A bead P slides on a frictionless semi-circular string ACB and is at point S at t=0; at this instant the horizontal component of its velocity is v. Another bead Q of the same mass as P is ejected from point A at t=0 along the horizontal string AB with the speed v. Friction between the beads and the…

Question

A bead P slides on a frictionless semi-circular string ACB and is at point S at t=0; at this instant the horizontal component of its velocity is v. Another bead Q of the same mass as P is ejected from point A at t=0 along the horizontal string AB with the speed v. Friction between the beads and the respective strings may be neglected in both cases. Let t_P and t_Q be the respective times taken by beads P and Q to reach the point B. Then the relation between t_P and t_Q is

Accepted Answer

(4) t_P<t_Q. Idea: compare the horizontal progress of the two beads, not their path lengths. Q runs along the horizontal string AB. The string is horizontal and frictionless, so gravity is balanced by the normal force and Q keeps the constant speed v the whole way. P drops below the line AB on the arc. Gravity does positive work on the way down, so P speeds up, and its horizontal component of velocity is larger than v over most of the journey. On the way back up P slows again, returning to horizontal component v at B by symmetry, but in between it has been running ahead. So P covers the horizontal span AB in less time: t_P<t_Q. This is the same reason a ball on a dipped track beats one on a straight track over the same horizontal distance.

Step-by-step solution

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