Inelastic Collision Calculator

Find final velocity when objects stick together using m₁v₁ + m₂v₂ = (m₁+m₂)v_f.

💢 Collision 📐 Perfectly Inelastic 💨 Momentum

Objects merge after collision. Momentum is conserved; kinetic energy is not.

Mass 1 (m₁) kg
Velocity 1 (v₁) m/s
Mass 2 (m₂) kg
Velocity 2 (v₂) m/s
⚠️ Enter valid positive numbers.

Understanding Perfectly Inelastic Collision

A perfectly inelastic collision is one where the colliding objects stick together and move as one combined mass after impact. Momentum is conserved, but kinetic energy is not — the lost energy converts to heat, sound, and deformation. This type of collision produces the maximum kinetic energy loss possible while still conserving momentum.

The formula is v_f = (m₁v₁ + m₂v₂)/(m₁+m₂), derived directly from conservation of momentum. The merged object's velocity is the momentum-weighted average of the initial velocities. If both objects have the same initial momentum but opposite directions, they cancel and the merged mass is stationary.

Common examples: two railway cars coupling, a bullet embedding in a block, cars crashing and locking bumpers, two balls of clay merging. In all cases, the energy lost (ΔKE = KE_before − KE_after) converts to internal energy — deforming metal, creating heat, generating sound.

Perfectly inelastic collisions are important in forensics (crash reconstruction), materials science (impact testing), and ballistics (bullet stopping power). The fraction of KE lost depends on the mass ratio: equal masses lose 50% of KE; a very light projectile hitting a massive target loses nearly all its KE.

Formula Reference Table

Solve ForFormulaNotes
Final velocityv_f = (m₁v₁ + m₂v₂) / (m₁+m₂)Objects stick together
Momentum beforep = m₁v₁ + m₂v₂Conserved in collision
KE lostΔKE = ½(m₁v₁²+m₂v₂²) − ½(m₁+m₂)v_f²Always ≥ 0
KE fraction lostΔKE/KE₀ = m₁m₂(v₁−v₂)² / [(m₁+m₂)(m₁v₁²+m₂v₂²)]Depends on mass ratio
Equal massesv_f = (v₁+v₂)/2Simple average
Coeff. of restitutione = 0Defining feature of perfectly inelastic

3 Worked Examples

Example 1
Car Crash — Objects Stick Together

1,200 kg car at 15 m/s rear-ends stationary 1,500 kg truck. They lock together.

  • v_f = (1200×15 + 1500×0)/(1200+1500)
  • v_f = 18,000/2,700 = 6.67 m/s
  • KE before = ½×1200×225 = 135,000 J; KE after = ½×2700×44.5 = 60,075 J; ΔKE = 74,925 J lost
✓ Final velocity = 6.67 m/s; KE lost = 74.9 kJ
Example 2
Bullet Embedding in Block

10 g bullet at 300 m/s embeds in 2 kg wooden block.

  • v_f = (0.01×300 + 2×0)/(0.01+2)
  • v_f = 3/2.01 = 1.49 m/s
  • KE before = ½×0.01×90,000 = 450 J; KE after = ½×2.01×2.22 = 2.23 J; lost = 447.8 J (99.5%)
✓ Final velocity = 1.49 m/s; 99.5% of KE lost to heat/deformation
Example 3
Train Cars Coupling

Two railway cars: 5,000 kg at 8 m/s and 3,000 kg at −2 m/s (opposite direction). Find final velocity.

  • p_total = 5000×8 + 3000×(−2) = 40,000 − 6,000 = 34,000 kg·m/s
  • v_f = 34,000/(5,000+3,000) = 4.25 m/s (in direction of heavier car)
✓ Final velocity = 4.25 m/s forward

Real-World Applications

🚗
Crash Reconstruction
Accident investigators use perfectly inelastic collision equations to back-calculate impact speeds from skid marks and final resting positions. v_f is measured; m₁, m₂ are known; solve for initial velocities.
🔫
Ballistics
Bullet-block experiments measure bullet velocity: embed bullet in pendulum block, measure swing angle, calculate v_f, back-solve for bullet speed. This is the ballistic pendulum method used for centuries.
🚂
Railway Engineering
Train coupler design must withstand perfectly inelastic coupling impacts. The impulse (FΔt) and resulting deformation energy inform buffer spring specifications.
🌍
Asteroid Impacts
Large asteroid impacts are perfectly inelastic: the impactor vaporizes and merges energy with Earth. Energy calculations use KE = ½(m_asteroid)v² — explaining why even small asteroids at cosmic speeds release enormous energy.
🏈
Football Tackles
A tackle is a perfectly inelastic collision. A 100 kg defender at 5 m/s stopping a 90 kg running back at 6 m/s: v_f = (100×5−90×6)/190 = (500−540)/190 = −0.21 m/s (defender barely wins).

