JEE Advanced
Physics
Motion In Two Dimensions
2023
JEE Advanced 2023 (Paper 1)
JEE Advanced Physics Question (2023) — Solution
Question
A slide with a frictionless curved surface, which becomes horizontal at its lower end, is fixed on the terrace of a building of height 3 h from the ground, as shown in the figure. A spherical ball of mass m is released on the slide from rest at a height h from the top of the terrace. The ball leaves the slide with a velocity u → 0 = u 0 x ∧ and falls on the ground at a distance d from the building making an angle θ with the horizontal. It bounces off with a velocity v → and reaches a maximum height h 1 . The acceleration due to gravity is g and the coefficient of restitution of the ground is 1 3 . Which of the following statement(s) is(are) correct?
Options
- A. u → 0 = 2 g h x ∧
- B. v → = 2 g h x ∧ - z ∧
- C. θ = 60 o
- D. d h 1 = 2 3
Step-by-step solution
Using energy conservation, m g h = 1 2 m u 0 2 ⇒ u 0 = 2 g h u 0 is the horizontal component of the velocity throughout the flight. For vertical component of the velocity(just before collision with the ground), we can use equation of motion v z 2 = 0 2 + 2 - g - 3 h ⇒ v z = 6 g h Angle θ can be written as, tan θ = v z u = 3 ⇒ θ = 60 o Horizontal distance covered just before the collision will be, d = u 0 T = u 0 2 3 h g = 2 g h 2 3 h g = 2 3 h After collision, only velocity along z -direction changes. Therefore, v 1 = e v z = 2 g h and hence v → = u 0 i ∧ + v 1 k ∧ = 2 g h i ∧ + k ∧ = 2 g h x ∧ + z ∧ Now, maximum height achieved after collision will be, h 1 = v 1 2 2 g = h . Finally, u 0 = 2 g h ,   θ = 60 o ,   d h 1 = 2 3 .
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Related: Physics — Motion In Two Dimensions · All PYQ Banks