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Elastic vs Inelastic Collision: Which Cases Conserve Kinetic Energy?

The first record is perfectly inelastic because the objects stick and kinetic energy falls from 27 J to 24 J. Records 2 and 3 conserve both momentum and kinetic energy, so they are perfectly elastic.

Question

Take rightward velocity as positive. For each one-dimensional collision record, calculate the total kinetic energy before and after the collision. Then identify which collisions are perfectly elastic.

  1. A 4.0 kg4.0\,\mathrm{kg} block at +3.0 m/s+3.0\,\mathrm{m/s} and an 8.0 kg8.0\,\mathrm{kg} block at +1.5 m/s+1.5\,\mathrm{m/s} move together afterward at +2.0 m/s+2.0\,\mathrm{m/s}.
  2. A 3.0 kg3.0\,\mathrm{kg} block at +6.0 m/s+6.0\,\mathrm{m/s} and a 5.0 kg5.0\,\mathrm{kg} block at +4.0 m/s+4.0\,\mathrm{m/s} move afterward at +3.5 m/s+3.5\,\mathrm{m/s} and +5.5 m/s+5.5\,\mathrm{m/s}, respectively.
  3. A 3.0 kg3.0\,\mathrm{kg} block at +3.0 m/s+3.0\,\mathrm{m/s} and a 2.0 kg2.0\,\mathrm{kg} block at −2.0 m/s-2.0\,\mathrm{m/s} move afterward at −1.0 m/s-1.0\,\mathrm{m/s} and +4.0 m/s+4.0\,\mathrm{m/s}, respectively.

Answer

Use K=12mv2K=\tfrac12mv^2 for every object and add the values for the whole system.

Record 1

Ki=12(4.0)(3.0)2+12(8.0)(1.5)2=27 J.K_i=\tfrac12(4.0)(3.0)^2+\tfrac12(8.0)(1.5)^2=27\,\mathrm{J}.

Because the blocks move together afterward,

Kf=12(12.0)(2.0)2=24 J.K_f=\tfrac12(12.0)(2.0)^2=24\,\mathrm{J}.

Momentum is 24 kg m/s24\,\mathrm{kg\,m/s} before and after, but 3 J3\,\mathrm{J} of translational kinetic energy is lost. This is a perfectly inelastic collision.

Record 2

Ki=12(3.0)(6.0)2+12(5.0)(4.0)2=94 J,K_i=\tfrac12(3.0)(6.0)^2+\tfrac12(5.0)(4.0)^2=94\,\mathrm{J}, Kf=12(3.0)(3.5)2+12(5.0)(5.5)2=94 J.K_f=\tfrac12(3.0)(3.5)^2+\tfrac12(5.0)(5.5)^2=94\,\mathrm{J}.

Momentum is 38 kg m/s38\,\mathrm{kg\,m/s} on both sides. This collision is perfectly elastic.

Record 3

Ki=12(3.0)(3.0)2+12(2.0)(−2.0)2=17.5 J,K_i=\tfrac12(3.0)(3.0)^2+\tfrac12(2.0)(-2.0)^2=17.5\,\mathrm{J}, Kf=12(3.0)(−1.0)2+12(2.0)(4.0)2=17.5 J.K_f=\tfrac12(3.0)(-1.0)^2+\tfrac12(2.0)(4.0)^2=17.5\,\mathrm{J}.

Momentum is 5.0 kg m/s5.0\,\mathrm{kg\,m/s} on both sides. This collision is also perfectly elastic.

The key distinction is that total momentum is conserved in every isolated collision, while total kinetic energy is conserved only in a perfectly elastic collision.

Evidence boundary

The calculations treat each record as an isolated one-dimensional system and include only translational kinetic energy. They classify the stated before-and-after states; they do not model deformation, heat, sound, rotation, or measurement uncertainty. ‘Perfectly inelastic’ is used only for the first record because the blocks stick together.

Sources

These references support the concepts and methods used in the explanation above.

Elastic vs Inelastic Collision Kinetic Energy | Verla