A body is falling freely under the action of gravity alone in vacuum. Which of the following quantities remain

NCERT Exemplar · Q5

Q. A body is falling freely under the action of gravity alone in vacuum. Which of the following quantities remain constant during the fall?

(a) Kinetic energy.

(b) Potential energy.

(d) Total linear momentum.

Assam AhsecMCQ· 1mImportance★★★★★est

48% · 40/83 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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 …

Start your 14-day free trial to unlock the full solution →

Gravity is a conservative force acting alone in vacuum, so mechanical energy is conserved throughout the fall. The answer is (C).

When a body falls freely under gravity in vacuum, only one force acts: the gravitational force. This is a conservative force, meaning the work it does depends only on the initial and final positions, not on the path taken. For conservative forces acting in isolation, the total mechanical energy—the sum of kinetic and potential energy—remains constant.

Let's examine each quantity during the fall:

Kinetic energy changes continuously.

As the body falls, its speed increases. The kinetic energy $KE = \frac{1}{2}mv^2$ grows larger with every passing moment. At the start of the fall (if released from rest), $KE = 0$ ; by the time it has fallen through height $h$ , $KE = mgh$ . Clearly not constant.
Potential energy decreases continuously.

Taking the usual convention where gravitational potential energy $PE = mgh$ (with $h$ measured upward from some reference), as the body falls and $h$ decreases, so does $PE$ . The potential energy is being converted into kinetic energy. Again, not constant.
Total mechanical energy is conserved.

The total mechanical energy is $E = KE + PE$ . At any instant during the fall:

E = \frac{1}{2}mv^2 + mgh

Because gravity is conservative and no non-conservative forces (like air resistance) are present, the work done by gravity exactly equals the change in kinetic energy, which in turn exactly compensates the loss in potential energy:

$\Delta KE = -\Delta PE$ …

Start your 14-day free trial to unlock the full solution →

←Previous questionQ4 — A bicyclist comes to a skidding stop in 10 m. During this process, the…→Next questionQ6 — During inelastic collision between two bodies, which of the following …

Q.A body is falling freely under the action of gravity alone in vacuum. Which of the following quantities remain constant during the fall?

Concept understanding — Conservation Of Mechanical Energy

Conservation of Mechanical Energy

The Intuition First

The Precise Statement

What This Means in Practice

The Two Critical Conditions

A Simple Example to Cement It …

Conservative Force and Free Fall

Concept: Conservation of Mechanical Energy Under a Single Conservative Force

Q.A body is falling freely under the action of gravity alone in vacuum. Which of the following quantities remain constant during the fall?

🎓Unlock everything free for 14 days

Unlock everything free for 14 days