A slide with a frictionless curved surface, which becomes horizontal at its lower end, is fixed on the terrace of a building of height 2h 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=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 12. Which of the following statement(s) is (are) correct ?

By conserving energy mgh=12mv02V0=2ghV0→=2ghx^After loosing contact with wedge Vy2−Vy2=2as Vy2−0=6gh Vy=6ghtan⁡θ=VyVx=3 θ=600 d=V0t d=2gh2(3h)g=23hAfter collision, Vy1Vy=e⇒Vy1=136ghVy1=2ghh1=Vy22g=h ∴dh1=23

A slide with a frictionless curved surface, which becomes horizontal at its lower end, is fixed on the terrace of a building of height 2h 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=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 12. Which of the following statement(s) is (are) correct ?

By conserving energy mgh=12mv02V0=2ghV0→=2ghx^After loosing contact with wedge Vy2−Vy2=2as Vy2−0=6gh Vy=6ghtan⁡θ=VyVx=3 θ=600 d=V0t d=2gh2(3h)g=23hAfter collision, Vy1Vy=e⇒Vy1=136ghVy1=2ghh1=Vy22g=h ∴dh1=23

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

A slide with a frictionless curved surface, which becomes horizontal at its lower end, is fixed on the terrace of a building of height 2h 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 $\frac{1}{\sqrt{2}}$ . Which of the following statement(s) is (are) correct ?

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$θ = 60 °$

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

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

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

answer is A, B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

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, $\frac{V_{y}^{1}}{V_{y}} = e \Rightarrow V_{y}^{1} = \frac{1}{\sqrt{3}} \sqrt{6 gh} V_{y}^{1} = \sqrt{2 gh}$
$h_{1} = \frac{V_{y}^{2}}{2 g} = h ∴ \frac{d}{h_{1}} = 2 \sqrt{3}$