# Two protons are released from rest when they are 0.730 nm apart. A) What is the maximum speed...

## Question:

Two protons are released from rest when they are 0.730 nm apart.

A) What is the maximum speed they will reach?

B) What is the maximum acceleration they will achieve?

## Energy Conservation:

When we make a system isolated from its environment, the energy it already possessed when we did this becomes its total energy because no additional energy can flow in or out. So when the total energy is constant, we can use the conservation of energy principle to determine kinetic and potential energies. This is because here, the sum of the kinetic and potential energy at any point must always equal the constant total. We can thus calculate changes in kinetic energy if we just know changes in potential energy and vice-versa.

Given:

• {eq}\displaystyle r = 0.73\ nm = 7.3\ \times\ 10^{-10}\ m {/eq} is the distance between the protons

(a) If we use the conservation of energy principle and assign K as the kinetic energy and U as the potential energy, we can write:

{eq}\displaystyle K_1 + U_1 = K_2 + U_2 {/eq}

Before the protons are released, there is no kinetic energy. And when the protons are infinitely far from each other, then there is no potential energy. We thus reduce this to:

{eq}\displaystyle U_1 = K_2 {/eq}

We expand these terms as:

{eq}\displaystyle \frac{kq_1 q_2}{r} = \frac{1}{2} mv^2 + \frac{1}{2} mv^2 {/eq}

We note that due to repulsive forces, both protons will move away from each other at the same speed and hence we have to divide the total energy between them. We thus have:

{eq}\displaystyle mv^2 = \frac{kq_1 q_2}{r} {/eq}

We isolate the speed v:

{eq}\displaystyle v = \sqrt{\frac{kq_1 q_2}{mr}} {/eq}

Now for a proton, we have:

• {eq}\displaystyle m = 1.67\ \times\ 10^{-27}\ kg {/eq}
• {eq}\displaystyle q = 1.602\ \times\ 10^{-19}\ C {/eq}

We substitute:

{eq}\displaystyle v = \sqrt{\frac{(9\ \times\ 10^9\ Nm^2/C^2)(1.602\ \times\ 10^{-19}\ C)^2}{(1.67\ \times\ 10^{-27}\ kg)(7.3\ \times\ 10^{-10}\ m)}} {/eq}

We thus get:

{eq}\displaystyle \boxed{v = 13,765\ m/s} {/eq}

(b) In general, electric force between the two protons is given as:

{eq}\displaystyle F_E = \frac{kq^2}{r^2} {/eq}

So here we use Newton's second law to expand the force as:

{eq}\displaystyle ma = \frac{kq^2}{r^2} {/eq}

We isolate the acceleration as:

{eq}\displaystyle a = \frac{kq^2}{mr^2} {/eq}

So here let us recall that due to the inverse square law, the value of a is highest at the lowest r which is their release point. We thus substitute:

{eq}\displaystyle a = \frac{(9\ \times\ 10^9\ Nm^2/C^2)(1.602\ \times\ 10^{-19}\ C)^2}{(1.67\ \times\ 10^{-27}\ kg)(7.3\ \times\ 10^{-10}\ m)^2} {/eq}

We get:

{eq}\displaystyle \boxed{a = 2.6\ \times\ 10^{17}\ m/s^2} {/eq} 