Newton's First Law states: every body continues in its state of rest or uniform motion in a straight line unless compelled to change that state by an external unbalanced force (NCERT Class 11 Physics Chapter 4, page 2). This property of a body — its resistance to any change in velocity — is called inertia.
The trap NEET exploits: Students conflate Newton's First Law with Newton's Third Law. A book on a table is in equilibrium — net force is zero — so it stays at rest (First Law). But when asked "are the weight of the book and the normal force from the table an action-reaction pair?", many answer yes. They are not. Both forces act on the same body (the book). An action-reaction pair under the Third Law always acts on two different bodies. The weight's true reaction partner is the gravitational pull the book exerts on the Earth; the normal's true partner is the force the book pushes down on the table.
Inertia is not a force. A body at rest does not need a force to "keep it at rest." It stays at rest because no net external force acts on it. Similarly, an object moving at constant velocity on a frictionless surface needs no force to maintain that motion — First Law directly.
Connection to Newton's Second Law. When F_net = 0, the Second Law gives a = 0 — the body's velocity is constant (possibly zero). The First Law is sometimes called the "zero-force special case" of the Second Law, but it carries a deeper role: it defines what an inertial reference frame is. A frame where a free body (no net force) has zero acceleration is inertial. Non-inertial frames (accelerating bus, rotating platform) violate this and require pseudo-forces.
Watch out: NEET questions on the First Law are mostly recall or conceptual-application. They test whether you understand what inertia means, which reference frames are inertial, and whether you can correctly identify action-reaction pairs versus equilibrium force pairs.
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
Newton's First Law of motion defines:
A passenger standing in a bus lurches forward when the bus suddenly brakes. This is a direct illustration of:
Which of the following is an inertial reference frame?
A book lies at rest on a table. A student claims the weight of the book and the normal reaction from the table form a Newton's Third Law action-reaction pair. This claim is:
A hockey puck slides on a perfectly frictionless ice surface at constant speed. The net force acting on the puck is:
A body of mass 2 kg moves due east at 5 m/s. A net force acts on it for a brief interval, after which it moves due north at 5 m/s. The direction of the net force during that interval was:
A horizontal force of 20 N is applied to block A (mass 3 kg) which pushes block B (mass 2 kg) on a frictionless surface. The contact force that block A exerts on block B is:
Newton's First Law is sometimes called the law of inertia. Which statement about inertia is correct?
Given
A 0.50 kg ball moves due south at 4.0 m/s. After hitting a wall, it moves due west at 4.0 m/s. The collision lasts 0.010 s.
Required
Direction and magnitude of the average net force on the ball during the collision.
Concept
Newton's Second Law: F = Δp/Δt. The force is in the direction of the change in momentum, not the direction of final motion.
Formula
F = m(v_f − v_i) / Δt
Substitution
Choose south as +y, west as +x. v_i = 4.0 ŷ (due south) v_f = 4.0 x̂ (due west) Δv = v_f − v_i = 4.0 x̂ − 4.0 ŷ = (4.0, −4.0) m/s |Δv| = √(4.0² + 4.0²) = √32 = 4√2 m/s
Calculation
|F| = m|Δv|/Δt = 0.50 × 4√2 / 0.010 = 0.50 × 5.657 / 0.010 = 2.828 / 0.010 = 283 N Direction of Δv: components (+4.0 west, −4.0 north-to-south reversed = toward north from south). In our coordinate system, Δv = (4.0 x̂ − 4.0 ŷ) points toward north-west — specifically 45° north of west, i.e. toward the direction that is simultaneously west (gaining) and away from south (losing southward momentum). More precisely: with south = +y and west = +x, Δv = (+4.0, −4.0). This is in the fourth quadrant of our coordinate system (positive x, negative y) = west and anti-south = west and north. The angle from the +x (west) axis is arctan(4.0/4.0) = 45° toward the −y (north) direction. So the force points north-west, bisecting west and north. Note: The mass 0.50 kg and time 0.010 s are exact problem-defined values; they do not limit significant figures. The speeds 4.0 m/s have 2 significant figures, so the final answer is reported to 2 significant figures.
Final answer
Average force ≈ 2.8 × 10² N, directed north-west (45° from both north and west).
