1D and 2D Collisions: Avoiding Double-Counted Signs
Collision problems are among the most reliably error-prone in Further Mechanics — not because the physics is hard, but because of a single, avoidable sign mistake. Students often learn Newton's Experimental Law as a statement about unsigned speeds, and then try to use it alongside signed velocities. Mixing the two frameworks causes a negative sign to be counted twice. Use signed velocities throughout and the algebra handles the signs automatically.
The two equations — always in velocity form
Choose a positive direction and express every velocity as a signed quantity in that direction. Then write both equations using velocities, not speeds:
The NEL in this form is always correct. You never need to decide whether the balls are approaching or separating before writing it down — the algebra produces the right sign automatically.
Where the double-count comes from
Many textbooks state NEL as a ratio of unsigned speeds:
This is correct, but it requires you to compute the approach and separation speeds by hand — and that computation is where the sign error enters.
Suppose ball A moves right and ball B moves left toward it. A student assigns:
- uA = 3 (moving right, positive)
- uB = −2 (moving left, negative)
The speed of approach is 3 + 2 = 5. So far so good. Now the double-count: the student writes speed of approach = uA + uB = 3 + (−2) = 1. They have used uB = −2 in the sum rather than |uB| = 2. The negative sign in uB has already been applied once (by assigning it a negative value), and has then been applied again (by adding it instead of subtracting), cancelling itself out and producing the wrong answer.
The velocity form uA − uB = 3 − (−2) = 5 gives the correct approach speed automatically, with no separate reasoning required.
1D worked example
Ball A (mass 2 kg) moves right at 3 m s−1. Ball B (mass 1 kg) moves left at 2 m s−1. They collide directly. The coefficient of restitution is e = 0.5. Find the velocities after the collision.
Step 1 — define the positive direction and assign signed velocities
Take right as positive. Then:
Step 2 — write the two equations
Step 3 — solve
From (2): vB = vA + 2.5. Substituting into (1):
Both are positive, so both balls move right after the collision. Ball A has slowed; ball B has reversed direction and sped up. Check:
The same example done wrong
A student treats uA = 3 and uB = 2 as unsigned speeds, correctly writes momentum as:
So far identical. But for NEL they write speed of separation = e × speed of approachand decide that since A moves right and B left, separation means vB to the right, so they write:
This happens to give the same equation — but only because they were careful about which speed to add. If they had instead written uA − uBwith the unsigned values (3 − 2 = 1) they would obtain 0.5 × 1 = 0.5, which is wrong. The velocity form uA − uB with signed velocities (3 − (−2) = 5) always gives the right answer without any case analysis.
A quick check: impulse and Newton's third law
The impulse on A from B is mA(vA − uA), and by Newton's third law the impulse on B from A is equal and opposite. This gives a free check: the two impulse magnitudes must be equal.
Equal and opposite impulses, as expected. This check costs almost nothing and immediately catches a wrong sign.
2D oblique collisions
When two smooth spheres collide at an angle, the collision force acts along the line joining their centres (the line of centres). This means:
- Along the line of centres: apply momentum conservation and the velocity form of NEL to the components along this line. This is identical to the 1D problem.
- Perpendicular to the line of centres: no force acts, so the tangential velocity component of each sphere is unchanged.
The sign convention advice is the same as in 1D — define a positive direction along the line of centres, assign signed components, and use vB − vA = e(uA − uB) for those components only.
The physical meaning of e
The coefficient of restitution e must satisfy 0 ≤ e ≤ 1. The extremes are:
| e | Name | What happens |
|---|---|---|
| e = 1 | Perfectly elastic | Kinetic energy conserved; balls bounce apart at the same relative speed |
| 0 < e < 1 | Inelastic | Some kinetic energy lost; balls separate at reduced relative speed |
| e = 0 | Perfectly inelastic | Maximum energy lost; balls move together after collision |
Note: kinetic energy is never gained in a collision (e ≤ 1), and the balls never pass through each other (the separation speed is always at most the approach speed). If your answer gives vB − vA and uA − uB with opposite signs, the balls have passed through each other — a sure sign of a sign error in the working.