A person driving a car suddenly applies the brakes on seeing a child on the road ahead. If he is not wearing s

NCERT Exemplar · Q19

Q. A person driving a car suddenly applies the brakes on seeing a child on the road ahead. If he is not wearing seat belt, he falls forward and hits his head against the steering wheel. Why?

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Concept understanding — Newton's First Law

Newton's First Law: The Law of Inertia

Imagine you're sitting in a bus that's stopped at a signal. The bus suddenly lurches forward. What happens to you? You jerk backwards against the seat. Now imagine the bus is moving at a steady speed and the driver slams the brakes. You lurch forwards toward the front.

Why? Your body was trying to keep doing what it was already doing — staying still when the bus was still, and moving forward when the bus was moving. That instinct is the heart of Newton's First Law.

The Intuition: Objects Are Lazy

Things don't change their motion on their own. A ball sitting on the ground will stay sitting forever unless something pushes or pulls it. A rolling ball will keep rolling forever in a straight line — unless friction, air resistance, or a wall stops it.

This "laziness" of objects is called inertia. The more massive an object, the more inertia it has — a truck is much harder to start moving or stop than a bicycle.

The Precise Statement

Newton's First Law (Law of Inertia): An object at rest stays at rest, and an object in motion stays in motion with the same speed and in the same direction, unless acted upon by an unbalanced external force.

Let's break that down:

"At rest stays at rest" — A book on a table won't slide off unless you push it or the table shakes.
"In motion stays in motion" — A hockey puck on ice keeps gliding because friction is very low. On rough ground, it stops quickly — but that's because friction (a force) is acting on it.
"Unbalanced external force" — If you push a box and someone else pushes it equally from the opposite side, the forces cancel (balanced). The box doesn't move. Only when the net force is non-zero does motion change.

Watch out

A common mistake: students think a moving object needs a force to keep moving. That's false. A moving object needs a force only to change its motion — speed up, slow down, or turn. In the absence of forces, it keeps moving forever.

Why This Law Matters

Newton's First Law defines what a force is: anything that changes an object's state of motion. It also introduces the idea of inertial reference frames — if you're in a smoothly moving train with no windows, you can't tell you're moving at all. All physics works the same as if you were at rest.

Important

The First Law is really a special case of the Second Law ( $F = ma$ ). If net force $F = 0$ , then acceleration $a = 0$ , meaning velocity is constant — either zero (rest) or some steady value (uniform motion). But Newton listed it first because it's the foundation: it tells us what happens when no forces act.

Real-Life Examples

| Situation | What happens | Why |

|-----------|--------------|-----| …

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The driver's body continues moving forward at the car's original speed (Newton's first law) while the car decelerates beneath him, causing him to lurch forward into the steering wheel.

When the brakes slam on, two very different things happen to the car and to the driver's body. The car experiences a large backward force from the road through its wheels—friction between brake pads and discs converts kinetic energy into heat, and the entire vehicle slows down rapidly. But the driver's body, which was moving forward at the same speed as the car just a moment ago, receives no such direct force.

This is Newton's first law in action: an object in motion stays in motion with the same velocity unless acted upon by an external force. The driver's body wants to keep moving forward at the original speed. Without a seat belt, there is nothing to apply the necessary backward force to decelerate the body along with the car.

Let me walk through exactly what happens:

Before braking: Both car and driver move forward together at velocity $v$ . The driver is at rest relative to the car because the seat pushes forward on his back, maintaining his speed.
Brakes applied: The car experiences a large friction force $F_{\text{brake}}$ backward from the wheels. This gives the car a negative acceleration $a = -\frac{F_{\text{brake}}}{m_{\text{car}}}$ , and it begins to slow down.
The driver's body responds differently: The only contact points are the seat (beneath him) and his hands on the wheel. The seat can only push upward (normal force) and cannot pull backward. His hands exert a small force, but nowhere near enough to decelerate his entire body mass at the same rate as the car.
Relative motion develops: Because the car decelerates but the driver's torso does not (or decelerates much more slowly), the driver moves forward relative to the car. From the car's reference frame—which is now a non-inertial, decelerating frame—it looks like a forward "pseudo-force" throws the driver ahead. From an inertial frame outside, it's simply that the driver keeps moving while the car slows beneath him. …

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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-2026-05-11-AN1 markMCQ
$\dfrac{2}{3}$
›Reveal solutionSolution
Using conservation of mechanical energy, the maximum height reached by a body thrown upward with speed $\frac{2}{3}v_{\text{esc}}$ is $5120$ km above the Earth’s surface. The correct option is (B).

