JEE Advanced 2023 Paper 1 Q01 Physics Mechanics Projectile Motion Hard

JEE Advanced 2023 Paper 1 · Q01 · Projectile Motion

A slide with a frictionless curved surface, which becomes horizontal at its lower end, is fixed on the terrace of a building of height $3h$ 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 $\vec{u_0} = u_0 \hat{x}$ and falls on the ground at a distance $d$ from the building making an angle $\theta$ with the horizontal. It bounces off with a velocity $\vec{v}$ and reaches a maximum height $h_1$. The acceleration due to gravity is $g$ and the coefficient of restitution of the ground is $1/\sqrt{3}$. Which of the following statement(s) is(are) correct?

  1. A. $\vec{u_0} = \sqrt{2gh}\,\hat{x}$
  2. B. $\vec{v} = \sqrt{2gh}\,(\hat{x} - \hat{z})$
  3. C. $\theta = 60°$
  4. D. $d/h_1 = 2\sqrt{3}$
JEE Advanced multi-correct — pick every correct option, then check.
Reveal answer + step-by-step solution

Correct answer:A, C, D

Solution

Energy conservation on the slide: $\dfrac{1}{2}mu_0^2 = mgh \Rightarrow u_0 = \sqrt{2gh}$ — (A) correct. Vertical velocity just before impact $= \sqrt{2g(3h)} = \sqrt{6gh}$. So $\tan\theta = \sqrt{6gh}/\sqrt{2gh} = \sqrt{3} \Rightarrow \theta = 60°$ — (C) correct. After collision: horizontal component unchanged $= \sqrt{2gh}$; vertical component scaled by $e = 1/\sqrt{3}$: $v_z' = \sqrt{6gh}/\sqrt{3} = \sqrt{2gh}$. So $\vec{v} = \sqrt{2gh}(\hat{x} + \hat{z})$ — (B) shows minus sign so INCORRECT. $h_1 = v_z'^2/(2g) = h$. Range during fall: $d = u_0 \cdot t = \sqrt{2gh}\sqrt{6h/g} = 2h\sqrt{3}$. So $d/h_1 = 2\sqrt{3}$ — (D) correct.

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