NTA Abhyas JEE Main2020PhysicsLaws of MotionPractice
A particle is given a horizontal velocity u , from the point P on a smooth horizontal floor which has a vertical circular track at the end. A B C is a semicircular track of radius r in the vertical plane. If the path length P A is x = 3 r and the particle returns to point P , then the initial speed of the particle is
Options
- Au = 5 g r
- Bu = 5 2 g r
- Cu = 4 g r
- Du = 3 2 g r
Correct answer
B. u = 5 2 g r
Step-by-step solution
v C 2 = u 2 - 2 g × 2 r v C 2 = u 2 - 4 g r … 1 x = v C 4 r g … 2 or, x 2 = v C 2 4 r g as, x = 3 r ∴ 9 r 2 = v C 2 4 r g v C 2 = 9 4 g r From equation (1) u 2 - 4 g r = 9 4 g r ⇒ u 2 = 25 4 g r or u = 5 2 g r