Elastic and Inelastic Collisions

An elastic collision occurs when kinetic energy is conserved, that is the initial kinetic energy = the final kinetic energy of the system. 

  • In elastic collisions the kinetic energy of the system is conserved.
  • In inelastic collisions the kinetic energy of the system is not conserved.
  • In all collisions the total momentum of the system is conserved.

When an inelastic collision occurs there will be a loss of kinetic energy. The kinetic energy lost is transformed into other forms of energy (heat, light, sound). The objects in the system may also become deformed (bent, crushed, broken), or the objects may collide and stick together (train engine coupling with carriage).

Example 1:

A 250g toy car travelling at 5\cfrac { m }{ s } east, collides with an identical toy car travelling in the same direction at 2.5\cfrac { m }{ s } . After the collision the first car continues in the same direction at 1.5\cfrac { m }{ s } . The second car continues in the same direction after the collision at 4\cfrac { m }{ s } . Is this collision elastic or inelastic?

{ m }_{ 1 } = 250g

{ u }_{ 1 } = 5\cfrac { m }{ s } to the east (note: east as positive)

{ v }_{ 1 } = 1.5\cfrac { m }{ s } east 

{ m }_{ 2 } = 250g

{ u }_{ 2 } = 2.5\cfrac { m }{ s } east

{ v }_{ 2 } = 4\cfrac { m }{ s } east

We need to compare the initial kinetic energy to the final kinetic energy of the system:

KE_{ i }={ KE }_{ 1 }+{ KE }_{ 2 }

KE_{ i }=\cfrac { 1 }{ 2 } mu^{ 2 }(car1)+\cfrac { 1 }{ 2 } mu^{ 2 }(car2)

KE_{ i }=\cfrac { 1 }{ 2 } \times 0.25\times { 5 }^{ 2 }+\cfrac { 1 }{ 2 } \times 0.25\times { 2.5 }^{ 2 }

KE_{ i }=3.125+0.781

KE_{ i }=3.91J

KE_{ f }={ KE }_{ 1 }+{ KE }_{ 2 }

KE_{ f }=\cfrac { 1 }{ 2 } mv^{ 2 }(car1)+\cfrac { 1 }{ 2 } mv^{ 2 }(car2)

KE_{ f }=\cfrac { 1 }{ 2 } \times 0.25\times { 1.5 }^{ 2 }+\cfrac { 1 }{ 2 } \times 0.25\times { 4 }^{ 2 }

KE_{ f }=0.281+2

KE_{ f }=2.28J

Therefore; KE_{ i } does not = KE_{ f } and the collision was inelastic.