A bullet of mass $m$ fired at $30^\circ$ to the horizontal leaves the barrel of the gun with a velocity $v$. The bullet hits a soft target at a height $h$ above the ground while it is moving downward and emerges out with half the kinetic energy it had before hitting the target. Which of the following statements are correct in respect of bullet after it emerges out of the target? (Note: more than one of the given options may be correct.) (a) The velocity of the bullet will be reduced to half its initial value. (b) The velocity of the bullet will be more than half of its earlier velocity. (c) The bullet will continue to move along the same parabolic path. (d) The bullet will move in a different parabolic path. (e) The bullet will fall vertically downward after hitting the target. (f) The internal energy of the particles of the target will increase.

A bullet of mass m fired at 30^° to the horizontal leaves the barrel of the gun with a velocity v. The bullet

NCERT Exemplar · Q20

m

(a) The velocity of the bullet will be reduced to half its initial value.

(b) The velocity of the bullet will be more than half of its earlier velocity.

(d) The bullet will move in a different parabolic path.

(e) The bullet will fall vertically downward after hitting the target.

(f) The internal energy of the particles of the target will increase.

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Concept understanding — Work Energy Theorem

The Work-Energy Theorem: From Intuition to Precision

Imagine pushing a heavy box across the floor. The harder you push and the farther it slides, the faster it moves when you let go. That connection — the push (force) over a distance (displacement) changing the box's speed — is exactly what the Work-Energy Theorem captures.

The Intuition First

Think of work as the "currency" that buys motion. When you do work on an object, you transfer energy to it. That energy shows up as kinetic energy — the energy of motion. The more work you do, the more the object's kinetic energy changes.

If you push a stationary ball, it starts moving. If you push a moving ball in the same direction, it speeds up. If you push against its motion, it slows down. In every case, the work done equals the change in the ball's kinetic energy.

Note

Work is done by a force on an object. The object's kinetic energy changes by exactly that amount (assuming no other forces do work).

The Precise Statement

W_{\text{net}} = \Delta K = K_f - K_i

Where:

$W_{\text{net}}$ is the net work done on the object (the total work from all forces combined)
$K_f$ is the final kinetic energy
$K_i$ is the initial kinetic energy

And kinetic energy is defined as:

K = \frac{1}{2}mv^2

So the theorem can also be written as:

W_{\text{net}} = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2

Why "Net" Work Matters

This is the most common point of confusion. The theorem uses net work — the work done by the net force (the vector sum of all forces). If you push a box and friction opposes it, the net work is the work you do minus the work friction does. Only that net amount changes the kinetic energy.

Watch out

If you push a box at constant speed, your work is positive, but friction does equal negative work. The net work is zero, so kinetic energy doesn't change — the box keeps moving at the same speed. Your work didn't "disappear"; it was dissipated as heat by friction.

A Simple Derivation (for constant force)

Consider a constant net force $F_{\text{net}}$ acting on an object of mass $m$ over a displacement $s$ . From Newton's second law:

F_{\text{net}} = ma

From kinematics (constant acceleration):

v_f^2 = v_i^2 + 2as

Multiply both sides by $\frac{1}{2}m$ :

\frac{1}{2}mv_f^2 = \frac{1}{2}mv_i^2 + mas

But $mas = F_{\text{net}} s = W_{\text{net}}$ , so:

\frac{1}{2}mv_f^2 = \frac{1}{2}mv_i^2 + W_{\text{net}}

Rearranging:

W_{\text{net}} = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2 = \Delta K

Tip

The theorem holds even for variable forces and curved paths — the derivation uses calculus then, but the result is the same.

What It Tells You (and What It Doesn't) …

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The Work–Energy Theorem directly links the loss in kinetic energy to the work done by the target. Halving the kinetic energy reduces the speed by a factor of $\frac{1}{\sqrt{2}}$ , not $\frac12$ , and the change in speed changes the projectile’s path. The correct options are (B), (D), and (F).

1. The core idea: Work–Energy Theorem

The bullet loses kinetic energy because the target does negative work on it — that work is the force of the target multiplied by the distance over which it acts. The Work–Energy Theorem says:

The net work done on an object equals its change in kinetic energy.

So if the bullet emerges with half its original kinetic energy, the target has removed exactly half of that energy. No other forces (like gravity) do work during the extremely brief collision, because the displacement during the hit is horizontal and gravity acts vertically — so the work done by gravity during the collision is negligible.

