L-3: Laws of Motion
Physics — Class 12 · NIOS Code 312 · Module 1 · Source: 312_Physics_Eng_Lesson3.pdf
Why Do Things Move? — Starting Point
In the previous lesson you learnt to describe motion using displacement, velocity, and acceleration. But a deeper question remains: what makes an object move? Why does a ball rolling on the ground eventually stop? From everyday experience we know that pushing or pulling is needed to change an object's position. A football must be kicked, a cricket ball must be hit hard — muscular activity is involved and its effect is visible.
Yet many causes of motion are not visible. What makes raindrops fall? What makes Earth orbit the Sun? In this lesson you discover that force causes motion. Newton showed that force and motion are intimately connected. The laws of motion are fundamental — they help us understand walking, driving, collisions, rocket propulsion, and countless everyday phenomena.
NIOS learning objectives: explain inertia; state and illustrate Newton's three laws; explain conservation of momentum; understand equilibrium of concurrent forces; define coefficients of friction; distinguish static, kinetic, and rolling friction; suggest methods of reducing friction; and apply Newton's laws using free body diagrams.
3.1 Concepts of Force and Inertia
Stationary objects remain where placed unless forced to move. Similarly, an object moving with constant velocity must be forced to change its state of motion. The property by which an object resists change in its state of rest or uniform motion in a straight line is called inertia. Think of it as laziness built into matter — objects prefer to keep doing what they are already doing.
Mass is the measure of inertia. A heavy truck at low speed is harder to stop than a light cricket ball at high speed because the truck has far more mass. Inertia is a remarkable property. Without it, your books could mingle with your sibling's notes and your wardrobe could drift to a friend's house. However, rest and uniform motion are not absolute: an object at rest for one observer may appear in motion to another (Lesson 1). Observations confirm that change in velocity occurs only when a net force acts on the body.
3.1.1 Force and Its Effects
Force is familiar from daily life — pulling, pushing, kicking, hitting. Though invisible, its effects are seen. Forces can: (a) change the shape and size of an object (a balloon deforms); (b) influence motion — start motion, stop motion, or change direction or speed; (c) rotate a body about an axis (covered in Lesson 7).
Force is a vector quantity. When several forces act simultaneously, a net equivalent force is found by vector addition (Lesson 1). Motion is characterised by displacement and velocity. Velocity may increase (free fall) or decrease (ball on horizontal surface). A net non-zero force is required to change state of motion.
- If net force is parallel to velocity → speed increases.
- If net force is opposite to velocity → speed decreases.
- If net force is perpendicular to velocity → speed stays constant but direction changes (uniform circular motion).
Student tip: A body does NOT always move in the direction of net force. If it already has velocity, force may only bend the path. Acceleration always aligns with net force; velocity may not.
3.1.2 Galileo's Experiments and the First Law
When you roll a marble on a smooth floor it stops — friction reduces velocity to zero. To keep it moving at constant velocity you must continuously apply force. Similarly, a trolley at constant velocity must be continuously pushed or pulled to balance friction.
Galileo showed that without external force, a body continues in rest or uniform straight-line motion. On an inclined plane moving downward, a body accelerates; moving upward, it retards. On a horizontal plane with no friction, it would move with uniform speed forever.
In his thought experiment, a ball rolls down plane PQ and rises on plane RS to nearly the same height. As RS becomes less inclined, the ball travels farther. When RS is horizontal, the ball keeps moving indefinitely if friction is absent.
Newton's First Law: A body continues in a state of rest or of uniform motion in a straight line unless acted upon by a net external force.
Because motion is relative, measurements must be made with respect to a frame of reference. An inertial frame is one in which a body in translatory motion has constant velocity when no net external force acts. A frame fixed to Earth is inertial for practical purposes.
3.2 Concept of Momentum
A fielder struggles to stop a fast cricket ball (small mass, large velocity) but also struggles with a slow truck (large mass, small velocity). Both mass and velocity matter when studying force and motion.
