What am I doing wrong? Potential Difference

[kamehamehaaa]

Homework Statement

On attachment. Part B only. I got C.

Homework Equations

mass of proton is 1.673*10^-27
q0 = 1.60*10^-19

The Attempt at a Solution

For part B I thought I could divide the intial speed by 2 and plug it back into the energy conservation equation:

.5mv^2 + q0V1 = .5mv^2 + q0V2 and solve for (V1 - V2) which would be the potential difference but it doesn't seem to work.

 

[Doc Al]

That should work just fine. (Show what you actually did.)

Another way to look at it is: By what fraction must the total energy be decreased to bring the speed to half its initial value? Compare that with the answer you got for part A.

 

[Moetasim]

First of all, it's great that you are actively trying to find a solution for your homework problem. However, it seems like you may have made a mistake in your approach. Let's break down the problem and see where the issue may lie.

The problem states that you need to find the potential difference (V1 - V2) for part B, which involves using the conservation of energy equation .5mv^2 + q0V1 = .5mv^2 + q0V2. This equation is used to relate the kinetic energy of a particle (in this case, a proton) to its potential energy at two different points, V1 and V2.

Your attempt at a solution involves dividing the initial speed by 2 and plugging it back into the energy conservation equation. However, this doesn't make sense because the initial speed (v) is already accounted for in the equation. In other words, you cannot simply divide the initial speed by 2 and substitute it into the equation.

To solve for the potential difference, you need to first determine the kinetic energy of the proton at the two different points, V1 and V2. This can be done by using the mass of the proton (1.673*10^-27 kg) and its charge (q0 = 1.60*10^-19 C). From there, you can rearrange the energy conservation equation to solve for (V1 - V2).

In summary, it seems like the mistake you made was trying to substitute the initial speed into the equation instead of first calculating the kinetic energy at the two different points. I would suggest reviewing the energy conservation equation and making sure you understand how to use it to solve for potential difference. Good luck!