Electric Potential Energy Question With two Similarly charged Objects

[HarleyM]

Homework Statement

Two frictionless pucks are placed on a level surface as shown, at an initial distance of 20m.

Puck 1 has a mass of 0.8 kg and a charge of + 3x10^-4 while puck 2 has a mass of 0.4 kg and a charge of +3 x10^-4.

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 min distance between the two pucks

Homework Equations

Pt=Pt'
M1V1+M2V2=(M1+M2)V
Ek=1/2mv²

The Attempt at a Solution

Conservation of momentum is used to find the velocity of closest approach, since it must be constant for both.

Pt=Pt'
M1V1+M2V2=(M1+M2)V
(0.8)(12)+(0.4)(8)=(1.2)V (Should I set East or west as +, and therefore should 1 velocity be negative?)

V=10.7 m/s when pucks reach point of closest approach

Ek1+Ek2= Ee +Ek
1/2M1v1² +1/2M2v2²=(kq1q2/r) + 1/2(M1+M2)V

1/2(0.8)(12)²+1/2(0.4)(8)²= [(9x10^9)*(3x10^-4)²/r] + [1/2(1.2)*(10.7)]

(70.4-6.42)r=(9x10⁹)(3x10^-4)²
r= 12.66 m

SO my question is, is 12.7 m the closest they reach or would it be 20-12.7?


I think I put all the variables into the equations correctly and accounted for conservation of energy and momentum, any help would be great. I am quite confused.

Thanks everyone and merry Christmas, happy holidays and a happy winter solstice

 

[SammyS]

Homework Statement

Two frictionless pucks are placed on a level surface as shown, at an initial distance of 20m.

Puck 1 has a mass of 0.8 kg and a charge of + 3x10^-4 while puck 2 has a mass of 0.4 kg and a charge of +3 x10^-4.

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 min distance between the two pucks

Homework Equations

Pt=Pt'
M1V1+M2V2=(M1+M2)V
Ek=1/2mv²

The Attempt at a Solution

Conservation of momentum is used to find the velocity of closest approach, since it must be constant for both.

Pt=Pt'
M1V1+M2V2=(M1+M2)V
(0.8)(12)+(0.4)(8)=(1.2)V (Should I set East or west as +, and therefore should 1 velocity be negative?)

V=10.7 m/s when pucks reach point of closest approach

Ek1+Ek2= Ee +Ek
1/2M1v1² +1/2M2v2²=(kq1q2/r) + 1/2(M1+M2)V²

1/2(0.8)(12)²+1/2(0.4)(8)²= [(9x10^9)*(3x10^-4)²/r] + [1/2(1.2)*(10.7)]

(70.4-6.42)r=(9x10⁹)(3x10^-4)²
r= 12.66 m

SO my question is, is 12.7 m the closest they reach or would it be 20-12.7?I think I put all the variables into the equations correctly and accounted for conservation of energy and momentum, any help would be great. I am quite confused.

Thanks everyone and merry Christmas, happy holidays and a happy winter solstice

The pucks travel in opposite directions, so, yes, their initial velocities should have opposite sign. You pick what direction is positive. That is often East, but it also makes sense to have the sign of the one with greater speed. In this case, that's also East.

It's also possible to work this out in the center of mass reference frame, but that's not necessary.

Added in Edit:

The initial potential energy is not zero. you need to include it of the left side of your equation:

(1/2)M₁v₁² +(1/2)M₂v₂² = (kq₁q₂/r) + (1/2)(M₁+M₂)V²​

 

[HarleyM]

DO you have any insight as to whether I am close with my answer of 12.7 m?

I can't help but wonder if it should be 20-12.7m but I can't really explain why.

 

[cupid.callin]

I have another thought about the question ...
if we work in the frame of COM of the pucks ... it has no external force acting so it is in inertial frame ...

So simply using energy conservation in that case will give a solution because final velocities will be zero of both pucks ... (i guess correct me if I'm wrong) ...

 

[HarleyM]

Im still working at this question, 12.7 M seems like way to big of a distance to be the minmum distance these pucks come between each other. Am I doing something wrong in my calculations/equations?

 

[HarleyM]

I am also wondering is there electric potential energy at 20 m away so it should be

E_E + E_ΔK=E_E

This gives me an answer of 7.3 m, a little better but that still seems huge.. I also get 11.5 m sometimes as well.

Im really confused here.

 

[SammyS]

I am also wondering is there electric potential energy at 20 m away so it should be

E_E + E_ΔK=E_E

This gives me an answer of 7.3 m, a little better but that still seems huge.. I also get 11.5 m sometimes as well.

I'm really confused here.

How much force do the pucks exert on each other at those distances ?

Added in Edit:

I get an answer of about 8.6 meters for the minimum distance.

 

[HarleyM]

how much force do the pucks exert on each other at those distances ?

40.5 n?

 

[SammyS]

40.5 n?

