[Quantum Mechanics]Uncertainty principle

[pezzang]

I have a question for Heisenberg's uncertainty principle. Here is the question:
You are driving your car when you race thru a stop sign and get subsequently pulled over. The cop measured your speed within a meter of the stop sign. You decide to use the uncertainty principle as your defense, claiming that your measured speed differed from zero by less than the uncertainty principle would allow for. Therefore, your speed could well have been zero and you should not be ticketed. How fast could you have gone for this to be a valid defense? (Mass of the car=1000kg, h=6x10^-34)
I used m delta(v) x delta(s) >or= h/(4pi)
solving for delta(v), i got delta(v) >= 4.77x10^-38 m/s. What does this value represent? How do I go from here to find the maximum speed the car could have gone under the valid defense condition? I guess I am not exactly grasping the meaning of this equation. Please let me know.
Also, another question:
Not having evaded a ticket in the past problem, you switch to a super-light quantum car wtiht eh mass of an electron. The cop now measures your speed when you are within a micron of the stop sign. How fast can you go now, in km/hr? (mass of electron = 10^-30kg, 1 micron = 10^-6m)
Again, I used the uncertainty principle equation to solve for delta(v), and I got delta(v) >= 1.33x10^-5 km/hr. Does this mean that the maximum is 1.33x10^-5km/hr? Please please help this stupid person out. Thanx :!)

 

[thepatientmental]

The uncertainty principle is a fundamental concept in quantum mechanics that states that there is a limit to how precisely certain pairs of physical quantities, such as position and momentum, can be known simultaneously. In this case, the uncertainty principle can be applied to the measurement of your car's speed and position at the time of the traffic stop.

The equation you used, mΔvΔs ≥ h/(4π), represents the uncertainty principle in terms of the mass of the object (your car), the uncertainty in velocity (Δv), and the uncertainty in position (Δs). The value you calculated, 4.77x10^-38 m/s, represents the minimum uncertainty in your car's velocity that is allowed by the uncertainty principle. This means that your car's measured speed could have been as low as 4.77x10^-38 m/s, but it could have also been higher. It is not possible to determine the exact speed with perfect precision.

To find the maximum speed your car could have gone for this to be a valid defense, you would need to use the mass of your car and the uncertainty in position (which is not specified in the question) to solve for the uncertainty in velocity. This would give you an upper limit for your car's speed, but it would not necessarily be the exact speed at the time of the traffic stop.

For the second part of the question, where the car has the mass of an electron and the uncertainty in position is given as 1 micron, you would use the same equation to calculate the minimum uncertainty in velocity. However, since the mass of the car is much smaller, the minimum uncertainty in velocity is much larger, which is why you get a larger value in km/hr. This means that the maximum speed your car could have gone for this to be a valid defense is 1.33x10^-5 km/hr, but it could have also been higher. Again, it is not possible to determine the exact speed with perfect precision.

I hope this helps clarify the concept of the uncertainty principle and how it applies in this situation. Keep in mind that the uncertainty principle is a fundamental concept in quantum mechanics and may not necessarily be applicable in a real-world scenario such as this.