The product of mass m and velocity v is linear momentum p:
m = Mass (kilograms, kg)
v = Velocity (m·s⁻¹)
Momentum is a vector — direction same as velocity.
Momentum changes when magnitude, direction, or both change. Example: a freely falling 2 kg object has zero momentum at t = 0; at t = 1 s, v = 9.8 m/s downward so p = 19.6 kg·m/s downward; at t = 2 s, p = 39.2 kg·m/s downward. Gravitational force causes this continuous increase.
When a 0.2 kg ball strikes a wall at 10 m/s and rebounds at 10 m/s, magnitude of momentum is unchanged but direction reverses. If initial momentum is +2 kg·m/s, final is −2 kg·m/s, so change Δp = −4 kg·m/s. The wall exerts force on the ball.
3.3 Newton's Second Law of Motion
A body at constant velocity has constant momentum; the first law implies no net external force. For a freely falling body, momentum increases with time — connecting force, time, and momentum change.
Second Law (momentum form): The rate of change of momentum is directly proportional to the net force, and change occurs in the direction of the net force.
Δp = Change in momentum (kg·m·s⁻¹)
Δt = Time interval (seconds, s)
SI unit of force: kg·m·s⁻² = newton (N)
Writing p = mv and assuming constant mass:
m = Mass (kilograms, kg)
a = Acceleration (m·s⁻²)
Hover tip: Mass resists acceleration. The more mass, the more force needed for the same acceleration.
Example: A 0.4 kg ball rolling at 20 m/s stops in 10 s. |F| = m|Δv/Δt| = 0.4 × 20/10 = 0.8 N opposite to motion.
Example: A 10 kg body at 10 m/s is stopped by 50 N opposing force. t = m(v₀ − v)/F = 10 × 10/50 = 2 s.
Exam note: For rockets where mass changes with time, use F = d(mv)/dt — the more general form. F = ma alone is for constant mass only.
3.4 Forces in Pairs — Third Law and Impulse
When Earth pulls an object, does the object pull Earth? When you push an almirah, does it push back? Actions between two bodies are always mutual. Forces exist in pairs — both are real interaction forces.
Newton's Third Law: When two objects interact, the force exerted by one on the other is equal in magnitude and opposite in direction to the force exerted by the second on the first.
F₂₁ = Force on object 2 due to object 1
Equal magnitude, opposite direction. Act on DIFFERENT bodies — they do NOT cancel on one body.
A book on a table exerts downward force F₁ = mg on the table; the table exerts upward F₂ on the book. F₁ and F₂ do not cancel because they act on different bodies. Action and reaction exist simultaneously, not one after the other.
Impulse
Δt = Time duration (s)
Δp = Change in momentum (kg·m·s⁻¹)
SI unit: N·s. Impulse is a vector. Explains why fielders draw hands back when catching.
Action-Reaction Concept Map
NEWTON'S THIRD LAW
|
+--------------+--------------+
| | |
Forces exist Equal Opposite
in PAIRS magnitude direction
| | |
v v v
+---------------+ F₁₂ = -F₂₁ Act on DIFFERENT
| Examples: | bodies (no cancel)
+---------------+
| Man kicks | Action: foot on ball
| football | Reaction: ball on foot
+---------------+
| Earth-Moon | Action: Earth pulls moon
| | Reaction: moon pulls Earth
+---------------+
| Ball hits | Action: ball on wall
| wall | Reaction: wall on ball
+---------------+
| High jumper | Action: jumper pushes ground
| | Reaction: ground pushes jumper UP
+---------------+
3.5 Conservation of Momentum
When two bodies interact and mutual interaction is the only force, the vector sum of their momenta remains unchanged. For a closed (isolated) system — no external forces — total momentum is constant. Individual momenta may change due to internal mutual forces.
