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# Could you help me with Physics II?

Two protons are fixed and cannot move on the x axis. One at x =a and the other at x = -a. Let a = 1.5 * 1-^-14

An alpha [article which is a helium nucleus (q = +2e, m = 6.64 *10^-27) is moving straight down along the y axis. When the protons are very far from each other the alpha particle had a speed of 2.5 * 10^6m/s

Solve for the closest y value (y min) and the closest y axis distance the alpha article could get from the origin?

### 1 Answer

- 2 months agoFavorite Answer
This question is most easily solved using conservation of energy.

First, we can check if the alpha particle can actually make it all the way to the origin. To do this, we first need to calculate its initial kinetic energy. They give us the mass and velocity, so we just use K = 1/2 mv^2.

K = 1/2 mv^2 = 1/2 * 6.64*10^-27 kg * (2.5 * 10^6 m/s)^2 = 2.075*10^-24 J

Now we can use the formula for voltage to determine the energy in a state where the alpha particle got exactly to the origin and then stopped moving. Because the setup is symmetrical, we can just find the voltage for one proton and then double it:

V = 2 * (9*10^9 Nm^2/C^2)(1.602*10^-19 C)/(1.5*10^-14 m) = 1.922*10^5 V

Recall the definition of voltage: U = qV. Use this to find the potential energy in the final state:

U = 2*(1.602*10^-19 C)*(1.922*10^5 J/C) = 6.158*10^-14 J

The potential energy in the final state is less than the initial kinetic energy, the alpha particle does not have enough energy to get to the origin, so it must stop somewhere along the way.

In order to see where exactly it stops, we need to find the potential energy at any given y value. First, we need to get the distance from either proton at any given y distance.

We can construct a right triangle here with the x distance on one leg and the unknown y distance on the other. We know that the x distance from the alpha particle to either proton is equal to a, so the total distance is r = sqrt(a^2 + y^2).

Plugging this into our voltage formula gives us

V = 2 * (9*10^9 Nm^2/C^2)(1.602*10^-19 C)/r = 2.884*10^-9/sqrt(a^2+y^2) Nm^2/C

Now multiplying by the charge of the alpha particle gives us this:

U = 2*(1.602*10^-19 C)*V = 9.239*10^-28/sqrt(a^2+y^2) J*m

Now, equate the final energy U to the initial energy K and solve for y:

9.239*10^-28/sqrt(a^2+y^2) J*m = 2.075*10^-24 J

sqrt(a^2 + y^2) = 9.239*10^-28 J*m/2.075*10^-24 J = 3.360*10^-4 m

y = sqrt((3.360*10^-4 m)^2 - (1.5*10^-14 m)^2) = 3.360*10^-4 m

(a is very small compared to 3.360 * 10^-4 here, so its effect is negligible)

The alpha particle will be at rest when y = 3.36*10^-4 m, so after it gets that far it will turn around and start accelerating back up the y-axis.