A particle slides down a frictionless parabolic track shaped like $y = x^2$. It starts from rest at point A on the upper-left arm of the parabola, slides down through point B at the vertex (the lowest point of the parabola), and continues up the right arm to point C, which is at a height less than that of A. At C the particle leaves the track and moves freely through the air as a projectile, launched at some speed at an angle above the horizontal, and reaches the highest point of its flight at P. Given this, which of the following are correct? (Note: more than one option may be correct.) (a) KE at P = KE at B (b) height at P = height at A (c) total energy at P = total energy at A (d) time of travel from A to B = time of travel from B to P.

A particle slides down a frictionless parabolic track shaped like y = x². It starts from rest at point A on th

NCERT Exemplar · Q12

y = x^2

(a) KE at P = KE at B

(b) height at P = height at A

(d) time of travel from A to B = time of travel from B to P.

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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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No friction acts and only gravity does work, so the total mechanical energy is the same at every instant — hence (C) is correct. At the top of its flight the projectile still moves horizontally, so it retains kinetic energy there; that makes P lower than A and its KE smaller than at B, so (A) and (B) fail. The two travel times in (D) have no reason to match. Answer: (C) only.

Concept

The parabolic track is frictionless and, after C, the particle is in free projectile motion. In both phases the only forces are the (normal, does no work) track reaction and gravity (conservative). Therefore total mechanical energy $E = KE + PE$ is conserved throughout. Take B (the vertex) as the reference level, and let $h_A$ and $h_C$ be the heights of A and C above B, with $h_A > h_C$ .

Working through each option

(C) total energy at P = total energy at A — TRUE. With no energy dissipation, $E$ is constant, so $E_P = E_A$ . This is the safe consequence of energy conservation.

(A) KE at P = KE at B — FALSE. At B all of A's potential energy has become kinetic: $KE_B = mgh_A$ . At the apex P the vertical velocity is zero but the horizontal velocity $v_x = v_0\cos\theta$ survives, so $KE_P = \tfrac12 m v_0^2\cos^2\theta$ . Since $\tfrac12 m v_0^2 = mg(h_A-h_C)$ , we get $KE_P = mg(h_A-h_C)\cos^2\theta < mgh_A = KE_B$ .

(B) height at P = height at A — FALSE. By energy conservation $mgh_A = KE_P + mgh_P$ , so

$h_P = h_A - \frac{KE_P}{mg} = h_A - (h_A-h_C)\cos^2\theta < h_A,$ …

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Concept: Settle Each Claim by Contradiction, Using Only "Horizontal Velocity Never Changes"

Method: Proof by Contradiction for (A), (B), (D); Direct Energy Conservation for (C)

Rather than computing $h_P$ , $KE_P$ explicitly in terms of the launch angle at C, this method assumes each false statement is true and shows it forces an impossible conclusion -- using only one physical fact that's true throughout the entire flight, on the track and in the air: gravity and the (frictionless) normal force never touch the particle's horizontal momentum, so the horizontal component of velocity, once the particle leaves the track at C, is fixed for the rest of the flight and equal to whatever it was at C. Since C is on the curved, rising part of the track (not its flat vertex), that horizontal speed at C is strictly positive.

Setting up the one fact everything rests on

Once airborne, the only force is gravity (vertical) -- the standard projectile setup -- so $v_x(t) = v_{Cx} > 0$ for the entire flight after C, including at the apex P. ( $v_{Cx}>0$ because C sits on the sloped right arm of the parabola, where the track's own tangent has a nonzero horizontal component -- only at the vertex B would the tangent be perfectly vertical, which C is not.)

(C): total energy at P = total energy at A -- proved directly, not by contradiction

The track (frictionless) and the air (gravity only) both do work only via a conservative force, so mechanical energy $E=KE+PE$ is conserved across the entire journey A $\to$ B $\to$ C $\to$ P with no exceptions. Hence trivially $E_P=E_A$ . (C) is true.

