How momentum conservation principle and Newton third law are related
        Assuming constancy of inertial mass of an object, the most popular form of Newton second law   m a = F  can be can be changed to another form originally introduced by Newton himself. Namely, for any motion along a straight line we may write

m a = m (Dv / Dt) = D(mv) / Dt = Dp / Dt ,                (1)

where a new magnitude p called a momentum is defined as p = mv. Newton called this magnitude an amount of motion. Indeed, we may say that if two bodies have the same velocity v , the one with a greater mass carries more motion. Consequently the most popular form of Newton second law can be replaced by

Dp / Dt = F  .                            (2)

This new form turns out to be correct also for objects with variable inertial mass. Please notice that both forms are equivalent only if object's inertial mass stays constant. Consequently the last form is more general.
       If total force  F  acting upon the object equals to zero then a momentum rate of change is also equal to zero, and the momentum itself is conserved. It means, the momentum stays constant. But this implies a constant velocity only for objects with fixed inertial mass. Otherwise, if mass changes, velocity changes too, to keep the product  mv  constant. Practically, however, cases of objects with variable masses are very rare. So, you may think , it is not worth to bother with the more general form of Newton second law and related to it idea of momentum just for sake of a few rare cases. But this is not a case, and you will see soon why the idea of momentum is so important.
        Let us concentrate for a moment on two point like objects interacting with each other via gravitational forces. As you remember the magnitudes of these attracting forces is given by the following formula

F_G  =  G m₁ m₂ / R ² .                (3)

This means that these forces have the same magnitudes and opposite directions  as it is shown in Fig. 1 below.     

Introducing momenta for both objects  p₁ = m₁ v₁ ,  p₂ = m₂ v₂  we can write for these objects

Dp₁ / Dt = F₂₁ ,  Dp₂ / Dt = F₁₂ .          (4)

Adding these equations leads to the following result

Dp₁ / Dt + Dp₂ / Dt = D(p₁ + p₂) / Dt = F₂₁ + F₁₂ = 0,          (5)

because the gravity forces there have the same magnitudes but opposite directions. Introducing a total momentum of the two particle system as  p = p₁ + p₂, we have

Dp / Dt = 0,                      (6)

which means that the total momentum of this system is conserved. This conservation is assured by two facts. Namely, a lack of external forces acting from outside on the particles of the system and the gravity forces acting between these two particles have the same magnitudes but opposite directions, so they add up to a zero force.
       Now a more general question can be asked. If instead of gravity forces there are other forces acting between these two particles, will it still be true that they have the same magnitudes but opposite directions?  An answer can come only from experiment. But how to arrange experiments to test it? Imagine yourself two railway cars colliding head on along a leveled and straight track. As long as the cars are in a good technical condition and there is no wind their velocities before collision do not change practically because friction forces in their axles and air resistance are negligible. As they collide, they act upon each other with certain forces and if their locking mechanisms are not set up they will usually depart. If these forces have the same magnitudes and opposite directions then a total momentum of both cars should be conserved. Thus investigating this momentum we may find out if the forces have the properties that were described above.
       Using the applet from this section you can study several types of collisions. These collisions are similar to collisions of two railway cars described above. Let us start with nonelastic collisions. This means that the kinetic energy of the system after collision will be smaller than before collision. The kinetic energy lost in the process of collision is transformed into another types of energy. For example, real collisions around us are usually noisy. This noise, of course, is a part of the initial kinetic energy that was transformed into sound energy in the collision process.
       Select in the applet nonelastic collisions, set up initial velocities (v_rⁱⁿ, v_bⁱⁿ) and mass ratio for colliding objects as they are given in the first row in the table below. A fraction of lost kinetic energy is randomly generated for each collision by the computer. This ratio is kept unchanged as long as you operate only with three buttons on the bottom of the applet. Using the selection devices in the upper part of the applet automatically changes this fraction. Prepare your stopwatch and make measurements needed for calculation of final (after collision) velocities (v_r ^f, v_b ^f) of both red and black object. You do not need to make all these measurements in a single applet run because with help of the bottom buttons you can repeat exactly the same experiment. Assuming that the mass of the red object is 1 kg calculate total momentum of the system before and after collision (pⁱⁿ, p ^f) as well as the ratio of final kinetic energy and initial kinetic energy.

