00:01
In this problem, we are given that there are two protons and they are initially separated by a distance of 0 .75 nanometers, which is equal to 0 .75 times 10 raised to minus 9 meters.
00:18
And they are initially at rest and they are released.
00:23
And we are required to determine the maximum speed, what they will reach, and along with that, we have to also determine the maximum acceleration which they will attain.
00:36
So this is part b, where we will compute the maximum acceleration, and in part a, we will compute the maximum speed which they reach.
00:45
And we see that these two protons, they are containing similar charges because both are positive, so they will experience repulsive force here.
00:54
And we apply energy conservation, and we see that if the maximum speed, that's attained by each of these particles is v, then the total kinetic energy, it will be because of the electrostatic potential energy which is stored when the system is at rest.
01:12
And we use this expression to get the electrostatic potential energy of a system of two charges, and this will be transformed into kinetic energy that will be carried by each of these particles.
01:25
So k times the square of the charge carried by the proton divided by the initial, the initial, separation this will be equal to two times the kinetic energy carried by each of these particles.
01:38
And now let's substitute the values here...