(A): KE at P = KE at B -- refuted by contradiction

Assume $KE_P = KE_B$ . Since $E$ is conserved (from (C)) and $E = KE+mgh$ , equal $KE$ at two points forces equal height too: $h_P=h_B=0$ (taking B as the reference level). But P is the highest point of a projectile launched with $v_{Cx}>0$ -- for that to have zero height above B, C itself would already have to be at height $0$ (i.e. $C=B$ ), contradicting that C is a distinct point higher up the right arm. Contradiction $\Rightarrow$ (A) is false.

(B): height at P = height at A -- refuted by contradiction

Assume $h_P=h_A$ . By conservation of energy, this forces $KE_P=KE_A=0$ (the particle started from rest at A). But $KE_P \geq \tfrac12 m v_{Cx}^2 > 0$ , since the horizontal velocity $v_{Cx}$ never vanishes during the flight (Step 1) -- the particle cannot have zero kinetic energy at P while still carrying nonzero horizontal speed. Contradiction $\Rightarrow$ (B) is false.

(D): time A $\to$ B equals time B $\to$ P -- refuted by a structural mismatch, not a computation …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
$Q. A ball of mass 0.5 kg is dropped freely from a point A which is at a height of 10 m from the ground. Between second and third collisions with the ground, the linear momentum of the ball becomes zero at a point B. If the coefficient of restitution between the ball and the ground is 0.5, then the percentage loss of the potential energy of the ball when it reaches point B is (A) 23.25 (B) 83.75 (C) 6.25 (D) 93.75$
›Reveal solutionSolution
The ball loses energy with each bounce due to the coefficient of restitution. Between the second and third collisions, its momentum becomes zero at the highest point of the second bounce. The percentage loss of potential energy relative to the initial drop is 93.75%, so the correct option is (D).

Concept and Intuition

When a ball is dropped from a height, it gains kinetic energy as it falls. On hitting the ground, it rebounds with a speed reduced by the coefficient of restitution $e$ . This means the height of each successive bounce is $e^2$ times the previous height. The question asks: Between the second and third collisions, when does the linear momentum become zero? That happens at the highest point of the second bounce — because at the peak, the ball momentarily stops before falling again. At that point, its kinetic energy is zero, so all its mechanical energy is gravitational potential energy. We need to find what percentage of the original potential energy (from point A) is lost by the time the ball reaches that peak.

Step-by-step solution

Initial potential energy The ball is dropped from height $h_0 = 10 \, \text{m}$ , mass $m = 0.5 \, \text{kg}$ . Initial potential energy:

$U_0 = m g h_0 = 0.5 \times 9.8 \times 10 = 49 \, \text{J}$
(We can keep $g$ symbolic; it will cancel.)

Height after first bounce Coefficient of restitution $e = 0.5$ . After the first impact, the rebound speed is $e$ times the impact speed. Since height is proportional to the square of speed, the height after the first bounce is:

$h_1 = e^2 h_0 = (0.5)^2 \times 10 = 0.25 \times 10 = 2.5 \, \text{m}$

Height after second bounce The same factor applies again: after the second impact, the ball rebounds to a height:

$h_2 = e^2 h_1 = (0.5)^2 \times 2.5 = 0.25 \times 2.5 = 0.625 \, \text{m}$

Where does momentum become zero between second and third collisions? The ball is in the air between the second and third collisions. Its momentum becomes zero at the top of the second bounce, i.e., at height $h_2$ . At that point, the ball has only potential energy:
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TG EAPCET 2025Set ap-2025-04-30-FN1 markMCQ
$= 2 \times 10^{11} \ \mathrm{Nm}^{-2}$
›Reveal solutionSolution

For a beam supported at both ends and loaded at its centre, the sag is $\delta = \dfrac{WL^3}{4bd^3 Y} = 1\ \text{cm}$ .

Given: length $L = 1\ \text{m}$ , breadth $b = 2.5\ \text{cm} = 2.5\times10^{-2}\ \text{m}$ , thickness $d = 5\ \text{mm} = 5\times10^{-3}\ \text{m}$ , central load $W = 25\ \text{N}$ , and $Y = 2\times10^{11}\ \text{N m}^{-2}$ .

