The four 1.0 g spheres shown in FIGURE P25.42 are released sim...

Question

The four 1.0 g spheres shown in FIGURE P25.42 are released simultaneously and allowed to move away from each other. What is the speed of each sphere when they are very far apart?

Accepted Answer

Hey, everyone. Let's go through this practice problem. Four spheres each with a mass of 2 g are simultaneously released from rest and allowed to move apart without any external forces acting on them. What will each sphere's speed be when they have moved a significant distance away from each other? Option A 2.8 m per second. Option B 5.49 m per second. Option C 0.13 m per second and option D 0.31 m per second. We are also given a diagram showing the arrangement of the four charges as well as the horizontal and vertical distances between each pair of particles. So this problem can be kind of tricky in the fact that it's not immediately obvious how we're supposed to tackle this problem. But there are a few clues here since we talked about the spheres being released from rest and we're looking for their final speed after they've moved some distance away from each other due to the electric forces. And what we can kind of determine is that we can figure this out using energy concepts. So recall the formula for energy conservation with mechanical energy, which states that the initial potential energy plus the initial kinetic energy. So U of I and K sub I is equal to the final potential energy plus the final kinetic energy. So what we can do with this problem is find terms and and quantities for the initial potential energy and the initial kinetic energy and the final potential energy and the final kinetic energy and then use the final kinetic energy to figure out what the final speeds should be. So in this case, initial represents when the particles are first placed released from rest and final represents after they have moved a significant distance away from each other. So let's begin this problem by setting up the how we're going to find the use of I the initial potential energy. Now recall that energy is a scaler, which means we can just find different energy components and then add them up in order to get the total energy at that point for the case of the initial potential energy, it is going to be electric potential energy due to the electrostatic forces between the particles, which means this can get a little bit complicated because the total potential energy from this arrangement of four particles is going to be due to the electrostatic force between every single possible pair of these four particles. So the initial potential energy is going to be equal to the potential energy between particles A and B. So I'm gonna call it U sub A B but plus the potential energy between particles B and C. So U sub BC, but we're not done yet. We had a lot more to go plus the potential energy from particles C and D, the potential between those two plus the potential energy between D and A and you may think we're all done because you've gone the full circuit. But no, there was also the diagonal forces to consider because there's also the potential energy between particles A and C got this kind of diagonal thing going on here. And also the part, the potential energy between particles B and D. So we've got six different terms that all make up this total potential energy term. And so that's where things can start to get a little tricky just because of the, the nature of these electrostatic force problems. Luckily, for our brains, though this doesn't actually make the problem that much harder, just more time consuming and a little more annoying to do. So recall that the formula for electric potential energy you is equal to the Coolum constant K multiplied by the product of the two relevant charges. So Q sub one multiplied by Q sub two, all divided by the distance between the two particles. And there are a few things to know about how we will be applying this formula to the diagram. First off. Since every single particle has the same magnitude of charge, we can more simply write this product, this product of charges as just Q squared where one of those cues just represents the charge on any of the particles. The R. However, the distance between the two particles is going to be a little more tricky for us because depending on which pair of particles we're looking at, there are different distances because we can see that the diagram tells us that our sub A B and R sub CD are clearly given in the diagram as being four centimeters or 0.04 m. Similarly, our sub BC and R sub D A are both equal to two centimeters or 0.02 m. And there's also the matter of the diagonal distances which can be found using the Pythagorean theorem since it's not given to us in the diagram. And that's R sub AC which is also R sub BD. Like I said, this can be found using the Pythagorean theorem. So we can find it this distance by taking the square root of the two distances we were given. So four centimeters squared plus two centimeters squared, which if you put that into a calculator is equal to 4.5 centimeters or alternatively 0.045 m. OK. So now that we have all those R values out of the way and we've got them sorted out. Now let's actually set up a formula to find the initial potential energy. And like I mentioned earlier, in order to find this, we're gonna have to add up all six of these terms. But luckily for us, there are a few things that actually makes this a little easier and, and less writing some things we can simplify more because the ones that are symmetrical, the pairs of charges that are symmetrical to one another and have the same distances can be written as being equal to one another. So for example, because all of the charges have the exact same magnitude and because particles A and B have the same distance apart from one another as particles C and D, that means that U sub A B and U sub CD have the exact same magnitude. So instead of writing out tediously, U sub A B plus U sub CD, we could just more simply write out two multiplied by U sub A B because that makes it more simple