They collide when $\dfrac{\vec r_1-\vec r_2}{|\vec r_1-\vec r_2|}=\dfrac{\vec v_2-\vec v_1}{|\vec v_2-\vec v_1|}$, after time $t=\dfrac{|\vec r_1-\vec r_2|}{|\vec v_2-\vec v_1|}$.
(ii) Problem Statement
"Two particles, 1 and 2, move with constant velocities $\vec v_1$ and $\vec v_2$. At the initial moment, their radius vectors are equal to $\vec r_1$ and $\vec r_2$. How must these four vectors be interrelated for the particles to collide?"
Reference Answer Verification (from the supplied solution image):
Let the particles meet at point A with position vector $\vec r_3$. Then $\vec r_3=\vec r_1+\vec v_1t=\vec r_2+\vec v_2t$, so $\vec r_1-\vec r_2=(\vec v_2-\vec v_1)t$. This gives $t=\dfrac{|\vec r_1-\vec r_2|}{|\vec v_2-\vec v_1|}$ and $\dfrac{\vec r_1-\vec r_2}{|\vec r_1-\vec r_2|}=\dfrac{\vec v_2-\vec v_1}{|\vec v_2-\vec v_1|}$. Our work matches.
(iii) The Big Idea in Plain Words
A vector is an arrow. It has a length and a direction. A radius vector $\vec r$ is an arrow from the origin O to a particle. A velocity $\vec v$ is an arrow showing how fast and which way a particle moves.
To collide, both particles must be at the same place at the same time. Call that place A and that time $t$. Each particle gets to A by walking straight along its own path.
This means the starting gap between them, $\vec r_{12}=\vec r_1-\vec r_2$, must be closed by their relative velocity, $\vec v_{21}=\vec v_2-\vec v_1$ (how particle 2 moves as seen by particle 1). The gap closes in a straight line. So the gap arrow and the relative-velocity arrow must point the same way. A unit vector, written with a hat, is an arrow of length 1 that shows only direction. So the two unit vectors must be equal.
(iv) The Solution Viewed in Each Reference Frame
1. Ground or Lab Frame (Stationary Observer)
Origin O fixedWhy: at the collision both particles are at the same point A, with position vector $\vec r_3$.
$$\vec r_3=\vec r_1+\vec v_1t=\vec r_2+\vec v_2t$$
Why: moving the terms shows the gap equals the relative velocity times the time.
$$\vec r_1-\vec r_2=(\vec v_2-\vec v_1)\,t\qquad(1)$$
Why: take the length (magnitude) of both sides of (1) to find $t$. The time is positive.
$$t=\frac{|\vec r_1-\vec r_2|}{|\vec v_2-\vec v_1|}\qquad(2)$$
Why: divide (1) by its own length. Then $t$ cancels, leaving only directions.
$$\frac{\vec r_1-\vec r_2}{|\vec r_1-\vec r_2|}=\frac{\vec v_2-\vec v_1}{|\vec v_2-\vec v_1|}$$
With the default sliders: $\vec r_1-\vec r_2=(-5,\,4)$ m and $\vec v_2-\vec v_1=(-2.5,\,2)$ m/s. Both have unit vector $(-0.78,\,0.62)$. So $t=\dfrac{6.40}{3.20}=2.0$ s, and A is at $(5,\,6)$ m.
2. Centre-of-Mass (CM) Frame
Origin at the centre of massWhy: the CM frame needs the masses $m_1$ and $m_2$. The problem does not give them, and the answer does not depend on them. The sliders only place the CM.
The CM position is $\vec R=\dfrac{m_1\vec r_1+m_2\vec r_2}{m_1+m_2}$. Its velocity is $\vec V=\dfrac{m_1\vec v_1+m_2\vec v_2}{m_1+m_2}$. No force acts, so $\vec V$ is constant. The frame does not accelerate, so no pseudo-force (a "fake" force) is needed.
Why: subtract the CM motion from each particle to get their motion about the CM.
$$\vec r_1'=\frac{m_2}{M}\,\vec r_{12},\quad \vec r_2'=-\frac{m_1}{M}\,\vec r_{12},\quad \vec v_1'=-\frac{m_2}{M}\,\vec v_{21},\quad \vec v_2'=\frac{m_1}{M}\,\vec v_{21}$$
Here $M=m_1+m_2$. Total momentum is $m_1\vec v_1'+m_2\vec v_2'=-\dfrac{m_1m_2}{M}\vec v_{21}+\dfrac{m_1m_2}{M}\vec v_{21}=0$. The telemetry shows $|\vec P|=0$ in this frame.
Why: in this frame both particles move along one straight line through the origin.
They collide only if both head straight at the origin. That means $\vec r_1'$ points opposite to $\vec v_1'$. This is the same as $\hat r_{12}=\hat v_{21}$. The time is $\dfrac{|\vec r_1'|}{|\vec v_1'|}=\dfrac{|\vec r_{12}|}{|\vec v_{21}|}$. They meet at the CM.
3. Medium Frame (Observer Moving at Steady Velocity $\vec u$)
Origin moves at $\vec u$Why: a medium (like wind or a river) moving at $\vec u$ shifts every velocity by the same vector.
