The Concepts of Energy and Momentum Conservation

©Fernando Caracena 17 September 2012

The concept of energy and its conservation is among the key concepts in physics. If you want to hear a great lecture on the subject, check out Richard P. Feynman's  lecture on the Great Conservation Principles.


Newton handed engineers an important tool: equations for developing prototypes as mathematical models rather than building scale models. The Industrial Revolution became possible because calculations based on dynamical models were cheaper to carry out than building expensive prototypes, and this is before the invention of computers. Nowadays, with increasing computer power, it becomes feasible to do most of the designing of mechanical and electronic devices by way of the computer.

The first application that drew on Newtonian Mechanics was to define quantitatively the effort necessary to produce useful motion, such as in a steam engine. The basic idea was that no matter how much force was involved, it had to accomplish some motion, or it was useless. It did nothing. To describe the useful application of a force, physicists defined a quantity called work as the product of a component of force (F) acting in the direction of a displacement (dr) that it produced in an object:

dW = F•dr.                                (1)

Not that this expression represents an infinitesimal amount of work, the work done by a small part of the application of a variable force, which therefore, lends itself to the methods of calculus (see below). Calculus simplifies the mathematics in favor of clearly presenting the physics.

Application of Newton's Laws

When a force accelerates an object, that force does work on that object. Now, let us do some calculations to see how work operates in this case. First, recall how Newton's laws apply to the force on an object of mass m:

                            F = d(m v)/dt ,                            (2a)

which is the most general form of Newton's equations of motion that applies to systems of variable mass, such as rockets and comets. Most of the time, we are interested in objects of constant mass, which simplifies the equations of motion of an object of constant mass, m, to

                              F = m a  .                                    (2b)

The acceleration (a) produced by the vector force (F) is in the same direction as the force and is a variable as the force. For example, the acceleration of gravity (g) is a vector that points straight down, more or less toward the center of the Earth, and the mass times the acceleration of gravity (m g) is equal to the force of gravity.

Using the methods of calculus, we can solve (1) in a case where a force acts on an otherwise free mass. First, consider the following differential relations:

dv = a dt ,                                     (3a)

a vector equation, which represents a set of three equations,

dvi=ai*dt, i=1,2,3                          (3aa)


d r = v dt .                                     (3b)

Inserting (3) in (1), we have

dW = Fv dt.                                 (4a)

We now specialize the equation for a fixed mass object (2b), which substituting for the force in (4), results in the following:

dW = m dvv .                                  (4b)

An integration of (4b) yields

                            W-W0 = (½) m (v2 - v02) .                       (5a)

Note that v2 is a very compact way of indicating the sum of the square of the x, y, z, components of the velocity (labeled as v1,v2,v3 , respectively),

                     v2 =v•v,                                          (5b)


                      v2 = v12+v22+v32  .                              (5c)

The terms on the RHS of (5) represent the change in kinetic energy of the mass, m. The definition of the kinetic energy of a mass is suggested by the reasoning that led up to (5).

Note that W0 in (5) is the work required to get the mass accelerated to the initial kinetic energy,  (½) m v02 , from an initial  value of zero, and that the difference,   W-W0 , is additional work required to boost the kinetic energy up to (½) m v2. This means that W  is the work required to increase the kinetic energy energy from zero to a final value of (½) m v2.

Work is a measure the amount of energy transfer by a force. There is a piece of the energy puzzle that we have not yet discovered here. If the kinetic energy of an object of mass, m, receives energy transfer, how do we describe the source of that energy? In a collision, there is a transfer of energy from one mass to another. One mass loses kinetic energy, and another gains it. But what about when you drop a stone? Were does its kinetic energy come from? We shall deal with this question that discusses potential energy.

Potential Energy

Fig. 1 Forces on car on an incline.

Consider the problem of a roller coaster cart released from rest at the top of an incline (height, H, above the ground). The problem is to find its kinetic energy anywhere along the incline, given that frictional losses of energy are negligible.

