A slide with a frictionless curved surface, which becomes horizon… - Vidyadip

MCQ

A slide with a frictionless curved surface, which becomes horizontal at its lower end,, is fixed on the terrace of a building of height $3 h$ from the ground, as shown in the figure. A spherical ball of mass $\mathrm{m}$ 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 $\overrightarrow{\mathrm{v}}$ and reaches a maximum height $h_l$. 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?

($AV$) $\vec{u}_0=\sqrt{2 g h} \hat{x}$ ($B$) $\vec{v}=\sqrt{2 g h}(\hat{x}-\hat{z})$ ($C$) $\theta=60^{\circ}$ ($D$) $d / h_1=2 \sqrt{3}$

✓
$A,C,D$
B
$A,C,B$
C
$A,C$
D
$A,D$

✓

Answer

Correct option: A.

$A,C,D$

a
$\overrightarrow{\mathrm{v}}_1=\sqrt{2 g h} \hat{\mathrm{i}}-\sqrt{2 g 3 \mathrm{~h}} \hat{\mathrm{k}}$

$\overrightarrow{\mathrm{v}}=\sqrt{2 g h} \hat{\mathrm{i}}+\sqrt{2 g 3 \mathrm{~h}} \times \frac{1}{\sqrt{3}} \hat{\mathrm{k}}$

$=\sqrt{2 g h} \hat{\mathrm{i}}+\sqrt{2 g h} \hat{\mathrm{k}}$

$\tan \theta=\frac{\sqrt{2 g 3 \mathrm{~h}}}{\sqrt{2 g h}}=\sqrt{3} \quad \theta=60^{\circ}$

$\mathrm{h}_1=\frac{\mathrm{v}_{1 y}^2}{2 \mathrm{~g}}=\frac{2 g h}{2 \mathrm{~g}}=\mathrm{h}$

$\mathrm{d}=\mathrm{v}_x \mathrm{t}=\sqrt{2 g h} \times \sqrt{\frac{2 \times 3 \mathrm{~h}}{\mathrm{~g}}}$

$=\sqrt{2 g h} \sqrt{\frac{6 \mathrm{~h}}{\mathrm{~g}}}=2 \sqrt{3 \mathrm{~h}}$

$=\frac{\mathrm{d}}{\mathrm{h}_1}=2 \sqrt{3}$

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