## Law of Conservation of Momentum

**In general**, the **law of conservation of momentum** or principle of momentum conservation states that the momentum of an isolated system is a **constant**. The vector sum of the momenta (momentum is equal to the mass of an object multiplied by its velocity) of all the objects of a system cannot be changed by interactions within the system. In classical mechanics, this law is implied by **Newton’s laws**. This principle is a direct consequence of Newton’s third law.

Let us assume the **one dimensional elastic collision** of two billiard balls,, the ball A and the ball B. We assume the net external force on this system of two balls is zero—that is, the only significant forces during the collision are the forces that each ball exerts on the other. According to **Newton’s third law**, the two forces are always equal in magnitude and opposite in direction. Hence, the impulses that act on the two balls are equal and opposite, and the changes in momentum of the two balls are equal and opposite.

These two balls are moving with velocities **v**** _{A}** and

**v**

**along the x axis before the collision. After the collision, their velocities are**

_{B}**v’**

**and**

_{A}**v’**

**. The**

_{B}**conservation of the total momentum**demands that the total momentum before the collision is the same as the total momentum after the collision.

**The law of conservation of linear momentum** in words:

*If no net external force acts on a system of particles, the total linear momentum of the system cannot change.*

## Conservation of Momentum and Energy in Collisions

The use of the **conservation laws for momentum and energy** is very important also in **particle collisions**. This is a very powerful rule because it can allow us to determine the results of a collision without knowing the details of the collision. The law of **conservation of momentum** states that in the collision of two objects such as billiard balls, the **total momentum is conserved**. The assumption of conservation of momentum as well as the conservation of kinetic energy makes possible the calculation of the **final velocities** in two-body collisions. At this point we have to distinguish between two types of collisions:

## Elastic Collisions

**perfectly elastic collision**is defined as one in which there is

**no net conversion of kinetic energy**into other forms (such as heat or noise). For the brief moment during which the two objects are in contact, some (or all) of the energy is stored momentarily in the form of

**elastic potential energy**. But if we compare the total kinetic energy just before the collision with the total kinetic energy just after the collision, and they are found to be the same, then we say that the

**total kinetic energy is conserved**.

- Some large-scale interactions like the
**slingshot type gravitational interactions**(also known as a planetary swing-by or a gravity-assist manoeuvre) between satellites and planets are**perfectly elastic**. - Collisions between
**very hard spheres**may be**nearly elastic**, so it is useful to calculate the limiting case of an elastic collision. - Collisions in
**ideal gases**approach perfectly elastic collisions, as do scattering interactions of sub-atomic particles which are deflected by the electromagnetic force. **Rutherford scattering**is the elastic scattering of charged particles also by the electromagnetic force.- A
**neutron-nucleus scattering reaction**may be also elastic, but in this case the neutron is deflected by the strong nuclear force.

**one dimensional elastic collision**of two objects, the object A and the object B. These two objects are moving with velocities

**v**and

_{A}**v**along the x axis before the collision. After the collision, their velocities are

_{B}**v’**and

_{A}**v’**. The

_{B}**conservation of the total momentum**demands that the total momentum before the collision is the same as the total momentum after the collision. Likewise, the

**conservation of the total kinetic energy**, which demands that the total kinetic energy of both objects before the collision is the same as the total kinetic energy after the collision. Both law may be expressed in equations as:

**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**.

## Inelastic Collisions

**An inelastic collision**is one in which part of the

**kinetic energy is changed**to some other form of energy in the collision. Any macroscopic collision between objects will convert some of the kinetic energy into

**internal energy**and other forms of energy, so

**no large scale impacts are perfectly elastic**. For example, in collisions of common bodies, such as two cars, some energy is always transferred from

**kinetic energy**to other forms of energy, such as

**thermal energy**or

**energy of sound**. The inelastic collision of two bodies always involves a loss in the kinetic energy of the system. The greatest loss occurs if the bodies stick together, in which case the collision is called a

**completely inelastic collision**. Thus, the

**kinetic energy**of the system is

**not conserved**, while the

**total energy is conserved**as required by the general principle of conservation of energy.

**Momentum is conserved in inelastic collisions**, but one cannot track the kinetic energy through the collision since some of it is converted to other forms of energy.

**In nuclear physics**, an inelastic collision is one in which the incoming particle causes the nucleus it strikes to become excited or to break up. Deep inelastic scattering is a method of probing the structure of subatomic particles in much the same way as Rutherford probed the inside of the atom (see Rutherford scattering).

