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

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 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 $θ$ 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 [2023]

Smart Achievers · Accepted Answer

(1, 3, 4)

$From v^{2} - u^{2} = 2 g h$

$0^{2} - u_{0}^{2} = 2 g h$ $∴ u_{0} = \sqrt{2 g h}$ $∴ {\vec{u}}_{0} = \sqrt{2 g h} \hat{x}$

$and v_{z} = \sqrt{2 g (3 h)}$

$∴$ $\tan θ = \frac{v_{z}}{u_{0}} = \frac{\sqrt{2 g (3 h)}}{\sqrt{2 g h}} = \sqrt{3}$ $∴$ $θ = 60 °$

$Distance d = u_{0} t = \sqrt{2 g h} \times \sqrt{\frac{2 \times 3 h}{g}} = 2 \sqrt{3} h$

After collision only velocity along z-direction will change

$v_{1} = e v_{z} = \frac{1}{\sqrt{3}} \times \sqrt{2 g (3 h)} = \sqrt{2 g h}$

$∴$ $\vec{v} = v_{1} \hat{k} + u_{0} \hat{i} = \sqrt{2 g h} \hat{k} + \sqrt{2 g h} \hat{i} = \sqrt{2 g h} [\hat{i} + \hat{k}]$

$Height h_{1} = \frac{v_{1}^{2}}{2 g} = \frac{{(\sqrt{2 g h})}^{2}}{2 g} = h$

$∴$ $\frac{d}{h_{1}} = \frac{2 \sqrt{3} h}{h} = 2 \sqrt{3}$

Therefore options (1, 3, 4) are correct.

Solution

1 ${\vec{u}}_{0} = \sqrt{2 g h} \hat{x}$

2 $\vec{v} = \sqrt{2 g h} (\hat{x} - \hat{z})$

3 $θ = 60 °$

4 $\frac{d}{h_{1}} = 2 \sqrt{3}$

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Solution

1 u→0=2gh x^

2 v→=2gh (x^-z^)

3 θ=60°

4 dh1=23

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1 ${\vec{u}}_{0} = \sqrt{2 g h} \hat{x}$

2 $\vec{v} = \sqrt{2 g h} (\hat{x} - \hat{z})$

3 $θ = 60 °$

4 $\frac{d}{h_{1}} = 2 \sqrt{3}$