🏐 1D Collision Solver

Object 1

Object 2

The Law of Conservation of Momentum

For any system of interacting bodies, the total momentum remains constant, provided no external resultant force acts on the system. This principle is derived directly from Newton's Third Law: for every internal action force, there is an equal and opposite reaction force.

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

1. Elastic vs. Inelastic Collisions

In every collision, momentum is conserved. However, kinetic energy (KE) behavior defines the collision type:

  • Perfectly Elastic: Kinetic energy is fully conserved (KE_{initial} = KE_{final}). This is common for gas molecules but rare for macroscopic objects.
  • Inelastic: Some KE is transformed into heat, sound, or deformation energy. Most real-world collisions are inelastic.
  • Perfectly Inelastic: The objects stick together after the collision, moving with a common velocity. This results in the maximum possible KE loss.

2. Impulse and Force

Impulse (J) is the change in momentum. It can be caused by a large force acting over a short time, or a small force acting over a long time. In A Level exams, this is often represented as the area under a Force-Time graph.

Impulse (J) = FΔt = Δp = m(v - u)
Exam Tip: Direction is EVERYTHING. Momentum is a vector. If you define "Right" as positive, make sure any object moving "Left" has a negative velocity in your m_1u_1 + m_2u_2 calculation. Failing to do this is the #1 reason for incorrect answers.

Deep Dive: Worked Examples

✅ Example 1: Perfectly Inelastic (Stuck Together)

A 2.0 kg trolley moving at 6.0 m/s hits a stationary 4.0 kg trolley. They stick together. Find the final common velocity.

Step 1: Conservation of Momentum (2.0 × 6.0) + (4.0 × 0) = (2.0 + 4.0) × v
Step 2: Simplify 12.0 = 6.0v
Step 3: Solve v = 2.0 m/s

✅ Example 2: Impulse on a Wall

A 0.5 kg ball hits a wall at 12 m/s and rebounds at 8 m/s in the opposite direction. The contact time is 0.05s. Find the average force.

Step 1: Change in Momentum (Define right as +) Δp = m(v - u) = 0.5 × (-8 - 12) = 0.5 × (-20) = -10 kg·m/s
Step 2: Apply Force = Δp / Δt F = -10 / 0.05
Step 3: Solve F = -200 N (The negative sign means the force acts away from the wall).

✅ Example 3: Explosion (Recoil)

A 5.0 kg rifle fires a 0.02 kg bullet at 400 m/s. Calculate the recoil velocity of the rifle.

Step 1: Initial Momentum is Zero 0 = (5.0 × V_recoil) + (0.02 × 400)
Step 2: Rearrange -5.0V = 8.0
Step 3: Solve V = -1.6 m/s

✅ Example 4: Kinetic Energy Loss

In Example 1, calculate the total Kinetic Energy lost during the collision.

Step 1: Initial KE KE_i = ½(2.0)(6.0)² = 36.0 J
Step 2: Final KE (v = 2.0 m/s) KE_f = ½(2.0 + 4.0)(2.0)² = ½(6.0)(4.0) = 12.0 J
Step 3: Loss Loss = 36.0 - 12.0 = 24.0 J

✅ Example 5: Rocket Propulsion

A rocket of mass 10,000 kg ejects 50 kg of gas per second at a velocity of 2000 m/s relative to the rocket. Find the thrust.

Step 1: Resultant Force = Rate of change of momentum F = (Δm / Δt) × v
Step 2: Plug in values F = 50 kg/s × 2000 m/s
Step 3: Solve F = 100,000 N (100 kN)