Hi,
This is a challenge question for my intro chemistry class and I'd really like to do it. Its in 2 parts.
1. Consider a 2790-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?
∆v≧???mph
2. If the speed limit is 75 mph, could the driver of the car reasonably evade a speeding ticket by invoking the Heisenberg uncertainty principle?
yes or no?
The first one seems like an equation but I don't know which one to pick. The chapter we had to read for our homework packet plus this challenge question was kinda confusing. And then the second question also confuses me because it seems like it should be a calculation too but the only options are yes or no.
To solve the first part of the question, we need to determine the uncertainty in the velocity of the car. This can be done using the concept of the Heisenberg uncertainty principle applied to position and momentum.
The Heisenberg uncertainty principle states that it is impossible to know both the exact position and momentum (or velocity) of a particle at the same time with absolute certainty. The uncertainty in position and momentum is related by the equation:
∆x * ∆p ≥ h/2
where ∆x is the uncertainty in position, ∆p is the uncertainty in momentum, and h is the Planck constant (h = 6.626 × 10^-34 J·s).
In this case, the uncertainty in position (∆x) is given as 5.0 feet. To find the uncertainty in velocity (∆v), we need to relate it to the uncertainty in momentum (∆p). The momentum of an object can be calculated using the equation:
p = mv
where p is momentum, m is mass, and v is velocity.
Since we are given the mass of the car as 2790 lb, we need to convert it to the SI unit of mass (kilograms). 1 lb is approximately equal to 0.4536 kg. Therefore:
m = (2790 lb) * (0.4536 kg/lb) ≈ 1265.0 kg
Now we can relate the uncertainty in position to the uncertainty in momentum and solve for ∆v:
∆x * ∆p ≥ h/2
∆x * (m * ∆v) ≥ h/2
(5.0 ft) * (1265.0 kg * ∆v) ≥ (6.626 × 10^-34 J·s)/2
Notice that we need to convert the units of distance and mass to a consistent unit system. Let's use meters and kilograms:
(1.524 m) * (1265.0 kg * ∆v) ≥ (6.626 × 10^-34 J·s)/2
Multiplying the values:
1.524 * 1265.0 * ∆v ≥ 3.313 × 10^-34 J·s
Dividing both sides by (1.524 * 1265.0):
∆v ≥ (3.313 × 10^-34 J·s) / (1.524 * 1265.0)
Simplifying:
∆v ≥ 1.368 × 10^-37 m/s
Since we were given the initial velocity in mph, we need to convert the uncertainty in velocity to mph:
∆v = (1.368 × 10^-37 m/s) * (2.23694 mph/m/s)
∆v ≈ 3.063 × 10^-37 mph
Therefore, the uncertainty in velocity (∆v) of the car is approximately 3.063 × 10^-37 mph.
Now let's move on to the second question.
To determine if the driver of the car could reasonably evade a speeding ticket by invoking the Heisenberg uncertainty principle, we need to consider the speed limit and the uncertainty in velocity.
The speed limit is given as 75 mph. If the uncertainty in velocity (∆v) of the car exceeds the speed limit, then it would be possible to argue that the driver could evade a speeding ticket by invoking the Heisenberg uncertainty principle.
Comparing the uncertainty in velocity (∆v) calculated earlier (approximately 3.063 × 10^-37 mph) to the speed limit (75 mph), we can conclude that the uncertainty in velocity is extremely small compared to the speed limit. Therefore, the driver cannot reasonably evade a speeding ticket by invoking the Heisenberg uncertainty principle.
In conclusion, the answer to the first part of the question is ∆v ≥ 3.063 × 10^-37 mph, and the answer to the second part is no, the driver cannot reasonably evade a speeding ticket by invoking the Heisenberg uncertainty principle.