Two smooth (frictionless) inclined tracks share the same top point A and rise to the same vertical height. One track is a gentle slope making a small angle with the horizontal (it meets the ground at B, far from the foot of the vertical); the other is a steep slope making a large angle with the horizontal (it meets the ground at C, close to the foot). Both tracks start at A at the same height $h$ above the ground. Two identical stones are released from rest at A — stone I slides down the gentle track and stone II slides down the steep track. Which of the following statements is correct? (a) Both stones reach the bottom at the same time but not with the same speed. (b) Both stones reach the bottom with the same speed and stone I reaches the bottom earlier than stone II. (c) Both stones reach the bottom with the same speed and stone II reaches the bottom earlier than stone I. (d) Both stones reach the bottom at different times and with different speeds.

Two smooth (frictionless) inclined tracks share the same top point A and rise to the same vertical height. One

NCERT Exemplar · Q7

h

(a) Both stones reach the bottom at the same time but not with the same speed.

(b) Both stones reach the bottom with the same speed and stone I reaches the bottom earlier than stone II.

(d) Both stones reach the bottom at different times and with different speeds.

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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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Both tracks are frictionless and drop the same height $h$ , so by energy conservation both stones reach the bottom with the same speed $v=\sqrt{2gh}$ . But the steeper track gives a larger acceleration ( $g\sin\theta$ ) over a shorter length, so stone II (steep) arrives earlier. Correct option: (C).

Speeds

Both surfaces are frictionless, so mechanical energy is conserved. Dropping through the same height $h$ :

\tfrac{1}{2}mv^2 = mgh \implies v = \sqrt{2gh}

This is independent of the slope, so both stones reach the bottom with equal speed. (This rules out (A) and (D), which claim different speeds.)

Times

On an incline of angle $\theta$ , the acceleration along the slope is $a=g\sin\theta$ and the slope length is $L=h/\sin\theta$ . Using $L=\tfrac{1}{2}at^2$ : …

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