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Irodov 1.5Trickykinematicsvectorsrelative-motioncross-product

Collision Condition for Two Particles

Problem 1.5
Two particles, 1 and 2, move with constant velocities v⃗1\vec{v}_1 and v⃗2\vec{v}_2. At the initial moment their radius vectors equal r⃗1\vec{r}_1 and r⃗2\vec{r}_2. How must these four vectors be interrelated for the particles to collide?
Answer: (r⃗1−r⃗2)×(v⃗1−v⃗2)=0⃗(\vec{r}_1 - \vec{r}_2) \times (\vec{v}_1 - \vec{v}_2) = \vec{0}, with particles approaching
Solution Path
For collision, the initial separation must be parallel to the relative velocity. In particle 2's frame, particle 1 must aim directly at particle 2. The condition is (r1 - r2) x (v1 - v2) = 0, with the particles approaching each other.
01Problem Restatement
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Two particles are moving through space, each with a constant velocity. At time t=0t = 0, particle 1 is at position r⃗1\vec{r}_1 and particle 2 is at position r⃗2\vec{r}_2. Particle 1 moves with velocity v⃗1\vec{v}_1 and particle 2 moves with velocity v⃗2\vec{v}_2.Given:
- r⃗1\vec{r}_1 = initial position vector of particle 1
- r⃗2\vec{r}_2 = initial position vector of particle 2
- v⃗1\vec{v}_1 = constant velocity of particle 1
- v⃗2\vec{v}_2 = constant velocity of particle 2We need to find: the condition on these four vectors that guarantees the two particles collide (occupy the same point at the same time).
Find the collision condition on r⃗1,r⃗2,v⃗1,v⃗2\vec{r}_1, \vec{r}_2, \vec{v}_1, \vec{v}_2
02Physical Picture and Strategy
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Picture two cars on a vast flat field, each driving in a straight line at constant speed. Car 1 starts at one location and drives northeast. Car 2 starts at another location and drives northwest. Will they crash into each other?For them to collide, their straight-line paths must intersect, AND they must reach the intersection point at exactly the same time. Two paths can cross without a collision if the timing is wrong.Each particle's position at time tt is given by uniform motion:r⃗1(t)=r⃗1+v⃗1 t,r⃗2(t)=r⃗2+v⃗2 t\vec{r}_1(t) = \vec{r}_1 + \vec{v}_1\, t, \qquad \vec{r}_2(t) = \vec{r}_2 + \vec{v}_2\, tStrategy: We will set up the collision equation r⃗1(t)=r⃗2(t)\vec{r}_1(t) = \vec{r}_2(t) and find what conditions on the four vectors allow a real, positive solution for tt. Then we will re-interpret the result geometrically using the relative-motion frame of particle 2.
Position at time tt: r⃗(t)=r⃗0+v⃗ t\vec{r}(t) = \vec{r}_0 + \vec{v}\,t
03Step 1: Write the Collision Equation
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For a collision, both particles must be at the same point at the same instant t∗t^*. Using the equation of uniform motion for each particle:r⃗1+v⃗1 t∗=r⃗2+v⃗2 t∗\vec{r}_1 + \vec{v}_1\, t^* = \vec{r}_2 + \vec{v}_2\, t^*Rearranging by collecting all terms with t∗t^* on one side:r⃗1−r⃗2=(v⃗2−v⃗1) t∗\vec{r}_1 - \vec{r}_2 = (\vec{v}_2 - \vec{v}_1)\, t^*This is a vector equation. The left side, r⃗1−r⃗2\vec{r}_1 - \vec{r}_2, is a fixed vector (the initial separation). The right side is the vector (v⃗2−v⃗1)(\vec{v}_2 - \vec{v}_1) scaled by the scalar t∗t^*.This means: the initial separation vector must equal a scalar multiple of the relative velocity vector. For two vectors to be related by a scalar, they must be parallel.
r⃗1−r⃗2=(v⃗2−v⃗1) t∗\vec{r}_1 - \vec{r}_2 = (\vec{v}_2 - \vec{v}_1)\,t^*
04Step 2: Switch to Particle 2's FrameKEY INSIGHT
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The algebra becomes obvious when we look at it from particle 2's reference frame.In particle 2's frame, particle 2 is stationary at the origin. Particle 1 starts at position (r⃗1−r⃗2)(\vec{r}_1 - \vec{r}_2) and moves with relative velocity (v⃗1−v⃗2)(\vec{v}_1 - \vec{v}_2).For particle 1 to hit particle 2 (which is sitting at the origin), particle 1's velocity must point directly at the origin. That means the relative velocity (v⃗1−v⃗2)(\vec{v}_1 - \vec{v}_2) must be anti-parallel to the relative position (r⃗1−r⃗2)(\vec{r}_1 - \vec{r}_2).Anti-parallel means the vectors point in exactly opposite directions. If v⃗1−v⃗2\vec{v}_1 - \vec{v}_2 pointed in any other direction, particle 1 would sail past particle 2 and miss.This means: the relative velocity must aim directly at the target. Any sideways component causes a miss.
