Guest Roop Posted November 16, 2009 Share Posted November 16, 2009 I need your help with a ridiculous centre of mass problem. Help appreciated. Link to comment Share on other sites More sharing options...
cams Posted November 16, 2009 Share Posted November 16, 2009 I need your help with a ridiculous centre of mass problem. Help appreciated. Graduated 97 2:1 Applied Physics, but fecked if i can remember anything Link to comment Share on other sites More sharing options...
Victorian Posted November 16, 2009 Share Posted November 16, 2009 yes, may i help you? Link to comment Share on other sites More sharing options...
Guest Roop Posted November 16, 2009 Share Posted November 16, 2009 Graduated 97 2:1 Applied Physics, but fecked if i can remember anything Ta. I don't know whether to assume you can help, then. I'm going to try and look for a Physics forum.. Link to comment Share on other sites More sharing options...
cams Posted November 16, 2009 Share Posted November 16, 2009 PM the question Link to comment Share on other sites More sharing options...
milky_26 Posted November 16, 2009 Share Posted November 16, 2009 whats the question, i may be able to help Link to comment Share on other sites More sharing options...
Guest Roop Posted November 16, 2009 Share Posted November 16, 2009 Two titanium spheres approach each other head-on with the same speed and collide elastically. After the collision, one of the spheres, whose mass is 300 g remains at rest. 1) What is the mass of the other sphere? 2) Find the speed of the two-sphere center of mass both before and after the collision, if the initial speed of each sphere was 2 ms-1. Check each explicitly. If you take my word for it that I've calculated part 1) to be 0.1Kg (check it if you like but it's incredibly long and boring), you've now got all the values relating to the collision. It's part 2 I don't really get. I think you give yourself an arbitrary distance between the 2 particles and then work out what happens 1 second later? How do you check that? What do you even do?! Link to comment Share on other sites More sharing options...
Beverley Posted November 16, 2009 Share Posted November 16, 2009 how quickly do you need to know as i actually do know a physicist, quantum i think. he works on that bosson, god particle thing that was gonna kill us all, and turn us into a big black hole Link to comment Share on other sites More sharing options...
Guest Roop Posted November 16, 2009 Share Posted November 16, 2009 how quickly do you need to know as i actually do know a physicist, quantum i think. he works on that bosson, god particle thing that was gonna kill us all, and turn us into a big black hole Hand-in is on Wednesday morning, Bev. Link to comment Share on other sites More sharing options...
The Gasman Posted November 16, 2009 Share Posted November 16, 2009 how quickly do you need to know as i actually do know a physicist, quantum i think. he works on that bosson, god particle thing that was gonna kill us all, and turn us into a big black hole No offence Bev, but if your pal's working on that big ass haydron colider thing, I'd rather you don't disturb them ! Link to comment Share on other sites More sharing options...
Bigsmak Posted November 16, 2009 Share Posted November 16, 2009 how quickly do you need to know as i actually do know a physicist, quantum i think. he works on that bosson, god particle thing that was gonna kill us all, and turn us into a big black hole Hand-in is on Wednesday morning, Bev. Well with any luck he'll kill us all on Tuesday night and you won't have to submit it! Link to comment Share on other sites More sharing options...
davemclaren Posted November 16, 2009 Share Posted November 16, 2009 how quickly do you need to know as i actually do know a physicist, quantum i think. he works on that bosson, god particle thing that was gonna kill us all, and turn us into a big black hole I've always had a lot of affection for the w vector boson since my student days... Link to comment Share on other sites More sharing options...