Common trap
Students pick "due west" (direction of final motion) as the force direction. The force is along Δp = p_f − p_i, which has components in BOTH the west and north directions. Drawing the vector subtraction triangle is essential.
Similar NEET-style question
A 0.20 kg ball moving east at 3.0 m/s is struck and moves north at 3.0 m/s. Find the magnitude of the impulse on the ball. *Approach: J = mΔv. Δv = 3.0 ŷ − 3.0 x̂, |Δv| = 3√2. J = 0.20 × 3√2 = 0.60√2 ≈ 0.85 kg·m/s, directed north-west.* ---
Aristotle held that an external force is required to keep a body in motion. This view is intuitively appealing but flawed. Galileo's frictionless-plane argument and Newton's first law overturn it: a body with NO net external force continues in its state of rest or uniform motion in a straight line.
-- NCERT Class 11 Physics, Ch. 4, p. 2If no external force acts on a body, the body in motion continues to move uniformly in a straight line, and the body at rest continues to be at rest. Friction had been masking this principle in everyday observation.
-- NCERT Class 11 Physics, Ch. 4, p. 2Every body continues in its state of rest, or of uniform motion in a straight line, unless compelled by an external force to change that state. This identifies inertia as the property by which a body resists change in its state of motion.
-- NCERT Class 11 Physics, Ch. 4, p. 3An object moving in a circle of radius r at constant speed v has acceleration of magnitude v^2/r (or equivalently omega^2 * r) directed toward the centre. This is centripetal (radially inward), not tangential.
| Symbol | Quantity | SI Unit |
|---|---|---|
| a_c | Centripetal acceleration | m/s^2 |
| v | Tangential speed | m/s |
| r | Radius of circle | m |
| omega | Angular speed | rad/s |
On a road banked at angle theta from horizontal with tyre-road friction coefficient mu_s, this is the maximum speed for safe negotiation of a curve of radius r.
| Symbol | Quantity | SI Unit |
|---|---|---|
| v_max | Maximum safe speed | m/s |
| g | Gravitational acceleration | m/s^2 |
| r | Radius of the curve | m |
| mu_s | Coefficient of static friction | (dimensionless) |
| theta | Banking angle | rad/deg |
The net inward force required to keep a body of mass m moving in a circle of radius r at speed v is m*v^2/r. This 'centripetal' force is NOT a new fundamental force — it is whichever real force (tension, friction, gravity, etc.) provides the inward acceleration.
| Symbol | Quantity | SI Unit |
|---|---|---|
| F_c | Centripetal force | N |
| m | Mass of the body | kg |
| v | Tangential speed | m/s |
| r | Radius of the circle | m |
| omega | Angular speed | rad/s |
If the net external force on a system of particles is zero, the total linear momentum of the system is conserved (vector equality of total p before and after any internal interaction). Basis of all collision and recoil analysis.
| Symbol | Quantity | SI Unit |
|---|---|---|
| F_ext | Net external force on system | N |
| p_total | Sum of m_i*v_i over all particles | kg*m/s |
Impulse equals the product of force and the time interval over which it acts. By Newton's Second Law, impulse equals the change in linear momentum during that interval.
| Symbol | Quantity | SI Unit |
|---|---|---|
| J | Impulse (vector) | N*s = kg*m/s |
| F | Force (treated as average) | N |
| Delta_t | Time interval | s |
| Delta_p | Change in momentum | kg*m/s |
When two surfaces slide relative to each other, kinetic friction opposes the motion with magnitude proportional to the normal force.
| Symbol | Quantity | SI Unit |
|---|---|---|
| f_k | Kinetic friction force | N |
| mu_k | Coefficient of kinetic friction | (dimensionless) |
| N | Normal force | N |
Static friction is the only force available to provide centripetal acceleration on a level road. Setting mu_s*N = m*v^2/r and N = m*g gives this maximum-safe speed bound.
| Symbol | Quantity | SI Unit |
|---|---|---|
| v_max | Maximum safe speed (no skid) | m/s |
| mu_s | Coefficient of static friction (tyre vs road) | (dimensionless) |
| g | Gravitational acceleration | m/s^2 |
| r | Radius of the circular path | m |
The net external force on a body equals the rate of change of its linear momentum. For a body of constant mass, this reduces to F = m*a — net force equals mass times acceleration. Both F and a are vectors; the acceleration is in the direction of the net force.
