
#1
Feb2212, 11:39 AM

P: 1

1. The problem statement, all variables and given/known data
Two frictionless pucks are placed on a level surface with an initial distance of 20m. Puck 1 has a mass of 0.8 kg and a charge of + 3x10^4 C while puck 2 has a mass of 0.4 kg and a charge of +3 x10^4 C. The initial velocity of puck 1 is 12 m/s [E] and the initial velocity of puck 2 is 8 m/s [W]. Find the minimum separation of the two pucks (the minimum distance between the two pucks). 3. The attempt at a solution This is my attempt at a solution I first used the law of conservation of momentum to find the velocity of each mass at minimum separtion. The 2 masses have the same velocity at this point. Note: [east] is positive Pt=Pt' m1v1+m2(v2)=(m1+m2)v' isolating for v' v'=(0.80)(12)+(0.40)(8)/0.80+0.40 v'=5.33 m/s [east] Next i used the law of conservation of energy: Ek=kinetic energy Ee=electric potential energy Ek1+Ek2=Ee+Ek' 1/2m1v1^2 + 1/2m2(v2^2) = kq^2/r + 1/2(m1+m2)(v')^2 inserting each value 1/2(0.80)(12)^2 + 1/2(0.40)(8)^2 = (9.0X10^9)(+3.0X10^4)^2/r + 1/2(0.80+0.40)(5.33)^2 isolating for r (the distance) I get: r=15.2 m so the minimum distance between the 2 pucks is 15.2 m Is this correct? I am not feeling too confident with my answer of 15.2 m it just seems too big a separation. If i made a mistake or forgot to include something or included something i shouldn't have. please let me know. Also I am not sure if i am correct in assuming that there is no electric potential energy when the pucks are 20 m apart. It seems that they are too far apart to have any electric potential energy between them. Can someone confirm this with me also? It took me some time to write this up so can someone please look over my answer and the questions that i have. It would be most appreciated! 



#2
Jan2913, 10:05 AM

P: 91

bump? i have the same problem and got the same answer  was wondering if it is correct




#3
Jan2913, 01:38 PM

Mentor
P: 11,410

Perhaps you should make your own, fresh attempt. 



#4
Jan2913, 02:11 PM

P: 91

Find the minimum separation between 2 pucks
Its negative because its travelling in the opposite direction. I've attached a picture of the situation.
Here's my work (let's ignore, for now, what's done): Let East [E] = (+) d_{i}=20m m_{1}=0.8kg q_{1}=+3*10^4 C V_{i1} = 12m/s m_{2}=0.4kg q_{2}=+3*10^4 C v_{i2}= 8m/s r_{min}=? v_{f}' = v_{f1} = v_{f2} p_{total}=p'_{total} m_{1}v_{1}+m_{2}v_{2}=(m_{1}+m_{2})v' 9.63.2=1.2v' 6.4/1.2 = v' v' = 5.333 m/s [E] I didn't include initial electrical potential energy in this next part because its negligible at that distance (20m). E_{K1}+E_{K2} = E_{E}+E'_{K} 0.5(0.8)(12)^2 + (0.5)(0.4)(8)^2 = (k(3*10^4)^2)/r +0.5(0.8)(0.4)(5.33)^2 57.6+12.8=k(9*10^8)/r + 17.04534 53.46566=k(9*10^8)/r 53.46566r=k(9*10^8) r = 15.18142933 bon? 



#5
Jan2913, 02:30 PM

Mentor
P: 11,410





#6
Jan2913, 02:56 PM

P: 91

E_{E}+E_{K1}+E_{K2} = E'_{E}+E'_{K}
(k(3*10^4)^2 )/20 + 0.5(0.8)(12)^2 + (0.5)(0.4)(8)^2 = (k(3*10^4)^2)/r +0.5(0.8)(0.4)(5.33)^2 40.5+57.61.6=810/r' +17.0645334 r'=10.19695653m I was wrong about the initial electrical potential energy. I thought that at that range, its magnitude would be minuscule. OH WELL. How's that look? 



#7
Jan2913, 03:14 PM

Mentor
P: 11,410





#8
Jan2913, 03:22 PM

P: 91

Damnit. When I wrote the calculation down on paper, I forgot to include the square on the (8).
E_{E}+E_{K1}+E_{K2} = E'_{E}+E'_{K} (k(3*10^4)^2 )/20 + 0.5(0.8)(12)^2 + (0.5)(0.4)(8)^2 = (k(3*10^4)^2)/r +0.5(0.8)(0.4)(5.33)^2 40.5+57.6+12.817.0645334=810/r' 93.8354666r'=810 r'=8.632130572m 



#9
Jan2913, 03:30 PM

Mentor
P: 11,410

That looks much better
Be sure to round your final results to the appropriate number of significant figures. 



#10
Jan2913, 04:19 PM

P: 91

thanks!




#11
Apr314, 06:03 PM

P: 2

I ended up with the same steps and answer, wondering if it ended up being correct in the end? 


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