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

Solution

Physics sir · 5 steps

Gravity cannot change P's horizontal speed. Only the string's inward push can, and it speeds P up.

1. Bead Q. On the straight horizontal string, no force acts along the motion, so Q keeps speed v. It covers AB = 2R, so t_Q = 2R/v.
2. What pushes P sideways. Gravity is vertical, so only the normal force from the string can change P's horizontal velocity. Below AB, gravity has a part pointing away from the centre, so the string must pull the bead toward the centre to keep it moving on the circle.
3. Left half, right half. On the left half (S to C), that inward force has a component to the right, so P's horizontal speed rises above v. On the right half it points left, so the horizontal speed falls. At the mirror point S' (45° on the right), energy gives the same speed as at S and the tangent is again at 45°, so the horizontal component is back to v.
4. Compare distances. P starts at S, already ahead of A, so it needs only R + R cos 45° ≈ 1.71R of horizontal travel, against 2R for Q. Over the long stretch S to S' its horizontal speed is above v, and only the short last stretch near B is slower.
5. So. P covers less horizontal distance at a larger average horizontal speed, so it reaches B first: t_P < t_Q, option (a).

The idea to keep. For a bead on a curved wire, gravity never changes the horizontal velocity. Only the normal force does, so look at which way the normal force points along each part of the path.

Answer from the NTA final answer key. Solution by PrepWiser.