$A$ 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 $e = 1 / \sqrt{3}$. Which of the following statement$(s)$ is(are) correct?
$(A)$ $\vec{u}_0 = \sqrt{2gh} \hat{x}$
$(B)$ $\vec{v} = \sqrt{2gh} \hat{x} + \sqrt{2gh} \hat{z}$
$(C)$ $\theta = 60^{\circ}$
$(D)$ $d / h_1 = 2\sqrt{3}$

  • A
    $A, C, D$
  • B
    $A, C, B$
  • C
    $A, C$
  • D
    $A, D$

Explore More

Similar Questions

Two bodies of masses $0.1\, kg$ and $0.4\, kg$ move towards each other with the velocities $1\, m/s$ and $0.1\, m/s$ respectively. After collision they stick together. In $10\, s$,the combined mass travels ............ $m$.

Two cars,both of mass $m$,collide and stick together. Prior to the collision,one car had been traveling north at speed $2v$,while the second was traveling at speed $v$ at an angle $\phi$ south of east (as indicated in the figure). The magnitude of the velocity of the two-car system immediately after the collision is:

Difficult
View Solution

$A$ ball of mass $m_1$ falls from height $h_1$ from rest to strike a spring of force constant $K$,which forces another ball of mass $m_2$ to jump on a horizontal floor at a height $h_2$ below it. Find the horizontal distance at which the ball of mass $m_2$ strikes the floor from the position of the start. [Assume the spring is fixed and the collision is elastic].

This question has Statement-$1$ and Statement-$2$. Of the four choices given after the statements, choose the one that best describes the two statements.
Statement-$1$: $A$ point particle of mass $m$ moving with speed $u$ collides with a stationary point particle of mass $M$. If the maximum energy loss possible is given as $f \left( \frac{1}{2} m u^2 \right)$, then $f = \left( \frac{m}{M + m} \right)$.
Statement-$2$: Maximum energy loss occurs when the particles get stuck together as a result of the collision.

$A$ student skates up a ramp that makes an angle $30^{\circ}$ with the horizontal. He/she starts (as shown in the figure) at the bottom of the ramp with speed $v_0$ and wants to turn around over a semicircular path $xyz$ of radius $R$ during which he/she reaches a maximum height $h$ (at point $y$) from the ground as shown in the figure. Assume that the energy loss is negligible and the force required for this turn at the highest point is provided by his/her weight only. Then ($g$ is the acceleration due to gravity):
$(A)$ $v_0^2 - 2gh = \frac{1}{2} gR$
$(B)$ $v_0^2 - 2gh = \frac{\sqrt{3}}{2} gR$
$(C)$ The centripetal force required at points $x$ and $z$ is zero.
$(D)$ The centripetal force required is maximum at points $x$ and $z$.

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo