Applied Maths: Impacts and Collisions (Strand 6) — Higher Level Notes

1. Impulse

Impulse is a force multiplied by the time it acts, and that product is exactly the change in momentum:

J = Ft = mv - mu

Which makes its unit the newton second, N s. Not the joule — that is a newton metre and belongs to work. Impulse is a vector, so direction and sign matter.

2. The two conservation rules, and how they differ

• Momentum is conserved in every collision. Always. m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
• Kinetic energy is conserved only when the collision is perfectly elastic, meaning e = 1.

This is the distinction the chapter is built on. Assuming both are always conserved will wreck a question from the first line.

3. Newton's law of restitution

e = (v₂ - v₁) / (u₁ - u₂)

In words: the relative speed of separation is e times the relative speed of approach. Since e lies between 0 and 1, the spheres can never separate faster than they approached.

• e = 1 — perfectly elastic. No kinetic energy is lost.
• e = 0 — the particles coalesce and move on together. This is where the energy loss is greatest.

Momentum, note, is conserved for every value of e. Only the energy depends on it.

4. Direct collisions: the standard method

1. Choose a positive direction and mark every velocity with its sign.
2. Write the conservation of momentum equation.
3. Write the restitution equation.
4. Solve the two simultaneously for the two unknown velocities.

Then check: if e came out as 1, the kinetic energy before and after should match exactly. That takes fifteen seconds and catches most errors.

5. Oblique collisions and smooth surfaces

When a sphere strikes a smooth fixed wall, resolve its velocity parallel and perpendicular to the wall:

v_parallel = u_parallel   |   v_{perpendicular} = e × u_{perpendicular} (reversed)

Because the wall is smooth it exerts no force along itself, so the parallel component passes through untouched. Applying e to both components is a common and expensive error.

A consequence worth stating: since the perpendicular component shrinks while the parallel one does not, the path flattens, so the angle the path makes with the wall gets smaller after the bounce. It is not like light off a mirror.

6. Projectiles that bounce

Ground is a horizontal smooth surface, so the same rule applies: horizontal velocity unchanged, vertical velocity reversed and multiplied by e. Each hop is therefore e times the previous one in vertical speed, which makes each successive range e times the one before.

For a ball dropped onto a floor, the rebound height follows:

h' = e²h