**In nuclear reactors**, inelastic collisions are of importance in **neutron moderation** process. An inelastic scattering plays an important role in slowing down neutrons especially **at high energies and by heavy nuclei**. Inelastic scattering occurs above **a threshold energy**. This threshold energy is higher than the energy the first excited state of target nucleus (due to the laws of conservation) and it is given by following formula:

**E**_{t}** = ((A+1)/A)* ε**_{1}

where **E**** _{t}** is known as the

**inelastic threshold energy**and

**ε**

**is the energy of the first excited state. Therefore especially scattering data of**

_{1}

^{238}**U**, which is a major component of nuclear fuel in commercial power reactors, are one of the most important data in the neutron transport calculations in the reactor core.

**A ballistic pendulum** is a device for measuring the velocity of a projectile, such as a **bullet**. The ballistic pendulum is a kind of “transformer,” exchanging the high speed of a light object (the bullet) for the low speed of a massive object (the block). When a bullet is fired into the block, its momentum is transferred to the block. The bullet’s momentum can be determined from the** amplitude of the pendulum swing**.

When the bullet is embedding itself in the block, it occurs so quickly that the block does not move appreciably. The supporting strings remain nearly vertical, so negligible external horizontal force acts on the bullet–block system, and the** horizontal** component of **momentum is conserved**. **Mechanical energy is not conserved** during this stage, however, because a **nonconservative force** does work (the force of friction between bullet and block).

In the second stage, the bullet and block move together. The only forces acting on this system are gravity (a conservative force) and the string tensions (which do no work). Thus, as the block swings, **mechanical energy is conserved**. **Momentum is not conserved during this stage**, however, because there is a net external force (the forces of gravity and string tension don’t cancel when the strings are inclined).

**Equations governing the ballistic pendulum**

**Equations governing the ballistic pendulum**

In the first stage **momentum is conserved** and therefore:

where** v** is the initial velocity of the projectile of mass **m _{P}**.

**v’**is the velocity of the block and embedded projectile (both of mass

**m**) just after the collision, before they have moved significantly.

_{P}+ m_{B}In the second stage **mechanical energy is conserved**. We choose y = 0 when the pendulum hangs vertically, and then y = h when the block and embedded projectile system reaches its maximum height. The system swings up and comes to rest for an instant at a height y, where its kinetic energy is zero and the potential energy is **(m _{P} + m_{B})gh**. Thus we write the law of conservation of energy:

which is the initial velocity of the projectile and our final result.

**When we use some realistic numbers:**

- m
_{P}= 5 g - m
_{B}= 2 kg - h = 3 cm
- v = ?

**then we have:**

## Conservation Laws in Nuclear Reactions

**nuclear reaction**is considered to be the process in which two nuclear particles (two nuclei or a nucleus and a nucleon) interact to produce two or more nuclear particles or ˠ-rays (gamma rays). Thus, a

**nuclear reaction**must cause a transformation of at least one nuclide to another. Sometimes if a nucleus interacts with another nucleus or particle without changing the nature of any nuclide, the process is referred to a

**nuclear scattering**, rather than a nuclear reaction.

In analyzing nuclear reactions, we apply the **many conservation laws**. **Nuclear reactions** are subject to classical **conservation laws for charge, momentum, angular momentum, and energy** (including rest energies). Additional conservation laws, not anticipated by classical physics, are are **electric charge**, **lepton number and baryon number**. Certain of these laws are obeyed under all circumstances, others are not. We have accepted conservation of energy and momentum. In all the examples given we assume that the number of protons and the number of neutrons is separately conserved. We shall find circumstances and conditions in which this rule is not true. Where we are considering non-relativistic nuclear reactions, it is essentially true. However, where we are considering relativistic nuclear energies or those involving the weak interactions, we shall find that these principles must be extended.

Some conservation principles have arisen from theoretical considerations, others are just empirical relationships. Notwithstanding, any reaction not expressly forbidden by the conservation laws will generally occur, if perhaps at a slow rate. This expectation is based on quantum mechanics. Unless the barrier between the initial and final states is infinitely high, there is always a **non-zero probability** that a system will make the transition between them.

For purposes of this article it is sufficient to note four of the fundamental laws governing these reactions.

**Conservation of nucleons**. The total number of nucleons before and after a reaction are the same.**Conservation of charge**. The sum of the charges on all the particles before and after a reaction are the same**Conservation of momentum**. The total momentum of the interacting particles before and after a reaction are the same.**Conservation of energy**. Energy, including rest mass energy, is conserved in nuclear reactions.

**nuclear reactions**is enormous, nuclear reactions can be sorted by types. Most of nuclear reactions are accompanied by gamma emission. Some examples are:

**Elastic scattering**. Occurs, when no energy is transferred between the target nucleus and the incident particle.

** 208Pb (n, n) 208Pb**

**Inelastic scattering**. Occurs, when energy is transferred. The difference of kinetic energies is saved in excited nuclide.