Concept and Intuition

When a body is thrown upward from Earth’s surface, its total mechanical energy (kinetic + gravitational potential) is conserved because only gravity (a conservative force) acts. The escape velocity $v_{\text{esc}} = \sqrt{\frac{2GM}{R}}$ is the minimum speed needed for the body to reach infinity with zero kinetic energy. Here, the launch speed is $\frac{2}{3} v_{\text{esc}}$ , so the body will not escape; it will rise to a finite maximum height $h$ where its kinetic energy becomes zero. We equate the total energy at launch to the total energy at the highest point, using the correct gravitational potential energy formula (not $mgh$ , because $h$ is comparable to Earth’s radius).

Step-by-step solution

Write the escape velocity formula Escape velocity from Earth’s surface:

$v_{\text{esc}} = \sqrt{\frac{2GM}{R}}$
where $G$ is the gravitational constant, $M$ is Earth’s mass, and $R = 6400$ km is Earth’s radius.

Given launch speed

$v = \frac{2}{3} v_{\text{esc}} = \frac{2}{3} \sqrt{\frac{2GM}{R}}$

Conservation of mechanical energy At the surface (point 1):

$E_1 = \frac{1}{2} m v^2 - \frac{GMm}{R}$
At maximum height $h$ (point 2): velocity = 0, distance from Earth’s center = $R + h$ :
$E_2 = 0 - \frac{GMm}{R+h}$
Set $E_1 = E_2$ :
$\frac{1}{2} m v^2 - \frac{GMm}{R} = - \frac{GMm}{R+h}$

Cancel $m$ and substitute $v^2$

$\frac{1}{2} \left( \frac{4}{9} \cdot \frac{2GM}{R} \right) - \frac{GM}{R} = - \frac{GM}{R+h}$
Simplify the first term:
$\frac{1}{2} \cdot \frac{8GM}{9R} = \frac{4GM}{9R}$
So the equation becomes:
$\frac{4GM}{9R} - \frac{GM}{R} = - \frac{GM}{R+h}$

Combine the left side

$\frac{4GM}{9R} - \frac{9GM}{9R} = -\frac{5GM}{9R}$
Thus:
$-\frac{5GM}{9R} = - \frac{GM}{R+h}$
Cancel the minus signs and $GM$ :

$\frac{5}{9R} = \frac{1}{R+h}$ …
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TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQ
$W$
›Reveal solutionSolution
The rope is the only thing connecting the body to the helicopter, so the tension must supply exactly the net force needed to give the body the same acceleration as the helicopter — both vertically ( $g$ upward) and horizontally ( $g$ ). Combining this with the body's weight acting downward gives a tension of $T = W\sqrt{5}$ , option (C).

The body has mass $m = W/g$ . Since it moves rigidly with the helicopter, its acceleration components are
$a_x = g \ \text{(horizontal)}, \qquad a_y = g \ \text{(upward)}.$
Only two forces act on the body: the tension $T$ (along the rope, at some angle $\theta$ to the vertical) and its weight $W$ acting straight down.

Step-by-step reasoning

Required net force.

$F_{\text{net},x} = ma_x = \frac{W}{g}\cdot g = W, \qquad F_{\text{net},y} = ma_y = \frac{W}{g}\cdot g = W.$

Resolve the tension. Horizontally, only the tension acts:

$T\sin\theta = W$
Vertically, tension acts up and weight acts down, and their net must equal $F_{\text{net},y} = W$ (upward):
$T\cos\theta - W = W \implies T\cos\theta = 2W$

Solve for the tension. Squaring and adding:

$(T\sin\theta)^2 + (T\cos\theta)^2 = W^2 + (2W)^2$ $T^2(\sin^2\theta + \cos^2\theta) = W^2 + 4W^2 = 5W^2$ $T = W\sqrt{5}$

Cross-check with vectors. The net force vector required is $\vec{F}_{\text{net}} = W\hat{i} + W\hat{j}$ . The two actual forces are $\vec{T}$ and the weight $-W\hat{j}$ , so …
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TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
$2.8~\mathrm{ms^{-2}}$
›Reveal solutionSolution
The apparent weight is the normal reaction from the lift floor. With upward retardation (deceleration), the net acceleration is downward, so apparent weight = $m(g - a) = 60(9.8 - 2.8) = 420~\mathrm{N}$ , which corresponds to option (D).

Concept & Intuition

“Apparent weight” is not your true weight ( $mg$ ), but the force you feel through the floor — the normal reaction $N$ . When a lift accelerates, your body “wants” to keep moving at constant velocity (Newton’s first law), so the floor must push harder or less hard to change your motion.

