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Question
Two particles, 1 and 2, move with constant velocities v1 and v2. At the initial moment their radius vectors are equal to r1 and r2. The relation between these four vectors for the particles to collide is (→r1−→r2)|→r1−→r2|=x(→v2−→v1)|→v2−→v1|. Find x.
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Solution
Considering the diagram given,
Let the particles collide at the point A (shown in figure below), whose position vector is →r3 (say). If t be the time taken by each particle to reach at point A, from triangle law of vector addition:
→r3=→r1+→v1t=→r2+→v2t
so, →r1−→r2=(→v2−→v1)t (1)
therefore, t=|→r1−→r2||→v2−→v1| (2)
From equations (1) and (2)
→r1−→r2=(→v2−→v1)|→r1−→r2||→v2−→v1|
or, →r1−→r2|→r1−→r2|=→v2−→v1|→v2−→v1|, which is the sought relationship.
Let the particles collide at the point A (shown in figure below), whose position vector is →r3 (say). If t be the time taken by each particle to reach at point A, from triangle law of vector addition:
→r3=→r1+→v1t=→r2+→v2t
so, →r1−→r2=(→v2−→v1)t (1)
therefore, t=|→r1−→r2||→v2−→v1| (2)
From equations (1) and (2)
→r1−→r2=(→v2−→v1)|→r1−→r2||→v2−→v1|
or, →r1−→r2|→r1−→r2|=→v2−→v1|→v2−→v1|, which is the sought relationship.
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