Total momentum before = total momentum after
Applies to collisions, explosions, gun recoil, rockets, nuclear reactions.
Derivation: From second law Δp = FΔt. For bodies A and B: F_AB = −F_BA, so Δp_A/Δt = −Δp_B/Δt, hence Δp_A + Δp_B = 0.
(a) Gun recoil: mv₁ + Mv₂ = 0, so v₂ = −(m/M)v₁. Since m ≪ M, recoil is much slower than the bullet.
(b) Collision: Two trolleys (mass m each) at v hit three stationary trolleys. 2mv = 5mv′ → v′ = (2/5)v.
(c) Explosion: Bomb at rest has zero momentum; fragments fly opposite ways so vector sum stays zero.
(d) Rocket propulsion: Escaping gases provide thrust via momentum conservation.
3.5.3 Equilibrium of Concurrent Forces
Forces acting simultaneously at one point are concurrent forces. They are in equilibrium when their resultant is zero: F₁ + F₂ + F₃ = 0. Graphically, the vector sum of any two must equal and oppose the third. For a point object in static equilibrium, the vector sum of all forces must be zero.
3.6 Friction
When a batsman hits a ball along the ground, it eventually stops — friction opposes motion and changes momentum. Friction is a contact force, parallel to surfaces and opposite to attempted or actual motion. Friction is a necessary evil: it wastes energy yet lets us walk, drive, and brake.
3.6.1 Static and Kinetic Friction
A block on a horizontal surface does not move until applied force F_ext exceeds a limit. While at rest, static friction f_s matches F_ext up to maximum f_s(max). Beyond this, the block slides and kinetic friction f_k acts. Starting motion needs more force than maintaining it: f_s(max) > f_k.
μ_s = Coefficient of static friction (dimensionless)
F_N = Normal force (N)
While f_s < f_s(max): f_s = F_ext. Independent of contact area.
μ_k = Coefficient of kinetic friction
F_N = Normal force (N)
Generally μ_s > μ_k. On horizontal surface F_N = mg.
Example: 2 kg block, μ_s = 0.25 → f_s(max) = 0.25 × 2 × 9.8 = 4.9 N.
Example: 5 kg block, μ_k = 0.1, pulled by 10 N. f_k = 4.9 N; F_net = 5.1 N; a = 1.02 m/s².
3.6.2 Rolling Friction and Reducing Friction
Rolling friction is much smaller than sliding friction (~1/100 for steel on steel). Methods to reduce friction: wheels and ball bearings; lubricants (grease, oil); compressed air between surfaces; streamlined shapes for fluid friction. Cars are most fuel-efficient around 40–45 km/h due to air resistance.
3.7 Free Body Diagram Technique
Newton's laws become easier with free body diagrams (FBD) — diagrams showing all forces on an isolated object.
HOW TO SOLVE FBD PROBLEMS
=========================
|
v
[1] Draw neat diagram of system
|
v
[2] ISOLATE the object of interest
(this is the "free body")
|
v
[3] Mark ALL external forces with arrows
touching the body; show line of action
|
v
[4] Apply Newton's 2nd Law:
ΣF = ma (or ΣFx = max, ΣFy = may)
|
v
[5] Count unknowns vs independent equations
Need equal number for complete solution
|
v
[6] Solve + CHECK extreme cases
(e.g. m1 = m2, m1 >> m2 for pulleys)
Example — Two blocks on smooth surface: Blocks m₁ and m₂ connected by string; m₂ pulled by F. a = F/(m₁ + m₂). Tension T = m₁F/(m₁ + m₂).
Example — Atwood machine: m₁ > m₂ over frictionless pulley. a = (m₁ − m₂)g/(m₁ + m₂); T = 2m₁m₂g/(m₁ + m₂).
Example — Trolley with friction: M = 10 kg, m = 2 kg, μ_k = 0.02 gives a = 1.47 m/s², T = 16.66 N.