to write and we can continue this pattern. Particles B and C have the same distance apart as A and D. So we can write that as, as just two multiplied by U sub BC and particles A and C have the same distance apart as particles B and D. So we can just write that as two multiplied by U sub ac. So we've made this a lot more simple to write, but let's keep going and include our formula for the electric potential energy. So that's two multiplied by the potential energy formula we discussed earlier. So that is the Coolum constant K multiplied by Q squared all divided by R sub A B. And then we're adding on to this, on adding on to this two multiplied by K Q squared. The same term except this time divided by R sub EC. Then we're adding two. And then for AC, it's the same formula, but it's two KQ squared all divided by R sub AC. So now from here, things are starting to look pretty simple. Uh We can simplify the formula even further so that we have less, fewer things to type in our calculator by factoring out all the twos and the KS and the Q squared. So we can more simply write this as two KQ squared and then multiplied by, in parentheses, one divided by R sub A B plus one divided by R sub BC plus one divided by R sub AC. So that is our formula for the initial potential energy. So now actually quantifying and finding the magnitude of that is as simple as putting our values into the calculator. So that's two multiplied by the Coolum constant 8. multiplied by 10 to the power of Newton meters squared per Coolum squared multiplied by the square of the charges. And the magnitude of one charge is given to us in the problem as nano coons, negative 15 nano coons, we can disregard the negative sign because when we're dealing with energies, those tend to be irrelevant so we could, you could just put an absolute value sign over that if you want, but it's 15, multiplied by 10, the power of negative nine S and then multiplied by the sum of our distances. And don't forget to square the charge. So that's one divided by the distance between R and B. So 0.04 m plus one divided by R sub BC. So 0.02 m plus one divided by R sub ac or 0.045 m. And if you put all of that into your calculator, then we find an initial potential energy of approximately 3. multiplied by 10 rays to the power of negative four jewels. So that is our value for the initial, the initial potential energy. Now let's take another look at the energy formula. We discussed all the way at the beginning of the problem and discuss what we have next. So next, we want to find the initial kinetic energy. But since the problem tells us that the particles are all released from rest, that means that the initial kinetic energy is zero. So that term can be ignored. Next, let's look at the final potential energy. Uh The problem specifically tells us to find the speed when they have moved a significant distance away from each other. Now, what exactly significant means in this context is of course ambiguous. But what the general implication is is that they've moved apart at some distance that the electric potential energy between them becomes negligible as it gets smaller and smaller and closer to zero as they move further and further apart. So we can assume that U sub F is also equal to zero. So lastly, let's look at K sub F the final kinetic energy. This is the thing we're actually most interested in finding because it involves the speed, recall that kinetic energy is equal to one half multiplied by the mass multiplied by the square of the speed of the particle. So this is going to be very pertinent to what we're looking at. So K sub F the problem asks us to find the speed of just one of the particles, but there are four of them. So the total final kinetic energy of the system is going to be four multiplied by the kinetic energy formula we discussed earlier. That's four multiplied by one half MB squared. I'm gonna specifically label this as V sub F just for efficiency and completeness. We can simplify this a little bit down by multiplying the four by the one half. So that just divides the four by two. So we're left with two multiplied by M multiplied by V sub F squared. And we know from the problem that the mass of each of these particles are g, that's two multiplied by 2 g or two multiplied by 10 to the power of negative 3 kg V sub F squared. All right. So now all that's left for us to do to solve for V sub F is put all of this into our full energy equation. So as we discussed earlier, the initial potential energy that's 3.9375 multiplied by 10 to the power of negative four Jews plus zero Jews of kinetic energy equals zero Jews of final potential energy plus two multiplied by. And we could just simplify that down. Actually, it was just four multiplied by 10 to the power of a negative three kilograms multiplied by V sub F squared. And to solve for the sub F to solve for the final speed, we simply divide both sides of the equation by 4 g and then take the square root of the whole thing. So that's 3.9375 multiplied by 10 to the power of negative four Jews all divided by four multiplied by 10 to the power of negative 3 kg. And all of that is underneath a square root. And if we put this into our calculator, then we find a final speed for each of the particles of approximately 0.31 m per second. So that is our final answer for the speed of the particles. And if we look at our multiple choice options, we can see that this agrees with option D. So option D is the correct answer to this problem. And that is it for the problem. I hope this video helped you out if you would like more practice. Please consider checking out some of our other tutoring videos which will give you more experience with these types of problems, but that's all for now and I hope you all have a lovely day. Bye bye.

The four 1.0 g spheres shown in FIGURE P25.42 are released simultaneously and allowed to move away from each other. What is the speed of each sphere when they are very far apart?

Key Concepts

Conservation of Energy

Kinetic Energy

Electrostatic Force

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