       Repeat the same kind of measurements and calculations for the next four rows of the table or any other your own choices of four sets of initial data. If for some of these choices measurements are difficult because after collision the objects are moving too slow or too fast, just make another choice.
       If your measurements and calculations are accurate and correct you should learn two things. Kinetic energy for these collisions is not conserved, whereas momentum is conserved.  Your results, of course, are not exact because measurements are always  carrying certain experimental errors. From the momentum conservation for the investigated collisions you also can deduce that the forces acting between the colliding objects must have the same magnitudes and opposite directions. Otherwise according to the relations (5) the momentum could not be conserved. Consequently for gravitational forces and forces involved in investigated collisions you have rediscovered Newton third law. This law states that if body #1 acts on body #2 with certain force  F₁₂  then body #2 acts on body #1 with the force  F₂₁  having the same magnitude but opposite direction. Mathematically it means that  F₁₂ + F₂₁ = 0. In Newtonian mechanics this law is generally valid for any two interacting bodies. If we take a set containing any number of such bodies with no external forces (forces from outside of this set) acting on them, then a sum of all internal forces acting between of these bodies must add up due to Newton third law to zero. Consequently a total momentum of such system is conserved. Thus for two body system restricted to a straight line motion we can write

m₁v₁ⁱⁿ + m₂v₂ⁱⁿ = m₁v₁ ^f + m₂v₂ ^f  ,         (7)

where the subscripts 1 and 2 mark properties of bodies #1 and #2, whereas the superscripts in and f are refering to initial (before collision) and final (after collision) magnitudes of velocities. If  such a system is restricted to a planary motion its momentum is conserved when x and y components of momentum are conserved, that is when

 m₁v_1xⁱⁿ + m₂v_2xⁱⁿ = m₁v_1x ^f + m₂v_2x ^f  ,         (8)

m₁v_1yⁱⁿ + m₂v_2yⁱⁿ = m₁v_1y ^f + m₂v_2y ^f   .         (9)

      An interesting thing about the momentum conservation conditions for both rectilinear (7) and planar collisions  (8),(9)  is that they do not provide us with full information  about final velocities of the objects after collision. From mathematical point of view the condition (7),  if initial velocities and masses are known, represents a single equation with two unknowns that are the final velocities. Thus, to find the final velocities we have to look for an additional physical condition  relating the final velocities to initial velocities and masses. For planar collisions we need two such additional conditions. 

      Perfectly inelastic collision
      If after collision the two railway cars described above are lock together and travel with a final velocity  v ^f , then the additional condition must be v₁ ^f = v₂ ^f = v ^f . Inserting this condition into the momentum conservation condition (7) we obtain

v ^f = (m₁v₁ⁱⁿ + m₂v₂ⁱⁿ) / (m₁ + m₂).             (10)

We already know from "experiment" that kinetic energy does not need to be conserved in a collision process. Let us find out how much of it is lost in this type of collision:

Loss in KE = K_in - K_f = (1/2){m₁(v₁ⁱⁿ)² + m₂(v₂ⁱⁿ)² - (m₁ + m₂)(v ^f)²} =

m₁ m₂ (v₁ⁱⁿ - v₂ⁱⁿ)² / {2(m₁ + m₂)}.             (11)

To obtain the last result  v ^f   was replaced by the right side of (10) and a simple but slightly tedious algebra was executed. 
      Looking at the final result for Loss in KE  we realize that it is never equal to zero. Thus for this type of collision some kinetic energy is always lost. Considering all possible kinds of rectilinear collisions of two bodies with fixed masses and initial velocities we find that for this particular case a maximal amount of kinetic energy is lost. This is why a kind of collision resulting with two bodies traveling finally together is called a perfectly inelastic collision. Notice that if both colliding masses are identical (m₁ = m₂ = m) and their velocities are exactly opposite (v₁ⁱⁿ = - v₂ⁱⁿ = vⁱⁿ)  then in the collision process all kinetic energy is lost and both bodies stop. Indeed,   Loss in KE = m(vⁱⁿ)²  which is exactly a total initial kinetic energy of the system.
      Now using the applet investigate the perfectly inelastic collisions described in the table below. Find "experimental" values for final velocities, calculate initial and final kinetic energies assuming that a smaller mass is equal to 1 kg, and both, "experimental" Loss in KE  and theoretical (Loss in KE )^th  values of dissipated kinetic energy.   The (Loss in KE) t^h values should be calculated from the formula (11) .

Table 2

vrin   m/s	vbin     m/s	mass    ratio	K in       J	v f        m/s	K f       J	K       J	Kth       J	pin  kg m/s	p f    kg m/s
0.50	-0.50	1
0.50	0.00	1
0.30	-0.50	2
0.50	0.20	0.5
0.20	0.50	2

 

      Perfectly elastic collisions
      If in a process of collision kinetic energy is conserved such collision is called a perfectly elastic. Then using momentum conservation (7) and kinetic energy conservation we can predict (calculate) final velocities of colliding bodies. A derivation of formulae for these final velocities is not difficult but certainly tedious. Doing it in a well organized fashion helps to simplify these calculations. We shall start with both conservation laws:

m₁(v₁ⁱⁿ)² + m₂(v₂ⁱⁿ)² = m₁(v₁ ^f)² + m₂(v₂ ^f)²  

m₁v₁ⁱⁿ + m₂v₂ⁱⁿ = m₁v₁ ^f + m₂v₂ ^f .