$\vec v_1'=\vec v_1-\vec u$ and $\vec v_2'=\vec v_2-\vec u$. The start positions $\vec r_1,\vec r_2$ are unchanged at $t=0$.
Why: only differences matter for a collision. Check what happens to them.
$\vec v_2'-\vec v_1'=\vec v_2-\vec v_1$ and $\vec r_1'-\vec r_2'=\vec r_1-\vec r_2$. Nothing changes, so the condition and $t$ are the same. Only the meeting point moves: $\vec r_3'=\vec r_3-\vec ut$.
What Changes and What Stays the Same Across Frames
- What changes: each particle's own position and velocity, and the place where they meet.
- What stays the same: $\vec r_1-\vec r_2$, $\vec v_2-\vec v_1$, the collision condition, the time $t$, and the closest distance. The graph at the bottom of the simulation is identical in every frame.
(v) The Solution by Each Method
Method 1: Relative-Motion Method
Fastest- Why: if you ride on particle 2, it stands still. Then only particle 1 moves.
Particle 1 starts at $\vec r_{12}=\vec r_1-\vec r_2$ from you and moves with velocity $-\vec v_{21}$. - Why: to hit a particle that is standing still, you must move straight at it.
So $-\vec v_{21}$ must point opposite to $\vec r_{12}$. That is $\hat r_{12}=\hat v_{21}$, and $t=\dfrac{|\vec r_{12}|}{|\vec v_{21}|}$.
Method 2: Graph Method (Triangle of Vectors)
- Why: arrows add tip to tail, so each particle's path ends at A.
From O draw $\vec r_1$, then add $\vec v_1t$. From O draw $\vec r_2$, then add $\vec v_2t$. Both ends must land on A, with $\vec r_3$ drawn from O to A. - Why: the two routes to A form a closed four-sided figure (a parallelogram-like shape, as in the figure).
Going round it: $\vec r_1-\vec r_2=(\vec v_2-\vec v_1)t$. - Why: the same condition follows from the lengths and directions.
$\hat r_{12}=\hat v_{21}$ and $t=\dfrac{|\vec r_{12}|}{|\vec v_{21}|}$.
Method 3: Calculus Method (Closest Approach)
- Why: the gap between the particles changes with time as a straight-line motion.
$\vec D(t)=\vec r_{12}-\vec v_{21}\,t$. They collide if $\vec D=0$ at some $t>0$. - Why: the smallest gap occurs where the slope of $|\vec D|^2$ is zero.
$\dfrac{d|\vec D|^2}{dt}=0\Rightarrow t^*=\dfrac{\vec r_{12}\cdot\vec v_{21}}{|\vec v_{21}|^2}$. We need $t^*>0$, meaning the dot product is positive (they approach). - Why: the smallest gap must be zero.
$d_{min}=\dfrac{|\vec r_{12}\times\vec v_{21}|}{|\vec v_{21}|}=0$, so the cross product is zero (the vectors are parallel). With the dot product positive, this is $\hat r_{12}=\hat v_{21}$, and $t^*=\dfrac{|\vec r_{12}|}{|\vec v_{21}|}$.
Methods 4 and 5: Force and Energy Methods
Force Method: not suited. The velocities are constant, so no forces act before the collision.
Energy Method: not suited. Energy tells us nothing about where and when the particles meet.
Fastest: Method 1 (relative motion). Most reliable in an exam: Method 3 (calculus), since the cross product and dot product give a clear yes or no test. All give $\hat r_{12}=\hat v_{21}$ and $t=\dfrac{|\vec r_{12}|}{|\vec v_{21}|}$.
(vi) Parameter Table
| Symbol | Meaning | Unit | Live value |
|---|---|---|---|
| $\vec r_1,\ \vec r_2$ | start positions (radius vectors) | m | |
| $\vec v_1,\ \vec v_2$ | constant velocities | m/s | |
| $\vec r_{12}$ | gap, $\vec r_1-\vec r_2$ | m | |
| $\vec v_{21}$ | relative velocity, $\vec v_2-\vec v_1$ | m/s | |
| $t$ | time to collide, $|\vec r_{12}|/|\vec v_{21}|$ | s | |
| $d_{min}$ | closest approach | m | |
| $m_1,\ m_2,\ \vec u,\ \rho$ | masses (CM frame only), medium velocity, contact radius (sim only) | kg, m/s, m |
(vii) Common Exam Traps
- Thinking crossing paths means a collision. The particles must reach the crossing at the same time.
- Writing $\vec r_1-\vec r_2$ with $\vec v_1-\vec v_2$. The signs must match: $\vec r_1-\vec r_2$ goes with $\vec v_2-\vec v_1$.
- Only checking that the vectors are parallel. They must also point the same way. Opposite directions mean the particles move apart.
- Using the masses. They do not matter for whether the particles meet.
- Real particles have size. Point particles need $d_{min}=0$ exactly. The sim's contact radius $\rho$ is only to show the effect.
Verified: the live numbers, all frames and all methods give the same condition and time ($t=2.0$ s, A at $(5,6)$ m by default), with units in m, s and m/s.