** 40Ca (α, α’) 40mCa**

**Capture reactions**. Both charged and neutral particles can be captured by nuclei. This is accompanied by the emission of ˠ-rays. Neutron capture reaction produces radioactive nuclides (induced radioactivity).

** 238U (n, ˠ) 239U**

**Transfer Reactions**. The absorption of a particle accompanied by the emission of one or more particles is called the transfer reaction.

**4He (α, p) 7Li**

**Fission reactions**. Nuclear fission is a nuclear reaction in which the nucleus of an atom splits into smaller parts (lighter nuclei). The fission process often produces free neutronsand photons (in the form of gamma rays), and releases a large amount of energy.

**235U (n, 3 n) fission products**

**Fusion reactions**. Occur when, two or more atomic nuclei collide at a very high speed and join to form a new type of atomic nucleus.The fusion reaction of deuterium and tritium is particularly interesting because of its potential of providing energy for the future.

**3T (d, n) 4He**

**Spallation reactions**. Occur, when a nucleus is hit by a particle with sufficient energy and momentum to knock out several small fragments or, smash it into many fragments.

**Nuclear decay**(**Radioactive decay**). Occurs when an unstable atom loses energy by emitting ionizing radiation. Radioactive decay is a random process at the level of single atoms, in that, according to quantum theory, it is impossible to predict when a particular atom will decay. There are many types of radioactive decay:**Alpha radioactivity**. Alpha particles consist of two protons and two neutrons bound together into a particle identical to a helium nucleus. Because of its very large mass (more than 7000 times the mass of the beta particle) and its charge, it heavy ionizes material and has a very short range.

**Beta radioactivity**. Beta particles are high-energy, high-speed electrons or positrons emitted by certain types of radioactive nuclei such as potassium-40. The beta particles have greater range of penetration than alpha particles, but still much less than gamma rays.The beta particles emitted are a form of ionizing radiation also known as beta rays. The production of beta particles is termed beta decay.

**Gamma radioactivity**. Gamma rays are electromagnetic radiation of an very high frequency and are therefore high energy photons. They are produced by the decay of nuclei as they transition from a high energy state to a lower state known as gamma decay. Most of nuclear reactions are accompanied by gamma emission.

**Neutron emission.**Neutron emission is a type of radioactive decay of nuclei containing excess neutrons (especially fission products), in which a neutron is simply ejected from the nucleus. This type of radiation plays key role in nuclear reactor control, because these neutrons are delayed neutrons.

## Conservation of Momentum in Nuclear Reactions – Neutron Moderation

See also: Neutron Inelastic Scattering

See also: Neutron Moderators

It is known the fission neutrons are of importance in any chain-reacting system. All neutrons produced by fission are born as **fast neutrons** with high kinetic energy. Before such neutrons can efficiently cause additional fissions, they must be slowed down by collisions with nuclei in the moderator of the reactor. The probability of the fission U-235 becomes very large **at the thermal energies** of slow neutrons. This fact implies increase of multiplication factor of the reactor (i.e. lower fuel enrichment is needed to sustain chain reaction).

The neutrons released during fission with an average energy of **2 MeV** in a reactor on average undergo a **number of collisions** (elastic or inelastic) before they are absorbed. During the scattering reaction, a fraction of the neutron’s kinetic energy **is transferred to the nucleus**. Using the laws of **conservation of momentum and energy** and the analogy of collisions of billiard balls for elastic scattering, it is possible to derive the following equation for the mass of target or moderator nucleus (M), energy of incident neutron (E_{i}) and the energy of scattered neutron (E_{s}).

where A is the atomic mass number.In case of the **hydrogen (A = 1)** as the target nucleus, the incident neutron **can be completely stopped**. But this works when the direction of the neutron is completely reversed (i.e. scattered at 180°). In reality, the direction of scattering ranges from 0 to 180 ° and the energy transferred also ranges from 0% to maximum. Therefore, the average energy of scattered neutron is taken as the average of energies with scattering angle 0 and 180°.

Moreover, it is useful to work **with logarithmic quantities** and therefore one defines **the logarithmic energy decrement per collision (ξ)** as a key material constant describing energy transfers during a neutron slowing down. ξ is not dependent on energy, only on A and is defined as follows:For heavy target nuclei,** ξ** may be approximated by following formula:From these equations it is easy to determine the number of collisions required to slow down a neutron from, for example from **2 MeV to 1 eV**.

Example: Determine the number of collisions required for thermalization for the 2 MeV neutron in the carbon.

