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Physics · Motion in Plane

JEE Advanced 2023 — Paper 1 — Question 17

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 hh from the top of the terrace. The ball leaves the slide with a velocity u→0=u0x^∧\overrightarrow{\mathrm{u}}_{0}=\mathrm{u}_{0} \hat{\mathrm{x}}^{\wedge} and falls on the ground at a distance d from the building making an angle θ\theta with the horizontal. It bounces off with a velocity v⃗\vec{v} and reaches a maximum height h1h_{1}. The acceleration due to gravity is gg and the coefficient of restitution of the ground is 1/31 / \sqrt{ } 3. Which of the following statement(s) is(are) correct?

Question figure
  1. Option A:

    u→0=2ghx^\overrightarrow{\mathrm{u}}_{0}=\sqrt{2 \mathrm{gh}} \hat{\mathrm{x}}

    Correct
  2. Option B:

    v→=2gh(x^−z^)\overrightarrow{\mathrm{v}}=\sqrt{2 \mathrm{gh}}(\hat{\mathrm{x}}-\hat{\mathrm{z}})

  3. Option C:

    θ=60∘\theta=60^{\circ}

    Correct
  4. Option D:

    d/h1=23\mathrm{d} / \mathrm{h}_{1}=2 \sqrt{3}

    Correct

Answer: A, C, D

Step-by-step solution

mgh=12mu02\mathrm{mgh}=\frac{1}{2} \mathrm{mu}_{0}^{2} u0=2gh⇒u→0=2ghx^\mathrm{u}_{0}=\sqrt{2 \mathrm{gh}} \Rightarrow \overrightarrow{\mathrm{u}}_{0}=\sqrt{2 \mathrm{gh}} \hat{x} On ground horizontal component of velocity vx=2gh\mathrm{v}_{\mathrm{x}}=\sqrt{2 \mathrm{gh}} Vertical component, VZ=2 g×3 h=6gh\mathrm{V}_{\mathrm{Z}}=\sqrt{2 \mathrm{~g} \times 3 \mathrm{~h}}=\sqrt{6 \mathrm{gh}} tan⁡θ=VZVx=6gh2gh=3⇒θ=60∘\tan \theta=\frac{\mathrm{V}_{\mathrm{Z}}}{\mathrm{V}_{\mathrm{x}}}=\frac{\sqrt{6 \mathrm{gh}}}{\sqrt{2 \mathrm{gh}}}=\sqrt{3} \Rightarrow \theta=60^{\circ} v→=2ghx^+(6gh)13z^\overrightarrow{\mathrm{v}}=\sqrt{2 \mathrm{gh}} \hat{\mathrm{x}}+(\sqrt{6 \mathrm{gh}}) \frac{1}{\sqrt{3}} \hat{\mathrm{z}} =2gh(x^+z^)=\sqrt{2 \mathrm{gh}}(\hat{\mathrm{x}}+\hat{\mathrm{z}}) h1=(eVz)22 g=(1/3)6gh2 g=hh_{1}=\frac{\left(\mathrm{eV}_{\mathrm{z}}\right)^{2}}{2 \mathrm{~g}}=\frac{(1 / 3) 6 \mathrm{gh}}{2 \mathrm{~g}}=\mathrm{h} Time to hit ground after leaving slide t=(6 h g)\mathrm{t}=\sqrt{\left(\frac{6 \mathrm{~h}}{\mathrm{~g}}\right)} d=(2gh)(6 h g)=23 h\mathrm{d}=(\sqrt{2 \mathrm{gh}})\left(\sqrt{\frac{6 \mathrm{~h}}{\mathrm{~g}}}\right)=2 \sqrt{3} \mathrm{~h} dh=23hh=23\frac{d}{h}=\frac{2 \sqrt{3} h}{h}=2 \sqrt{3}.

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2023
Paper
Paper 1
Subject
Physics
Chapter
Motion in Plane
Topic
Oblique and Horizontal Projectile Motion