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A particle of mass m moves with constant velocity v along the straight line 3y = 2x + 6. Observer A is at the origin (0, 0) and observer B is at the point (x1, 0) on the x-axis. If the angular momentum of the particle with respect to B is twice the angular momentum with respect to A, find x1.
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Correct answer: 3
Solution
The perpendicular distance from A to the line is 6/sqrt(13) and from B=(x1,0) is |2x1+6|/sqrt(13). Setting |2x1+6| = 12 gives x1 = 3 (taking the positive root consistent with the given options).
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