A pesky 1.5-mg mosquito is annoying you as you attempt to study p...

Question

A pesky 1.5-mg mosquito is annoying you as you attempt to study physics in your room, which is 5.0 m wide and 2.5 m high. You decide to swat the bothersome insect as it flies toward you, but you need to estimate its speed to make a successful hit. (a) What is the maximum uncertainty in the horizontal position of the mosquito? (b) What limit does the Heisenberg uncertainty principle place on your ability to know the horizontal velocity of this mosquito? Is this limitation a serious impediment to your attempt to swat it?

Accepted Answer

Hey, everyone. Let's go through this problem. A hamster of mass 100 g lives in a cage of length, two m, width 20.75 m and height 0.5 m. Assume that the hamster runs along only a straight line that is parallel to the two m side of the cage. And the X axis part one of the problem asks us to calculate the maximum uncertainty in the hamster's position. And part two of the problem asks us to calculate the minimum uncertainty in the hamster's velocity. We're given four options to choose from option A for part one delta X equals 10.5 m and for part two delta V X equals 10.6 multiplied by 10 to the power of negative 32 m per second. Option B says for part one, delta X equals 10.75 m. And for part two delta V X equals seven point oh four multiplied by 10 to the power of negative 34 m per second. Option C says for part one delta X equals one m and for part two delta V X equals 5.28 multiplied by 10 to the power of negative 34 m per second. And for option D part one delta X equals two m. And for part two, delta V X equals 2.64, multiplied by 10 to the power of negative 34 m per second. Let's first talk about part one, which asks us to calculate the maximum uncertainty in the hamster's position. So the problem tells us directly that the hamster is only allowed to run in a straight line that is paralleled to the two m side of the cage. So it means that the hamster can't move across a distance greater than two m. So if we imagined a line, it had a length of two m, then even if the hamster was exactly at one end of the cage, for all, we know it could be on the other end if we don't know any better. So therefore, the maximum uncertainty of the hamster is going to be delta X equals 2.0 m because it can't be any further away from two m from whatever point we're looking at. So that's the answer to the first part of the problem. Part two asks us to calculate the minimum uncertainty in the hamster's velocity. Now, when we're trying to relate uncertainties in velocity to things like uncertainties in distance, the most useful equation we have is the equation for the Heisenberg uncertainty principle. And recall that it states that the uncertainty in an object's position multiplied by its uncertainty and momentum is greater than or equal to the plant constant divided by four pi and recalled, the plank constant has a value of 6.6, 26 multiplied by 10 to the power of negative 34 jeal seconds. We don't know anything about the hamster's momentum, but recall that for an object, its momentum is equal to its mass multiplied by its velocity. So we can use that in the Heisenberg Uncertainty principle equation, I'm gonna switch colors. So another way of writing this is by saying that the uncertainty in the position multiplied by the hamster's mass multiplied by its uncertainty and velocity is greater than or equal with the plan constant divided by four pi. Now, because the problem explicitly asks us for the minimum uncertainty, that means we want to find the smallest possible value for delta V sub X that'll make this equation still valid. And since the Heisenberg uncertainty principle is specifically meant for something that is greater than or equal to the quantity on the right, the H divided by four pi, that means that the minimum value can be found at the point where delta X multiplied by M delta V X is exactly equal to H divided by four pi. So this is the final equation we'll be using. So we want to solve this for delta V sub X. So we can find a formula for delta V sub X algebraically by dividing both sides of the equation by delta X multiplied by M. So delta V sub X is equal to H the plan constant divided by four pi multiplied by the hamster's mass multiplied by the uncertainty in the hamster's position. As we discussed earlier, the uncertainty, the maximum uncertainty in the hamster's position is going to be 2.0 m. And the mass of the hamster M is given to us in the problem as g. So converting this into kilograms, that's going to be 1000.100 kg. All that's left for us to do is to plug these values into our equation. So for H that's 6.626 multiplied by 10 to the power of negative 34 dual seconds, all divided by four pi multiplied by 0.100 kg multiplied by by delta X which is just two m. If we put this into a calculator, then we find value of about 2.64 multiplied by 10 to the power of negative 34 meters per second. And so this is the hamsters, the minimum uncertainty in the hamster's velocity. And if we look at our multiple choice options, we can see that option D includes for part one, a value of two m and for part B an uncertainty in the velocity of 2.64 multiplied by 10 to the power of negative 34 m per second. So this agrees with our answer. And therefore option B is the correct answer to this problem and that's it. I hope this video helped you out if it did and you'd like more practice. Please check out some of our other videos which will hopefully give us, give you some more experience with these types of problems, but that's all for now and I hope you all have a lovely day. Bye bye.

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