K_{\text{after}} = \frac12 K_{\text{before}} \quad\Longrightarrow\quad \frac12 m v_{\text{after}}^2 = \frac12 \left( \frac12 m v_{\text{before}}^2 \right)

2. Step-by-step reasoning

Step 1: Find the speed after the hit

Let the speed just before impact be $u$ . Then:

K_{\text{before}} = \frac12 m u^2

After emerging, $K_{\text{after}} = \frac12 K_{\text{before}} = \frac14 m u^2$ .

But $K_{\text{after}} = \frac12 m v_{\text{after}}^2$ , so:

\frac12 m v_{\text{after}}^2 = \frac14 m u^2 \quad\Longrightarrow\quad v_{\text{after}}^2 = \frac12 u^2

\boxed{v_{\text{after}} = \frac{u}{\sqrt{2}}}

This is not half the speed — it’s about $0.707\,u$ . So option (A) is false, and option (B) is true: the velocity is more than half of its earlier value.

Watch out

A common mistake is to think “half the kinetic energy means half the speed”. But kinetic energy depends on $v^2$ , so halving $K$ means dividing $v$ by $\sqrt{2}$ , not by $2$ .

Step 2: Does the path remain the same?

The bullet is a projectile. Its path is determined by the velocity vector (magnitude and direction) at the moment it leaves the target. The target changes both:

The magnitude drops from $u$ to $u/\sqrt{2}$ . …

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

Showing the 12 most recent of 42 on this concept.

AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
$Q. If the work done in increasing the velocity of a body by 25% is 900 J, then the work done in decreasing the velocity of the same body by 25% is (A) 700 J (B) 500 J (C) 225 J (D) 450 J$
›Reveal solutionSolution
Kinetic energy scales as $v^2$ , so a 25% increase and a 25% decrease change KE by different amounts (not symmetric) — the work values are 900 J and 700 J respectively.

Concept and Intuition

Because $KE\propto v^2$ , percentage changes in velocity do not translate into equal-and-opposite percentage changes in KE. Increasing $v$ by 25% multiplies KE by $(1.25)^2=1.5625$ , while decreasing $v$ by 25% multiplies KE by $(0.75)^2=0.5625$ — these are not mirror images, so the two work-done values differ.

Step-by-Step Solution

Let the initial KE be $K=\tfrac12mv^2$ .

Increasing velocity by 25%: $v'=1.25v$ , so $KE' = \tfrac12m(1.25v)^2 = 1.5625K$ .

Work done to increase $= KE'-K = 0.5625K = 900$ J $\Rightarrow K = \dfrac{900}{0.5625}=1600$ J.

Decreasing velocity by 25%: $v''=0.75v$ , so $KE''=\tfrac12m(0.75v)^2=0.5625K = 0.5625(1600)=900$ J. …
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AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
$\cos^{-1}(0.6)$
›Reveal solutionSolution
Applying $v^2=u^2+2as$ separately over the two half-lengths with their different friction coefficients gives a final speed of 5.6 m/s.

Concept and Intuition

Since the coefficient of kinetic friction changes halfway down the incline, the acceleration is piecewise-constant — the problem must be split into two stages, using the velocity at the end of stage one as the initial velocity for stage two.

Step-by-Step Solution

$\cos\theta=0.6\Rightarrow\sin\theta=\sqrt{1-0.36}=0.8$ (3-4-5 triangle).

Upper half, length $1.4$ m, $\mu_1=0.3$ : $a_1=g\sin\theta-\mu_1g\cos\theta=10(0.8)-0.3(10)(0.6)=8-1.8=6.2\ ms^{-2}$ .

Starting from rest: $v_1^2=0+2(6.2)(1.4)=17.36\ m^2s^{-2}$ . …
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AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
$20\ ms^{-1}$
›Reveal solutionSolution
The work-energy theorem, with both gravity and constant air resistance doing negative work on the way up, gives a height of 10 m.

Concept and Intuition

Going up, both gravity and air resistance oppose the motion (both act downward), so the loss in kinetic energy equals the sum of the work done against both — this is a direct energy-balance shortcut, avoiding a separate calculation of a modified "effective g."

Step-by-Step Solution

Initial KE at X: $\dfrac12mv^2=\dfrac12(5)(20)^2=1000$ J.

KE at Y: given as 400 J.

