Electric Potential Energy-minimum distance between two pucks

[chef99]

Homework Statement

Two frictionless pucks are placed on a level surface, as shown(see diagram) at an initial distance of 20.0m. The mass of puck 1 is 0.80kg and has a charge of +3.0 x10^-4 C, while puck two has a mass of 0.40 kg and a charge of +3.0 x10^-4 C. The initial velocity of puck 1 is 12.0m/s [E]. The initial velocity of puck 2 is 8.0m/s [W]. Find the minimum distance between the two pucks.

Homework Equations

E_to= E_tf

The Attempt at a Solution

E_to= E_tf

E_po + E_ko = E_pf + E_kf

0 + E_ko = E_pf + 0

1/2 m(v_1o)² + 1/2 m(v_2o)² = kq₁q₁/r

1/2(0.80kg)(12.0m/s)² + 1/2(0.40kg)(8.0m/s)² = (9.0 x10⁹)(3.0 x10^-4)((3.0 x10^-4) / r

70.4kgm/s = 810Nm / r

r = 810Nm / 70.4kgm/s

r = 11.5m

The minimum distance between the two puck will be 11.5m.

This is the first time I've tried to determine the minimum value of r (I have only determined the final velocity of the pucks) so I just tried what I thought made sense. I'm pretty sure I have the right idea but just want to make sure. Any input is greatly appreciated.

 

[tnich]

Homework Statement

Two frictionless pucks are placed on a level surface, as shown(see diagram) at an initial distance of 20.0m. The mass of puck 1 is 0.80kg and has a charge of +3.0 x10^-4 C, while puck two has a mass of 0.40 kg and a charge of +3.0 x10^-4 C. The initial velocity of puck 1 is 12.0m/s [E]. The initial velocity of puck 2 is 8.0m/s [W]. Find the minimum distance between the two pucks.

Homework Equations

E_to= E_tf

The Attempt at a Solution

E_to= E_tf

E_po + E_ko = E_pf + E_kf

0 + E_ko = E_pf + 0

1/2 m(v_1o)² + 1/2 m(v_2o)² = kq₁q₁/r

1/2(0.80kg)(12.0m/s)² + 1/2(0.40kg)(8.0m/s)² = (9.0 x10⁹)(3.0 x10^-4)((3.0 x10^-4) / r

70.4kgm/s = 810Nm / r

r = 810Nm / 70.4kgm/s

r = 11.5m

The minimum distance between the two puck will be 11.5m.

This is the first time I've tried to determine the minimum value of r (I have only determined the final velocity of the pucks) so I just tried what I thought made sense. I'm pretty sure I have the right idea but just want to make sure. Any input is greatly appreciated.

You are close. Think of it this way. For the total energy to remain constant
$ΔE_p + ΔE_k = 0$
You have accounted for the change in kinetic energy, but not for the change in potential energy. What is the potential energy of the starting configuration?

 

[TSny]

E_po + E_ko = E_pf + E_kf

0 + E_ko = E_pf + 0

As @tnich pointed out, the initial potential energy is not zero.
Also, the final kinetic energy of the system is not zero.

 

[tnich]

As @tnich pointed out, the initial potential energy is not zero.
Also, the final kinetic energy of the system is not zero.

You are right. I assumed the pucks would be at zero velocity at the minimum distance, but that is not true. It looks conservation of momentum has to be considered, too.

 

[chef99]

You are right. I assumed the pucks would be at zero velocity at the minimum distance, but that is not true. It looks conservation of momentum has to be considered, too.

Ok, so this is what I have come up with. I used the conservation of momentum to determine how the velocity of each mass is related to each other.

Using conservation of momentum,

P_to = P_tf

m₁v_1o - m₂v_2o = m₁v_1f - m₂v_2f

(0.80kg)(12.0m/s) - (0.40kg)(8.0m/s) = (0.80kg)(v_1f)(0.40kg)(v_2f)

9.6kgm/s - 3.2kgm/s = (0.80kg)(v_1f)(0.40kg)(v_2f)

v_1f = (6.4kgm/s)(0.40kg)(v_2f) / (0.80kg)

v_1f = 3.2v_2fNow using conservation of energy, to determine the final velocities of the pucks, in order to find the final kinetic energy.

E_E = E_k1 + E_k2

kq₁q₂ / r = 1/2 m₁(v_1f)² + 1/2 m₂(v_2f)²

(9.0 x10⁹)(3.0 x10^-4C)(3.0 x10^-4C) / (20.0m) = 1/2 (0.80kg)(3.2v<sub>2f)² + 1/2 (0.40kg)(v2f)²

40.5 J = 4.296(v2f)²

v2 = √9.427

v2 = 3.07m/sv1 = 3.2v2f

v1 = 3.2(3.1)

v1 = 9.92m/s

Now plugging the velocity values to solve for Ekf.

Ekf = 1/2 m1(v1f)² + 1/2 m2(v2f)²

Ekf = 1/2 (0.80kg)(9.92m/s)² + 1/2 (0.40kg)(3.1m/s)²

Ekf = 5.89 J

Now since the values of Ekf, Eko and Epf are known, the value of Epo can be determined:

Epo +Eko = Epf +Ekf

Epo + 70.4 J = 5.89 J + 70.4 J**

Epo + 70.4 J = 76.29 J

Epo = 76.29 J - 70.4 J

Epo = 5.89 J

**earlier determined that

70.4kgm/s = 810Nm/ r

r = 810Nm / 70.4kgm/s

= 11.5m. Using this;

810Nm/ 11.5m

= 70.4 J

This is what I figured, but I may be going way off here. Also is my calculation from the first attempt of r = 11.5 correct? it makes sense here but if that is incorrect then so is this. Thanks</sub>

Also, I believe the question just needs me to determine the value of r_f.