(v⃗1−v⃗2)(\vec{v}_1 - \vec{v}_2) must be anti-parallel to (r⃗1−r⃗2)(\vec{r}_1 - \vec{r}_2)
05Step 3: The Cross Product Condition
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From the rearranged collision equation:r⃗1−r⃗2=(v⃗2−v⃗1) t∗\vec{r}_1 - \vec{r}_2 = (\vec{v}_2 - \vec{v}_1)\,t^*For this to have a solution, the left and right sides must be parallel vectors. Using the property that the cross product of parallel vectors is zero:(r⃗1−r⃗2)×(v⃗1−v⃗2)=0⃗( \vec{r}_1 - \vec{r}_2 ) \times ( \vec{v}_1 - \vec{v}_2 ) = \vec{0}Additionally, the collision time must be positive (t∗>0t^* > 0). Since r⃗1−r⃗2=(v⃗2−v⃗1) t∗\vec{r}_1 - \vec{r}_2 = (\vec{v}_2 - \vec{v}_1)\,t^* and t∗>0t^* > 0, the vectors (r⃗1−r⃗2)(\vec{r}_1 - \vec{r}_2) and (v⃗2−v⃗1)(\vec{v}_2 - \vec{v}_1) must point in the same direction (not opposite).Equivalently, (r⃗1−r⃗2)(\vec{r}_1 - \vec{r}_2) must be anti-parallel to (v⃗1−v⃗2)(\vec{v}_1 - \vec{v}_2), meaning the particles must be approaching each other, not moving apart.This means: the cross product being zero ensures the vectors are parallel (collision is geometrically possible), and the direction constraint ensures the collision happens in the future, not the past.
(r⃗1−r⃗2)×(v⃗1−v⃗2)=0⃗(\vec{r}_1 - \vec{r}_2) \times (\vec{v}_1 - \vec{v}_2) = \vec{0}
06Verification and Summary
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Verification:Dimensional check: The cross product of (r⃗1−r⃗2)(\vec{r}_1 - \vec{r}_2) (dimension [L][\text{L}]) and (v⃗1−v⃗2)(\vec{v}_1 - \vec{v}_2) (dimension [L/T][\text{L/T}]) gives dimension [L2/T][\text{L}^2/\text{T}], which is a valid physical quantity (areal velocity). Setting it to zero is dimensionally consistent. ✓\checkmarkLimiting case: If both particles start at the same point (r⃗1=r⃗2\vec{r}_1 = \vec{r}_2), then (r⃗1−r⃗2)=0⃗(\vec{r}_1 - \vec{r}_2) = \vec{0}, so the cross product is automatically zero regardless of velocities. This makes sense: particles at the same point have already collided at t=0t = 0. ✓\checkmarkLimiting case: If both particles have the same velocity (v⃗1=v⃗2\vec{v}_1 = \vec{v}_2), they move in parallel and the separation never changes. Collision requires r⃗1=r⃗2\vec{r}_1 = \vec{r}_2 (already together). The formula gives (r⃗1−r⃗2)×0⃗=0⃗(\vec{r}_1 - \vec{r}_2) \times \vec{0} = \vec{0}, which is trivially satisfied, but t∗t^* would be undefined (division by zero), meaning no finite collision time unless they started together. ✓\checkmarkSummary: For two particles moving with constant velocities to collide, the initial separation vector (r⃗1−r⃗2)(\vec{r}_1 - \vec{r}_2) must be exactly parallel to the relative velocity vector (v⃗2−v⃗1)(\vec{v}_2 - \vec{v}_1). In particle 2's frame, this means particle 1 is heading straight toward particle 2 with no sideways drift. The mathematical condition is that the cross product (r⃗1−r⃗2)×(v⃗1−v⃗2)=0⃗(\vec{r}_1 - \vec{r}_2) \times (\vec{v}_1 - \vec{v}_2) = \vec{0}, with the particles approaching (not receding).
(r⃗1−r⃗2)×(v⃗1−v⃗2)=0⃗\boxed{(\vec{r}_1 - \vec{r}_2) \times (\vec{v}_1 - \vec{v}_2) = \vec{0}}
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