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5.(30 points) A 13 kg cat sits on the left end of a 4 m long; 187 kg boat while a 43 kg dog sits on the right end A) Calculate the center of mass of this system rel...

Question

5.(30 points) A 13 kg cat sits on the left end of a 4 m long; 187 kg boat while a 43 kg dog sits on the right end A) Calculate the center of mass of this system relative to the left end of the boat. The cat now attacks the defenseless running dog by to the right end of the boat and jumping on top of the dog: B) Where is the new center of mass of this system relative to the original coordinate system? C) How far has the boat shifted in the water and in which direction?

5.(30 points) A 13 kg cat sits on the left end of a 4 m long; 187 kg boat while a 43 kg dog sits on the right end A) Calculate the center of mass of this system relative to the left end of the boat. The cat now attacks the defenseless running dog by to the right end of the boat and jumping on top of the dog: B) Where is the new center of mass of this system relative to the original coordinate system? C) How far has the boat shifted in the water and in which direction?



Answers

Jack and Jill are standing on a crate at rest on the frictionless, horizontal surface of a frozen pond. Jack has mass 75.0 kg, Jill has mass $45.0 \mathrm{kg},$ and the crate has mass 15.0 $\mathrm{kg}$ . They remember that they must fetch a pail of water, so each jumps horizontally from the top of the crate. Just after each jumps, that person is moving away from the crate. with a speed of 4.00 $\mathrm{m} / \mathrm{s}$ relative to the crate. (a) What is the final speed of the crate if both Jack and Jill jump simultaneously and in the same direction? (Hint: Use an inertial coordinate system attached to the ground.) (b) What is the final speed of the crate if Jack jumps first and then a few seconds later Jill jumps in the same direction? (c) What is the final speed of the crate if Jill jumps first and then Jack, again in the same direction?

In this problem. We have a person on a boat walking from end to end on. We're told how much the boat moves relative to the water that it's in and were asked to determine the first the length of the boat and then also how much the center of mass of the system moves and were given. The mass of the boat is 130 kg, and the mass of our person is 80 kg and the boat moves 0.8 m. Eso what we'll do here is will use conservation of momentum, uh, to say that our system since we're told that our the system conserves momentum. So the center of mass of our system does not change eso when the boat moves by 0.8 m are center of mass of relative to the boat has to move by an equivalent 0.8 m. We could look at this from a point of view of velocities and distances, but it will be simpler just to look at it from the center of mass perspective. So our part a, uh is asking us what is the length of the boat. And so we're looking at Here are distance. So our boat, we're assuming our boat is uniform. So this 130 um, kilograms is acting at a distance of El over to. And so what we'll do, we'll say that are X. So we start with our center of mass at X initial, and we finish with our senator mass at X Final, where we're measuring relative to the boat on So we can say that X final minus R X initial equals this 0.8 since our center of mass is going to remain in the same position. So as our person is walking to the right, our boat has to move equivalently to the left. Right. Okay, so let's find our to, uh, positions for center of mass. So x initial and x one. Okay. Initially, our center mass is going to be 3100 and 30. So we're starting at the left end of the boat is zero. So we have 130 times l over to over 130 plus 80 and we can That will give us a 13 over 42 help, and then finally, Rx final. It's going to be again measuring from the left end of the boat relative to the boat, the left handed zero. So our center of mass is going to be 80 or person has moved to the right edge of the boat 80 times l and plus our mass of the boat acting at its center l A. To and divided by 1 30 plus 80. And that will give us 100 45 over 210 out. So relative to the boat, the center of Mass has moved by this distance. And then the boat has moved by this distance 0.8. So doing our location here 145 210 hour minus 13/42 eloquent zero 0.8. And if we just saw that for l get our AL equals to 2.1 meters and then our part B asked, asks us what happens to the center of mass of the system, right? And it, uh, it doesn't move. So what we can see what would happen here is are so are yeah. So we saw our center of math. Our boat moves one way our person moves this way on our boat. It is moving in the other direction, but the center of mass of the system is going to remain in the same position. Right? So if we're are looking from an outside perspective, right. So our boat, our boat moved 0.8 m to the left and our center of mass relative to the boat moved that much distance to the right.

