Two identical solid balls, one of ivory and the other of wet-clay, are dropped from the same height on the flo

NCERT Exemplar · Q23

Q. Two identical solid balls, one of ivory and the other of wet-clay, are dropped from the same height on the floor. Which one will rise to a greater height after striking the floor and why?

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Concept understanding — Conservation Of Mechanical Energy

Conservation of Mechanical Energy

The Intuition First

Imagine you're holding a heavy stone at shoulder height. Your arm is tired. That stone has potential energy — energy stored because of its position. Now let it go. As it falls, it speeds up. The potential energy is turning into kinetic energy — the energy of motion. Just before it hits the ground, all the original potential energy has become kinetic energy.

Now imagine the reverse: you throw a ball straight up. It leaves your hand fast (lots of kinetic energy), rises, slows down, stops for an instant at the top (zero kinetic energy), then falls back. At the top, all the kinetic energy you gave it has turned back into potential energy.

This back-and-forth transformation — potential ↔ kinetic — is the heart of the idea. Energy doesn't disappear; it just changes form. That's conservation.

The Precise Statement

Important

Conservation of Mechanical Energy: In an isolated system where only conservative forces (like gravity or an ideal spring) do work, the total mechanical energy of the system remains constant.

Total mechanical energy is the sum of kinetic energy ( $K$ ) and potential energy ( $U$ ):

E_{\text{mech}} = K + U

The law says:

K_{\text{initial}} + U_{\text{initial}} = K_{\text{final}} + U_{\text{final}}

Or, in symbols:

E_{\text{mech, initial}} = E_{\text{mech, final}}

What This Means in Practice

Let's go back to the falling stone. Suppose you hold it 5 metres above the ground. Its mass is 2 kg. Take $g = 10\ \text{m/s}^2$ .

At the top (initial):

$K_i = 0$ (not moving)

$U_i = mgh = 2 \times 10 \times 5 = 100\ \text{J}$

$E_{\text{mech}} = 0 + 100 = 100\ \text{J}$
Just before hitting ground (final):

$U_f = 0$ (height = 0)

$K_f = \frac{1}{2}mv^2$

Conservation says $K_f = 100\ \text{J}$ , so $\frac{1}{2} \times 2 \times v^2 = 100$ , giving $v = 10\ \text{m/s}$ .

You never needed to know the time of fall or acceleration. Energy conservation gave you the speed directly.

The Two Critical Conditions

Watch out

Mechanical energy is not always conserved. It is conserved only when:

No non-conservative forces (like friction, air resistance, or applied pushes/pulls) do work.
The system is isolated — no external forces transfer energy in or out.

If friction is present, some mechanical energy turns into heat (thermal energy). The total energy of the universe is still conserved, but mechanical energy alone is not.

A Simple Example to Cement It …

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The ivory ball will rise to a greater height because it undergoes a more elastic collision with the floor, losing less mechanical energy compared to the wet-clay ball, which experiences a highly inelastic collision.

When an object is dropped from a certain height, its potential energy is converted into kinetic energy as it falls. Upon striking the floor, this kinetic energy is then partially converted back into potential energy as the object rebounds and rises. The key to understanding which ball rises higher lies in how much mechanical energy is conserved during the collision with the floor.

Mechanical energy, which is the sum of an object's kinetic energy and potential energy, is conserved only when conservative forces (like gravity) are at play and no non-conservative forces (like air resistance or internal friction during deformation) do work. In a collision, some mechanical energy is almost always lost, primarily converted into heat, sound, and permanent deformation of the colliding objects. The extent of this energy loss depends on the material properties of the colliding objects, specifically their elasticity.

Here's a step-by-step breakdown:

Initial State and Impact Velocity: Both identical solid balls are dropped from the same height, let's call it $h$ . Assuming they start from rest, their initial potential energy is $mgh$ , where $m$ is the mass of each ball and $g$ is the acceleration due to gravity. Just before striking the floor, all this potential energy is converted into kinetic energy.

KE_{initial} = mgh

Since both balls have the same mass and are dropped from the same height, they will have the same kinetic energy and thus the same speed just before impact.

2. The Collision and Energy Transformation:

During the collision with the floor, the ball deforms, momentarily storing some of its kinetic energy as elastic potential energy within its material. This stored energy is then released, pushing the ball back up. However, not all of the initial kinetic energy is recovered as kinetic energy for the rebound. Some of it is irreversibly converted into other forms of energy:

* Heat: Due to internal friction within the material during deformation.

* Sound: Generated by the impact.

* Permanent Deformation: Energy used to permanently change the shape of the ball or the floor.

The amount of energy lost depends on the material's elasticity:
*   **Ivory Ball:** Ivory is a highly elastic material. When it deforms during impact, it stores a large fraction of the kinetic energy as elastic potential energy and then efficiently converts most of this stored energy back into kinetic energy for rebound. Very little energy is lost as heat, sound, or permanent deformation. Such a collision is considered **highly elastic**.
*   **Wet-Clay Ball:** Wet clay is a highly inelastic material. When it deforms during impact, a significant portion of the kinetic energy is used to permanently change its shape (the ball might flatten or dent). Much less of the stored energy is converted back into kinetic energy for rebound. A large amount of energy is lost as heat, sound, and permanent deformation. Such a collision is considered **highly inelastic**.

> [!IMPORTANT]
> In an ideal elastic collision, kinetic energy is conserved. In an ideal inelastic collision, kinetic energy is not conserved, and the maximum possible kinetic energy is lost (e.g., when objects stick together). Real-world collisions fall between these two extremes.

3. Kinetic Energy After Rebound: …

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Q.Two identical solid balls, one of ivory and the other of wet-clay, are dropped from the same height on the floor. Which one will rise to a greater height after striking the floor and why?

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