Head-On and Stuck — Choosing the Positive Direction
Two gliders ride an air track on a lab bench, with a big + → painted on the rail: right is positive. Glider A comes from the left and glider B from the right; hook-and-loop pads on the facing ends make them stick when they meet. Four sliders set the two masses and the two velocities, and a negative velocity means moving left. Press ▶ to watch the approach, the brief squeeze of the pads and the stuck pair moving off, four times slower than real life — the run plays once and stops, and the sign of the pair's velocity tells you which way it goes.
How to use this simulation
- Mass of A, mA — 0.5 to 3.0 kg, starting at 1.2 kg: the glider body is 0.50 kg, and each extra 100 g brass slotted mass appears on its pegs.
- Velocity of A, vA — −4.0 to +4.0 m/s, starting at +3.0 m/s; negative means moving left.
- Mass of B, mB — 0.5 to 3.0 kg, starting at 2.8 kg.
- Velocity of B, vB — −4.0 to +4.0 m/s, starting at −2.0 m/s. A change re-computes the collision and keeps the moment on screen; nothing moves until you press ▶.
- Play/pause, back to the start, a timeline you can drag to any moment, and ½×, ¼×, ⅛× slow motion (at 1× the scene already runs four times slower than real life, so ⅛× shows the pads squeezing 32 times slower).
- Drag anywhere on the scene to look around.
What to look for
- At the starting values the big read-out gives v′ = −0.50 m/s — 0.50 m/s to the left — and Σp is −2.0 kg m/s both before and after.
- On the number line the momenta of A (+3.6) and B (−5.6) are drawn tip to tail; their total, −2.0 kg m/s, matches the arrow for the stuck pair.
- At ⅛×, or dragging the timeline through the contact, watch the pads squeeze: A's velocity arrow flips from right to left while B slows down, and then the two move off together.
- Make the momenta equal and opposite — 1.4 kg at +2.0 m/s against 2.8 kg at −1.0 m/s — and the pair stops dead. With 2.0 kg at +3.0 m/s against 1.0 kg at rest, it moves right at +2.0 m/s.
- Set vA below vB and the card reads "no collision": the gliders never meet, and each keeps its own momentum.
The physics behind it
Choose the positive direction first — here, to the right — and write the momentum balance on one line: mA·vA + mB·vB = (mA + mB)·v′. At the starting values 1.2 × 3.0 + 2.8 × (−2.0) = 4.0 v′, so 3.6 − 5.6 = 4.0 v′ and v′ = −0.50 m/s. The answer is negative, so the pair moves left: B brought the larger momentum. The contact is modelled as two pads pressing and gripping, every push on A matched by an equal and opposite push on B, so the total momentum measured from the gliders' positions is −2.0 kg m/s before and after. The pair stops dead when the two momenta cancel — for example 1.4 kg at +2.0 m/s against 2.8 kg at −1.0 m/s. If A is not faster than B (vA ≤ vB), the gap never closes and there is no collision at all. The same one-line balance works for any collision in which the bodies stick: 2.0 kg at 3.0 m/s joining 1.0 kg at rest move off at 6.0/3.0 = 2.0 m/s.