Examples of Kinetic Energy

What is Kinetic Energy

The kinetic energy, K, is defined as the energy stored in an object because of its motion. An object in motion has the ability to do work and thus can be said to have energy. It is called kinetic energy, from the Greek word kinetikos, meaning “motion.”

The kinetic energy depends on the speed of an object and is the ability of a moving object to do work on other objects when it collides with them. On the other hand, the kinetic energy of an object represents the amount of energy required to increase the velocity of the object from rest (v = 0) to its final velocity. The kinetic energy also depends linearly on the mass, which is a numerical measure of object’s inertia and the measure of an object’s resistance to acceleration when a force is applied.

We define the quantity:

K = ½ mv2

to be the translational kinetic energy of the object. It must be added, it is called the “translational” kinetic energy to distinguish it from rotational kinetic energy.

Examples of Kinetic Energy

Kinetic energy of fission fragments
As can be seen when the compound nucleus splits, it breaks into two fission fragments. In most cases, the resultant fission fragments have masses that vary widely, but the most probable pair of fission fragments for the thermal neutron-induced fission of the 235U have masses of about 94 and 139.

The largest part of the energy produced during fission (about 80 % or about 170 MeV or about 27 picojoules) appears as kinetic energy of the fission fragments. The fission fragments interact strongly (intensely) with the surrounding atoms or molecules traveling at high speed, causing them to ionize. Creation of ion pairs requires energy, which is lost from the kinetic energy of the charged fission fragment causing it to decelerate. The positive ions and free electrons created by the passage of the charged fission fragment will then reunite, releasing energy in the form of heat (e.g. vibrational energy or rotational energy of atoms).

The range of these massive, highly charged particles in the fuel is of the order of micrometers, so that the recoil energy is effectively deposited as heat at the point of fission. This is the principle how fission fragments heat up fuel in the reactor core.

Kinetic energy of prompt neutrons
Prompt neutrons are emitted directly from fission and they are emitted within very short time of about 10-14 second. Usually more than 99 percent of the fission neutrons are the prompt neutrons, but the exact fraction is dependent on the nuclide to be fissioned and is also dependent on an incident neutron energy (usually increases with energy).

For example a fission of 235U by thermal neutron yields 2.43 neutrons, of which 2.42 neutrons are the prompt neutrons and 0.01585 neutrons (0.01585/2.43=0.0065=ß) are the delayed neutrons. Almost all prompt fission neutrons have energies between 0.1 MeV and 10 MeV. The mean neutron energy is about 2 MeV. The most probable neutron energy is about 0.7 MeV.

Most of this energy is deposited in the coolant (moderator), because the water have the highest macroscopic slowing down power (MSDP) of the materials that are in a reactor core (PWR). The range of neutrons in a reactor depends strongly on certain reactor type, in the case of PWRs it is usually of the order of centimeters.

Kinetic Energy in Elastic Nuclear Collision
A neutron (n) of mass 1.01 u traveling with a speed of 3.60 x 104m/s interacts with a carbon (C) nucleus (mC = 12.00 u) initially at rest in an elastic head-on collision.

What are the velocities of the neutron and carbon nucleus after the collision?

Solution:

This is an elastic head-on collision of two objects with unequal masses. We have to use the conservation laws of momentum and of kinetic energy, and apply them to our system of two particles.

We can solve this system of equation or we can use the equation derived in previous section. This equation stated that the relative speed of the two objects after the collision has the same magnitude (but opposite direction) as before the collision, no matter what the masses are.

The minus sign for v’ tells us that the neutron scatters back of the carbon nucleus, because the carbon nucleus is significantly heavier. On the other hand its speed is less than its initial speed. This process is known as the neutron moderation and it significantly depends on the mass of moderator nuclei.

Block sliding down a frictionless incline slope

The 1 kg block starts out a height H (let say 1 m) above the ground, with potential energy mgH and kinetic energy that is equal to 0. It slides to the ground (without friction) and arrives with no potential energy and kinetic energy K = ½ mv2. Calculate the velocity of the block on the ground and its kinetic energy.

Emech = U + K = const

=> ½ mv2 = mgH

=> v = √2gH = 4.43 m/s

=> K2 = ½ x 1 kg x (4.43 m/s)2 = 19.62 kg.m2.s-2 = 19.62 J

Pendulum

Assume a pendulum (ball of mass m suspended on a string of length L that we have pulled up so that the ball is a height H < L above its lowest point on the arc of its stretched string motion. The pendulum is subjected to the conservative gravitational force where frictional forces like air drag and friction at the pivot are negligible.

We release it from rest. How fast is it going at the bottom?

The pendulum reaches greatest kinetic energy and least potential energy when in the vertical position, because it will have the greatest speed and be nearest the Earth at this point. On the other hand, it will have its least kinetic energy and greatest potential energy at the extreme positions of its swing, because it has zero speed and is farthest from Earth at these points.

If the amplitude is limited to small swings, the period T of a simple pendulum, the time taken for a complete cycle, is:

where L is the length of the pendulum and g is the local acceleration of gravity. For small swings the period of swing is approximately the same for different size swings. That is, the period is independent of amplitude.

Example: Proton’s kinetic energy

A proton (m = 1.67 x 10-27 kg) travels at a speed v = 0.9900c = 2.968 x 108m/s. What is its kinetic energy?

According to a classical calculation, which is not correct, we would obtain:

K = 1/2mv2 = ½ x (1.67 x 10-27 kg) x (2.968 x 108m/s)2 = 7.355 x 10-11 J

With relativistic correction the relativistic kinetic energy is equal to:

K = (ɣ – 1)mc2

where the Lorentz factor

ɣ = 7.089

therefore

K = 6.089 x (1.67 x 10-27 kg) x (2.9979 x 108m/s)2 = 9.139 x 10-10 J = 5.701 GeV

This is about 12 times higher energy as in the classical calculation. According to this relationship, an acceleration of a proton beam to 5.7 GeV requires energies that are in the order different.

References:
Reactor Physics and Thermal Hydraulics:
1. J. R. Lamarsh, Introduction to Nuclear Reactor Theory, 2nd ed., Addison-Wesley, Reading, MA (1983).
2. J. R. Lamarsh, A. J. Baratta, Introduction to Nuclear Engineering, 3d ed., Prentice-Hall, 2001, ISBN: 0-201-82498-1.
3. W. M. Stacey, Nuclear Reactor Physics, John Wiley & Sons, 2001, ISBN: 0- 471-39127-1.
4. Glasstone, Sesonske. Nuclear Reactor Engineering: Reactor Systems Engineering, Springer; 4th edition, 1994, ISBN: 978-0412985317
5. Todreas Neil E., Kazimi Mujid S. Nuclear Systems Volume I: Thermal Hydraulic Fundamentals, Second Edition. CRC Press; 2 edition, 2012, ISBN: 978-0415802871
6. Zohuri B., McDaniel P. Thermodynamics in Nuclear Power Plant Systems. Springer; 2015, ISBN: 978-3-319-13419-2
7. Moran Michal J., Shapiro Howard N. Fundamentals of Engineering Thermodynamics, Fifth Edition, John Wiley & Sons, 2006, ISBN: 978-0-470-03037-0
8. Kleinstreuer C. Modern Fluid Dynamics. Springer, 2010, ISBN 978-1-4020-8670-0.
9. U.S. Department of Energy, THERMODYNAMICS, HEAT TRANSFER, AND FLUID FLOW. DOE Fundamentals Handbook, Volume 1, 2 and 3. June 1992.

Kinetic Energy