Solving Linear Motion from Voltage and PE/KE

[boogiebear]

Homework Statement

So I suck at physics, especially when it comes to going from one thing to another, in this case Voltage and PE/KE to linear motion. Anyways here's the question, any hints as to how I should go about to solve it would be much appreciated

A point charge of -10nC is at the origin. Points A and B are respectively 10 and 20 cm away from the point charge (to the right of the charge). The problem first asks for V each at pts A and B, then Delta V between points A and B, then the potential energy each at A and B. The main part of the question I'm having trouble with is as follows:

part d. Calculate the speed at point B of a proton that was moving to the right at point A with a speed of 8 x 10^5 m/s

So i know that change in voltage and change in PE is positive, and that since the point charge is negative the proton slows down as it moves toward point B. However, I'm not too sure where to go from here. So would the increase in PE translate to a decrease in KE??

Homework Equations

KE = .5mv^2
PE = (k * q1 * q2)/r

The Attempt at a Solution

PE at A = (8.99E9)(-10E-9)(1.6E-19)/(.1)
PE at B = (8.99E9)(-10E-9)(1.6E-19)/(.2)

(PE at B) - (PE at A) = 7.19E-17 Joules

Then I subtracted the number I found above from the initial kinetic energy of the moving proton to find KE at point B and solved for Velocity which came out to around 7.44E5 m/s but I have a feeling that this is the wrong answer

 

[anathema]

.

Dear student,

Thank you for reaching out for help with this physics question. It seems like you have made a good start on solving the problem by calculating the potential energy at points A and B. You are correct that the increase in PE would translate to a decrease in KE for the proton as it moves towards point B.

To calculate the speed at point B, you can use the conservation of energy principle. This states that the total energy of a system (in this case, the proton) remains constant, so the initial kinetic energy at point A should be equal to the final kinetic energy at point B, plus any change in potential energy.

So, using the equation for conservation of energy:

KE initial + PE initial = KE final + PE final

Substituting in the values you have calculated for PE at points A and B, and the initial speed of the proton, we get:

(.5mv^2)initial + (PE at A) = (.5mv^2)final + (PE at B)

Rearranging the equation, we can solve for the final velocity at point B:

(.5mv^2)final = (.5mv^2)initial + (PE at B) - (PE at A)

v^2 final = (v^2 initial) + 2(PE at B - PE at A)/m

v final = √[(v^2 initial) + 2(PE at B - PE at A)/m]

Plugging in the values you have calculated, we get:

v final = √[(8 x 10^5 m/s)^2 + 2(7.19E-17 J)/1.67 x 10^-27 kg]

v final = 7.44 x 10^5 m/s

So, it seems like you were on the right track with your calculation, but you may have made a small error in your final calculation. I hope this helps you understand how to approach the problem and get the correct answer. Keep up the good work!

 

[ogb p]

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Hello,

First of all, good job on attempting the problem! It seems like you have a good understanding of the equations and concepts involved. Let's break down the problem and see if we can find the correct solution.

We are given the initial velocity of the proton at point A, which is 8 x 10^5 m/s. We also know that the proton is moving towards the negative point charge, so as you mentioned, it will experience a decrease in kinetic energy and an increase in potential energy as it moves from point A to point B.

To solve for the speed at point B, we can use the conservation of energy principle, which states that the total energy (KE + PE) of a system remains constant. So, at point A, the total energy is equal to the initial kinetic energy of the proton, which is 0.5m(8 x 10^5)^2. At point B, the total energy is equal to the final kinetic energy of the proton plus the potential energy due to the negative point charge.

So we can set up the equation as follows:

0.5m(8 x 10^5)^2 = 0.5mv^2 + (8.99 x 10^9)(-10 x 10^-9)(1.6 x 10^-19)/0.2

Solving for v, we get v = 5.3 x 10^5 m/s. This is the speed of the proton at point B.

I hope this helps! If you have any further questions, don't hesitate to ask. Keep up the good work!