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 u0=u0x^ 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 h1. The acceleration due to gravity is 𝑔 and the coefficient of restitution of the ground is 13. Which of the following statement(s) is(are) correct?

θ = 60 ° , u 0 = 2 gh x ^ , v → = 2 gh ( x ^ − z ^ )

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 u0=u0x^ 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 h1. The acceleration due to gravity is 𝑔 and the coefficient of restitution of the ground is 13. Which of the following statement(s) is(are) correct?

By conserving energy mgh = 1 2 mv 0 2 V 0 = 2 gh V 0 → = 2 gh x ^ After loosing contact with wedge V y 2 − V y 2 = 2 as V y 2 − 0 = 6 gh V y = 6 gh tan ⁡ θ = V y V x = 3 θ = 60 0 d = V 0 t d = 2 gh 2 ( 3 h ) g = 2 3 h After collision, V y 1 V y = e ⇒ V y 1 = 1 3 6 gh V y 1 = 2 gh h 1 = V y 2 2 g = h ∴ d h 1 = 2 3

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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 $u_{0} = u_{0} \hat{x}$ 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 h₁. The acceleration due to gravity is 𝑔 and the coefficient of restitution of the ground is $\frac{1}{\sqrt{3}}$ . Which of the following statement(s) is(are) correct?

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$θ = 60 °$

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

$u_{0} = \sqrt{2 gh} \hat{x}$

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

answer is A, C, D.

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Detailed Solution

By conserving energy
$\begin{array}{l} mgh = \frac{1}{2} {mv}_{0}^{2} \\ V_{0} = \sqrt{2 gh} \\ \vec{V_{0}} = \sqrt{2 gh} \hat{x} \end{array}$
After loosing contact with wedge
$V_{y}^{2} - V_{y}^{2} = 2 as V_{y}^{2} - 0 = 6 gh V_{y} = \sqrt{6 gh}$
$\tan θ = \frac{V_{y}}{V_{x}} = \sqrt{3} θ = 60^{0} d = V_{0} t d = \sqrt{2 gh} \sqrt{\frac{2 (3 h)}{g}} = 2 \sqrt{3} h$
After collision, $\frac{V_{y}^{1}}{V_{y}} = e \Rightarrow V_{y}^{1} = \frac{1}{\sqrt{3}} \sqrt{6 gh} V_{y}^{1} = \sqrt{2 gh}$
$h_{1} = \frac{V_{y}^{2}}{2 g} = h ∴ \frac{d}{h_{1}} = 2 \sqrt{3}$

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