In the execution we have given to men's and both are sitting at the extreme of Both of two m long and Wait or 70 Kg in still water. We have to find that they come to the middle to the boat. Now we have to find it. How far does the boat move on the water during this process. So as we know that the position of center of moth will remain same because there is no external force. So we can say that the vision of center of mass that is given by which is RCO musical to disease and one R. One plus M. Two R. Two A bonded and one blood. I am too. But this is M. one r. 3 M three RT upon entry. So we can see that if we differentiate this. So this is the R. C. Um Is equal to disease and one into D. R. one plus this is I'm doing two D. R. Two Plus entry into D. R. three upon and one plus M. Two plus M. Three. So this really called zero because the position of center of march will remain same from here. We can say that and went into D. R. One vector plus I'm going to the er to vector plus entry into the artery, vector is about 20. No, if the boat moves by X. Distance so we can see that disease 17 2 X plus. This is 16. To change in position of, this is one plus X. Last the C 17 2 X minus one is equal to zero from here. We can find the value of this will be 70 X plus. This is 60 plus 60 X plus 70 X minus 70 Z equals to zero. What we can say Access equals 25 semi. So this is the shifting of board and this is our answer. So we can say for that correct answer. His option. B 35 for me the right word. So this is our answer. Thank you.

So we're told an 80 kilogram person stands at one end of a 130 kilograms boat. You gonna walk to the other and the boat so that the boat moves. 87 years was extracted bottom of the lake. 1st 1 to figure out what is the length of the boat. And how much did the center mass of the boat person system move when the person walked from one into the other? Um, well, we can try schematic. So here we have the person standing at one end of the boat, and then we assume the center of mass of the boat itself. Is that its geometric center here? And so the center of massive total system is going to be somewhere towards the back of the boat. And so we're gonna measure relative to the front of the boat where that is. And then when the person moves to the other side of the boat, um, the center of mass of the boat moves towards the front. We'll send of the mass of the boats person system moved towards the front so that we could calculate where that center of mass is again. With respect to the front of the boat. Now we can actually answer the second question first. And that is that the center of mass of the boat person system actually doesn't move it all because there is no external forces acting on this system. So that stays where it is. And that's why the boat moves relative to the bottom of the lake. When the person moves so we can figure these things out, figure out the length of the boat. And so we know that the center of mass here is the mass of the boat, times the distance from here to the center of mass cell over too, plus the mass of the person times distance from here to the where the person standing, which is l invited by the total mass of the system. And then, in this case, we have the person that's standing at zero. So there they don't contribute to the numerator here. And so we get that the mass of the boat, its center of Mass, is at El over to sew the mass of the boat times over to fight it by the total mass of the system gives us the center of mass relative to the front of the boat. Now, Now the difference here is is 80 centimeters. So given that as 80 centimeters. So that's how much the boat moved. Come with respect to wait fixed reference here. So this is 80 centimeters and that's ex cm one minus x e m, too. So we subtract this from this this term in this term cancel out. And so we're left with the mass of the person times the length of the boat divided by the total mass of the system. And we can Then you know, these two things we can solve for l. And we find out that l is, um 120 centimeters. So that was that's how alarmed the boat is 2.1 meters. So then you asked about Like I said, the second part of the question says, You know where? How does the center of mass in the boat move of the boat person system moved again? It doesn't move because there are no external forces acting upon this system as a whole. There are forces internal from the person on the boat and the boat on the person. But externally, here we're assuming that there is nothing no Net forces acting to cause the boat to move, um, from the water or something so the center of mass stays globally at the same position.

Okay, so during Chapter nine problems 77 here. So we have two people of Mass 85 55 kilograms sitting in a row boat, which has a massive 78 grams. And it says the boats initially at rest when they're sitting at opposite ends shown here and it says the boats three meters long. They then decided to switch positions and it moves about a distance. D such as this and we wouldn't figure out how far did the boat move. So to do this, we need to figure out we need to realize that the center of mass is not gonna change because the only forces are within your system. So the senator mass must stay constant. So if we take this is our ex system with our origin, be here at the original position of original left position of the rowboat. Then we can figure out the center of mass of the system as mass of the first person comes a position of the first times the mass of the boat, time to position of the boat, plus the mess of second person times the position of second person all over the total mess So it's do that well, The first position is your over. The first person to guys at zero boats sent 78 kilograms is at a distance of 1.5 years and the farm dude is 85 kilograms is at a position of three years. The total mass is 78 plus the five plus 55 grams. When we get a center of mass uh, 1.7 06 meters. Okay, so we know that's gonna stay constant. So now, once they have switched positions, it's moved some distance D so we can calculate what the senator masses. And that crazy, you know, that 1.706 meters is a center of mass and now we have 85 kilograms at a position D. So this is 85 kilograms times D plus 78 kilograms and now it's a 1.5 meters plus p. What if I plus B and we have 55 kilograms in a position of three dusty. Okay, miss. We divide this by the total which was 218 kilograms. If you add them all together and this becomes too 18 de kilogram years plus 2 82 kilogram years. All over to 18 in this equals 1.706 This is craps. Okay, so we just saw for this. Fine. What d is he becomes 0.412 years. This this is positive. It moves to the right. So now we know to the right and how we do this or towards the have your persons original spark.


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