $\frac{h}{4\pi }$

here uncertainty momentum ΔPx = mΔVx

and uncertainty velocity = ΔVx

and h = 6.626 × $10^{-34}$ Js

so put here all these value in equation 1

$10^{-3}$ × 0.0459 × ΔVx =  $\frac{6.626*10^{-34}}{4\pi }$

ΔVx = 1.15 × $10^{-30}$ m/s

and

so % of uncertainty in speed = ΔV / m

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

% of uncertainty in speed =  2.07 × $10^{-30}$ %