Depression of a centrally-loaded beam supported at both ends:
$\delta = \frac{WL^3}{4\,b\,d^3\,Y}$
Substitute the values:

$d^3 = (5\times10^{-3})^3 = 1.25\times10^{-7}\ \text{m}^3$ …
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TG EAPCET 2024Set ap-2024-05-08-FN1 markMCQ
$6~\mathrm{ms}^{-1}$
›Reveal solutionSolution
The work done by the variable force equals the change in kinetic energy. Integrating $F = -25x$ from $x=1.5$ to $2.5$ m and equating to $\frac12 m (4^2 - 6^2)$ gives $m = 5$ kg.

The key here is that the retarding force is not constant — it depends on position $x$ . When a force varies with position, you cannot use $F = ma$ with constant acceleration. Instead, the work–energy theorem is the natural tool: the net work done on the block equals its change in kinetic energy. That theorem holds for any force, constant or variable, as long as you can compute the work integral.

The force is given as $F = -25x$ , where the negative sign means it opposes the motion. The block moves from $x = 1.5$ m to $x = 2.5$ m, so the work done by this force is the integral of $F$ over that displacement.

Write the work–energy relation. The work done by the retarding force is

$W = \int_{x_i}^{x_f} F \, dx = \int_{1.5}^{2.5} (-25x) \, dx.$
This work equals the change in kinetic energy:
$W = \frac12 m v_f^2 - \frac12 m v_i^2.$
Here $v_i = 6$ m/s and $v_f = 4$ m/s.

Evaluate the integral.

$W = -25 \int_{1.5}^{2.5} x \, dx = -25 \left[ \frac{x^2}{2} \right]_{1.5}^{2.5}.$
Compute the bracket:
$\frac{(2.5)^2}{2} - \frac{(1.5)^2}{2} = \frac{6.25 - 2.25}{2} = \frac{4}{2} = 2.$
So $W = -25 \times 2 = -50$ J.

The negative sign makes sense — the force opposes motion, so it removes energy from the block.

Set up the kinetic energy change. …
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TG EAPCET 2024Set eng-2024-05-10-AN1 markMCQ
$100\, \mathrm{Nm}^{-1}$
›Reveal solutionSolution
The kinetic energy at a given displacement is the difference between the total mechanical energy and the elastic potential energy at that point. Using conservation of energy, the answer is 0.275 J, which corresponds to option (C).

The key idea here is conservation of mechanical energy in a spring-block system on a frictionless surface. Since there is no friction, the total energy (kinetic + elastic potential) remains constant. At the maximum displacement (amplitude), all energy is stored as spring potential energy. At any other position, the kinetic energy is simply the total energy minus the potential energy at that position.

Let’s work through it step by step.

Identify the given data and convert to SI units.

Spring constant: $k = 100 \, \text{N/m}$

Amplitude (maximum stretch from equilibrium): $A = 8 \, \text{cm} = 0.08 \, \text{m}$

Position of interest: $x = 3 \, \text{cm} = 0.03 \, \text{m}$ from the mean position.

Find the total mechanical energy of the system.

At the extreme position ( $x = A$ ), the block is momentarily at rest, so kinetic energy is zero. All energy is elastic potential energy:

$E_{\text{total}} = \frac{1}{2} k A^2 = \frac{1}{2} \times 100 \times (0.08)^2$ $E_{\text{total}} = 50 \times 0.0064 = 0.32 \, \text{J}$

Find the elastic potential energy at $x = 0.03 \, \text{m}$ .

$U = \frac{1}{2} k x^2 = \frac{1}{2} \times 100 \times (0.03)^2$
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TG EAPCET 2023Set eng-2023-05-13-FN1 markMCQ
$\dfrac{ML^2}{K}$
›Reveal solutionSolution
The block loses all its kinetic energy to the spring at maximum compression, then regains it symmetrically, so its maximum momentum after collision equals the initial momentum, which is $L\sqrt{MK}$ — option (C).

The key concept is energy conservation in a frictionless system. The block collides with the spring, compresses it, and then the spring pushes the block back. Since there's no friction, mechanical energy is conserved: the block's kinetic energy transforms entirely into spring potential energy at maximum compression, and then back into kinetic energy as the spring expands. The maximum momentum after the collision occurs when the block leaves the spring with its original speed (but opposite direction), so we just need to find the initial momentum from the given compression.