Notice that the first of these relations is the kinetic energy conservation relation already, for sake of simplicity, multiplied by 2. Rearranging these relations we can get

m₁{(v₁ⁱⁿ)² - (v₁ ^f)²} = m₂{(v₂ ^f)² -(v₂ⁱⁿ)²} 

m₁{v₁ⁱⁿ - v₁ ^f} = m₂{v₂ ^f - v₂ⁱⁿ}.

Dividing  the first of these relations by the second relation and applying the very well known identity a² - b² = (a - b)(a + b) that holds for any a and b we obtain

v₁ⁱⁿ + v₁ ^f = v₂ ^f + v₂ⁱⁿ.

Rearranging again the momentum conservation relation and this last relation we can write

  m₁v₁ ^f + m₂v₂ ^f  = m₁v₁ⁱⁿ + m₂v₂ⁱⁿ       (13)

v₁ ^{f  -}  v₂ ^f   =  v₂^{in  -}  v₁ⁱⁿ .      (14) 

This constitute a set of linear equations for v₁ ^f  and   v₂ ^f  that can be easily solved.  Multiplying the second of these equations by  m₂   and adding it to the first equation the solution for  v₁ ^f   is obtained

(m₁ + m₂) v₁ ^f = 2 m₂ v₂ⁱⁿ + (m₁ - m₂) v₁ⁱⁿ  

or

v₁ ^f  = {2 m₂ v₂ⁱⁿ + (m₁ - m₂) v₁ⁱⁿ} / (m₁ + m₂) .      (15) 

The solution for  v₂ ^f  can be obtained a similar way  multiplying the equation (14) by  m₁   and subtracting it from the equation (13).  There is, however, a simplest and more instructive way to find this solution. Notice that the equations (13) and (14) have an interesting symmetry. If the subscripts  1  and  2  are interchanged these equations stay equivalent to the initial set. It means that the new set can be rearranged with help of algebraic operations such a way that it will become identical with the original set. Try to show it, please. Then, the solutions must exhibit the same sort of symmetry. Consequently, to obtain the solution for v₂ ^f  it is enough to interchange the subscripts  1  and  2  in (15) and get

  v₂ ^f  = {2 m₁ v₁ⁱⁿ + (m₂ - m₁) v₂ⁱⁿ} / (m₁ + m₂) .      (16) 

      These general results do not look too simple, but it is still possible to derive from them some simple and instructive conclusions for specific cases. If masses of both bodies are the same (m₁ = m₂) then any perfectly elastic collision leads to interchange of velocities between the bodies because the solutions (15) and (16) are reduced to  v₁ ^f = v₂ⁱⁿ  and  v₂ ^f = v₁ⁱⁿ . Use the applet to study a few such cases setting the mass ratio equal to 1 and velocities as you wish. But do not forget to run at least one case with one initial velocity equal to zero. The other specific case is slightly more challenging. Let us assume that the mass of the second body is very big if compared with the mass of the first body, and that the second body stays initially at rest. In this case the ratio  (m₁ - m₂) / (m₁ + m₂) is practically equal to -1. Consequently  v₁ ^f  =  - v₁ⁱⁿ  and  v₂ ^f  = 0 . This is a model of perfectly elastic collision of a body with a standing wall. Use again the applet to study this kind of collisions also with a moving wall. Figure out differences if the wall moves toward the body and away from it.
      Finally study cases with other mass distributions specified in the Table 3.    

      Nonelastic collisions
      For nonelastic collisions kinetic energy conservation does not hold and must be replaced by a relation describing kinetic energy retention. This relation can be written in the following form

K^f  = r Kⁱⁿ ,

where  r  is a retention coefficient obviously limited by double inequality  0 < r  < 1  and additionally limited by initial velocities and mass ratio.  Analytical solutions for final velocities are possible, but they are much more complicated than for perfectly elastic collisions. 

      Some remarks about planar collisions
      For all kinds of planar collisions momentum conservation principle is described by the set of relations (8) and (9).  Dealing with perfectly inelastic collisions we can use this set to find components of final velocity of both objects because after collision they travel together. Situation for perfectly elastic collisions is, however. more complicated. Kinetic energy conservation principle adds only one relation. But there are four unknowns to look for. Thus we have to add one more relation between the components of final and initial velocities. To find four unknowns we must have four equations. But a fourth equation can be added only if we have additional information about a nature of the collision.

      Evaluation
        If :

• your results recorded in the tables 1, 2 and 3 are consistent with theoretical predictions
• you understand relation between Newton third law and momentum conservation
• you can derive relations (15) and (16)
• you can show that the interchange of subscripts 1 and 2 does not change the relations (13) and (14)
• you realize that the last two collisions described in the table 2 are really identical, and the same is true about the last two collisions in the table 3

the objectives of this lesson are fully achieved. If you have doubts try to read it once more concentrating on them, but do not try to memorize this text. Physics is not about memorizing, it is about understanding.