# Find the minimum separation between 2 pucks

1. Feb 22, 2012

### Helios12

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. Jan 29, 2013

### quicksilver123

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

3. Jan 29, 2013

### Staff: Mentor

Looks dubious to me. I just glanced at the workings and it doesn't look like the electrical PE is being handled properly --- there should be contributions at both initial and final separations. And for the initial KE, why is the square of the velocity of the second puck made negative???

Perhaps you should make your own, fresh attempt.

4. Jan 29, 2013

### quicksilver123

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] = (+)

di=20m

m1=0.8kg
q1=+3*10^-4 C
Vi1 = 12m/s

m2=0.4kg
q2=+3*10^4 C
vi2= -8m/s

rmin=?

vf' = vf1 = vf2

ptotal=p'total
m1v1+m2v2=(m1+m2)v'
9.6-3.2=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).

EK1+EK2 = EE+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?

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5. Jan 29, 2013

### Staff: Mentor

Kinetic energy has no direction, it's always a positive scalar value. KE from all particles sum to a total.
You'd better confirm that. Especially since your 'final' distance of around 15m isn't significantly different. What's the PE for the initial separation?
Check the potential energy contribution and redo.

6. Jan 29, 2013

### quicksilver123

EE+EK1+EK2 = 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+57.6-1.6=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. Jan 29, 2013

### Staff: Mentor

Still not quite right. Check your KE for m2. The magnitude of the value looks incorrect, and KE is never negative!

8. Jan 29, 2013

### quicksilver123

Damnit. When I wrote the calculation down on paper, I forgot to include the square on the (-8).

EE+EK1+EK2 = 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+12.8-17.0645334=810/r'
93.8354666r'=810
r'=8.632130572m

9. Jan 29, 2013

### Staff: Mentor

That looks much better

Be sure to round your final results to the appropriate number of significant figures.

10. Jan 29, 2013

### quicksilver123

thanks!

11. Apr 3, 2014