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Physics · Laws of Motion

Newton's Second Law of Motion

The first law tells you what happens when no net force acts. The second law handles every other case — the actual workhorse of mechanics. Built on momentum rather than acceleration, it gives you \(F = \dfrac{dp}{dt}\), reduces to the familiar \(F = ma\) for constant mass, and powers the impulse-momentum theorem behind airbags, follow-through, and NEET's favourite "instant-after" spring problems. This deep-dive walks through the NCERT derivation, the four interpretive points students miss, and two PYQ traps from 2021 and 2017.

From first to second law

The first law settles one question: what does a body do when no net force acts? It keeps doing whatever it was already doing. That covers exactly one situation. The second law steps in for everything else — every push, weight, tension and friction problem you will meet in NEET. It is the law that turns a force diagram into a prediction about motion.

Why momentum?

Before stating the law, NCERT builds intuition for one quantity — momentum. The point is that "how hard it is to change a body's motion" depends on more than just speed. NCERT motivates this with four everyday observations.

  1. A truck travelling at the same speed as a car needs a much larger push to start or stop. Mass matters.
  2. A heavy stone is harder to catch than a light one moving at the same speed. Mass matters, again.
  3. A bullet fired at high speed pierces wood; a bullet thrown by hand at low speed does not. Speed matters.
  4. A stone whirled at the end of a string at constant speed but changing direction still requires a force to keep it moving in a circle. Direction matters.

One quantity ties all four together — the product of mass and velocity. NCERT defines linear momentum as

\[ \vec{p} = m\,\vec{v}. \]

Momentum is a vector, pointing along the velocity. Its SI unit is \(\mathrm{kg\,m\,s^{-1}}\). Doubling the mass doubles momentum; doubling the speed doubles momentum; turning the velocity vector changes momentum even at constant speed. This single object captures every effect in observations (i)–(iv).

Statement of the second law

"The rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction in which the force acts."

Mathematically,

\[ \vec{F} = k\,\frac{d\vec{p}}{dt}. \]

For a body of fixed mass \(m\),

\[ \frac{d\vec{p}}{dt} = m\,\frac{d\vec{v}}{dt} = m\,\vec{a}, \]

so \(\vec{F} = k\,m\,\vec{a}\). The SI unit of force is chosen to make \(k = 1\), giving the textbook statement

\[ \vec{F} = \frac{d\vec{p}}{dt} = m\,\vec{a}. \]

The corresponding unit, the newton, is the force that gives a \(1\,\mathrm{kg}\) mass an acceleration of \(1\,\mathrm{m\,s^{-2}}\):

\[ 1\,\mathrm{N} = 1\,\mathrm{kg\,m\,s^{-2}}. \]

Four NCERT points

NCERT highlights four interpretive points before any worked example. Each shows up regularly in NEET distractors.

1. Consistency with the first law

Set \(\vec{F} = 0\) in \(\vec{F} = m\vec{a}\). Then \(\vec{a} = 0\), so velocity is constant. The first law is not an independent postulate — it is the \(F = 0\) special case of the second. The first law's separate status is historical and conceptual: it defines what an inertial frame is, the frame in which the second law takes the simple form above.

2. Vector character and the projectile

The law is a vector equation. In component form,

\[ F_x = m\,a_x, \qquad F_y = m\,a_y, \qquad F_z = m\,a_z. \]

Force and acceleration are parallel; force and velocity need not be. A force perpendicular to the velocity changes only the perpendicular component of velocity — it never alters the parallel component. The standard illustration is a projectile near the Earth's surface. The only force is gravity, vertically downward. The horizontal component of velocity therefore remains unchanged throughout the flight, while the vertical component picks up \(g\) every second. Circular motion at constant speed is the same idea pushed further: the centripetal force is always perpendicular to velocity, so it only turns the velocity vector and never speeds the body up or slows it down.

3. F means net external force

In every application of the second law, \(\vec{F}\) stands for the resultant of all forces acting on the body — and, when the law is applied to a system of particles or a rigid body, only the external forces. Internal forces between parts of a system always cancel pairwise by the third law and contribute nothing to the motion of the centre of mass. For a rigid body or a system, \(\vec{a}\) is the acceleration of the centre of mass:

\[ \vec{F}_{\text{ext}} = M\,\vec{a}_{\text{cm}}. \]

4. The law is local in time

Newton's second law is a local statement. The force acting at an instant determines the acceleration at that same instant — and only at that instant. The law has no memory. The moment a force ceases, the acceleration it produced ceases too; the body retains the velocity it had built up, not any "leftover" acceleration. NCERT's example: a stone dropped from an accelerating train. While the stone was in the passenger's hand, the hand exerted a horizontal force on it. Once it is released, the only force is gravity. The stone keeps the horizontal velocity it had at the instant of release, but it has zero horizontal acceleration afterwards — even though the train (still pushed by its engine) keeps accelerating.

