Question

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 $\vec{u} = u_0 \hat{x}$ 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 $1/\sqrt{3}$. Which of the following statement(s) is(are) correct?

• A. $\vec{u_0} = \sqrt{2ghx}$
• B. $\vec{v} = \sqrt{2gh(x-z)}$
• C. $\theta = 60^\circ$
• D. $d/h_1 = 2\sqrt{3}$

Answer 1

Answer: (A), (B), (C), (D)

All options are correct, assuming typos in options A and B are corrected to standard physics notation.

Answer 2

The problem involves three main parts: the motion of the ball on the slide, the projectile motion of the ball from the terrace to the ground, and the collision with the ground followed by another projectile motion.

1. Velocity $u_0$ when leaving the slide: The ball is released from rest at a height $h$ from the top of the terrace. Since the slide is frictionless and the lower end is horizontal, we can use the conservation of mechanical energy. Let the terrace level be the reference for potential energy. Initial potential energy = $mgh$ Initial kinetic energy = $0$ Final potential energy (at terrace level) = $0$ Final kinetic energy = $\frac{1}{2}mu_0^2$ By conservation of energy: $mgh = \frac{1}{2}mu_0^2$ $u_0^2 = 2gh$ $u_0 = \sqrt{2gh}$ The ball leaves the slide horizontally, so its velocity vector is $\vec{u} = u_0 \hat{x} = \sqrt{2gh} \hat{x}$.

• Checking Option (A): The option given is $\vec{u_0} = \sqrt{2ghx}$. This is dimensionally incorrect if 'x' is a variable. However, it is highly probable that 'x' is a typo for 'h' and the unit vector $\hat{x}$ is implied, so the intended option is $\vec{u_0} = \sqrt{2gh}\hat{x}$. Assuming this correction, statement (A) is correct.

2. Projectile motion (first flight) to the ground: The ball leaves the terrace at a height $H = 3h$ from the ground with an initial horizontal velocity $u_x = u_0 = \sqrt{2gh}$ and initial vertical velocity $u_z = 0$. The z-axis is defined as downwards.

• Time of flight (T): Using the equation for vertical motion: $H = u_z T + \frac{1}{2}gT^2$ $3h = 0 \cdot T + \frac{1}{2}gT^2$ $3h = \frac{1}{2}gT^2$ $T^2 = \frac{6h}{g}$ $T = \sqrt{\frac{6h}{g}}$

• Distance 'd': The horizontal distance covered is $d = u_x T$. $d = \sqrt{2gh} \cdot \sqrt{\frac{6h}{g}}$ $d = \sqrt{2gh \cdot \frac{6h}{g}} = \sqrt{12h^2} = 2h\sqrt{3}$

• Velocity components just before impact: Horizontal component: $V_x = u_x = \sqrt{2gh}$ Vertical component: $V_z = u_z + gT = 0 + g\sqrt{\frac{6h}{g}} = \sqrt{6gh}$ (downwards) The velocity vector before impact is $\vec{V}_{impact} = V_x \hat{x} + V_z \hat{z} = \sqrt{2gh}\hat{x} + \sqrt{6gh}\hat{z}$.

• Angle $\theta$ with the horizontal: The angle $\theta$ is given by $\tan\theta = \frac{|V_z|}{|V_x|}$. $\tan\theta = \frac{\sqrt{6gh}}{\sqrt{2gh}} = \sqrt{3}$ $\theta = 60^\circ$

• Checking Option (C): $\theta = 60^\circ$. This statement is correct.

3. Collision with the ground and subsequent motion: The coefficient of restitution is $e = 1/\sqrt{3}$. For a collision with a horizontal surface, the horizontal component of velocity remains unchanged, and the vertical component changes direction and magnitude.

• Velocity after bounce ($\vec{v}$): Horizontal component: $v_x = V_x = \sqrt{2gh}$ Vertical component (upwards): $v_z' = e V_z = \frac{1}{\sqrt{3}} \cdot \sqrt{6gh} = \sqrt{\frac{6gh}{3}} = \sqrt{2gh}$ Since the z-axis is defined downwards, the upward vertical velocity is in the $-\hat{z}$ direction. So, $\vec{v} = v_x \hat{x} - v_z' \hat{z} = \sqrt{2gh}\hat{x} - \sqrt{2gh}\hat{z} = \sqrt{2gh}(\hat{x}-\hat{z})$.

• Checking Option (B): The option given is $\vec{v} = \sqrt{2gh(x-z)}$. This is dimensionally incorrect. Assuming it's a typo and the intended option is $\vec{v} = \sqrt{2gh}(\hat{x}-\hat{z})$, then statement (B) is correct.

• Maximum height $h_1$ after bounce: The maximum height reached is determined by the initial vertical velocity after bounce $v_z'$. Using $v_z'^2 = 2gh_1$: $(\sqrt{2gh})^2 = 2gh_1$ $2gh = 2gh_1$ $h_1 = h$

• Checking Option (D): The ratio $d/h_1$. $d/h_1 = \frac{2h\sqrt{3}}{h} = 2\sqrt{3}$ Statement (D) is correct.

Conclusion: Assuming the typographical errors in options (A) and (B) are corrected to the standard physics notation as derived: (A) $\vec{u_0} = \sqrt{2gh}\hat{x}$ (B) $\vec{v} = \sqrt{2gh}(\hat{x}-\hat{z})$ All four statements (A), (B), (C), and (D) are correct.