A bead P sliding on a frictionless semi-circular string (ACB) and it is at point S at t = 0 and 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 tP and tQ be the respective times taken by beads P and Q to reach the point B, then the relation between tP and tQ is [figure]
- tP < tQ
- tP = tQ
- tP > 1.25tQ
- tP > tQ
Answer
(1) tP < tQ
Solution
Let R be the radius. Q moves with constant speed v along AB:
tQ = 2Rv
At S the radius makes 45 with AB, so the velocity of P (along the tangent) makes 45 with the horizontal:
speed of P at S = v 45 = 2v
From S to C and on to the same level on the other side, P is below S, so (energy conservation) its speed is more than 2v; only on the last 45 arc near B does it slow down a little.
Path of P from S to B = arc of 135 = 3 R4
tP 3 R/4 2v = 3 4 2 Rv 1.67 Rv
while tQ = 2 Rv
tP < tQ
Answer: (1)