Impulse and the impulse-momentum theorem

Rewriting the second law as \(\vec{F}\,dt = d\vec{p}\) and integrating over an interval \([t_1, t_2]\) gives the impulse-momentum theorem:

\[ \vec{J} = \int_{t_1}^{t_2} \vec{F}\,dt = \Delta \vec{p} = \vec{p}_f - \vec{p}_i. \]

The vector \(\vec{J}\) is called the impulse of the force. For a constant force, the integral collapses to \(\vec{J} = \vec{F}\,\Delta t\). For a varying force, the impulse is the area under the \(F\)–\(t\) graph between the chosen limits. The SI unit is the newton-second, \(\mathrm{N\,s}\), which is identical to \(\mathrm{kg\,m\,s^{-1}}\) — the same unit as momentum itself, as the theorem requires.

Impulsive forces — from collisions, hammer blows, ball impacts, explosions — are large and brief. NCERT stresses that they are not different in nature from ordinary forces; they simply act over short times. The impulse-momentum theorem replaces an unknown \(F(t)\) with a single algebraic relation between initial and final momenta.

Why softening the blow works

The impulse-momentum theorem explains a family of safety tricks. The incoming \(\Delta p\) is fixed by the situation. To bring you to rest the surface must absorb exactly \(\Delta p\) of impulse. Stretch the contact time \(\Delta t\), and the peak force \(F\) drops in inverse proportion.

  • Catching a cricket ball. A fielder pulls the hands back as the ball arrives, lengthening the stopping time from a few milliseconds to a tenth of a second. Same momentum change, average force on the palm an order of magnitude smaller.
  • Jumping onto sand vs concrete. Sand deforms and gives way under the foot, taking tens of milliseconds to stop you. Concrete stops you in a millisecond. Same \(\Delta p\), drastically larger \(F\) on concrete — which is why knees and ankles complain.
  • Airbags. The bag deforms over roughly a tenth of a second instead of the few milliseconds a rigid dashboard would allow. Peak deceleration force on the occupant is cut by a factor of ten or more.

In each case the theorem is the same — \(F\,\Delta t = \Delta p\) — and the design lever is \(\Delta t\).

Worked examples

Worked example 1 · NCERT 4.2

A bullet of mass \(0.04\,\mathrm{kg}\) moving at \(90\,\mathrm{m\,s^{-1}}\) enters a heavy wooden block and is brought to rest after penetrating \(0.60\,\mathrm{m}\). Find the average resistive force exerted by the block on the bullet.

Using \(v^2 = u^2 - 2as\) with \(v = 0\), the deceleration magnitude is

\[ a = \frac{u^2}{2s} = \frac{(90)^2}{2 \times 0.60} = \frac{8100}{1.2} = 6750\,\mathrm{m\,s^{-2}}. \]

The resistive force from \(F = ma\):

\[ F = 0.04 \times 6750 = 270\,\mathrm{N}, \]

directed opposite to the bullet's motion.

Worked example 2 · NCERT 4.4

A batsman strikes back a ball of mass \(0.15\,\mathrm{kg}\) straight along the line of approach. The ball comes in at \(12\,\mathrm{m\,s^{-1}}\) and leaves at the same speed. What is the impulse imparted to the ball?

Choose the bowler-to-batsman direction as positive. Incoming velocity \(v_i = -12\,\mathrm{m\,s^{-1}}\); rebound velocity \(v_f = +12\,\mathrm{m\,s^{-1}}\).

\[ J = \Delta p = m(v_f - v_i) = 0.15 \times (12 - (-12)) = 0.15 \times 24 = 3.6\,\mathrm{N\,s}, \]

directed from the batsman to the bowler.

Worked example 3 · NEET 2021 deep-dive

A ball of mass \(0.15\,\mathrm{kg}\) is dropped from a height of \(10\,\mathrm{m}\), strikes the ground and rebounds to the same height. Find the magnitude of the impulse imparted to the ball. Use \(g = 10\,\mathrm{m\,s^{-2}}\).

Speed at impact: \(v = \sqrt{2gh} = \sqrt{2 \times 10 \times 10} = \sqrt{200} = 10\sqrt{2}\,\mathrm{m\,s^{-1}}\), downward.

Since the ball returns to the same height, the rebound speed is also \(10\sqrt{2}\,\mathrm{m\,s^{-1}}\), now upward.