3.8 Inertial and Non-Inertial Frames
An inertial frame is stationary or moves at constant velocity; Newton's laws hold directly. A non-inertial frame accelerates — add a pseudo force (−ma). Centrifugal force appears in rotating frames. Water tilts in a starting train because effective gravity in the train frame is g − a.
Summary
- Inertia resists change in rest or uniform motion; mass measures inertia.
- First Law: Uniform motion or rest continues unless net external force acts.
- Second Law: F = Δp/Δt; for constant mass F = ma.
- Third Law: F₁₂ = −F₂₁ on different bodies.
- Conservation: Isolated system → total momentum constant.
- Friction: f_s(max) = μ_s F_N; f_k = μ_k F_N; rolling ≪ kinetic.
- FBD before every numerical — the most reliable exam strategy.
Q1. Which physical quantity is a measure of the inertia of a body?
Q2. The SI unit of force, one newton, is equivalent to:
Q3. Linear momentum p of a body of mass m moving with velocity v is defined as:
Q4. Newton's first law of motion is also known as the law of:
Q5. For a block on a horizontal surface, the maximum force of static friction is given by:
Q6. In an isolated system of colliding bodies with no external force, which quantity remains constant?
Q7. According to Newton's third law, when a book rests on a table:
Q8. A gun of mass M fires a bullet of mass m with velocity v₁. The recoil velocity v₂ is:
Q9. In the free body diagram technique, Newton's second law is applied as:
Q10. Rolling friction between steel wheels and steel rails is approximately:
PYQ — Previous Year Questions
Extracted from NIOS Physics (312) board exam papers in your PDF. Chapter L3 — Laws of Motion only. Use Model Answer for marking points; Explanation for concept clarity.
Chapter 3 — Laws of Motion (L3)
20 questions · Section A (MCQ & objective) + Section B (short/long) · Sources: 312/TUS/104A, 312/MAY/204A–C, 68/ESS/1-312-A
Section A — Multiple Choice (1 mark)
PYQ1. When a constant net external force acts on a body, which of the following physical quantities may not change?
Model Answer
Answer: (D) Acceleration
From Newton's second law F = ma, a constant net force gives constant acceleration. Position, speed and velocity change with time; acceleration remains unchanged.
Explanation
Under constant F_net, a = F/m is fixed. Velocity v = u + at changes; position changes. Acceleration is the quantity that stays constant (not zero — unchanged in magnitude and direction).
PYQ2. A force F acts on a body of mass m for t seconds. The change in its linear momentum will be — (A) Ft (B) Fm (C) F/t (D) F/m
Model Answer
Answer: (A) Ft
Explanation
Impulse–momentum theorem: Δp = FΔt when F is constant. Hence change in linear momentum = Ft. Units: N·s = kg·m·s⁻¹.
PYQ3. The mass of a body is 2 kg. Its weight is — (A) 19.6 N (B) 9.8 N (C) 10 N (D) 5 N
Model Answer
Answer: (A) 19.6 N
W = mg = 2 × 9.8 = 19.6 N
Explanation
Weight is gravitational force W = mg near Earth's surface. Distinct from mass (kg).
PYQ4. A body of mass 200 g falls through air with acceleration 6 m·s⁻². The air drag on the body is — (A) 1200 N (B) 1.2 N (C) 1.96 N (D) 0.76 N
Model Answer
Answer: (D) 0.76 N
Mg − F_a = ma → F_a = Mg − ma = 0.20×9.8 − 0.20×6 = 1.96 − 1.2 = 0.76 N
Explanation
Newton's second law with air resistance upward: net force = weight − drag = ma. Drag reduces acceleration below g.
PYQ5. A passenger in a moving bus is thrown forward when the bus suddenly stops. This is explained by — (A) Newton's first law (B) Newton's second law (C) Newton's third law (D) conservation of mass
Model Answer
Answer: (A) Newton's first law
Explanation
Inertia (first law): body tends to continue its state of motion. Bus stops but passenger's upper body keeps moving forward until seat belt or friction acts.
