Elastic and Inelastic Collisions
The colliding system's momentum is conserved when net external impulse is zero, and approximately conserved when it is negligible. An elastic collision also conserves total kinetic energy. Compare the kinetic-energy totals before and after to classify the collision.
What a learner can do afterwards
- Applies conservation of momentum in one dimension with signs that carry the directions
- Totals the kinetic energy before and after and classifies the collision from the comparison
- Explains how a decrease in kinetic energy can transfer energy to internal energy or sound without destroying total energy
1 · Read
As in your momentum work, use p = m × v and signed velocities. Choose both colliding trolleys as the system. Its total momentum is conserved if net external impulse is zero, and approximately conserved if it is negligible. External impulse is the combined effect of outside forces over the collision time. Equal and opposite contact forces cancel in the system total. Use kg and m/s for momentum in kg m/s.
Right is positive. A 2 kg trolley at +4 m/s meets a 1 kg trolley at -2 m/s. Total momentum is +8 - 2 = +6 kg m/s. Net external impulse is zero. After impact, the 2 kg trolley moves at +1 m/s and carries +2 kg m/s. The other trolley must carry +6 - 2 = +4 kg m/s. Divide by its 1 kg mass: its final velocity is +4 m/s.
Find each trolley's kinetic energy with KE = 0.5 × m × v², using kg and m/s to get joules. Squaring a negative velocity gives positive KE. Add both energies before impact and again after, in the same reference frame. Equal totals define an elastic collision; unequal totals mean inelastic. These questions ignore rotation and treat the stated values as exact.
Inelastic means that total kinetic energy changes. Usually it decreases, but a spring carried by a trolley can release stored energy during impact and increase it. Objects can bounce apart in an inelastic collision. If they stick together, it is perfectly inelastic. Sticking does not require the joined pair to stop. Seeing a bounce cannot replace comparing the energy totals.
When trolleys stick, divide total signed momentum by combined mass to get their shared final velocity. With zero net external impulse, a 1 kg trolley at +4 m/s joining a resting 3 kg trolley gives v = (1 × 4 + 3 × 0) ÷ 4 = +1 m/s. KE before is 0.5 × 1 × 4² = 8 J; after, use the joined mass: 0.5 × 4 × 1² = 2 J. Subtract after from before: 8 - 2 = 6 J transferred out of motion.
A kinetic-energy decrease is not destruction of total energy. It can increase internal energy through heating or deformation, or transfer energy away as sound. These are possible destinations, not a required list for every collision. A permanent dent is not needed. To find the decrease, subtract total KE after from total KE before.
Check the external impulse before conserving momentum. Compare total kinetic energy separately: unchanged means elastic; changed means inelastic.
Lesson, questions and flashcards written for this dataset. Physics references: OpenStax College Physics, chapter 8, and University Physics Volume 1, section 9.4.
Readings and flashcards written for this dataset. Physics references: OpenStax collision chapters. Linked videos are credited to Khan Academy; their publisher terms have not been verified.
More to read
Finding the other trolley's velocity
Choose both trolleys as the system and take right as positive. A 0.5 kg trolley at +8 m/s has momentum +4 kg m/s. A 1.5 kg trolley at -2 m/s has momentum -3 kg m/s. Together they have +1 kg m/s.
Suppose net external impulse during impact is zero. If the smaller trolley leaves at -4 m/s, its momentum is -2 kg m/s. The larger trolley must then carry +3 kg m/s, because -2 + 3 = +1. Divide its momentum by its mass: +3 ÷ 1.5 = +2 m/s. This calculation uses signed velocity throughout. Adding the two speeds would lose the direction information.
Two checks for an elastic collision
Consider a 3 kg trolley moving at +2 m/s toward a stationary 1 kg trolley. After impact, their velocities are +1 and +3 m/s. Ignore rotation. Initial momentum is 3 × 2 = +6 kg m/s. Final momentum is 3 × 1 + 1 × 3 = +6 kg m/s.
Now calculate kinetic energy separately. Before impact it is 0.5 × 3 × 2² = 6 J. After impact it is 0.5 × 3 × 1² + 0.5 × 1 × 3² = 1.5 + 4.5 = 6 J. Equal energy totals identify an elastic collision. The individual trolleys have changed speed, even though the system totals match. Exact data describe an ideal model; laboratory measurements may support only approximate agreement within their uncertainty.
Sticking does not mean stopping
A 2 kg trolley at +3 m/s collides with a 1 kg trolley at -3 m/s, and they stay joined. Net external impulse is zero. Initial momentum is +6 - 3 = +3 kg m/s. The combined mass is 3 kg, so their shared final velocity is +1 m/s.
Ignoring rotation, their initial kinetic energy is 9 + 4.5 = 13.5 J. Their final kinetic energy is 0.5 × 3 × 1² = 1.5 J. The decrease is 12 J, not the full initial energy. This perfectly inelastic collision leaves the joined trolleys moving. For fixed initial masses and velocities, sticking gives the smallest final kinetic energy consistent with momentum conservation. The energy removed from motion can increase internal energy or be transferred away as sound.
More kinetic energy after impact
Two 1 kg trolleys approach with velocities +1 and -1 m/s. A compressed spring carried by a trolley releases stored energy during impact. The trolleys leave at -3 and +3 m/s. Include the spring in the system and suppose net external impulse is zero.
Total momentum is zero before and after. Ignoring rotation, total kinetic energy rises from 0.5 + 0.5 = 1 J to 4.5 + 4.5 = 9 J. If no other energy transfers occur, the 8 J gain comes from an 8 J decrease in the spring's stored energy. Total energy is conserved, but kinetic energy is not. This is not an elastic collision. Here, the term inelastic includes collisions whose kinetic energy changes in either direction.
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