 

[TSny]

Ok, so this is what I have come up with. I used the conservation of momentum to determine how the velocity of each mass is related to each other.

Using conservation of momentum,

P_to = P_tf

m₁v_1o - m₂v_2o = m₁v_1f - m₂v_2f

When the distance between the two pucks is at a minimum, there is something special about the velocities v_1f and v_2f. You will need to use this.

 

[tnich]

You are getting closer.
You have dropped a minus here:

m₁v_1o - m₂v_2o = m₁v_1f - m₂v_2f

(0.80kg)(12.0m/s) - (0.40kg)(8.0m/s) = (0.80kg)(v_1f)(0.40kg)(v_2f)

Also, you have not correctly used conservation of energy here.

E_E = E_k1 + E_k2
kq₁q₂ / r = 1/2 m₁(v_1f)² + 1/2 m₂(v_2f)²

First of all $E = E_{k_1} + E_{k_2}$ is not the correct equation for conservation of energy, or even for the total energy.
Second, you have assumed that the potential energy at your point of closest approach is equal to the kinetic energy there. That is not true either. I suggest that you go back to the equation for conservation of energy that you correctly wrote as
$E_{p_0} +E_{k_0} = E_{p_f} +E_{k_f}$
and expand each term, including $E_{p_0}$. By the way, you have already written the correct expression for potential energy
$\frac {kq_1q_2}r$
and all the information you need to calculate $E_{p_0}$ is given in the problem statement.

@TSny alluded to another issue you need to address - you are asked to find the minimum distance between the pucks. Here is a hint about that. Let $|x_1(t) - x_2(t)|$ be the distance between puck 1 and puck 2. Minimize the distance wrt to time t. (Remember that when the distance is at a minimum, the squared distance is also at a minimum.) What does that tell you about the velocities of the pucks when the distance is at a minimum? You should also discover that there is a possibility that the pucks will collide and you will need to decide whether that can happen.

 

[chef99]

You are getting closer.
You have dropped a minus here:

Also, you have not correctly used conservation of energy here.

First of all $E = E_{k_1} + E_{k_2}$ is not the correct equation for conservation of energy, or even for the total energy.
Second, you have assumed that the potential energy at your point of closest approach is equal to the kinetic energy there. That is not true either. I suggest that you go back to the equation for conservation of energy that you correctly wrote as
$E_{p_0} +E_{k_0} = E_{p_f} +E_{k_f}$
and expand each term, including $E_{p_0}$. By the way, you have already written the correct expression for potential energy
$\frac {kq_1q_2}r$
and all the information you need to calculate $E_{p_0}$ is given in the problem statement.

@TSny alluded to another issue you need to address - you are asked to find the minimum distance between the pucks. Here is a hint about that. Let $|x_1(t) - x_2(t)|$ be the distance between puck 1 and puck 2. Minimize the distance wrt to time t. (Remember that when the distance is at a minimum, the squared distance is also at a minimum.) What does that tell you about the velocities of the pucks when the distance is at a minimum? You should also discover that there is a possibility that the pucks will collide and you will need to decide whether that can happen.

Thank you for pointing out the error in the conservation of momentum equation. I have been looking through my textbook and there is a practice question with an alpha particle and a proton moving towards each other. These are they equations they used, which seem to me to be similar to the ones you were talking about.

Use Conservation of momentum to find the velocity of each mass at minimum separation. The two masses must have the same velocity at the point of minimum separation.

Let right [East] be positive.

P_to = P_tf

m₁v_1o + m₂v_2o = (m₁ + m₂)v_fv_f = m₁v₁₀ + m₂v₂₀ / m₁ + m₂

v_f = (0.80kg)(-12.0m/s) + (0.40kg)(8.0m/s) / (0.80kg) + (0.40kg)

v_f = -5.33m/sNow use conservation of energy to determine the distance of minimum approach. E_ko + E_Po = E_Pf + E_kf

kq₁q₂ /r_o + (1/2 m₁v_1o²+1/2 m₂v_2o² = kq₁q₂ /r_f + 1/2 (m₁+m₂)(v_f)²

(9.0 x10⁹)(3.0 x10^-4)(3.0 x10^-4) / (20.0m) + {1/2(0.80)(-12.0m/s)² + 1/2 (0.40)(8.0m/s)²} = (9.0 x10⁹)(3.0 x10^-4)(3.0 x10^-4) / r_f + 1/2 (0.80 + 0.40)(-5.33)²

40.5 J + 70.4 J = 810 / r_f + 17.05 J

110.9 J - 17.05 J = 810 / r_f

93.85 J = 810 / r_f

r_f = 810 / 93.85 J

r_f = 8.63mThe minimum distance between the pucks is 8.63m.

This seems to make sense, the one thing I'm not entirely sure about is way E_kf is written in this. It makes sense if the final velocities are they same, but I'm not positive that's the case.

 

[tnich]

This seems to make sense, the one thing I'm not entirely sure about is way E_kf is written in this. It makes sense if the final velocities are they same, but I'm not positive that's the case.

I gave you a hint about finding the relationship between the final velocities. Have you tried it?

Let $|x_1(t) - x_2(t)|$ be the distance between puck 1 and puck 2. Minimize the distance wrt to time t. (Remember that when the distance is at a minimum, the squared distance is also at a minimum.) What does that tell you about the velocities of the pucks when the distance is at a minimum? You should also discover that there is a possibility that the pucks will collide and you will need to decide whether that can happen.