Loss in KE $=1000-400=600$ J, which equals work done against gravity plus air resistance over height $h$ : $600=(mg+F_{air})h=(5\times10+10)h=60h$ . …
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AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
$2\ \mathrm{Nm^{-1}}$
›Reveal solutionSolution
Work-energy theorem: the block's initial kinetic energy is dissipated by friction and stored as spring PE by the time it stops, giving $v=0.4\ \mathrm{ms^{-1}}$ — option (C).

Concept and Intuition

An impulse gives the block an initial velocity; as it slides, friction removes energy (as heat) and the spring stores energy (as elastic PE). Since the block eventually comes to rest, ALL the initial kinetic energy must have gone into these two channels — this is a direct application of the work-energy theorem over the whole sliding distance.

Step-by-Step Solution

Let $v$ be the speed just after the impulse. Initial KE $=\tfrac12mv^2$ .

Energy lost to friction over distance $d$ : $W_f=\mu mg\,d$ .

Energy stored in the spring at maximum stretch $d$ (block stops, so this is the full compression): $W_s=\tfrac12kd^2$ .

Energy conservation (block starts with KE, ends at rest with the energy fully transferred): $\tfrac12mv^2=\mu mgd+\tfrac12kd^2$ .

Substitute numbers: $\mu mgd = 0.1\times0.18\times10\times0.06=0.0108\ \mathrm J$ ; $\tfrac12kd^2=0.5\times2\times0.06^2=0.0036\ \mathrm J$ . …
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AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
$g = 10\, \mathrm{ms^{-2}}$
›Reveal solutionSolution
The 0.36 kg mass is the lighter side of an Atwood machine and rises; the string (tension) does positive work on it equal to $T\times s=8$ J in the first second — option (C).

Concept and Intuition

In an Atwood machine, the heavier mass falls and the lighter mass rises with the same magnitude of acceleration, connected by a single tension throughout the (light, inextensible) string. The work done by the string on either mass is simply tension times the distance that mass moves (positive if the tension and displacement are in the same direction, as they are here for the rising 0.36 kg mass).

Step-by-Step Solution

Let $m_1=0.36$ kg (lighter, rises) and $m_2=0.72$ kg (heavier, falls).

Acceleration of the system: $a=\dfrac{(m_2-m_1)g}{m_1+m_2}=\dfrac{(0.72-0.36)\times10}{1.08}=\dfrac{3.6}{1.08}=\dfrac{10}{3}\ \mathrm{ms^{-2}}$ .

Tension from the lighter mass's equation of motion ( $T-m_1g=m_1a$ ): $T=m_1(g+a)=0.36\left(10+\dfrac{10}{3}\right)=0.36\times\dfrac{40}{3}=4.8\ \mathrm N$ .

(Check with the heavier mass: $T=m_2(g-a)=0.72\times\dfrac{20}{3}=4.8\ \mathrm N$ — consistent.) …
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AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
$g=10\ ms^{-2}$
›Reveal solutionSolution
Mechanical energy is conserved as the body rises (no air resistance mentioned): the kinetic energy lost equals the gravitational PE gained. Setting the KE to half its initial value and solving for height gives $x=6\,m$ .

Concept and Intuition

As a body is thrown straight up, kinetic energy steadily converts into gravitational potential energy while total mechanical energy stays constant. So at any height $x$ : $KE(x) = KE_0 - mgx$ . We just need the height at which this has dropped to half the initial value — this is a direct energy-bookkeeping problem, no need to find the initial velocity explicitly (though we could).

Step-by-Step Solution

Given: $m = 2\,kg$ , $KE_0 = 240\,J$ , $g = 10\,ms^{-2}$ .

At height $x$ : $KE(x) = KE_0 - mgx = 240 - (2)(10)x = 240 - 20x$ .

We want $KE(x) = \dfrac{240}{2} = 120\,J$ . …
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AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQ
$m = 0.1\ \text{kg}$
›Reveal solutionSolution
The block’s initial gravitational potential energy converts to kinetic energy, which is then dissipated by friction over repeated trips across the rough segment, with elastic bounces at the wall. The total distance traveled on the rough surface is 40 m, plus the initial 1 m smooth segment gives 41 m, but the block also travels the smooth segment each time it passes — careful accounting shows the total horizontal distance is 59 m, so the correct option is (B).