| Symbol | Quantity | SI Unit |
|---|---|---|
| F | Net external (vector) force | N |
| m | Mass of the body | kg |
| a | Acceleration (vector) | m/s^2 |
| p | Linear momentum (= m*v) | kg*m/s |
| t | Time | s |
The maximum value of static friction between two surfaces in contact equals the coefficient of static friction times the normal force. Below f_s_max, static friction self-adjusts to whatever value is needed to prevent relative motion.
| Symbol | Quantity | SI Unit |
|---|---|---|
| f_s_max | Maximum static friction | N |
| mu_s | Coefficient of static friction | (dimensionless) |
| N | Normal force | N |
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
Category: Similar Terms
Student claims velocity is constant in uniform circular motion (it's not — direction changes).
Question asks 'in uniform circular motion at constant speed, which is also constant?'
In UCM: SPEED constant; KE constant. VELOCITY (vector) NOT constant. ACCELERATION (centripetal, magnitude v²/r) constant in MAGNITUDE but NOT in direction.
Category: Sign Convention
Student picks the direction of MOTION as the direction of the net force, instead of the direction of the change in momentum (Δp). When velocity changes direction at constant speed, the force is perpendicular to BOTH the initial and final velocity vectors (in the limit) — it's the direction of Δv.
Question describes a body changing direction (e.g. turning) and asks for the force direction or the direction of Δp.
F ∝ Δp = m Δv = m (v_f − v_i). Always draw v_i and v_f as vectors and subtract; the result (v_f − v_i) is the direction of net force.
Category: Unit Conversion
Student computes μmg (friction FORCE in Newtons) when asked for the friction-limited ACCELERATION. The two differ by a factor of m: a = μg, F = μmg.
Question gives μ, g, and a body's mass and asks for the maximum acceleration of the supporting surface OR the friction force on the body.
Read carefully: 'maximum acceleration of the vehicle so the body stays still' = μg (no mass). 'Friction force on the body' = μmg (mass present). Units expose the error: N for force, m/s² for acceleration.
Category: Sign Convention
Student forgets that velocity DIRECTION reverses on bounce; computes |v₁ - v₂| instead of |v₁ + v₂|.
Question describes ball dropped from height h₁, rebounding to height h₂; asks for impulse on ball.
Impulse J = Δp = m(v_after - v_before). Take down as positive: v_before = +v₁, v_after = -v₂. So J = m(-v₂ - v₁) = -m(v₁ + v₂); magnitude = m(v₁ + v₂).
Category: Sign Convention
Student writes a = g sin θ for a rough incline (which is the smooth-incline answer); forgets to subtract μ g cos θ.
Question contrasts rough vs smooth inclines, or asks for acceleration on a rough incline.
On a rough incline (block sliding down): a = g(sin θ - μ_k cos θ). On a rough incline (block sliding up): a = -g(sin θ + μ_k cos θ). Smooth case (μ = 0): just g sin θ.
Category: Overthinking
Student takes torque about the centre of mass (introducing all 4 forces with non-zero moment arms) instead of about the floor contact (where 2 forces have zero moment arm and the equation simplifies).
Question gives a uniform rod or ladder leaning against a wall; asks for friction coefficient or limiting condition.
Pick the pivot to ELIMINATE unknown forces from the torque equation. Floor-contact pivot: normal force and friction at floor contribute zero torque; only weight (mid-length) and wall normal (top) appear. Result: μ_min = 1 / (2 tan θ).
Category: Sign Convention
Student adds magnitudes of momenta of fragments instead of vectors; ignores cancellation when fragments fly perpendicular or in opposite directions.
Question describes a body at rest exploding into multiple fragments with given mass ratios and partial velocity info.
Total momentum is a VECTOR. Initial p = 0; therefore Σ m_i v_i = 0 as a vector equation. Decompose along chosen axes (often natural symmetry axes); sum = 0 in each.
Category: Overthinking
Student tries to apply F=ma to the system as a whole (using net force = (m1-m2)g and total mass m1+m2) but loses track of the tension. The correct approach writes Newton's Second Law SEPARATELY for each mass and treats T as an unknown in two simultaneous equations.