Taking upward as positive: \(v_i = -10\sqrt{2}\), \(v_f = +10\sqrt{2}\).

\[ J = m(v_f - v_i) = 0.15 \times \left(10\sqrt{2} - (-10\sqrt{2})\right) = 0.15 \times 20\sqrt{2} \approx 4.24\,\mathrm{kg\,m\,s^{-1}}. \]

Worked example 4 · seatbelts

A passenger of mass \(60\,\mathrm{kg}\) is travelling at \(20\,\mathrm{m\,s^{-1}}\) in a car that crashes. The seatbelt brings the passenger to rest in \(0.10\,\mathrm{s}\). Find the average force exerted by the belt.

\(\Delta p = m\Delta v = 60 \times (0 - 20) = -1200\,\mathrm{kg\,m\,s^{-1}}\), magnitude \(1200\,\mathrm{N\,s}\).

\[ F = \frac{\Delta p}{\Delta t} = \frac{1200}{0.10} = 12000\,\mathrm{N}. \]

If a rigid dashboard had stopped the passenger in \(0.01\,\mathrm{s}\) instead, the force would have been \(1.2 \times 10^5\,\mathrm{N}\) — the same momentum change, ten times the peak force.

Quick recap

Newton's second law in one glance

  • Momentum \(\vec{p} = m\vec{v}\); it captures mass, speed and direction in one vector.
  • \(\vec{F} = d\vec{p}/dt\); for constant mass this reduces to \(\vec{F} = m\vec{a}\), with \(1\,\mathrm{N} = 1\,\mathrm{kg\,m\,s^{-2}}\).
  • \(F\) is the net external force; for a system, \(\vec{a}\) is the centre-of-mass acceleration.
  • The law is local: force right now sets acceleration right now; no memory of past forces.
  • Impulse-momentum theorem: \(\vec{J} = \int \vec{F}\,dt = \Delta \vec{p}\). The area under an \(F\)–\(t\) graph is the change in momentum.
  • Stretching \(\Delta t\) lowers peak \(F\) for the same \(\Delta p\) — airbags, sand pits and pulled-back hands all exploit this.
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Try the chapter test

Two PYQs below test the locality and bouncing-ball traps. Take the full Laws of Motion chapter test for a 25-question diagnosis.

NEET PYQ Snapshot — Newton's Second Law

Two PYQs that test impulse and locality directly. NCERT Exercise 4.5 closes the set.

NEET 2021

A ball of mass \(0.15\,\mathrm{kg}\) is dropped from a height \(10\,\mathrm{m}\), strikes the ground and rebounds to the same height. The magnitude of impulse imparted to the ball is \((g = 10\,\mathrm{m\,s^{-2}})\) nearly

  1. \(1.4\,\mathrm{kg\,m\,s^{-1}}\)
  2. \(0\)
  3. \(4.2\,\mathrm{kg\,m\,s^{-1}}\)
  4. \(2.1\,\mathrm{kg\,m\,s^{-1}}\)
Answer: (3) 4.2 kg m s⁻¹

Why: Speed at the ground \(= \sqrt{2gh} = \sqrt{200} = 10\sqrt{2}\,\mathrm{m\,s^{-1}}\). After bouncing it leaves upward at \(10\sqrt{2}\,\mathrm{m\,s^{-1}}\) (same height = same speed). With upward positive, \(\Delta p = m(v_f - v_i) = 0.15 \times (10\sqrt{2} - (-10\sqrt{2})) = 0.15 \times 20\sqrt{2} \approx 4.24\,\mathrm{kg\,m\,s^{-1}}\). Distractor (4) 2.1 is exactly half — the trap where students forget to flip the sign on the rebound velocity and compute \(m(v - 0)\) or \(m\sqrt{2gh}\) on one leg only. (2) 0 catches students who subtract speed magnitudes. (1) 1.4 swaps in \(\sqrt{2}\) for \(2\sqrt{2}\).

NEET 2017

Two blocks A and B of masses \(3m\) and \(m\) respectively are connected by a massless and inextensible string. The whole system is suspended by a massless spring as shown in the figure. The magnitudes of acceleration of A and B immediately after the string is cut, are respectively

  1. \(g/3, g/3\)
  2. \(g, g/3\)
  3. \(g/3, g\)
  4. \(g, g\)
Answer: (3) g/3, g

Why: Before the cut the spring supports the total weight: spring force \(= (3m + m)g = 4mg\), upward. The instant the string is cut, the spring's length is still its equilibrium length, so by locality the spring force is unchanged at \(4mg\) upward. Block B is now in free fall — only gravity acts — so \(a_B = g\) downward. Block A feels \(4mg\) up from the spring and \(3mg\) down from gravity, net \(mg\) upward; acceleration \(a_A = mg/(3m) = g/3\) upward. The classic trap is to claim the spring "instantly relaxes" after the cut, which would give \(g, g\) — but a spring's length, and hence its force, cannot change discontinuously.

