Find the minimum separation between 2 pucks

In summary, two frictionless pucks with masses and charges are placed on a level surface with an initial distance of 20m. Using the law of conservation of momentum and energy, the minimum separation of the two pucks is calculated to be 8.632m, taking into account their initial velocities and potential energy.
  • #1
Helios12
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Homework Statement



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).


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!
 
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  • #2
bump? i have the same problem and got the same answer - was wondering if it is correct
 
  • #3
quicksilver123 said:
bump? i have the same problem and got the same answer - was wondering if it is correct

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
Its negative because its traveling 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
quicksilver123 said:
Its negative because its traveling in the opposite direction. I've attached a picture of the situation.
Kinetic energy has no direction, it's always a positive scalar value. KE from all particles sum to a total.
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).
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?
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?

Check the potential energy contribution and redo.
 
  • #6
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
quicksilver123 said:
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?

Still not quite right. Check your KE for m2. The magnitude of the value looks incorrect, and KE is never negative!
 
  • #8
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
That looks much better :smile:

Be sure to round your final results to the appropriate number of significant figures.
 
  • #10
thanks!
 
  • #11
quicksilver123 said:
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

quicksilver123
I ended up with the same steps and answer, wondering if it ended up being correct in the end?
 

1. What is the definition of "minimum separation"?

The minimum separation refers to the closest distance between two objects, in this case, two pucks.

2. How do you calculate the minimum separation between two pucks?

The minimum separation between two pucks can be calculated by finding the distance between their centers. This can be done using the Pythagorean theorem or by measuring the distance with a ruler.

3. What factors can affect the minimum separation between two pucks?

The minimum separation between two pucks can be affected by factors such as the size and shape of the pucks, their initial velocity, any external forces acting on them, and the surface they are on.

4. Why is it important to find the minimum separation between two pucks?

Finding the minimum separation between two pucks is important in determining the potential for collision or interference between the two objects. It can also help in predicting the motion and behavior of the pucks.

5. How does the minimum separation between two pucks relate to their kinetic energy?

The minimum separation between two pucks can affect their kinetic energy by determining the distance they have to travel before colliding. A shorter minimum separation would result in a higher likelihood of collision and therefore, a greater transfer of kinetic energy between the pucks.

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