Concept and Intuition

The block starts with gravitational potential energy $mgh$ at the top of the smooth curve. Since the curve and the first horizontal segment AB are frictionless, all that energy becomes kinetic energy by the time the block reaches point B. From B onward, the rough segment BC (length 2 m) has friction, which does negative work, slowing the block. When the block hits the vertical wall at C, the collision is perfectly elastic, so the block rebounds with the same speed (but opposite direction) — no energy is lost in the bounce. Then it travels back across the rough segment, losing more energy to friction, until it either stops or reaches the smooth segment again. Because the smooth segment AB has no friction, the block coasts across it without losing energy, then re-enters the rough segment from the left. This back-and-forth continues until all the initial energy is dissipated by friction. The key is to find the total distance traveled on the rough surface, then add the distances traveled on the smooth surface (which are traversed multiple times).

Step-by-Step Solution

Find the speed at point B (just before entering the rough patch). The block starts from rest at height $h = 4\ \text{m}$ . By conservation of mechanical energy on the smooth curve and smooth AB:

$mgh = \frac{1}{2} m v_B^2 \quad \Rightarrow \quad v_B = \sqrt{2gh} = \sqrt{2 \cdot 10 \cdot 4} = \sqrt{80} = 4\sqrt{5}\ \text{m/s}.$
So the kinetic energy at B is $\frac{1}{2} m v_B^2 = mgh = 0.1 \times 10 \times 4 = 4\ \text{J}$ .

Work done by friction per unit distance.

On the rough segment, the friction force is $f = \mu m g = 0.1 \times 0.1 \times 10 = 0.1\ \text{N}$ .

The work done by friction over a distance $d$ is $W_f = -f d = -0.1 d$ (in joules). This work removes kinetic energy from the block.

First trip from B to C (2 m).

The block travels from B to C, a distance of 2 m. Friction removes $0.1 \times 2 = 0.2\ \text{J}$ of energy.

Kinetic energy remaining at C: $4 - 0.2 = 3.8\ \text{J}$ .

Speed at C: $v_C = \sqrt{2 \times 3.8 / 0.1} = \sqrt{76} \approx 8.72\ \text{m/s}$ .

The block hits the wall and rebounds elastically — speed magnitude unchanged, direction reversed.

First return trip from C to B (2 m).

The block travels back from C to B, again 2 m on rough surface. Friction removes another $0.2\ \text{J}$ .

Energy at B (on return): $3.8 - 0.2 = 3.6\ \text{J}$ .

Since AB is smooth, the block coasts across AB (1 m) without losing energy, then reaches the left end of the rough segment (point B again) with the same 3.6 J. But note: the block has now traveled 2 m (B→C) + 2 m (C→B) = 4 m on rough surface so far.

Subsequent trips: each full round trip across the rough segment (B→C→B) is 4 m and removes 0.4 J of energy.

However, the block does not necessarily always make a full round trip — it may stop partway. Let’s track the energy after each pass:

After first B→C: energy = 3.8 J

After first C→B: energy = 3.6 J

After second B→C: energy = 3.4 J

After second C→B: energy = 3.2 J

… and so on.

Each full back-and-forth (4 m) reduces energy by 0.4 J. Starting from 4 J, the number of full 4 m cycles possible is $\lfloor 4 / 0.4 \rfloor = 10$ full cycles, which would use $10 \times 4 = 40\ \text{m}$ of rough travel and reduce energy to $4 - 10 \times 0.4 = 0\ \text{J}$ . But wait — does it exactly hit zero? Let’s check: after 10 cycles (20 half-trips of 2 m each), the energy would be exactly 0. That means the block stops exactly at point B after the 20th half-trip (i.e., after returning from C to B for the 10th time). So the total rough distance is exactly 40 m.

Now account for the smooth segment AB (1 m).

Each time the block goes from B to the left end of the smooth segment and back to B, it travels 2 m on smooth surface. But careful: the block starts at A (after coming down the curve) and goes to B (1 m smooth). Then after each return to B from the rough side, it coasts across the smooth segment to the left end and back to B. How many times does it cross the smooth segment?

First time: from A to B (1 m) — this is the initial smooth travel. …
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AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQ
$S = \dfrac{t^3}{3}$
›Reveal solutionSolution
Differentiating displacement gives velocity at t=2s as 4 m/s; the

work-energy theorem then gives work done = 24 J. Answer: (D).

Concept and Intuition

Rather than computing the force $F = m\,d^2S/dt^2$ and integrating $F\,dS$

over the interval, the work-energy theorem ( $W = \Delta KE$ ) lets us

shortcut straight to the answer once we know the velocity at the start and

end of the interval — since the body starts from rest ( $S=0$ at $t=0$ gives

$v=0$ too), the work done in the first 2 seconds is just the final kinetic

energy.