Here the lift is moving up but slowing down (retardation = $2.8~\mathrm{m/s^2}$ upward). Slowing down while going up means acceleration is downward. So you feel lighter, because the floor doesn’t need to push as hard — part of gravity is “used” to decelerate you.

Step-by-step reasoning

Choose a sign convention. Let upward be positive. The lift’s velocity is upward, but it is retarding, so its acceleration is downward:

$a = -2.8~\mathrm{m/s^2}.$

Draw the free-body diagram for the man.

Two forces act:

Weight $W = mg$ downward (negative).

Normal reaction $N$ from the floor upward (positive).

Newton’s second law:

$N - mg = m a.$

Plug in the numbers. $m = 60~\mathrm{kg}$ , $g = 9.8~\mathrm{m/s^2}$ , $a = -2.8~\mathrm{m/s^2}$ :

$N - (60)(9.8) = (60)(-2.8).$ $N - 588 = -168.$

Solve for $N$ . $N = 588 - 168 = 420~\mathrm{N}.$ …
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TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
$m:n$
›Reveal solutionSolution
The acceleration of the centre of mass of two unequal masses in an Atwood machine is the weighted average of their individual accelerations. For mass ratio $m:n$ , the centre-of-mass acceleration is $\left(\frac{m-n}{m+n}\right)^2 g$ , so the correct option is (B).

Concept & Intuition

When two masses are connected by a string over a pulley, the heavier mass accelerates downward and the lighter one upward. The centre of mass (COM) of the system does not move with constant acceleration equal to either mass’s acceleration — instead, it’s the mass-weighted average of their accelerations. Since the accelerations are opposite in direction, the COM acceleration ends up being much smaller than either individual acceleration. The key is to first find each block’s acceleration using Newton’s laws, then combine them.

Step-by-step solution

Set up the masses and forces

Let the masses be $m_1 = m$ and $m_2 = n$ (the ratio is $m:n$ , but we can treat them as actual masses for calculation; the ratio will appear naturally). Assume $m > n$ so that $m$ moves downward and $n$ upward. Tension $T$ is the same on both sides (light, inextensible string, frictionless pulley).

Write Newton’s second law for each block

For the heavier mass $m$ (downward positive):

$mg - T = m a$
For the lighter mass $n$ (upward positive):
$T - n g = n a$
Here $a$ is the magnitude of acceleration of each block (they move with the same magnitude but opposite directions).

Solve for the acceleration $a$ Add the two equations:

$(mg - T) + (T - ng) = m a + n a$ $(m - n)g = (m + n)a$ $a = \frac{m - n}{m + n}\, g$
This is the familiar Atwood machine acceleration.

Find the acceleration of the centre of mass The centre of mass acceleration is the mass-weighted average of the individual accelerations. Take upward as positive.

Block $m$ accelerates downward: $a_m = -a$

Block $n$ accelerates upward: $a_n = +a$

$a_{\text{COM}} = \frac{m a_m + n a_n}{m + n} = \frac{m(-a) + n(+a)}{m + n} = \frac{(n - m)a}{m + n}$
Substitute $a = \frac{m - n}{m + n}\, g$ :
$a_{\text{COM}} = \frac{(n - m)}{m + n} \cdot \frac{m - n}{m + n}\, g$
Notice $n - m = -(m - n)$ , so:
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TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
$= 10\, \mathrm{ms}^{-2}$
›Reveal solutionSolution
The problem splits into two phases: free fall under gravity for 20 m, then uniform motion at the final free‑fall speed for the remaining 1980 m. The total time is 2 s + 99 s = 101 s, so the correct option is (C).

Concept & Intuition

The jumper first accelerates downward at $g = 10\ \mathrm{m/s^2}$ for a short distance (20 m). During this phase, speed increases from zero. Once the parachute opens, the problem states the motion becomes uniform — meaning the speed stays constant at whatever value was reached at the end of free fall. So we need to:

Find the speed after falling 20 m.

Find the time to fall that 20 m.

Then find the time to cover the remaining distance at that constant speed.

Add the two times.

A common pitfall is forgetting that the free‑fall distance is only 20 m, not the whole 2 km, and that the constant speed is the final speed from free fall, not an average.

