Head-On Elastic Collisions

Predicting the outcome

So far you have used conservation of momentum to find one unknown velocity. But if two carts collide and you know only their masses and initial velocities, there are two unknowns: both final velocities. One equation is not enough.

For an elastic collision, you have a second equation: kinetic energy is conserved too. Two equations, two unknowns. In this lesson you will solve them once, in general, for a head-on (one-dimensional) collision, and get formulas that predict any elastic collision on a line. Then you will see what the formulas say about billiards, nuclear reactors and a famous bouncing-ball trick.

Solving the two equations

Take object 1 (mass m₁, velocity v₁) hitting object 2 (mass m₂) at rest. For an elastic collision:

• Momentum: m₁v₁ = m₁v₁′ + m₂v₂′ (1)
• Kinetic energy: ½m₁v₁² = ½m₁v₁′² + ½m₂v₂′² (2)

Move the m₁ terms to the left side:

• From (1): m₁(v₁ − v₁′) = m₂v₂′
• From (2), after multiplying by 2: m₁(v₁² − v₁′²) = m₂v₂′², and v₁² − v₁′² factors as (v₁ − v₁′)(v₁ + v₁′).

Divide the second by the first (assuming the objects actually collide, so v₁ ≠ v₁′):

v₁ + v₁′ = v₂′

Substitute this into (1) and solve. The results are:

v₁′ = [(m₁ − m₂) ÷ (m₁ + m₂)] v₁ and v₂′ = [2m₁ ÷ (m₁ + m₂)] v₁

The step v₁ + v₁′ = v₂′ can be rearranged as v₁ − 0 = −(v₁′ − v₂′). In words: in a head-on elastic collision, the relative velocity of the two objects reverses, with the same size. They separate exactly as fast as they approached. This is true even when both objects are moving.

The three special cases

Look at the formulas for a target at rest:

• Equal masses (m₁ = m₂): v₁′ = 0 and v₂′ = v₁. The velocities swap.
• Heavy hits light (m₁ ≫ m₂): v₁′ ≈ v₁ and v₂′ ≈ 2v₁. The heavy object hardly notices; the light one leaves at up to twice the incoming speed. This is why a heavy golf club head launches the much lighter ball faster than the club is moving.
• Light hits heavy (m₁ ≪ m₂): v₁′ ≈ −v₁ and v₂′ ≈ 0. The light object bounces straight back at almost its original speed, like a ball bouncing off a wall.

Application: slowing neutrons in a CANDU reactor

Remember from Grade 11: in the SPH3U unit Thermal and Nuclear Energy, you learned that the CANDU reactors in Ontario use heavy water as a moderator to slow down the fast neutrons released in fission, because slow neutrons are much more likely to cause the next fission.

Elastic collisions explain why the moderator must contain light nuclei. A neutron that hits a nucleus at rest head-on transfers the fraction 4m₁m₂ ÷ (m₁ + m₂)² of its kinetic energy (you can derive this in the Challenge).

• Hydrogen nucleus (mass ≈ the neutron's): the neutron can give up all its kinetic energy in one head-on hit.
• Deuterium, the hydrogen in heavy water (mass ≈ 2 neutrons): up to 8/9 ≈ 89% per head-on hit.
• Carbon-12 (mass ≈ 12 neutrons): only about 28% per head-on hit.
• Uranium-238: about 2%. The neutron just bounces off.

Ordinary water slows neutrons well, but its hydrogen also absorbs some of them. Heavy water absorbs far fewer, which is one reason CANDU reactors can run on natural (unenriched) uranium.