Common Mistakes to Avoid

⚠️
Using elastic collision formula instead

Perfectly inelastic = objects STICK TOGETHER. Use v_f = (m₁v₁+m₂v₂)/(m₁+m₂). If using elastic formulas (which give two separate final velocities), the answer will be wrong.

⚠️
Forgetting sign conventions

Velocities are signed (direction matters). A car moving left is negative. Head-on: p = m₁v₁ + m₂(−v₂). Drop the sign and you get wrong momentum and wrong final velocity.

⚠️
Expecting to conserve kinetic energy

KE is NOT conserved in inelastic collisions. Do not set ½m₁v₁² + ½m₂v₂² = ½(m₁+m₂)v_f². Only momentum is conserved.

⚠️
Mixing up inelastic and perfectly inelastic

Inelastic: objects separate after collision (KE lost but not all). Perfectly inelastic: objects stick together (maximum KE lost). Most collision problems specify which type; read carefully.

⚠️
Confusing mass units

m must be in kg for SI units. Mixing grams and kg in the same problem gives wrong momentum values and incorrect final velocity.

Frequently Asked Questions

What makes a collision 'perfectly inelastic'? ▾
Objects stick together after impact, moving as one mass. This maximizes kinetic energy loss (converted to heat, sound, deformation) while conserving total momentum. The coefficient of restitution e = 0. Examples: clay balls merging, cars crashing and locking, bullet embedding in wood.
Is momentum truly conserved even when objects deform? ▾
Yes — provided no external forces act during the brief collision time. Internal forces between the objects are Newton's 3rd law pairs (equal and opposite) that cancel in the total. Deformation affects kinetic energy but not momentum.
How much kinetic energy is lost? ▾
ΔKE = m₁m₂(v₁−v₂)²/[2(m₁+m₂)]. For equal masses at opposite velocities: ΔKE = 100%. For a heavy object absorbing a light one: ΔKE ≈ (m_light/m_total)×KE_light. The lighter the absorbed object relative to the total, the smaller the fractional KE loss.
What is the ballistic pendulum? ▾
A classic device for measuring bullet speed: bullet embeds in suspended block (perfectly inelastic). From swing angle, calculate v_f using energy conservation (ΔPE = KE of block). Back-calculate bullet speed using momentum: m_bullet×v_bullet = (m_bullet+m_block)×v_f.
How is this used in crash reconstruction? ▾
Police measure post-crash vehicle positions and speeds (from skid marks, damage). Using v_f = (m₁v₁+m₂v₂)/(m₁+m₂) in reverse — knowing v_f and masses — they calculate pre-impact speeds to determine fault and whether speed limits were exceeded.
Can two objects stop completely in an inelastic collision? ▾
Only if their momenta are equal and opposite: m₁v₁ = −m₂v₂ → v_f = 0. This requires specific mass-velocity combinations. Equal masses at equal and opposite speeds give v_f = 0, losing 100% of kinetic energy.
What happens in a partially inelastic collision? ▾
Objects separate (unlike perfectly inelastic) but with KE lost (unlike elastic). The coefficient of restitution 0 < e < 1 characterizes the degree of inelasticity. Most real collisions fall in this range. Two equations (momentum + restitution) give two unknowns (v₁', v₂').
How does this apply to Newton's cradle? ▾
Newton's cradle demonstrates near-elastic collisions, not inelastic. The balls bounce elastically through the chain via compression waves. If balls were clay (perfectly inelastic), pulling one ball and releasing would result in all balls moving together at 1/n the speed — very different from the observable behavior.

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Interpretation: This relationship connects motion, force, momentum, work or energy in a mechanical system. Assumption: Choose a consistent reference direction and unit system. The model may assume constant acceleration, rigid bodies, negligible losses or an isolated system.

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