Step‑by‑step solution

Free‑fall phase (first 20 m) Initial velocity $u = 0$ , acceleration $a = g = 10\ \mathrm{m/s^2}$ , distance $s_1 = 20\ \mathrm{m}$ . Use $v^2 = u^2 + 2as$ to find the speed just before the parachute opens:

$v^2 = 0 + 2 \cdot 10 \cdot 20 = 400 \quad\Rightarrow\quad v = 20\ \mathrm{m/s}.$
The time for this phase comes from $v = u + at$ :
$20 = 0 + 10\,t_1 \quad\Rightarrow\quad t_1 = 2\ \mathrm{s}.$

Uniform‑motion phase (remaining distance) Total height = 2 km = 2000 m. Already fallen 20 m, so remaining distance: $s_2 = 2000 - 20 = 1980\ \mathrm{m}.$ …
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TG EAPCET 2025Set ap-2025-04-30-FN1 markMCQ
$4\, \mathrm{ms}^{-2}$
›Reveal solutionSolution
The key is that friction is constant, so doubling the acceleration requires increasing only the net force. The required force increase is 60%, making option (C) correct.

The problem involves a body on a rough surface, so friction is present and opposes motion. The applied force must overcome friction before producing acceleration. Since the surface is rough and the body is already moving, kinetic friction acts with a constant magnitude (as long as the normal force and surface properties don't change). This means the net force is the applied force minus the constant frictional force. To double the acceleration, we need to double the net force, not the applied force — that's the central insight.

Let's work through it step by step.

Find the frictional force using the first condition. Mass $m = 3\ \text{kg}$ , applied force $F_1 = 20\ \text{N}$ , acceleration $a_1 = 4\ \text{m/s}^2$ . Newton's second law: $F_{\text{net}} = m a$ . The net force is $F_1 - f$ , where $f$ is the kinetic friction. So:

$20 - f = 3 \times 4 = 12$
Solving: $f = 20 - 12 = 8\ \text{N}$ .

Determine the net force needed for double the acceleration.

Desired acceleration $a_2 = 2 \times 4 = 8\ \text{m/s}^2$ .

Required net force: $F_{\text{net,2}} = m a_2 = 3 \times 8 = 24\ \text{N}$ .

Find the new applied force.

The friction is still $8\ \text{N}$ (constant). So:

$F_2 - 8 = 24 \quad \Rightarrow \quad F_2 = 32\ \text{N}$

Calculate the percentage increase in the applied force. Increase = $F_2 - F_1 = 32 - 20 = 12\ \text{N}$ . …
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TG EAPCET 2025Set ap-2025-04-30-FN1 markMCQ
$\dfrac{25GM}{144R^2}$
›Reveal solutionSolution
The gravitational force on a unit mass at the centre of the spherical shell comes only from the solid sphere (since the shell’s own field inside is zero), and the distance from the sphere’s centre to the shell’s centre is $3R$ , giving a force of $\frac{GM}{9R^2}$ .

The key idea here is a classic result from Newton’s shell theorem: a uniform spherical shell exerts zero net gravitational force on any mass placed inside it. That means the unit mass at the centre of the shell feels nothing from the shell itself — the only contribution comes from the solid sphere.

But the solid sphere is not centred at the same point. Its centre is offset because the sphere and shell touch each other. So we need to find the distance between the centre of the shell (where our test mass sits) and the centre of the solid sphere.

Locate the centres. The spherical shell has radius $4R$ , so its centre is at point O. The solid sphere has radius $R$ , and its surface touches the inner surface of the shell. Since the shell’s inner radius is $4R$ , the distance from O to the centre of the solid sphere (call it O') is the shell’s radius minus the sphere’s radius:

$OO' = 4R - R = 3R.$

Force from the shell. By the shell theorem, a uniform spherical shell exerts zero gravitational field at any point inside it. Our unit mass is at the centre O, which is certainly inside the shell. Therefore,

$F_{\text{shell}} = 0.$

Force from the solid sphere. For a point outside a uniform solid sphere, the sphere behaves as if all its mass were concentrated at its centre. The unit mass at O is at a distance $3R$ from the sphere’s centre O', so it is outside the sphere (since $3R > R$ ). Hence the force is simply …
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TG EAPCET 2024Set ap-2024-05-07-FN1 markMCQ
$10~\mathrm{ms}^{-1}$
›Reveal solutionSolution
The key idea is conservation of horizontal momentum because the man jumps vertically, so his initial horizontal momentum is zero. The boat’s speed decreases by $1~\mathrm{ms}^{-1}$ , making the correct option (C).

