ELASTIC AND INELASTIC COLLISIONS

• In elastic collision, total linear momentum and total kinetic energy of the colliding bodies are conserved.
• In inelastic collision, total linear momentum of the colliding bodies is conserved, but part or whole of the kinetic energy is lost in other forms of energy.
• In both, elastic as well as inelastic collisions, total energy ( energy of all forms, mechanical, internal, sound, etc. ) and total linear momentum are conserved.
Inelastic collision in one dimension: -  Suppose a sphere A of mass m1 moving with velocity v1 in X-direction collides with another sphere of mass m2 moving in the same direction with velocity v2.
( v1 > v2 ) Let v1’ and v2’ be their velocities in the same ( X ) direction after the collision which we
want to find. According to the law of conservation of linear momentum,
m1v1 + m2v2 = m1v1’ + m2v2’
or, m1 ( v1 - v1’ ) = m2 ( v2 - v2’ ) … … ( 1 )
Now, a parameter known as coefficient of restitution, e, is defined as under:
Co-efficient of restitution, e = (v2’ - v1’) / v1 - v2
                                        = velocity of separation after the collision / velocity of approach before the collision
For partly inelastic collision, 0 < e < 1.
therefore v2 - v = e ( v1 - v2 ) … … … … … ( 2 )
Solving equations ( 1 ) and ( 2 ) for v1’ and v2’, we get
v1’ = (m1 - m2e)v1  + ( 1 + e)m2 v2     and  v2’ = (1 + e)m1v1    +    (m1e - m2)v2
            m1 + m2            m1 + m2                           m1 + m2                 m1 + m2
Thus, v1’ and v2’ can be calculated from the above equations.
Completely inelastic collision: -
Putting e = 0 for completely inelastic collision in the above equations,
v1 = v2 ' = m1v1 + m2v2    =   v ( the common velocity of the spheres after collision )
Thus, in completely inelastic collision, the colliding bodies move jointly with a common velocity.
Elastic collision in one dimension: -
Putting e = 1 for elastic collision in equation ( 2 ), we get v2’ - v1’ = v1 - v2
∴ v1 + v1’ = v2 + v2’ … … … … … ( 3 )
Multiplying equations ( 1 ) and ( 3 ), we get
m1 ( v12 - v1’2 ) = m2 ( v22 - v2’2 ).
Special cases of elastic collision: -
( 1 ) For m1 >> m2, neglecting m2 as compared to m1 in the above equations, we get v1’ ≈ v1 and v2’ ≈ 2v1 - v2
This shows that the larger sphere continues to move with the same velocity, whereas the velocity of the smaller sphere increases. If the smaller sphere was at rest, it moves with twice the velocity of the larger sphere after the collision.
( 2 ) For m2 >> m1, neglecting m1 as compared to m2 in the above equations, we get v2’ ≈ v2 and v1’ ≈ 2v2 - v1
In this case also, the larger sphere continues to move with the same velocity, but the velocity of the smaller sphere changes.
( i ) If the smaller sphere were moving with twice the speed of the larger sphere, it becomes stationary after the collision.
( ii ) If the smaller sphere were moving with velocity less than twice the velocity of the larger sphere, it  continues to move in the same direction, but with decreased speed.
( iii ) If the smaller sphere were moving with velocity more than twice the velocity of the larger sphere, it rebounds and starts moving in the opposite direction with the velocity given as above.
If the larger sphere was stationary, it remains stationary. In this case, the smaller sphere rebounds and moves with the same speed in the reverse direction. Summarizing the above two cases, when one of the two spheres colliding is much more massive than the other, then the velocity of the larger sphere remains almost the same after the collision, whereas the velocity of the smaller sphere after the collision is almost twice the
velocity of larger sphere less the velocity of the smaller sphere before the collision.
( 3 ) If m1 = m2, then v1’ = v2 and v2’ = v1. Thus, in elastic collision of two spheres of equal mass, their velocities get exchanged
Elastic collision in two dimensions: -
Suppose a sphere of mass m1 moving in X-direction with velocity v1 collides elastically with a stationary (v2 = 0 ) sphere of mass m2 as
shown in the figure. After the collision, they move in the directions making angles θ1 and θ2 with the X-axis with velocities v1’ and  v2’ respectively.
According to the law of conservation of momentum,
m1v1 = m1v1’ + m2v2’
Equating the X- and Y-components of the momentum we get

m1v1 = m1v1’ cos θ1 + m2 v2’cos θ2 … ( 1 ) and 0 = m1v1’sin θ1 - m2v2’sin θ2 … ( 2 )
This equation shows that all the momentum are conseverd