The molecules in a six-particle gas have velocitiesv₁ = (20î ─ 30...

Question

The molecules in a six-particle gas have velocitiesv₁ = (20î ─ 30ĵ) m/sv₂ = (40î + 70ĵ) m/sv₃ = (─80î + 20ĵ) m/sv₄ = 30î m/sv₅ = (40î ─ 40ĵ) m/sv₆ = (─50î ─ 20ĵ) m/sCalculate (a) →vₐᵥ₉ , (b) vₐᵥ₉, and (c) vᵣₘₛ.

Accepted Answer

Hey, everyone. Let's go through this practice problem in a physics simulation. Four particles are moving with unique velocities as follows. The sub one is 10, I minus 20 J meters per second. So the I and the J are referring to the unit vectors V sub two is 30 I plus 40 J meters per second. V sub three equals negative 40 I plus 10 J meters per second and V sub four equals 20 I minus 30 J meters per second. Determine one, the average velocity V sub average two, the magnitude of the average velocity V suba and three, the root means square velocity V sub R MS of the particles. In the simulation, we have four multiple choice options to choose from. Option A V five I meters per second for part 1 36.4 m per second for part two and 24. m per second. For part three, option B five I meters per second for part 1 37. m per second, for part 2 38.7 m per second. For part three, option C five I meters per second for part 1 41. m per second for part two and 38.7 m per second for part three and option D eight I meters per second for part 1 37.4 m per second for part two and 38.7 m per second for part three. Ok. So this is a problem that has three parts to it. So it might look intimidatingly long. But fortunately none of the parts are that difficult on their own. So let's just take this one step at a time and go from there starting with step one which asks for the average velocity of the four particles. So because we're given all four velocities with their vector components, then finding the average velocity is a matter of just taking the average of each of those components. So for example, all four vectors are given to us with the I component and the J component, you could think of that as the X direction and the Y direction. So to find the average velocity of the four particles will want to take the average of both of those components. So for example, the average of the sub X, the average of the velocities in the X direction or in the I component is going to be the average of the I component for each vector. So that's 10 I for V sub 1 30 I for V, sub two negative 40 I for V sub three and 20 I for V sub four. So I'll take the average of those things. So that's 10 m per second plus 30 m per second, minus 40 m per second plus 20 m per second. And recall that the way we take the average is by adding them all together and then dividing by the number of items there are. So we just added four things together. So we have to divide by four. And if you do the Matheus or put it into your calculator, then we find an average in the X component of five m per second. Now let's do the same thing for the Y component. In other words, for all the components that have the J unit vector beside them. So V sub Y average is equal to negative 20 m per second plus 40 m per second plus 10 m per second minus 30 m per second. And once again, there are four items here. So we divide it all by four. And if we do the math on this, then we find that this is actually equal to zero. So there's zero m per second in the Y direction. In other words, the total average velocity, it's equal to five m per second in the eye direction. So five I hat plus zero m per second in the J direction or zero J hat, which could just be ignored entirely. And all this is with units of meters per second. So it could be more simply written as five I meters per second. And so that is the answer to the first part of the problem. And notice that we do have several options given that show this answer for part one, but we still need to figure out the other two parts of the problem. So now let's move on to part two, part two of the problem asks for the magnitude of the average velocity. Now the magnitude of the average velocity is different because if we have a vector and we're given the components of that vector, then the magnitude of that vector can be found by taking the square root of the sum of the squares of all those different components. So for example, the magnitude of vector one, the sub one is equal to the square root of the sum of the squares of the two components. The problem tells us the vector one vs sub one is 10, I minus 20 J. So B sub one as a magnitude then is going to be equal to the square or the square root of the sum of the square of m per second plus, then we can see the Y component is negative 20 m per second. It was 20 m per second squared. Technically, it's actually a negative 20 m per second. But since it's getting squared, the minus sign just goes away anyhow. But if we put this into a calculator, then we find that the magnitude of the sub one is equal to about 22 0.36 m per second. Now we're going to repeat the same process for all three other vectors and then take the average of all those magnitudes. So the magnitude for vector two V sub two mag is equal to the square root. Again, the sum of the squares of the components. So that's the square root of 30 m per second squared plus the square of 40 m per second. And if you put this into a calculator, we find that this is equal to exactly 50 m per second. Now let's do the same thing for V sub three mag. So again, the square root of the sum of the squares of the components to the square root of the square of the first component negative 40 m per second plus the square of the second component. 10 J 10 m per second. And if we put that into a calculator, we find a value of about 41. m per second. Lastly, let's do the same thing. Four of the sub four, the sub four mag is equal to the square root of 20 m per second squared plus negative 30 m per second squared. And again, put this into a calculator and we find a magnitude of 36.6 m per second. Now, all that's up to do is take the average of these four values and we'll find the answer to the second part of the problem. So take the average, we simply add the four numbers together and then divide them by four just like we did in the first part of the problem. So that is 22 0. m per second plus 50 m per second plus 41.23 m per second plus 36.6 m per second. And so we divide all of this by four, put it into a calculator and we find a value of about 37 0.4 m per second. So that is our answer for the second part of the problem. Finally, let's move on to the third part of the problem which asks for the R MS speed of the four particles. For this, we simply have to recall that the root meets the root means squared of a number of values is equal to the square root of the average of the squares of the values. So in other words, we're going to be taking the square root of the average of the squares of the speeds. So it's gonna be the square root of the sub one squared plus the sub two squared plus the sub three squared plus the sub four squared all divided by four. So now let's simply do that. And for the actual values for V sub one and V sub two, we'll use the magnitude values that we discovered while we were doing the work for part two of the problem. In other words, the sub RMS is equal to the square root of 22 0.36 m per second squared plus the square of m per second plus the square of 41.23 m per second plus the square of 36.6 m per second, all divided by four. And if we put this into a calculator, then we find a root mean squared value of about 38.7 m per second. And so 38.7 m per second is our answer to the third and final part of the problem. And if we look at our multiple choice options, we can see that one of our options agrees with all three parts we've discovered for part 15 I meters per second, for part 2 37.4 m per second and for part 38.7 m per second. So part B is the ultimate answer to this problem and that is it for this video. Hope this video helped you out. If you feel you need more practice, please check out some of our other videos which will give you more experience with these types of problems. But that's all for now. I hope you all have a lovely day. Bye bye.

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The molecules in a six-particle gas have velocitiesv₁ = (20î ─ 30ĵ) m/sv₂ = (40î + 70ĵ) m/sv₃ = (─80î + 20ĵ) m/sv₄ = 30î m/sv₅ = (40î ─ 40ĵ) m/sv₆ = (─50î ─ 20ĵ) m/sCalculate (a) →vₐᵥ₉ , (b) vₐᵥ₉, and (c) vᵣₘₛ.

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