JEE Challenger
More from Motion in a Plane
PhyHardAdv 2023 • Paper 1

Motion of Ball on Curved Slide Bouncing off Ground

A slide with a frictionless curved surface, which becomes horizontal at its lower end, is fixed on the terrace of a building of height 3h3h from the ground, as shown in the figure. A spherical ball of mass mm is released on the slide from rest at a height hh from the top of the terrace. The ball leaves the slide with a velocity u⃗0=u0x^\vec{u}_0 = u_0 \hat{x} and falls on the ground at a distance dd from the building making an angle θ\theta with the horizontal. It bounces off with a velocity v⃗\vec{v} and reaches a maximum height h1h_1. The acceleration due to gravity is gg and the coefficient of restitution of the ground is 1/31/\sqrt{3}. Which of the following statement(s) is(are) correct?

Question Diagram 1

Options

A

u⃗0=2ghx^\vec{u}_0 = \sqrt{2gh}\hat{x}

Correct
B

v⃗=2gh(x^−z^)\vec{v} = \sqrt{2gh}(\hat{x} - \hat{z})

C

θ=60∘\theta = 60^\circ

Correct
D

d/h1=23d/h_1 = 2\sqrt{3}

Correct

Topics & Concepts

Motion in a PlaneProjectile MotionCollisionWork-Energy Theorem

Step-by-Step Solution

To determine which of the given statements are correct, we analyze the motion of the ball in different stages:

1. Motion on the frictionless curved slide:

The ball is released from rest at a height hh from the top of the terrace. Applying the conservation of mechanical energy: mgh=12mu02  ⟹  u0=2ghmgh = \frac{1}{2} m u_0^2 \implies u_0 = \sqrt{2gh} Since the slide becomes horizontal at the exit, the initial velocity vector as it leaves the terrace is: u⃗0=2gh x^\vec{u}_0 = \sqrt{2gh}\,\hat{x} Thus, Option (A) is correct.

2. Projectile motion before hitting the ground:

The height of the building is 3h3h.

  • Vertical motion under gravity (az=−ga_z = -g): 3h=12gt2  ⟹  t=6hg3h = \frac{1}{2}gt^2 \implies t = \sqrt{\frac{6h}{g}}
  • Horizontal distance dd traveled before hitting the ground: d=u0t=2gh×6hg=12h2=23hd = u_0 t = \sqrt{2gh} \times \sqrt{\frac{6h}{g}} = \sqrt{12h^2} = 2\sqrt{3}h
  • Components of velocity just before striking the ground: vx0=u0=2ghv_{x0} = u_0 = \sqrt{2gh} vz0=−gt=−6ghv_{z0} = -gt = -\sqrt{6gh}

The angle θ\theta made with the horizontal is given by: tan⁡θ=∣vz0∣vx0=6gh2gh=3  ⟹  θ=60∘\tan\theta = \frac{|v_{z0}|}{v_{x0}} = \frac{\sqrt{6gh}}{\sqrt{2gh}} = \sqrt{3} \implies \theta = 60^\circ Thus, Option (C) is correct.

3. Bouncing off the ground:

The coefficient of restitution is e=13e = \frac{1}{\sqrt{3}}.

  • Horizontal component of velocity remains unchanged: vx=vx0=2ghv_x = v_{x0} = \sqrt{2gh}
  • Vertical component of velocity after the bounce: vz=e∣vz0∣=13×6gh=2ghv_z = e |v_{z0}| = \frac{1}{\sqrt{3}} \times \sqrt{6gh} = \sqrt{2gh}

Hence, the velocity vector v⃗\vec{v} after bouncing is: v⃗=vxx^+vzz^=2gh(x^+z^)\vec{v} = v_x \hat{x} + v_z \hat{z} = \sqrt{2gh}(\hat{x} + \hat{z}) Option (B) states v⃗=2gh(x^−z^)\vec{v} = \sqrt{2gh}(\hat{x} - \hat{z}), which is incorrect because the rebound velocity is in the +z^+\hat{z} direction. Thus, Option (B) is incorrect.

4. Maximum height h1h_1 and the ratio d/h1d/h_1:

The maximum height h1h_1 reached after the bounce is: h1=vz22g=(2gh)22g=2gh2g=hh_1 = \frac{v_z^2}{2g} = \frac{(\sqrt{2gh})^2}{2g} = \frac{2gh}{2g} = h

Now, calculating the ratio dh1\frac{d}{h_1}: dh1=23hh=23\frac{d}{h_1} = \frac{2\sqrt{3}h}{h} = 2\sqrt{3} Thus, Option (D) is correct.

Conclusion:

The correct statements are A, C, and D.

PreviousNext