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 $t_{P}$ 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
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Question

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 $t_{P}$ 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

Options

G1 Physics · Accepted Answer

The supplied source solution compares the horizontal motion of the two beads. For bead Q, the horizontal distance from A to B is greater than the corresponding horizontal distance traversed by bead P from S to B. The source treats the relevant horizontal component of velocity as the same value v for the comparison. Therefore, t(P) = x(P)/v, t(Q) = x(Q)/v. Since x(Q) > x(P), we obtain t(Q) > t(P). Hence, tP < tQ. Correct option: B.