PYQ6. The need of banking of roads is — (A) additional gravitational force (B) additional centrifugal force (C) additional centripetal force (D) additional electrostatic force
Model Answer
Answer: (C) To provide additional centripetal force for higher velocity
Explanation
Banking tilts normal reaction so its horizontal component supplies centripetal force mv²/r, allowing safe turning at higher speed without excessive friction.
PYQ7. A body of mass m just starts sliding on an incline when the plane makes 30° with the vertical. Coefficient of friction μ is — (A) 1/√3 (B) √3 (C) mg/√3 (D) √3·mg
Model Answer
Answer: (B) √3
Angle with horizontal θ = 90° − 30° = 60°. At limiting equilibrium: μ = tan θ = tan 60° = √3.
Explanation
Along incline: mg sin θ = μ mg cos θ at limiting friction → μ = tan θ. Read angle carefully — 30° with vertical means 60° with horizontal.
PYQ8. A 2 kg body moves at constant velocity 5 m·s⁻¹ under constant force 3 N. Power loss due to friction is — (A) Zero (B) 15 W (C) −15 W (D) 30 W
Model Answer
Answer: (B) 15 W
Constant v ⇒ net force = 0, so friction = 3 N opposite motion. P = F·v = 3 × 5 = 15 W.
Explanation
At uniform velocity, applied force balances friction. Power dissipated against friction P = f_k × v (L6 link: P = Fv).
PYQ9. Passage — Friction: F = μR. (i) Max static friction independent of — ? (ii) Unit of μ? (iii) Arrange μr, μk, μms ascending? (iv) Static friction on body at rest under 5 N applied force?
Model Answer
- (i) (c) area of contact
- (ii) (d) unitless
- (iii) (b) μr < μk < μms
- (iv) (b) 5 N (static friction equals applied force until limiting value)
Explanation
Limiting friction F = μs F_N depends on normal reaction and μ, not contact area. μ has no dimensions. Rolling friction is smallest; static max exceeds kinetic. Static friction is self-adjusting up to μs F_N.
PYQ10. Complete using [more, force, linear momentum, inertia, isolated, less]: (i) Total linear momentum of _______ system is conserved. (ii) Rate of change of momentum is higher when force is _______.
Model Answer
- (i) isolated
- (ii) large
- (iii) force (also accepted in scheme)
- (iv) linear momentum
Explanation
Conservation of momentum holds for isolated system (no net external force). Newton's second law: F = dp/dt — larger force ⇒ faster momentum change.
PYQ11. Passage — Newton's third law: (a) Why ignore intermolecular forces when finding net force? (b) Equilibrium under two forces? (c) Equilibrium under three forces? (d) Forces on a book on a table?
Model Answer
- (a) Internal forces occur in equal and opposite pairs (3rd law) — cancel in pairs.
- (b) Two forces: equal magnitude, opposite direction, same line of action.
- (c) Three forces: vector sum zero (closed polygon / Lami's theorem).
- (d) Weight mg downward; normal reaction N upward; both act on book.
Explanation
Only external forces change centre-of-mass motion. Translational equilibrium: ΣF_ext = 0. Weight and normal on the same body are not an action–reaction pair.
PYQ12. Match Column I with II: (i) Law of conservation of linear momentum — (ii) Expression for friction force — Options: (a) F = μR (b) F = ma (c) Ptotal = constant
Model Answer
(i) Conservation of linear momentum ↔ (c) Ptotal = constant
(ii) Friction force ↔ (a) F = μR
Explanation
F = ma is Newton's second law, not friction. Momentum conservation: total p unchanged in isolated collision/interaction.
PYQ13. A boy throws a ball vertically upward with velocity v₀ and catches it on return. What is the change in linear momentum of the ball?
