Formulaxis
EN
Start calculating

Physics · Mechanics · Motion & Energy

Perfectly inelastic collision — common velocity and kinetic-energy loss

Solve a one-dimensional collision in which two bodies stick together after impact. Velocities are signed according to the positive direction shown in the diagram.

Created v1.0

By Thibaut Grzelak, Mechanical Analysis Engineer

Mechanics · Motion & Energy
01Calculation inputsp_i = m₁ × u₁ + m₂ × u₂
+xim₁m₂u₁u₂fm₁m₂v_f
Before/after 1D collision with the +x convention shown: the two initial bodies become one joined pair moving at the signed common velocity v_f.
m₁ > 0.
Signed velocity: positive points in the +x direction shown.
m₂ > 0.
Signed velocity: negative points opposite the +x direction.
02

Results

Final common velocity
—

Signed velocity v_f of the joined bodies immediately after impact.

Initial total momentum
—

Signed system momentum p_i = m₁u₁ + m₂u₂, conserved for the closed system.

Kinetic-energy decrease
—

ΔK = K_i − K_f ≥ 0, converted to other forms of energy rather than destroyed.

Initial kinetic energy
—

Internal intermediate K_i used to document the energy balance.

Final kinetic energy
—

Internal intermediate K_f for the joined mass.

Relation and numerical substitutionp_i = m₁ × u₁ + m₂ × u₂

—

One-dimensional closed-system collision with no relevant external impulse during the impact interval. The two bodies remain joined after impact. Momentum is conserved; kinetic energy generally is not.

01Formulas and symbols

Formulas used

p_ip_i = m₁ × u₁ + m₂ × u₂Initial total momentum before impact.
v_fv_f = p_i / (m₁ + m₂)Common final velocity from conservation of momentum.
K_iK_i = ½ × m₁ × u₁² + ½ × m₂ × u₂²Total kinetic energy before impact.
K_fK_f = ½ × (m₁ + m₂) × v_f²Kinetic energy of the joined bodies after impact.
ΔKΔK = K_i − K_fDecrease in kinetic energy during the perfectly inelastic collision.
02Assumptions and limits

Scope of validity

  • One-dimensional closed-system collision with no relevant external impulse during the impact interval.
  • The two bodies remain joined after impact. Momentum is conserved; kinetic energy generally is not.
03Validation example

Reference numerical case

  1. For m₁ = 2 kg, u₁ = +3 m/s, m₂ = 1 kg and u₂ = −1 m/s: p_i = 5 kg·m/s and v_f = 5/3 m/s.
  2. K_i = 9.5 J, K_f = 25/6 J and ΔK = 16/3 J ≈ 5.333333 J.
04References

Technical references

  1. University Physics Volume 1, §9.4 Types of Collisions — OpenStax

FAQ

Why is momentum conserved but kinetic energy is not?

For a closed system, the bodies exchange internal forces so total momentum is conserved. In a perfectly inelastic impact, some initial kinetic energy becomes deformation, thermal, sound or other internal energy.

How should I enter opposite directions?

Use signed velocities. Enter motion along +x as positive and motion in the opposite direction as negative, following the convention shown in the diagram.

Does this calculate impact force or deformation?

No. The model resolves only the before/after 1D momentum and kinetic-energy balance for bodies that stick together; it does not model impact duration, contact force or deformation.

05Conserve momentum, then compare kinetic energy

First compute p_i = m₁u₁ + m₂u₂. Because the bodies remain joined, their final mass is m₁ + m₂ and their common velocity is v_f = p_i/(m₁ + m₂). Then compare K_i with K_f to obtain ΔK = K_i − K_f.

06Strict perfectly inelastic 1D model

This model requires a one-dimensional closed system over the impact interval and assumes the bodies stick together. It does not model rebound, impact force, collision duration or deformation history.

07Signs and energy interpretation

Do not discard velocity signs when bodies approach from opposite directions. Kinetic energy uses squared speeds and is not conserved here. A positive ΔK is energy converted to other forms, not energy destroyed.

Calculation path and sources

Continue on Formulaxis