Question contains a frictionless pulley with two unequal masses tied to a string. Asks for tension T or acceleration a (or both).
Draw a free-body diagram for EACH mass. Write F=ma per body, treating T as the same magnitude on both sides of the string. Solve simultaneously: a = (m1 - m2) g / (m1 + m2); T = 2 m1 m2 g / (m1 + m2).
Category: Negative Marking
Multi-mass pulley problem requires computing acceleration first, then tension. T = 2 m1 m2 g/(m1+m2). Sign errors in m1−m2 propagate.
Atwood machine or pulley system with multiple masses.
Compute a = (m1-m2)g/(m1+m2) first with m1 the heavier mass and downward as positive. Then T from F=ma on either mass. Always re-check by plugging back into both Newton's 2nd Law equations.
Category: Overthinking
Student applies the full external force F to a single block instead of recognising the system needs to be analysed for the contact (internal) force.
Question gives horizontal force F on block A which pushes block B; asks for contact force between A and B or acceleration.
System acceleration: a = F / (m_A + m_B). Contact force on B from A = m_B × a. The full F acts on the system, not on each block independently.
Category: Overthinking
Student computes P = Mgv (just lifting against gravity) and ignores the friction-opposing-motion term.
Question describes a lift moving at constant speed with explicit friction force on cable or guides.
At constant speed, net force = 0, so cable tension T = Mg + f_friction. Power = T × v = (Mg + f) × v. Always add friction when stated.
Root cause: concept gap
Action-reaction pairs ALWAYS act on DIFFERENT bodies. The pair to the book's weight is the gravitational pull the book exerts on the Earth. The pair to the normal force from table on book is the force the book exerts on the table. Equal-and-opposite forces on the SAME body are an equilibrium statement, not third-law statement.
Distractor labels two forces on the same body as a Newton's-third-law pair.
Root cause: concept gap
Centripetal force is NOT a new fundamental force. It is the NET inward radial component of the real forces (tension, friction, normal, gravity, etc.). On a free-body diagram, draw only the real forces; their net inward component must equal m*v^2/r.
Distractor sums tension + 'centripetal force' as separate inward forces.
Root cause: unit error
Impulse J has dimensions of momentum (kg*m/s) and equals F*Delta_t. Force has dimensions kg*m/s^2. Confusing them inflates or deflates an answer by a factor of seconds. Always check units before declaring an answer.
Distractor reports an answer in newtons where the correct answer is in N*s (or vice versa).
Root cause: concept gap
Linear momentum is conserved only when the net EXTERNAL force is zero. Gravity over a finite time changes momentum. For collision problems we use conservation because the collision happens over a brief time interval where external impulses are negligible compared to internal collision forces.
Distractor sets initial momentum = final momentum for a free-fall problem where gravity has acted for several seconds.
Root cause: formula misuse
Static friction is SELF-ADJUSTING: f_s exactly cancels the applied tangential force up to a ceiling f_s_max = mu_s * N. Below the ceiling, f_s = applied force. At the ceiling, motion is impending. Substituting mu_s * N too early over-estimates the friction.
Distractor uses mu_s * N for the static friction force in a no-slip scenario where the applied force is well below threshold.
treats velocity as scalar
Conflates speed (scalar) with velocity (vector)
scalar sum of momenta
Treating momentum as scalar; ignores vector cancellation
force along final velocity
Default to thinking force points in direction of motion
force along initial velocity
Newton-1 misread: object 'wants' to keep moving in original direction
subtracts v2 from v1 instead of adding
Forgetting velocity DIRECTION reversal at bounce
ignores mu cos theta term
Forgetting the friction-along-incline component
torque about wrong pivot
Default pivot at center of mass adds complexity
uses g where a belongs
Forgetting that the system accelerates, so weight is balanced by net force minus T
confuses tension with weight
Treating T = m·g for one of the masses (which is true only when a=0)
uses mu times g times mass
Confusing force (μ·N = μmg) with acceleration
treats each block with full F
Forgetting Newton's 3rd law internal force decomposition
forgets friction term
Ignoring opposing force
Test yourself on this topic with real past-paper questions:
Practice this topic →Get a structured 30-day Mechanics plan and a complete formula booklet — delivered to your inbox instantly.