Guest King Posted November 16, 2009 Share Posted November 16, 2009 Two titanium spheres approach each other head-on with the same speed and collide elastically. After the collision, one of the spheres, whose mass is 300 g remains at rest. 1) What is the mass of the other sphere? 2) Find the speed of the two-sphere center of mass both before and after the collision, if the initial speed of each sphere was 2 ms-1. Check each explicitly. If you take my word for it that I've calculated part 1) to be 0.1Kg (check it if you like but it's incredibly long and boring), you've now got all the values relating to the collision. It's part 2 I don't really get. I think you give yourself an arbitrary distance between the 2 particles and then work out what happens 1 second later? How do you check that? What do you even do?! I dont know how people can do subjects like this, you have my respect. This is why I hated maths and sciences and prefer arty farty English and history. Good luck Link to comment Share on other sites More sharing options...
cams Posted November 16, 2009 Share Posted November 16, 2009 Ill go for 4m/s after the collision m1= mass first object = 300g m2 = mass second object = 100 g u1 = initial velocity first object = 2m/s ( to the right) u2= initial velocity second object = -2m/s ( to the left) v1 = final velocity first object = 0 m/s v2 = final velocity second object = ?? m1u1 +m2u2 = m1v1 + m2v2 (300 x 2) + (100x -2) = (300 x0) + (100X V2) 600+(-200 ) = 0 + 100 x V2 400 = 100 V2 v2 = 4m/s therefore the lighter object moves at 4m/s to the right Link to comment Share on other sites More sharing options...
H2 Posted November 16, 2009 Share Posted November 16, 2009 Two titanium spheres approach each other head-on with the same speed and collide elastically. After the collision, one of the spheres, whose mass is 300 g remains at rest. 1) What is the mass of the other sphere? 2) Find the speed of the two-sphere center of mass both before and after the collision, if the initial speed of each sphere was 2 ms-1. Check each explicitly. If you take my word for it that I've calculated part 1) to be 0.1Kg (check it if you like but it's incredibly long and boring), you've now got all the values relating to the collision. It's part 2 I don't really get. I think you give yourself an arbitrary distance between the 2 particles and then work out what happens 1 second later? How do you check that? What do you even do?! There are two many unknown variables to answer this question. For example if man could go to moon the effect of gravity on the moon is only about 1/10 the acceleration rate on earth. The next problem is that the 300g sphere is at rest, but how long did it take to come to rest and how far way is it from the point of impact, then how far away did the other object finish. Although you know the initial speed of both sphere there is no inof about speed or deceleration, or acceleration of the objects, mayebe one was going up and the other coming down. But the real answer I have absolutely no idea. Link to comment Share on other sites More sharing options...
Guest Roop Posted November 16, 2009 Share Posted November 16, 2009 Ill go for 4m/s after the collision m1= mass first object = 300g m2 = mass second object = 100 g u1 = initial velocity first object = 2m/s ( to the right) u2= initial velocity second object = -2m/s ( to the left) v1 = final velocity first object = 0 m/s v2 = final velocity second object = ?? m1u1 +m2u2 = m1v1 + m2v2 (300 x 2) + (100x -2) = (300 x0) + (100X V2) 600+(-200 ) = 0 + 100 x V2 400 = 100 V2 v2 = 4m/s therefore the lighter object moves at 4m/s to the right That's just linear momentum again, though. I did exactly that and was told I didn't answer the question. It's the centre of mass, taking into account both particles given that you know all the values needed. I thiiiiiiink you use 2m/s with the centre of mass formula as your initial speed, u, and then do what you just did to get the final value of the final speed, v2, and do the centre of mass equation again to answer it? That makes vague sense Thanks very much! Still not sure how you'd "explicitly check" it either tbh.. Link to comment Share on other sites More sharing options...
Guest Roop Posted November 16, 2009 Share Posted November 16, 2009 I dont know how people can do subjects like this, you have my respect. This is why I hated maths and sciences and prefer arty farty English and history. Good luck Thanks There are two many unknown variables to answer this question. For example if man could go to moon the effect of gravity on the moon is only about 1/10 the acceleration rate on earth. The next problem is that the 300g sphere is at rest, but how long did it take to come to rest and how far way is it from the point of impact, then how far away did the other object finish. Although you know the initial speed of both sphere there is no inof about speed or deceleration, or acceleration of the objects, mayebe one was going up and the other coming down. But the real answer I have absolutely no idea. Well, if you put it that way, there are enough values for your variables to solve it. Two spheres collide on a frictionless, horizontal plane. They both move at fixed speeds so there's no acceleration, the deceleration isn't considered because the collision is elastic, and the distance between the spheres is the arbitrary number I was going on about earlier, I think it implies you set your own distances and work out the relevant speeds from there. I just don't know how. Link to comment Share on other sites More sharing options...