At maximum compression, the block is momentarily at rest, so all its initial kinetic energy has been converted into elastic potential energy of the spring. Let the initial speed of the block be $v_0$ . Then:

$\frac{1}{2} M v_0^2 = \frac{1}{2} K L^2$
Cancel the $\frac{1}{2}$ :
$M v_0^2 = K L^2$

Solve for the initial speed:

$v_0 = \sqrt{\frac{K L^2}{M}} = L \sqrt{\frac{K}{M}}$

The initial momentum is: $p_0 = M v_0 = M \cdot L \sqrt{\frac{K}{M}} = L \sqrt{MK}$ …
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TG EAPCET 2022Set eng-2022-07-19-AN1 markMCQ
$v = 10~\text{m/s}$
›Reveal solutionSolution

Momentum + energy conservation for equal masses give $v_A + v_B = 10$ and $v_A^2 + v_B^2 = 99$ ; the moving particle emerges at $\approx 0.05~\text{m/s}$ .

Both particles have equal mass $m$ . Take the initial speed of A as $10~\text{m/s}$ and B at rest.

Momentum conservation:
$m(10) = m v_A + m v_B \implies v_A + v_B = 10$
Kinetic energy (decreased by $1\%$ , so final $= 0.99$ of initial):
$\tfrac{1}{2}m v_A^2 + \tfrac{1}{2}m v_B^2 = 0.99\left(\tfrac{1}{2}m\,(10)^2\right) \implies v_A^2 + v_B^2 = 99$
From these, $2 v_A v_B = (v_A+v_B)^2 - (v_A^2+v_B^2) = 100 - 99 = 1$ , so $v_A v_B = 0.5$ .

Thus $v_A,\,v_B$ are roots of $t^2 - 10t + 0.5 = 0$ :

$t = \frac{10 \pm \sqrt{100-2}}{2} = \frac{10 \pm \sqrt{98}}{2} \approx 9.95 \text{ or } 0.05$ …
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TG EAPCET 2021Set ap-2021-08-10-AN1 markMCQ
$Q. A shell of mass 20 g is fired by a gun of mass 100 kg. If the muzzle speed of the shell is 6 km/min., what is the recoil speed of the gun? (A) 1 cm/s (B) 2 cm/s (C) 3 cm/s (D) 4 cm/s$
›Reveal solutionSolution
The recoil speed of the gun is found by applying conservation of momentum to the gun-shell system. Converting units carefully gives a recoil speed of 2 cm/s, so the correct option is (B).

Concept and Intuition

When a gun fires a shell, the total momentum of the gun + shell system is conserved if no external horizontal forces act (like friction). Before firing, both are at rest, so total momentum = 0. After firing, the shell moves forward and the gun recoils backward. Their momenta must be equal in magnitude and opposite in direction:
$m_{\text{shell}} v_{\text{shell}} = m_{\text{gun}} v_{\text{gun}}$
The key trap: units. The shell’s speed is given in km/min, but the answer choices are in cm/s. We must convert everything to a consistent system (e.g., kg, m, s or g, cm, s).

Step-by-step solution

Write down given data

Mass of shell: $m_s = 20\ \text{g}$

Mass of gun: $m_g = 100\ \text{kg} = 100\,000\ \text{g}$ (since 1 kg = 1000 g)

Muzzle speed of shell: $v_s = 6\ \text{km/min}$

Convert shell speed to cm/s

$1\ \text{km} = 1000\ \text{m} = 100\,000\ \text{cm}$

$1\ \text{min} = 60\ \text{s}$

So $v_s = 6 \times \frac{100\,000\ \text{cm}}{60\ \text{s}} = 6 \times \frac{100\,000}{60} = 6 \times \frac{10\,000}{6} = 10\,000\ \text{cm/s}$ (Check: $6 \times 100\,000 / 60 = 600\,000 / 60 = 10\,000$ )

Apply conservation of momentum

Initial momentum = 0.

Final momentum: $m_s v_s + m_g v_g = 0$ (taking forward as positive, so $v_g$ will be negative).