ξ_{CARBON} = 0.158

N(**2MeV → 1eV**) = ln 2⋅10^{6}/ξ =14.5/0.158 = **92**

For a mixture of isotopes:

**A neutron (n)**of mass

**1.01 u**traveling with a speed of

**3.60 x 10**interacts with a

^{4}m/s**carbon (C)**nucleus (

**m**) initially at rest in an

_{C}= 12.00 u**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.

## Momentum of a Photon

**A photon, the quantum of electromagnetic radiation**, is an elementary particle, which is the force carrier of the electromagnetic force. The modern photon concept was developed (1905) by

**Albert Einstein**to explain of the photoelectric effect, in which he proposed the existence of discrete energy packets during the transmission of light.

In 1916, Einstein extended his concept of light quanta (photons) by proposing that a **quantum of light has linear momentum**. Although a photon is **massless**, it has momentum, which is related to its energy E, frequency f, and wavelength by:

Thus, when a photon interacts with another object, **energy and momentum are transferred**, as if there were a collision between the photon and matter in the classical sense.

**10**(let λ = 650 nm) emitted per second from the

^{19}photons**100 W lightbulb**. Suppose all photons are focused onto a piece of black paper and absorbed. Assume that the momentum of a photon changes

**from p = h/λ to zero**.

**Calculate the momentum** of one photon and calculate the **force** all these photons could exert on the paper.

**Solution:**

We use the formula of momentum of a single photon:

## Momentum of a Photon – Compton Scattering

The Compton formula was published in 1923 in the Physical Review. Compton explained that the X-ray shift is caused by particle-like momentum of photons. Compton scattering formula is the mathematical relationship between the shift in wavelength and the scattering angle of the X-rays. In the case of Compton scattering the photon of frequency *f* collides with an electron at rest. Upon collision, the photon bounces off electron, giving up some of its initial energy (given by Planck’s formula E=hf), While the electron gains momentum (mass x velocity), the **photon cannot lower its velocity**. As a result of momentum conservation law, the photon must lower its momentum given by:

So the decrease in photon’s momentum must be translated into **decrease in frequency** (increase in wavelength Δ**λ = λ’ – λ**). The shift of the wavelength increased with scattering angle according to **the Compton formula**:

where **λ** is the initial wavelength of photon **λ’** is the wavelength after scattering, **h **is the Planck constant = 6.626 x 10^{-34} J.s, **m _{e}** is the electron rest mass (0.511 MeV)

**c**is the speed of light

**Θ**is the scattering angle. The minimum change in wavelength (

*λ′*−

*λ*) for the photon occurs when Θ = 0° (cos(Θ)=1) and is at least zero. The maximum change in wavelength (

*λ′*−

*λ*) for the photon occurs when Θ = 180° (cos(Θ)=-1). In this case the photon transfers to the electron as much momentum as possible. The maximum change in wavelength can be derived from Compton formula:

The quantity h/m_{e}c is known as the **Compton wavelength** of the electron and is equal to **2.43×10 ^{−12}**

**m**.

## Conservation of Momentum in Fluid Dynamics

**In general**, the

**law of conservation of momentum**or principle of momentum conservation states that the momentum of an isolated system is a

**constant**. The vector sum of the momenta (momentum is equal to the mass of an object multiplied by its velocity) of all the objects of a system cannot be changed by interactions within the system. In classical mechanics, this law is implied by

**Newton’s laws**.

**In fluid dynamics**, the analysis of motion is performed in the same way as in solid mechanics – by use of **Newton’s laws of motion**. As can be seen moving fluids are exerting forces. The lift force on an aircraft is exerted by the air moving over the wing.

A jet of water from a hose exerts a force on whatever it hits. But here it is not clear **what mass of moving fluid** we should use and therefore it is necessary to use a different form of the equation.

**Newton’s 2nd Law can be written:**

The rate of change of momentum of a body is equal to the resultant force acting on the body, and takes place in the direction of the force.

**steady**and

**incompressible**. To determine the rate of change of momentum for a fluid we will consider a streamtube (

**control volume**) as we did for the Bernoulli equation. In this

**control volume**any change in momentum of the fluid within a control volume is due to the action of external forces on the fluid within the volume.

As can be seen from the picture the **control volume method** can be used to analyze the law of conservation of momentum in fluid. Control volume is an **imaginary surface** enclosing a volume of interest. The control volume can be fixed or moving, and it can be rigid or deformable. In order to determine all forces acting on the surfaces of the control volume we have to solve the conservation laws in this control volume.

