Two protons are released from rest when they are 0.785 nm apart. (A) What is the maximum speed...

Question:

Two protons are released from rest when they are 0.785 nm apart.

(A) What is the maximum speed they will reach?

(B) What is the maximum acceleration they will achieve?

Proton:

The proton is the particle that carries one unit of positive charge. Now as compared to the electron, the proton is heavier by four orders of magnitude and carries positive charge instead of the electron's negative charge. Moreover, protons are not fundamental particles because they are actually made up of three quarks, namely two "up" quarks and one "down" quark.

Answer and Explanation:


Given:

  • {eq}\displaystyle r = 0.785\ nm = 7.85\ \times\ 10^{-10}\ m {/eq} is the distance between the protons


(a) The use of the conservation of energy principle with assignments of the variables K for kinetic energies and U for potential energies allows us to set up the equation:

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

The kinetic energy is zero before the protons are released and the potential energies are zero when the protons are infinitely far from one another. We can thus reduce this equation 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)(7.85\ \times\ 10^{-10}\ m)}} {/eq}

We thus get:

{eq}\displaystyle \boxed{v = 13,274\ m/s} {/eq}


(b) Here we start with the Coulomb force:

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

And we expand the force using Newton's definition as:

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

We isolate the acceleration as:

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

Since a is inversely proportional to r, then this means that the lowest r value gives our highest a value. This r value is our release point. We 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)(7.85\ \times\ 10^{-10}\ m)^2} {/eq}

We get:

{eq}\displaystyle \boxed{a = 2.24\ \times\ 10^{17}\ m/s^2} {/eq}


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