NCERT Exercise 4.5

A constant retarding force of \(50\,\mathrm{N}\) is applied to a body of mass \(20\,\mathrm{kg}\) moving initially with a speed of \(15\,\mathrm{m\,s^{-1}}\). How long does the body take to stop?

  1. \(3\,\mathrm{s}\)
  2. \(6\,\mathrm{s}\)
  3. \(9\,\mathrm{s}\)
  4. \(12\,\mathrm{s}\)
Answer: (2) 6 s

Why: Deceleration \(a = F/m = 50/20 = 2.5\,\mathrm{m\,s^{-2}}\). Using \(v = u + at\) with \(v = 0\), time to stop \(t = u/a = 15/2.5 = 6\,\mathrm{s}\). Alternatively, by the impulse-momentum theorem \(F\Delta t = \Delta p = mu\), so \(\Delta t = mu/F = (20 \times 15)/50 = 6\,\mathrm{s}\) — exactly the same answer through a one-line shortcut.

FAQs — Newton's Second Law

Short answers to the questions NEET aspirants ask most about \(F = ma\) and momentum.

Is F = ma always true, or is F = dp/dt the more fundamental form?
F = dp/dt is the fundamental statement. F = ma is the special case for constant mass, obtained by writing dp/dt = m(dv/dt) + v(dm/dt) and dropping the second term. For systems where mass changes with time, such as rockets or conveyor belts, the v(dm/dt) term cannot be dropped, and F = ma alone gives wrong answers.
What is the difference between impulse and force?
Force is an instantaneous push or pull measured in newtons. Impulse is the cumulative effect of a force over a time interval, measured in newton-seconds, and equals the change in momentum. A small force acting for a long time can produce the same impulse as a large force acting briefly — that is why follow-through in cricket and airbags in cars both work.
Why do airbags and cricket fielders pulling their hands back reduce injury?
Both increase the time over which momentum is brought to zero. Since impulse F·Δt = Δp is fixed by the incoming momentum, lengthening Δt by even a few hundredths of a second slashes the peak force F. An airbag stretching contact from 0.01 s to 0.1 s cuts the deceleration force by a factor of ten.
Does Newton's second law apply to internal forces inside a system?
No. When the second law is applied to a system as a whole, F stands only for the net external force. Internal forces between parts of the system cancel in pairs by Newton's third law. A passenger pushing the dashboard from inside a car cannot accelerate the car — only an external force such as friction, engine thrust transmitted through tyres, or air drag can.
What does locality of the second law mean?
Locality means the acceleration at any instant is determined entirely by the force acting at that same instant — there is no memory of past forces. A stone dropped from an accelerating train falls with horizontal velocity equal to the train's velocity at the moment of release, but with zero horizontal acceleration thereafter, because no horizontal force acts on it once it is in the air.
How does the second law apply to a rigid body or extended system?
For an extended body, F = ma describes the motion of the centre of mass, with m the total mass and a the acceleration of the centre of mass. Individual particles inside the body may accelerate differently — for example, points on a rolling wheel — but the centre of mass obeys F_external = M a_cm exactly.
Related Topics
Laws of Motion — Parent chapter All three laws, friction, circular motion, and PYQ-loaded chapter overview. Inertia & Newton's First Law Galileo's experiment, the F = 0 case, inertial frames, and kinds of inertia. Newton's Third Law — Action & Reaction Pairs always act on different bodies; common rocket and recoil distractors. Conservation of Linear Momentum Derived from second and third laws together; collisions, recoils, explosions. Equilibrium of a Particle Concurrent forces, Lami's theorem, and three-force balance problems. Free-Body Diagrams Isolating bodies, identifying contact forces, and avoiding ghost forces. Friction — Static, Kinetic, Rolling Coefficient ranges, limiting friction, and rolling vs sliding distinction. Pulley & Connected-Body Problems Constraint relations, tensions on either side, Atwood-machine PYQs. Circular Motion & Centripetal Force Why a force perpendicular to velocity only turns it, never speeds it up. Banking of Roads Maximum safe speed with and without friction; tan θ = v²/rg derivation. Apparent Weight in Lifts Accelerating lifts, weighing-machine readings, and free-fall edge cases.