Question

Consider a 2250-lb automobile clocked by law-enforcement radar at a speed of 85.5 mph (miles/hour). if the position of the car is known to within 5.0 feet at the time of the measurement, what is the uncertainty in the velocity of the car?

Answer 1

Answer:

dV =3.39x10^-38 m/s

Explanation:

Consider a 2250-lb automobile clocked by law-enforcement radar at a speed of 85.5 mph (miles/hour). if the position of the car is known to within 5.0 feet at the time of the measurement, what is the uncertainty in the velocity of the car?

According to the Heisenberg uncertainty law  which states tat te

product of uncertainty in position and uncertainty in momentum will be constant

\Delta x . \Delta P = \frac{h}{4\pi}

\Delta x . m \Delta v = \frac{h}{4\pi}

Mass(automobile)= 2150 lbs

1lbs=0.453592kg so:

2250 lbs= 1020.3 kg

Uncertainty(delta X)= 5ft

1ft= 0.3048m so:

5ft= 1.524m

h= Planck's constant = 6.62606957 × 10-34 m2 kg / s

(delta V) = (delta P)/ Mass

Solving the original equation we get:

(delta V) = h/( (4pi)(Mass)(delta X) )

= (6.626X10^-34)/( (4pi)( 1020.3 kg )(1.524m)

dV =3.39x10^-38 m/s

the uncertainty in the velocity of the car is dV =3.39x10^-38 m/s

Answer 2

As per law of Heisenberg uncertainty law

product of uncertainty in position and uncertainty in momentum will be constant

$\Delta x . \Delta P = \frac{h}{4\pi}$

$\Delta x . m \Delta v = \frac{h}{4\pi}$

now plug in all data

$(5\times 0.3048). (2250 \times 0.454) \Delta v = \frac{6.6 \times 10^{-34}}{4\pi}$

$\Delta v = 3.37 \times 10^{-38} m/s$

So above is the uncertainty in velocity of the object