In Fig. 1 the force of gravity,  -m g, is decomposed into two orthogonal components, one acting parallel to the inclined plane, -mg sin(A), and another acting normal to the incline, which does no work. The work done by gravity, when the car descends from rest from a initial height, H, a distance s along the incline is m*g* sin(A)*s. Also note that

s*sin(A) = H-z,                  (6)

so that the work done by gravity on the incline is just

W(z)=mg(h-z).                  (7)

Assuming that friction is negligible, the cart accelerates by an amount

a=g sin(A).                              (8)

From kinematics we can compute the speed of the cart at the indicated position on the incline by the following

                                v2 = 2*a*s,


                                v2 = 2*g*sin(A)*s.                  (7a)

Substituting from (6) in (7a) results in the following expression,

                               v2 = 2*g*(H-z).                         (7b)

To convert (7b) to an energy equation, multiply (both sides) by the mass of the cart and divide by 2,

½ m v2 = m*g*(H-z).                  (7c)

Rearranging terms, (7c) is rewritten as

½ m v2 +  m*g* z = m*g*H,       (8a)


½ m v2 +  m*g* z =  Etotal.            (8b)

Now here is something amazing, throughout its motion down the incline, a frictionless cart will move with a kinetic energy and another term that is a function of position (its potential energy), which sum to a constant value, equal to the function of position evaluated at the release point, where the cart was at rest. Further, any horizontal distance travelled cancels out, so that the function of position depends only on the height of the cart above the ground.  In this case, we have an example of the idea of energy and conservation of energy. The first term on the LHS of (8) is the kinetic energy; the second term on the LHS of (8) is  the potential energy, and that on the RHS of (8), the total energy, which remains fixed throughout the motion.

The above presented as a Calculus Derivation

Repeating (1) but in integral form, we have

Wif = ∫if F•ds,            (9)

where ds is an infinitesimal displacement parallel to the curved path at every point traversed. Imagine for example that in Fig. 1 the line is tangent to an underlying curve of roller coaster tracks. Now, the direction of F is straight down,

                              F = m g                     (10a)

                              F = - e3 m g.                (10b)

Note that

                             e3ds = dz,

or the differential change in height. In this case

Wif =-m g ∫idz',            (11)

where zi = H and zf = z. The integral (11) evaluates to give the expression

Wif =-m g (z-H),


Wif =m g(H-z).                 (12)

We see from the above derivation that the energy transfer to a cart following a curved trajectory, such as a roller coaster track, from rest at a height, H, is independent of the shape of the rack, but depends only on the amount of distance it has descended vertically. Setting the kinetic energy to the work done in that descent, we arrive at the energy relation (8) by a more general argument than that presented previously,

½ m v2 +  m*g* z = m*g*H.       (8)

In this case the second term on the LHS of (8) is the potential energy that the cart has. Beginning from rest, its total energy is the potential energy on the RHS of (8) that it had at the initial height. Thereafter, the total energy remains the same, but splits up between kinetic and potential energy.

Physicists have found that energy in the general form (13) is very useful in analysing complex motions,

KE + PE =Etotal.                            (13)

Momentum conservation

Rewrite the momentum of a particle as a new variable, p, as follows:

                                      p=mv ,                         (14)

in which case (2a) becomes,

F=dp/dt .                      (2aa)

Consider an isolated cluster of particles interacting by forces acting between every pair of particles. That force alters the momentum of the ith interacting particle by the jth  particle by an amount given by the integral of (2aa), which is called an impulse,

Δjpi =∫o Fji dt                         (15a)

and for the other interacting particle of the pair,

                                   Δipj =∫o Fij dt ,                     (15b)

where Δjpiis the change in momentum produced on the ith particle by the jth particle, and Δipj, the change in momentum of the jth particle produced by the ith particle.

According to Newton's Third Law of motion: for every force acting on an object, the object reacts back with an equal but opposite force. That means that in (15a) and (15b)

Fji = -Fij                                   (16)

As a result, adding (15a) and (15b) gives zero on the RHS,

Δjpi + Δipj =0 .                          (17)

The interactions between pairs of particles contribute changes in momenta in the pair particles that are equal and opposite.  The interaction of each particle with each of the other particles produce zero net changes in the momentum of the whole cluster. The result is that the momentum of the cluster of particles is not changed by the interactions of all the particles. It can be changed only by a force acting from beyond the cluster. But if you expand your range of analysis by including the outside body responsible for the impressed force, you will find that the total momentum of the expanded system does not change as a result of all the forces acting between the various bodies.

The momentum and energy of an isolated cluster of interacting bodies cannot be altered by the interactions of its constituents. Momentum and energy are therefore known as conserved quantities in Newtonian mechanics. Throughout the whole history of physics, physicists have realized that momentum and energy and energy are conserved in every conceivable context. They are part of the great conservation laws.


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