Test Index

JEE Advanced 2023 Paper 1 Solutions

Section: Physics

© examsnet.com

Question : 1 of 51

Marks: +1, -0

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

\frac{1}{\sqrt{3}}

. Which of the following statement(s) is(are) correct?

[JEE Adv 2023 P1]

$\vec{u}_0=\sqrt{2gh} \hat{x}$
$\vec{v}=\sqrt{2gh}(\hat{x}-\hat{z})$
$\theta=60^{\circ}$
$\frac{d}{h_1}=2\sqrt{3}$

Solution:

u_0=\sqrt{2gh}

v_z=\sqrt{2g(3h)}

\tan\theta=\;\frac{v_z}{u}=\sqrt{3}

\theta=60^{\circ}

d=u_0 T=u_0 \sqrt{2\left(\frac{3h}{g}\right)}=\sqrt{2gh}\sqrt{2\left(\frac{3h}{g}\right)}

Velocity after collision, only velocity along z-direction change

v_1=e v_z=\sqrt{2gh}

\vec{v}=v_1 \hat{k}+u_0 \hat{i}

=\sqrt{2gh}[\hat{i}+\hat{k}]

h_1=\;\frac{v_1^2}{2g}=h

Finally,

u_0=\sqrt{2gh}, \theta=60^{\circ}, \;\frac{d}{h}=2\sqrt{3}

© examsnet.com

Go to Question:

Next Question