In any collision or explosion in a closed system, total momentum is always conserved, and so is total energy. Kinetic energy, however, may not be conserved, because some can be transferred to other stores.• Elastic collision: total kinetic energy is conserved. Collisions between gas molecules are elastic, and collisions between snooker balls are very nearly elastic.
• Inelastic collision: total kinetic energy decreases. Some is transferred to the thermal energy of the objects, to sound and to work done deforming them.
• Perfectly (totally) inelastic collision: the objects stick together and move off with a common velocity. This loses the most kinetic energy allowed by conservation of momentum.
• Explosions: total kinetic energy increases, because energy from another store (chemical, elastic or nuclear) is transferred to kinetic energy.To decide which type a collision is, calculate the total kinetic energy before and after. Ek = p2/2m is often quick when momenta are known.Worked example: a 2.0 kg trolley moving at 3.0 m s-1 hits a stationary 1.0 kg trolley and they stick together.
momentum before = 2.0 × 3.0 = 6.0 kg m s-1, so after: v = 6.0 / 3.0 = 2.0 m s-1
Ek before = ½ × 2.0 × 3.02 = 9.0 J
Ek after = ½ × 3.0 × 2.02 = 6.0 J
Kinetic energy has fallen by 3.0 J, so the collision is inelastic (in fact perfectly inelastic).
• Inelastic collision: total kinetic energy decreases. Some is transferred to the thermal energy of the objects, to sound and to work done deforming them.
• Perfectly (totally) inelastic collision: the objects stick together and move off with a common velocity. This loses the most kinetic energy allowed by conservation of momentum.
• Explosions: total kinetic energy increases, because energy from another store (chemical, elastic or nuclear) is transferred to kinetic energy.To decide which type a collision is, calculate the total kinetic energy before and after. Ek = p2/2m is often quick when momenta are known.Worked example: a 2.0 kg trolley moving at 3.0 m s-1 hits a stationary 1.0 kg trolley and they stick together.
momentum before = 2.0 × 3.0 = 6.0 kg m s-1, so after: v = 6.0 / 3.0 = 2.0 m s-1
Ek before = ½ × 2.0 × 3.02 = 9.0 J
Ek after = ½ × 3.0 × 2.02 = 6.0 J
Kinetic energy has fallen by 3.0 J, so the collision is inelastic (in fact perfectly inelastic).