Magnitude relation:

$m_s v_s = m_g |v_g|$
Solve for $|v_g|$ :
$|v_g| = \frac{m_s v_s}{m_g} = \frac{20\ \text{g} \times 10\,000\ \text{cm/s}}{100\,000\ \text{g}}$

Simplify …
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TG EAPCET 2021Set ap-2021-08-10-AN1 markMCQ
$k = 10^4$
›Reveal solutionSolution
The problem is solved by equating the car’s kinetic energy to the spring’s elastic potential energy at maximum compression. The result is $\frac{10}{\sqrt{10}}$ m, which matches option (A).

The key idea is energy conservation: when the car collides with the spring and comes to rest, all its kinetic energy has been transferred into the spring’s stored elastic potential energy. No energy is lost (ideal spring, no friction), so we can set $\frac{1}{2} m v^2 = \frac{1}{2} k x^2$ and solve for $x$ , the maximum compression.

Write the energy conservation equation The car’s kinetic energy just before impact is

$KE = \frac{1}{2} m v^2$
The spring’s potential energy at maximum compression $x$ is
$PE = \frac{1}{2} k x^2$
Because the car comes to rest, all kinetic energy becomes spring potential energy:
$\frac{1}{2} m v^2 = \frac{1}{2} k x^2$

Cancel the common factor $\frac{1}{2}$ This simplifies to

$m v^2 = k x^2$

Solve for $x$

$x^2 = \frac{m v^2}{k} \quad \Rightarrow \quad x = v \sqrt{\frac{m}{k}}$

Plug in the given values $m = 1000$ kg, $v = 10$ m/s, $k = 10^4$ N/m …
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TG EAPCET 2021Set ap-2021-08-10-FN1 markMCQ
$k = 10^4$
›Reveal solutionSolution
The car’s kinetic energy is completely converted into the spring’s elastic potential energy. Using energy conservation, the maximum compression is $x = \sqrt{\frac{m v^2}{k}} = \sqrt{\frac{1000 \cdot 10^2}{10^4}} = \sqrt{10}$ m, which matches option (A).

Concept & Intuition

When the car hits the spring, it slows down as the spring compresses. If we ignore friction and other losses, the car’s kinetic energy just before impact is entirely stored as elastic potential energy in the spring at the moment of maximum compression (when the car momentarily stops). This is a classic conservation of mechanical energy problem:
$\frac{1}{2} m v^2 = \frac{1}{2} k x^2$
where $x$ is the maximum compression. The mass $m$ , speed $v$ , and spring constant $k$ are given, so we can solve directly for $x$ .

Step-by-step solution

Write the energy conservation equation The car’s kinetic energy before collision:

$KE = \frac{1}{2} m v^2$
The spring’s potential energy at maximum compression:
$PE = \frac{1}{2} k x^2$
Setting them equal (no energy loss):
$\frac{1}{2} m v^2 = \frac{1}{2} k x^2$

Cancel the common factor $\frac{1}{2}$

$m v^2 = k x^2$

Solve for $x$

$x^2 = \frac{m v^2}{k} \quad \Rightarrow \quad x = \sqrt{\frac{m v^2}{k}}$

Plug in the numbers …
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TG EAPCET 2021Set eng-2021-08-05-FN1 markMCQ
$200\,\text{g}$
›Reveal solutionSolution

Spring energy $\to$ block's kinetic energy gives launch speed $v=\sqrt{2.5}~\text{m/s}$ ; projectile fall from $2~\text{m}$ takes $\sqrt{0.4}~\text{s}$ , so horizontal range $= v t = 1.0~\text{m}\approx 0.99~\text{m}$ , option (A).

Launch speed from the spring

Convert the stored spring energy to kinetic energy ( $m=0.2~\text{kg}$ , $k=50~\text{N/m}$ , $x=0.10~\text{m}$ ):
$\frac{1}{2}kx^2 = \frac{1}{2}mv^2 \ \Rightarrow\ v^2 = \frac{kx^2}{m} = \frac{50\times(0.10)^2}{0.2} = \frac{0.5}{0.2} = 2.5,$ $v = \sqrt{2.5} \approx 1.58~\text{m/s}.$
Projectile off the edge of the slab

The block leaves the slab horizontally at height $h=2~\text{m}$ . Time to fall: …
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Concept

Working through each option

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