Explosions, Recoil, and 2D Collisions

In an explosion or recoil, a system initially at rest splits into parts that move apart, and since the total momentum of the isolated system remains zero, the parts have equal and opposite momenta; in a two-dimensional collision, momentum is conserved separately along each axis. In brief: for a gun and bullet, m_gunv_gun = −m_bulletv_bullet, so the lighter part moves faster. Kinetic energy increases in explosions, because stored chemical or elastic energy is released. In 2D, ΣP_x and ΣP_y are each the same before and after.

Rockets, guns, fireworks and a swimmer pushing off a wall use recoil, while billiards and car crashes at intersections are two-dimensional collisions. This guide explains recoil and explosions, shows how to treat collisions in a plane using components, compares the cases in a table, works through numerical examples and corrects common mistakes.

What happens in explosions and recoil?

Before an explosion, a system (a rifle with a bullet, a firework shell, a spacecraft separating from its rocket stage) is at rest, so the total momentum is zero. The internal forces during the explosion are equal and opposite and act for the same time, so the parts get equal and opposite impulses. After the explosion the vector sum of momenta is still zero: m₁v₁ + m₂v₂ = 0, so v₂ = −(m₁ ÷ m₂)v₁. The lighter part moves faster. The kinetic energies are not equal: the energy of each part is p² ÷ 2m, so the lighter part gets the larger share. The total kinetic energy increases, from zero to a positive value, because chemical or elastic energy is converted.

The same principle applies to the recoil of a gun (the kick), a boat when a person jumps off, a thrown ball while standing on ice, and rocket propulsion. A rocket in space moves forward by expelling mass backward. The center of mass of an isolated system does not accelerate during the explosion: for a firework in the air, the center of mass of the fragments follows the same parabola as the original shell, until the pieces hit the ground.

How do 2D collisions work?

Momentum is a vector, so for a collision in a plane we write two equations:

• x-direction: Σm_iv_ix (before) = Σm_iv_ix (after)
• y-direction: Σm_iv_iy (before) = Σm_iv_iy (after)

Each velocity is resolved into components using v cos θ and v sin θ. For an elastic collision, an additional equation for the conservation of kinetic energy is available. For an elastic glancing collision of a moving object with an identical stationary object, the two objects move off at right angles to each other (90°). Typical applications: billiards, accident reconstruction at intersections, and nuclear and particle scattering experiments.

One-dimensional and two-dimensional problems compared

Worked example: a rifle and billiard balls

1. Recoil of a rifle. A 4.0 kg rifle fires a 0.010 kg bullet at 400 m/s. Total momentum is zero, so 4.0 × v_rifle + 0.010 × 400 = 0, giving v_rifle = −1.0 m/s. The rifle recoils at 1.0 m/s, opposite to the bullet.
2. Energies. Bullet: ½ × 0.010 × 400² = 800 J. Rifle: ½ × 4.0 × 1.0² = 2.0 J. The light bullet takes 99.75 percent of the energy.
3. Explosion into two parts. A 3.0 kg object at rest explodes into 1.0 kg and 2.0 kg pieces. If the 1.0 kg piece moves at 12 m/s east, then the 2.0 kg piece moves at v = 1.0 × 12 ÷ 2.0 = 6.0 m/s west.
4. 2D collision. A billiard ball of mass 0.17 kg moving at 5.0 m/s strikes an identical stationary ball. After the collision, ball A moves at 4.0 m/s at 37° above the original direction, and ball B moves at 3.0 m/s at 53° below it.
   • x-momentum after (per unit mass): 4.0 cos 37° + 3.0 cos 53° = 3.19 + 1.81 = 5.0, equal to the initial 5.0 ✓.
   • y-momentum after: 4.0 sin 37° − 3.0 sin 53° = 2.41 − 2.40 ≈ 0 ✓.
   • Kinetic energy (per unit of ½m): 4.0² + 3.0² = 25 = 5.0², so the collision is elastic. The angle between the paths is 37° + 53° = 90°, as predicted.

5. Perpendicular cars. A 1,000 kg car moving east at 20 m/s and a 1,500 kg car moving north at 10 m/s collide and stick together. p_x = 20,000, p_y = 15,000 kg·m/s, magnitude = 25,000 kg·m/s. v = 25,000 ÷ 2,500 = 10 m/s, at an angle tan⁻¹(15,000 ÷ 20,000) = 37° north of east.

How was recoil put to use?

The principle of rocket propulsion was used in Chinese fire arrows in the 13th century. Konstantin Tsiolkovsky derived the rocket equation in 1903, and Robert Goddard launched the first liquid-fueled rocket in 1926. Newton showed in the Principia how recoil follows from the third law, and Wren’s and Huygens’s experiments with colliding spheres in the 1660s tested the conservation of momentum in two dimensions.

Which mistakes are common?

Mistake 1: Thinking that equal momenta mean equal speeds

The lighter part moves faster. Momenta are equal in magnitude, not speeds.

Mistake 2: Saying that energy is conserved in an explosion

Kinetic energy increases, since other forms of energy are converted.

Mistake 3: Adding momentum magnitudes in 2D

Resolve into components and add each direction separately.

Mistake 4: Forgetting the recoil of the larger object

Both objects move. The heavier object has a smaller speed, but not zero.