Miller Jambo 60 Posted November 16, 2009 Share Posted November 16, 2009 Thanks Well, if you put it that way, there are enough values for your variables to solve it. Two spheres collide on a frictionless, horizontal plane. They both move at fixed speeds so there's no acceleration, the deceleration isn't considered because the collision is elastic, and the distance between the spheres is the arbitrary number I was going on about earlier, I think it implies you set your own distances and work out the relevant speeds from there. I just don't know how. Imagine the mups on Hibs net attempting a sum like that, makes me proud to be a jambo. Yes we are kings of Edinburgh:2thumbsup: Doug. Link to comment Share on other sites More sharing options...
H2 Posted November 16, 2009 Share Posted November 16, 2009 Thanks Well, if you put it that way, there are enough values for your variables to solve it. Two spheres collide on a frictionless, horizontal plane. They both move at fixed speeds so there's no acceleration, the deceleration isn't considered because the collision is elastic, and the distance between the spheres is the arbitrary number I was going on about earlier, I think it implies you set your own distances and work out the relevant speeds from there. I just don't know how. I'm guessing, but could it be as simple as Force = Mass x acceleration, but as there is no acceleration it will be 1? Link to comment Share on other sites More sharing options...
wibble Posted November 16, 2009 Share Posted November 16, 2009 Loads of google hits on your question. http://www.google.com.au/search?rlz=1C1GGLS_enAU352AU352&sourceid=chrome&ie=UTF-8&q=Two+titanium+spheres+approach+each+other+head-on+with+the+same+speed+and+collide+elastically. Link to comment Share on other sites More sharing options...
redjambo Posted November 16, 2009 Share Posted November 16, 2009 Check this out: http://www.cramster.com/physics-answers-5-515727-cpi0.aspx Link to comment Share on other sites More sharing options...
gadgey55 Posted November 16, 2009 Share Posted November 16, 2009 Exactly what I was gonna say! Link to comment Share on other sites More sharing options...
Guest Roop Posted November 16, 2009 Share Posted November 16, 2009 Why didn't I think to Google it? Cheers folks, Kickback's great sometimes. Link to comment Share on other sites More sharing options...
Guest Roop Posted November 16, 2009 Share Posted November 16, 2009 Quite funny how wrong I was as well. Link to comment Share on other sites More sharing options...
redjambo Posted November 16, 2009 Share Posted November 16, 2009 Quite funny how wrong I was as well. Aye, but you got the first bit correct. Link to comment Share on other sites More sharing options...
Ulysses Posted November 16, 2009 Share Posted November 16, 2009 Why didn't I think to Google it? Or you could have just dropped a PM to Gadgey. Link to comment Share on other sites More sharing options...
The People's Chimp Posted November 16, 2009 Share Posted November 16, 2009 Good to see that exam/coursework questions are regurgitated the world over. It wouldn't do for those in charge to put together their own problems. Link to comment Share on other sites More sharing options...
trotter Posted November 17, 2009 Share Posted November 17, 2009 Link to comment Share on other sites More sharing options...
Auld Reekin' Posted November 17, 2009 Share Posted November 17, 2009 Imagine the mups on Hibs net attempting a sum like that, makes me proud to be a jambo.Yes we are kings of Edinburgh:2thumbsup: Doug. They're all economists though, aren't they...??? Link to comment Share on other sites More sharing options...
Moriarty Posted November 17, 2009 Share Posted November 17, 2009 haha. the answer is 5 btw. ( it think...) Link to comment Share on other sites More sharing options...
H2 Posted November 17, 2009 Share Posted November 17, 2009 haha. the answer is 5 btw. ( it think...) Show your calculations and you might get a mark ! Link to comment Share on other sites More sharing options...
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