Well 40 Newtons produces a fairly large acceleration on objects with masses of 0.4 kg and 0.8 kg.; 100m/s² and 50 m/s² respectively.

However, I get a force of more like 10 Newtons at 8.6 meters separation. Even that gives pretty big accelerations on the pucks.

 

[HarleyM]

How did you get to 8.6 m seperation?this equation?

E_E+E_ΔK=E_E+E_K

Did you also calculate the mass and momentum as one singular object i.e (M1+M2)V² for the right side of the equation ?I really want to understand this question so I appreciate any help you can offer a lot!

 

[SammyS]

How did you get to 8.6 m separation?

this equation?

E_E+E_ΔK=E_E+E_K

Did you also calculate the mass and momentum as one singular object i.e (M1+M2)V² for the right side of the equation ?

I really want to understand this question so I appreciate any help you can offer a lot!

Actually, I did it in the center of mass system, but that doesn't change the answer.

(M1+M2)V² is not momentum. Velocity should not be squared. (M₁+M₂)V is the momentum of the system when the pucks are at minimum separation. Conservation of momentum gives that V=(20/3) m/s @ min. separation. Remember, v₂ is negative, initially.

The Kinetic Energy of the system is (1/2)(M₁+M₂)V² at minimum separation.

What is the K.E. at 20 m separation?

What is the Potential Energy at 20 m separation?

Conservation of energy then gives the P.E. at minimum separation. Solve this for r .

 

[HarleyM]

Should it look like this?

K_E (@20 m) + E_E(@ 20 m) = K_E(@ distance of closest approach) + E_E (solve for r)

and K_E= (M1+M2)V² ( on both sides of the equation)?

 

[HarleyM]

so I get 10.67 m/s as velocity at minimum separation M1v1+M2V2=(M1+M2)V

v=10.67 m/s

E_K1+E_K2+E_E=E_E+(M1+M2)(10.67)

1/2m1v1²+1/2m2v2² + Kq²/20= kq²/r+1/2(M1+M2)V ( I didn't square V here)

R= 7.75 m

What am I doing wrong?

 

[HarleyM]

OK I found out how to get 8.6 I was just confusing myself, now how are you getting 10 N?

I used

(9x10^9)(3x10^-4)²/8.6

E_E= 94.18 N

 

[SammyS]

so I get 10.67 m/s as velocity at minimum separation M1v1+M2V2=(M1+M2)V

v=10.67 m/s

E_K1+E_K2+E_E=E_E+(M1+M2)(10.67)

1/2m1v1²+1/2m2v2² + Kq²/20= kq²/r+1/2(M1+M2)V ( I didn't square V here)

R= 7.75 m

What am I doing wrong?

The velocity at minimum separation (which is also the velocity of the center of mass) is (16/3) m/s. I had the wrong value in my previous post.

Remember, v₁ and v₂ have opposite signs.​

What is momentum, (M₁+M₂)(10.67), doing in the following equation?

E_K1+E_K2+E_E=E_E+(M1+M2)(10.67)​

Take that out & you may get the right answer.

 

[HarleyM]

OK I found out how to get 8.6 I was just confusing myself, now how are you getting 10 N?

I used

(9x10^9)(3x10^-4)²/8.6

E_E= 94.18 N

Did you see this?

 

[lightgrav]

Harley, the initial (smallest) Force is q kQ/r^2 , with r starting at 20m => 2.025 N.
... IF it would get to 8.6m distance, then (9E9)(3E-4)^2/(8.6)^2 = 10.95 N.

Your post #13 is pretty much right, except that you did not square V(closest)
... which doesn't matter because (using c.o.mass frame) V(closest) is zero.
use velocities relative to c.o.m. and you'll get 19m as closest approach.
(WTF?)
the big mass starts off 1.333 m/s faster than it is at closest approach,
the small mass starts off -2.666 m/s different than it is at closest approach.
2 N causes the big mass to accelerate a=F/m = 2.5 m/s/s ,
so the closest approach occurs in about dv/a=dt= .53 seconds.
the big one's _avg_ speed relative to c.o.m (10.67 m/s) was 0.666 m/s
so it went about 0.333 m relative to the c.o.m. in that time.
small mass moves twice as far (0.666 m) relative to the center-of-mass
before they're both going 10.666 m/s.

the center-of-mass moves 5.6 meters before the closest approach, so a LOT of Energy gets transfered!

 

[SammyS]

OK I found out how to get 8.6 I was just confusing myself, now how are you getting 10 N?

I used

(9x10^9)(3x10^-4)²/8.6

E_E= 94.18 N

You need to square the 8.6 .

That will then give you 10.95 N .

 

[HarleyM]

I don't think R is squared in electric potential energy , only in electric force and field?

So are you guys calculating electric force? Its unneeded for the answer ( I just need distance of closest approach) I am just curious

thanks everyone for the help!