Consider a 1dimensional collision between
two bodies in which no external forces act, as shown below. 



F_{AB} is the force exerted by body A
on body B. 

F_{BA} is the force exerted by body B
on body A. 

Newton's third law states that 



Suppose that the bodies remain in contact for a short time ∆t, then the
change in momentum of A is given by 



and the change in momentum of B is given by 



so the total change in momentum 

is clearly equal to zero 




This shows that the principle of conservation of linear momentum is an
inevitable consequence of Newton's second and third laws of motion. 



Elastic collisions: An Example 

Two bodies A
and B, have an elastic collision during which no external
forces act (this can easily be arranged by having magnets with
similar poles facing each other, as shown below). 



Mass of A = m_{A} = 2kg 

Mass of B = m_{B} = 3kg 

Calculate the magnitudes and senses of the velocities of the
bodies after the collision. 



Assuming there are no external forces, the principle of
conservation of momentum can be used, so we have 



which, putting in the values for masses and velocities before collision,
gives us 


Equation 1 

As the
collision is elastic we can use the fact that the total K.E.
is the same before and after the collision. 

This can be expressed as follows 


Equation 2 

These two equations could be used to find the values of v_{A} and v_{B}
however, the calculations can be simplified if we remember that the relative
velocity of approach is equal to the relative velocity of separation, for an
elastic collision (see here for
proof). 



We can therefore use the following equation (in combination with Equation 1
above) to find our answers 



From the information given near the diagram we can see that 



therefore 



Combining this last equation with Equation 1 above
allows us to calculate v_{A} and v_{B}
easily. 



The results are: v_{A} =
1.6ms^{1} and v_{B}
= 3.6ms^{1} 