Concept & Intuition

When the man jumps vertically down from the bridge, he has no horizontal velocity relative to the ground at the moment he leaves the bridge. The boat is moving horizontally at $10~\mathrm{m/s}$ . As the man lands on the boat, the only horizontal forces are internal (between man and boat), so the total horizontal momentum of the man–boat system is conserved. The boat slows down because it must share its horizontal momentum with the man.

Define the system and initial conditions

Mass of man: $m = 60~\mathrm{kg}$

Mass of boat: $M = 540~\mathrm{kg}$

Initial speed of boat: $v_b = 10~\mathrm{m/s}$ (horizontal)

Initial speed of man (horizontal): $v_m = 0$ (he jumps straight down)

Total initial horizontal momentum:

$p_i = M v_b + m \cdot 0 = 540 \times 10 = 5400~\mathrm{kg\,m/s}$

After the man lands on the boat

The man and boat move together with a common horizontal speed $v_f$ .

Total mass: $M + m = 540 + 60 = 600~\mathrm{kg}$

Final horizontal momentum:

$p_f = (M+m) v_f = 600\,v_f$

Apply conservation of horizontal momentum

$p_i = p_f \quad \Rightarrow \quad 5400 = 600\,v_f$
$v_f = \frac{5400}{600} = 9~\mathrm{m/s}$ …
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TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
$5\,\mathrm{kg}$
›Reveal solutionSolution
The key idea is that the water jet exerts a force on the block equal to the rate of change of momentum of the water. Using $F = \frac{\Delta p}{\Delta t} = 2 \rho A v^2$ , we find $F = 25\,\mathrm{N}$ , so acceleration $a = F/m = 5\,\mathrm{m/s^2}$ . The correct option is (D).

Concept & Intuition

When a stream of water hits a block and rebounds with the same speed, the water’s momentum changes direction. That change in momentum per unit time is the force the water exerts on the block. Since the surface is smooth (no friction), this force is the net force on the block, giving it an acceleration. The key is to realize that the water’s velocity reverses, so the change in velocity is twice the original speed, not just once.

Step-by-step solution

Identify the relevant physics The force on the block comes from the water jet. For a fluid stream hitting a surface and rebounding, the force is given by the rate of change of momentum of the fluid:

$F = \frac{\Delta p}{\Delta t}$
Here, each water molecule’s velocity changes from $+v$ (toward the block) to $-v$ (away), so the change in velocity for each molecule is $\Delta v = v - (-v) = 2v$ .

Find the mass flow rate of water The water comes from a pipe of cross-sectional area $A = 5\,\mathrm{cm}^2 = 5 \times 10^{-4}\,\mathrm{m}^2$ with speed $v = 5\,\mathrm{m/s}$ . The volume of water hitting the block per second is $A v$ . Density $\rho = 1\,\mathrm{g/cc} = 1000\,\mathrm{kg/m}^3$ . So the mass flow rate (mass per second) is:

$\frac{\Delta m}{\Delta t} = \rho A v = 1000 \times (5 \times 10^{-4}) \times 5 = 2.5\,\mathrm{kg/s}$

Calculate the force on the block The momentum change per second for the water is: $F = \frac{\Delta m}{\Delta t} \times (\text{change in velocity}) = (2.5\,\mathrm{kg/s}) \times (2 \times 5\,\mathrm{m/s}) = 2.5 \times 10 = 25\,\mathrm{N}$ …
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TG EAPCET 2022Set ap-2022-07-31-FN1 markMCQ
$\sqrt{4gR_E}$
›Reveal solutionSolution

Use energy conservation between the surface and infinity; the launch speed exceeds escape speed, so it arrives at infinity with $v_\infty=\sqrt{2gR_E}$ .

Energy conservation from the surface to infinity (where potential energy is zero):
$\frac{1}{2}v_0^2 - \frac{GM}{R_E} = \frac{1}{2}v_\infty^2$
Using $\dfrac{GM}{R_E}=gR_E$ and the launch speed $v_0=\sqrt{4gR_E}$ (so $v_0^2=4gR_E$ ): …
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Q.A person driving a car suddenly applies the brakes on seeing a child on the road ahead. If he is not wearing seat belt, he falls forward and hits his head against the steering wheel. Why?

Concept understanding — Newton's First Law

Newton's First Law: The Law of Inertia

The Intuition: Objects Are Lazy

The Precise Statement

Why This Law Matters

Real-Life Examples

Concept: Newton's First Law (inertia) applied to the driver's body, not the car

Concept and Intuition

Step-by-step solution

Q.A person driving a car suddenly applies the brakes on seeing a child on the road ahead. If he is not wearing seat belt, he falls forward and hits his head against the steering wheel. Why?

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