A R C H I V E S
Madinat al-Muslimeen Islamic Message Board
| anyone smart in physics? |
|---|
| Issma |
| 04/17/02 at 11:57:43 |
| [slm] Asalamualaykum, ummm i need help with a physics problem :(...any of you guys good in the topic of momentum and (frictionless) collision, ??? this is the q: a 0.4 kg (m1) bead slides down a frictionless wire(that is on a slight slant, like a V, but without the pointed ends, curves are smooth.), starting at the height 1.5m (y initial) at the base of the wire is another bead, o.6 kg (m2). both beads start at rest (Vi1 = Vi2 = 0) i need to know how high the second bead travels up the second fold of wire. my attempts included using this formula: mVi(1) + mVi(2) = mVf(1) + mVf(2) and i split it into 2 phases, i considered vf(1) in stage 1 as Vi(1) in stage 2, so that when they collide, i can use this formula. then i used Vi(1) - Vi(2) = Vf(2 - Vf(1) then i sub this one into the first but my answer comes out as Vf(2) = 4.378 i put this as Vi in this formula Vf(squared) = Vi (squared) + 2 g y to find y i get 0.2 but i at the back of the book its says its sposed to be yf(2) = 0.96m please please please if some one can help me out. :'( u'll get hasanaats cos in the long term u'd be helping me get into medecine, and helping muslim sisters have lady doctors :-) instead of having to go to non muslim men for help(well that my plan anyways inshallah) [wlm] issma |
| Re: anyone smart in physics? |
|---|
| Abu_Hamza |
| 04/17/02 at 13:45:20 |
| [slm] Ok, this is how I think you do the problem: This is given: m1 = 0.4 kg v1 = 0 m/s m2 = 0.6 kg v2 = 0 m/s y = 1.5 m g = 9.8 m/s(squared) First you need to find out what the speed of m1 will be right before it collides with m2 at the bottom of the incline. Call that speed vf. We need to use this equation that you gave: vf (squared) = vi (squared) + 2gy We're looking for vf. We know vi ... 0 m/s. We know g ... 9.8 m/s (squared) We know y ... 1.5 m. So vf comes out to be 5.422 m/s. Now, we use m1v1 = m2v2 to find out the initial speed of m2 right *after* the collision. Assuming all momentum is transferred from m1 to m2, this is what we do ... We're looking for v2. We know m1 ... 0.4kg We know v1 ... 5.422 m/s We know m2 ... 0.6 kg So v2 comes out to be 3.615 m/s. This is the initial velocity of m2. We now want to know how far up it will travel, y. We use the first equation again ... vf (squared) = vi (squared) + 2gy We know vf ... 0 m/s. We know vi ... 3.615 m/s. We know g ... -9.8 m/s (squared) [the value is negative because the ball is now travelling against gravity) We're looking for y. The value you get for y is the final answer for this problem. Now, the trouble is, I get 0.67 m for y, which is not the same value that's in your book :/ So either I'm calculating something wrong, or the equation that you gave is not correct. But I'm pretty sure the logic that I've explained is correct. So try and see if you can get anywhere using this logic :) Sorry if I confused more than you were before :( Wassalamu alaikum wa rahmatullah |
| 04/17/02 at 13:46:05 |
| Abu_Hamza |
| Re: anyone smart in physics? |
|---|
| Mehak |
| 04/17/02 at 15:42:02 |
| [slm], Sister, we are doing the same topic in my physics class and i had a similiar problem. What I figured out was that if u have ur second bead at origin(assume ur x-y plane with m2 at origin), u can apply the formula Xcom=(m2 * d)/ (m1 + m2) I dont know what book u r using but the one from Halliday and Resnick,this is the eq 9-1. The beads are going to travel upto their center of mass and then perhaps stop or not depending on if there is a force applied or not which I belive is not, in this case. Thats why u jsut find out the center of mass. If I am wrong,someone please correct me but thats the way I figured out mine ;D . And so when u apply the formula ,u get 0.9m as ur answer. Br AbuHamza, this problem is a monmentum problem.U cannot apply those eqs bcuz they can only be used with constant acceleration. Sister, I am not the best person to reply u back but if u or someone else thinks I maybe wrong please correct me so I can correct my own problems too. ;) Hope that helps, [wlm] |