The first conservation equation we have to consider in the control volume is the continuity equation (the law of conservation of matter). In the simplest form it is represented by following equation:

∑ṁ_{in} = ∑ṁ_{out}

Sum of mass flow rates entering per unit time = Sum of mass flow rates leaving per unit time

**The second conservation equation we have to consider in the control volume is the momentum formula.**

## Momentum Formula

**momentum formula**can be represented by following equation:

### Choosing a Control Volume

**A control volume can be selected as any arbitrary volume through which fluid flows. This volume can be static, moving, and even deforming during flow. In order to solve any problem we have to solve basic conservation laws in this volume. It is very important to know all relative flow velocities to the control surface and therefore it is very important to define exactly the boundaries of the control volume during an analysis.**

**An elbow**(let say of primary piping) is used to deflect water flow at a velocity of

**17 m/s**. The piping diameter is equal to

**700 mm**. The gauge pressure inside the pipe is about

**16 MPa**at the temperature of 290°C. Fluid is of constant density

**⍴ ~ 720 kg/m**(at 290°C). The angle of the elbow is

^{3}**45°**.

Calculate the **force on the wall** of a deflector elbow (i.e. calculate vector F_{3}).

**Assumptions:**

- The flow is steady.
- The frictional losses are negligible.
- The weight of the elbow is negligible.
- The weight of the water in the elbow is negligible.

We take the elbow as the **control volume**. The control volume is shown at the picture. The momentum equation is a vector equation so it has three components. We take the x- and z- coordinates as shown and we will solve the problem separately according to these coordinates.

First, let us consider the component in the **x-coordinate**. The conservation of linear momentum equation becomes:

Second, let us consider the component in the **y-coordinate**. The conservation of linear momentum equation becomes:

The final force acting on the wall of a deflector elbow will be:

**A stationary plate**(e.g. blade of a watermill) is used to deflect water flow at a velocity of

**1 m/s**and at an angle of

**90°**. It occurs at atmospheric pressure and the mass flow rate is equal to

**Q =1 kg/s**.

**Calculate the pressure force.****Calculate the body force.****Calculate the total force.****Calculate the resultant force.**

**Solution**

- The
**pressure force**is zero as the pressure at both the inlet and the outlets to the control volume are atmospheric. - As the control volume is small we can ignore the
**body force**due to the weight of gravity. **F**= ρ.Q.(w_{x}_{1x}– w_{2x}) = 1000 . 1 . (1 – 0) =**1000 N**

**F**=_{y}**0**

**F**=**(1000, 0)**- The
**resultant force**on the plane is the same magnitude but in the opposite direction as the total force**F**(friction and weight are neglected).

The water jet exerts on the plate the force of 1000 N in the x-direction.

**Nuclear and Reactor Physics:**

- J. R. Lamarsh, Introduction to Nuclear Reactor Theory, 2nd ed., Addison-Wesley, Reading, MA (1983).
- J. R. Lamarsh, A. J. Baratta, Introduction to Nuclear Engineering, 3d ed., Prentice-Hall, 2001, ISBN: 0-201-82498-1.
- W. M. Stacey, Nuclear Reactor Physics, John Wiley & Sons, 2001, ISBN: 0- 471-39127-1.
- Glasstone, Sesonske. Nuclear Reactor Engineering: Reactor Systems Engineering, Springer; 4th edition, 1994, ISBN: 978-0412985317
- W.S.C. Williams. Nuclear and Particle Physics. Clarendon Press; 1 edition, 1991, ISBN: 978-0198520467
- Kenneth S. Krane. Introductory Nuclear Physics, 3rd Edition, Wiley, 1987, ISBN 978-0471805533
- G.R.Keepin. Physics of Nuclear Kinetics. Addison-Wesley Pub. Co; 1st edition, 1965
- Robert Reed Burn, Introduction to Nuclear Reactor Operation, 1988.
- U.S. Department of Energy, Nuclear Physics and Reactor Theory. DOE Fundamentals Handbook, Volume 1 and 2. January 1993.

**Advanced Reactor Physics:**

- K. O. Ott, W. A. Bezella, Introductory Nuclear Reactor Statics, American Nuclear Society, Revised edition (1989), 1989, ISBN: 0-894-48033-2.
- K. O. Ott, R. J. Neuhold, Introductory Nuclear Reactor Dynamics, American Nuclear Society, 1985, ISBN: 0-894-48029-4.
- D. L. Hetrick, Dynamics of Nuclear Reactors, American Nuclear Society, 1993, ISBN: 0-894-48453-2.
- E. E. Lewis, W. F. Miller, Computational Methods of Neutron Transport, American Nuclear Society, 1993, ISBN: 0-894-48452-4.