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Intersecting paths and particle collisions

Distinguish crossing paths from simultaneous collisions. Solve parametric position equations with separate parameters or a shared clock, check start delays and domains, and calculate separation.

Before you startParametric motion, simultaneous equations and distance between points

01 / A collision needs the same place at the same time

A drawing of crossing paths is not enough.

Two particles can pass through the same location at different times. Use separate parameters to find where paths intersect. To decide whether the particles collide, compare their positions at the same physical time.

Path intersection: equal coordinates. Collision: equal coordinates at one shared time.

In the model, explore the two paths independently, then switch to one shared clock. Delaying B changes whether the particles meet even though the paths still cross.

Two paths, two clocks?Explore

Clock modeShared timeB starts after

A: (2t, t). B after departure: (6 − s, s), in metres.

Shared time 0 s; both particles depart immediately.

A = (0, 0) m; B = (6, 0) m.

Separation 6 m.

Both paths pass through (4, 2); with no delay both arrive there at time 2 s.

The ring marks the path intersection. The dashed segment joins the currently selected positions. In shared-clock mode, B waits at (6, 0) until its stated departure, then uses s = shared time − delay. Independent mode selects two path positions without asserting simultaneity. Axis scales differ.

02 / Use separate parameters for path intersections

One path parameter does not automatically equal the other.

Path A is (2t, t); path B is (6 − s, s). Find their intersection.Worked example

2t = 6 − s and t = s

Equate x and y coordinates.

3t = 6, so t = 2 and s = 2

Solve the simultaneous pair.

Intersection (4, 2)

Substitute back into either path.

The equality t = s here is a result of the equations. It was not assumed merely because both curves were described by parameters.

03 / Changing the speed can prevent a collision

The same path can have different arrival times.

A has position (2t, t). B has position (6 − 2s, 2s). Their clocks both start at the same physical instant. Do they collide for non-negative time?Worked example

2t = 6 − 2s and t = 2s

Use separate parameters to locate the path crossing.

t = 2, s = 1; intersection (4, 2)

A arrives after 2 s; B arrives after 1 s.

They do not collide at that crossing

The arrival times differ.

Using one clock τ: y equality requires τ = 2τ, so τ = 0

At τ = 0 their x-coordinates are 0 and 6, so no collision anywhere.

Finding different arrival times rules out collision at that particular crossing. If paths have several crossings, check them all or solve the full shared-time equations.

Watch two particles reach a crossing at different times

Pause, replay or seek freely. The notes explain the same idea and stay in view.

04 / Solve both coordinates with one physical clock

A common x-coordinate alone does not prove contact.

A = (2τ, τ) and B = (6 − τ, τ), both starting at τ = 0. Find any collision.Worked example

2τ = 6 − τ

The x-coordinates agree when τ = 2.

The y-coordinates are both τ

They agree at that same time.

Collision at τ = 2, position (4, 2)

The time is non-negative.

For A = (2τ, τ), B = (6 − 2τ, 2τ), x agrees at τ = 1.5. But their y-coordinates are 1.5 and 3, so they are still separated.

05 / Translate local elapsed time to the shared clock

An object launched later has a different time origin.

A = (2τ, τ) starts at τ = 0. B follows (6 − 2s, 2s) and departs 1 s later. Does it collide with A?Worked example

s = τ − 1 for τ ≥ 1

B’s elapsed travel time is one second less.

B = (8 − 2τ, 2τ − 2)

Substitute into both B coordinates.

2τ = 8 − 2τ gives τ = 2

Check y: 2 = 2(2) − 2.

Collision at (4, 2), shared time 2 s

B has travelled for only 1 s.

Before departure, state what B is doing. If it waits at its start, use that fixed position; do not apply a negative elapsed travel time as though B had already begun moving.

06 / Find a parameter that makes a collision possible

One coordinate can determine the time; the other then determines the constant.

A = (2τ, τ). B = (6 − kτ, kτ), for shared time τ ≥ 0 and constant k > 0. Find k for collision.Worked example

2τ = 6 − kτ and τ = kτ

Both coordinate conditions are required.

τ = 0 is impossible because the starting x-values differ

Do not divide by τ before checking this case.

Hence k = 1

Use (1 − k)τ = 0 with τ ≠ 0.

Then τ = 2, at (4, 2)

Substitute the constant into the other equation.

Do not choose a constant from only one coordinate and assume the second condition will follow.

07 / Check time restrictions on both paths

An algebraic crossing outside a permitted interval is not part of the model.

Use paths A = (2t, t), 0 ≤ t ≤ 1, and B = (6 − s, s), 0 ≤ s ≤ 4.Worked example

The unrestricted crossing uses t = s = 2

This follows from the simultaneous equations.

A’s allowed parameter stops at 1

The crossing lies beyond its modelled segment.

The two restricted paths do not intersect

Do not extend A’s line silently.

Open endpoints matter too: a meeting at the excluded finish time is not included. A physical collision between moving objects must also fall after both departures and before either model ends.

08 / Find separation at a specified common time

Subtract simultaneous positions, then use Pythagoras.

A = (2τ, τ), B = (6 − 2τ, 2τ). Find their separation at τ = 1.5.Worked example

A = (3, 1.5); B = (3, 3)

Use the same time in both rules.

B − A = (0, 1.5)

The horizontal separation is zero.

Distance = 1.5 m

They are vertically separated despite the equal x-coordinate.

Distance at shared time τ = √((x_B(τ) − x_A(τ))² + (y_B(τ) − y_A(τ))²).

For these rules, separation squared is (6 − 4τ)² + τ². A collision would require this non-negative sum to be zero, so both components must vanish together.

09 / Check every possible meeting

Curved paths may cross more than once.

A = (t, t²), 0 ≤ t ≤ 3; B = (s, 2s), 0 ≤ s ≤ 3, with clocks aligned.Worked example

t = s and t² = 2s

Path-intersection conditions.

t(t − 2) = 0

Candidate parameters are 0 and 2.

Meetings at (0, 0) and (2, 4)

Both occur at equal non-negative times.

A solution gives two different local travel times for the same position. What extra information is needed before deciding whether it is a collision?

Hint

Local clocks may start at different moments.

Worked solution

The relationship between the clocks, including departure times and time units, is needed. Convert both local arrival times to one physical clock and check they agree.

11 / Recap

Equal coordinates must be tied to the right clocks.

Use separate parameters to locate intersections of paths.
Use a single physical time to test collisions.
Compare both coordinates at that same time.
Account for delayed departures and restricted domains.
Check every crossing if there is more than one.
Calculate separation from simultaneous positions, not unrelated points on the paths.

Section 1 of 11 · A collision needs the same place at the same time