| Re: anyone smart in physics? |
|---|
| Abu_Hamza |
| 04/18/02 at 02:32:06 |
| [slm] :) :) :) Mehak, this *is* a momentum problem. Read the first line of bint Louay's post. Acceleration is constant in the problem. The only acceleration that acts upon the two masses is gravity. Center of mass has nothing to do with this problem. I'm trying to figure out what center of mass has to do with this problem (Asim?) My guess is that it's nothing but a wild coincidence that that equation gives you an answer which is close to the right answer. The answer of 0.9 is still wrong btw. The correct answer is supposed to be 0.96 Wassalamu alaikum wa rahmatullah :) |
| Re: anyone smart in physics? |
|---|
| eleanor |
| 04/18/02 at 11:15:14 |
| [slm] :o :o :o boggle boggle boggle :o :o :o [i]said eleanor, in direct violation of the Madina Constitution[/i] ;) wasalaam eleanor :-* |
| Re: anyone smart in physics? |
|---|
| M.F. |
| 04/18/02 at 14:29:45 |
| I used to know this stuff, I KNOW I did!! Where did it go? How do you all keep your knowledge and skills updated? I feel so un-intelligent now that I'm not in college anymore (well distance classes don't count). |
| Re: anyone smart in physics? |
|---|
| Mehak |
| 04/18/02 at 15:16:39 |
| [slm], Okay I asked one of the graduate students and he said that my way of doing it is wrong. :( Obviously the problem calls for taking into consideration the initial and final velocities(thats why they are given!). So, in our book we havent studied momentum related problems invloving gravity. Even for my problem he told me to use : m1v1= m2v2 0.4 * 1.5 = 0.6 * (x-1.5) 0.5m which isnt the right answer for this problem since it doesnt give the right answer so obviously I was wrong. I will try to ask the instructor , InshaAllah since now I am interested in knowing how to solve this too :-) Thanks for pointing it out Br Ars....err Abu Hamza. [wlm] |
| 04/18/02 at 15:18:02 |
| Mehak |
| Re: anyone smart in physics? |
|---|
| Dawn |
| 04/19/02 at 04:44:53 |
| OK, getting around to this a little late and getting the same answer as Abu_Hamza -- two-thirds. Just to check, my husband and I both did it, and we both got two-thirds. Is there any chance the answer in the book is a typo? |
| Re: anyone smart in physics? |
|---|
| inti |
| 04/20/02 at 06:34:15 |
| [slm] Let me try ... From what you describe, here are the eqns: (1) v1i*v1i = 2*g*h1 (2) m1*v1i + m2*v2i = m1*v1f + m2*v2f (3) 0.5*m1*v1i*v1i + 0.5*m2*v2i*v2i = 0.5*m1*v1f*v1f + 0.5*m1*v1f*v1f (4) v2f*v2f = 2*g*h2 m1=0.4, m2=0.6, h1=1.5 From (1) v1i=5.422 v2i=0 (given) (2) => 0.4*v1f + 0.6*v2f= 2.17 (3) => 0.2*v1f*v1f + 0.3*v2f*v2f = 5.88 gives: v2f=4.34 Using (4), h2=0.96 m ===================== <<<. Assuming all momentum is transferred from m1 to m2, this is what we do ... >>> That is wrong! [wlm] |
| Re: anyone smart in physics? |
|---|
| Abu_Hamza |
| 04/20/02 at 16:00:59 |
| [slm] Ah! I see. So much for Grad school! ::) Right Dawn? :) |
| Re: anyone smart in physics? |
|---|
| Dawn |
| 04/20/02 at 16:41:04 |
| HeHe, I didn't touch Physics after undergrad! And I can't say as I really missed it! :D (Of course, it would also explain why I got tripped up, too. ::) ) |
Madinat al-Muslimeen Islamic Message Board |