Step-by-Step Solution

Given $S(t) = \dfrac{t^3}{3}$ m.

Velocity: $v(t) = \dfrac{dS}{dt} = t^2$ .

At $t=0$ : $v=0$ (starts from rest). At $t=2$ s: $v = 2^2 = 4\ \text{ms}^{-1}$ . …
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AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
$t=0$
›Reveal solutionSolution
This tests the work–energy theorem applied directly from a velocity–time graph — total work equals the change in kinetic energy, regardless of the path velocity takes in between.

Concept and Intuition

The work–energy theorem states that the net work done by all forces on a body equals its change in kinetic energy, $W = \Delta KE = \tfrac12 m v_f^2 - \tfrac12 m v_i^2$ . This holds true no matter how the velocity varies in between (linear decrease, reversal of direction, etc.) — only the initial and final speeds matter. Here the body starts at $+10\ \mathrm{ms^{-1}}$ and ends at $-20\ \mathrm{ms^{-1}}$ ; even though the velocity passes through zero and reverses direction, kinetic energy depends only on $v^2$ , so the sign reversal doesn't subtract — it's the magnitude that counts in $v^2$ .

Step-by-Step Solution

From the graph: at $t=0$ , $v_i = 10\ \mathrm{ms^{-1}}$ ; at $t=5\ \mathrm{s}$ , $v_f = -20\ \mathrm{ms^{-1}}$ .

Initial KE $= \tfrac12(4)(10)^2 = 200\ \mathrm{J}$ . …
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AP EAPCET 2026Set ap-2026-05-20-AN1 markMCQ
$\dfrac{-2}{9}\mu MgL$
›Reveal solutionSolution
Integrate the (shrinking) friction force over the length the chain slides, from $x=L/3$ to $x=L$ ; the result is $-\dfrac{2}{9}\mu MgL$ . Answer: (A).

Concept and Intuition

As the chain slips, the length resting on the table (and hence the normal force and friction force) keeps decreasing, so friction is NOT constant — you must integrate. The friction force at any moment is $\mu$ times the weight of only the portion still on the table, since the hanging portion contributes no normal force on the table.

Step-by-Step Solution

Let $x$ = length of chain hanging off the table at some instant; it starts at $x_0 = L/3$ and ends at $x_1 = L$ (whole chain off).

Mass on table at that instant: $m_{table} = M\dfrac{L-x}{L}$ .

Friction force (opposing the slipping motion): $f(x) = \mu m_{table}\, g = \mu\dfrac{Mg}{L}(L-x)$ .

Work done by friction (force opposes displacement $dx$ ): $W_f = -\displaystyle\int_{L/3}^{L} \mu\dfrac{Mg}{L}(L-x)\,dx$ . …
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AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
$12\ \mathrm{m\,s^{-1}}$
›Reveal solutionSolution
This tests the work–energy theorem for a ball accelerated over a known distance; the net force equals the kinetic energy gained divided by the displacement, giving 20 N.

Concept and Intuition

The hand exerts a net force on the ball while it is still in contact, over the distance the hand travels forward. Since there is no other agent doing work on the ball during this phase (we're told the net force), the work done by that net force converts entirely into the kinetic energy the ball has when it leaves the hand. This is the work–energy theorem: $W_{net} = \Delta KE$ .

Step-by-Step Solution

The ball starts at rest in the hand and leaves with speed $v = 12\ \mathrm{m\,s^{-1}}$ .

Kinetic energy gained: $\Delta KE = \tfrac12 m v^2 = \tfrac12(0.25)(12)^2 = \tfrac12(0.25)(144) = 18\ \mathrm{J}$ . …
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AP EAPCET 2025Set ap-2025-05-19-FN1 markMCQ
$6\ \mathrm{kg}$
›Reveal solutionSolution
At uniform speed the net work done is zero, so the work done against friction simply equals (friction force) × (distance) = 12 J.

Concept and Intuition

When a body moves at constant velocity, its kinetic energy does not change, so by the work–energy theorem the net work done on it is zero. This means the work done by whatever force is pushing/pulling the body forward is exactly cancelled by the (negative) work done by friction — in other words, all the driving work is "spent" fighting friction. So the magnitude of work done against friction is simply the friction force times the distance moved.

Step-by-Step Solution

Convert distance: $200\ \text{cm} = 2\ \text{m}$ . …
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1. The core idea: Work–Energy Theorem

2. Step-by-step reasoning

Step 1: Find the speed after the hit

Step 2: Does the path remain the same?

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