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Two masses are attached to a 1 m long massless bar. Mass 1 is 3 kg and is attached to the far left side of the bar. Mass 2 is 5 kg and is attached to the far right side of the bar. If a third mass that is 2 kg is added to the middle of the bar, how does the center of mass of the system change?(A) The center of mass shifts to the left by 0.025 m.(B) The center of mass shifts to the right by 0.025 m.(C) The center of mass shifts to the left by 0.075 m.(D) The center of mass shifts to the right by 0.075 m.

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Chapter 14

Practice Test 4

Section 1

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So we have I thought. And this rod, let's say it is of a meter. And the two monsters are attached At the two ends. So this mass is three. Katie had. The other moss attached is a white kitchen. Okay, now. Okay, A 3rd masses addict. So where the third month is at it And in the middle, that means if I think this is the reference point, so exactly at .5 m from this end, the third mass is attached to the centre and it is a do pizza. Now, before this thermos was added and after this third mass is added, we need to figure out the difference the center of moss difference in central mask that needs to be computed. So let us first compute the center of mass in this situation. So when this mass is added, when the third master is two cages at it, the new center of mass. That will be next dash. So we use here effectively X. S. Submission. Am I excited over submission? Mhm. So the new center of mass will be three times its position. So this is the reference point ticket and then two times .5 Plus five times 1. And this will be divided by three plus four plus five. And this gives the new center of process one plus 55 Then and this is .6 m. So now the center of mass is at .6 m and earlier When only the two masses were there. So we don't need to include the two KK bus. So in this case the center of boss that is X will be three times 0 was five times 1 over the two masses at least to treat this fight. So that will be five x 8 and this will be .625 m. So now we have the center of mass when the thermoses at it and the center of mass when only two months is were present. So we take the difference in central mass that is x minus extension, and that comes out to be 0.65 m minus 0.6 m, and this is point zero june five m. So this is the difference in central Mosque. After This, two KK mass is added to this year system.

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