Model Answer
Initial momentum p_i = mv₀ (upward). On return, velocity = −v₀ (downward before catch). Final p_f = −mv₀.
Δp = p_f − p_i = −mv₀ − mv₀ = −2mv₀
Magnitude of change = 2mv₀.
Explanation
Momentum is vector. Same speed but reversed direction ⇒ change is not zero. Sign shows downward impulse delivered by hands when catching.
PYQ14. Give any two methods of reducing friction between two surfaces.
Model Answer
Any two from L3 notes:
- Applying lubricants (oil/grease) between surfaces
- Using ball bearings or rollers (rolling friction < sliding)
- Polishing/smoothing surfaces
- Streamlining (for fluids)
- Using wheels instead of dragging
Explanation
Friction arises from surface irregularities and adhesion. Lubrication separates surfaces; rollers convert sliding to rolling with smaller μ.
PYQ15. Give any two examples of conservation of linear momentum.
Model Answer
- Recoil of a gun when bullet is fired
- Two ice skaters pushing apart and moving in opposite directions
- Collision of two billiard balls on a smooth table
- Rocket propulsion (ejecting gases backward)
Explanation
If net external force is zero, total momentum before = total momentum after. Internal forces cancel in pairs.
PYQ16. Identify the action–reaction forces on a book lying on a table, explaining how these forces are developed.
Model Answer
Pair 1: Book presses table downward → table presses book upward (normal contact forces).
Pair 2: Earth attracts book (weight) → book attracts Earth with equal and opposite gravitational force.
Forces develop from deformation at contact and gravitational interaction.
Explanation
Action and reaction act on different bodies, same line, equal magnitude, opposite direction. Weight and normal on the book are NOT a third-law pair (both on book).
PYQ17. A boy throws a ball of mass m upward with speed v; it returns to his hands at the same speed. Find the change in momentum.
Model Answer
Same as PYQ13: |Δp| = 2mv (direction of momentum reverses).
Explanation
Take upward positive: Δp = m(−v) − m(v) = −2mv.
PYQ18. Calculate limiting friction: (a) 2 kg block, μs = 0.3, g = 10 m·s⁻² (b) 5 kg block, μ = 0.1 (c) 3 kg block, μs = 0.4
Model Answer
flimit = μs F_N = μs mg
- (a) 0.3 × 2 × 10 = 6 N
- (b) 0.1 × 5 × 10 = 5 N
- (c) 0.4 × 3 × 10 = 12 N
Explanation
On horizontal surface F_N = mg. Limiting (maximum static) friction is threshold before sliding begins: fs,max = μs mg.
PYQ19. A 1 kg body at rest explodes into three fragments of mass ratio 1 : 1 : 3. The two equal fragments fly off perpendicular to each other at 30 m·s⁻¹ each. Find the velocity of the heavier fragment.
Model Answer
Masses: m₁ = m₂ = 0.2 kg, m₃ = 0.6 kg. Conservation of momentum (initial = 0):
0.2×30 î + 0.2×30 ĵ + 0.6 v₃ = 0
|v₃| = √(6² + 6²)/0.6 = 6√2/0.6 = 10√2 ≈ 14.1 m·s⁻¹
Direction: opposite to resultant of the two lighter fragments (135° from each).
Explanation
Explosion — internal forces only ⇒ total momentum conserved. Vector addition of perpendicular momenta gives magnitude 6√2 kg·m·s⁻¹; divide by 0.6 kg for heavy piece speed.
PYQ20. Explain how to determine impulse of a force when the force is (a) constant and (b) variable.
Model Answer
(a) Constant F: Impulse J = FΔt (area of rectangle under F–t graph). Equals change in momentum Δp.
(b) Variable F: J = ∫ F dt from t₁ to t₂ (area under F–t curve). Measure F at intervals, sum FΔt or integrate if F(t) known.
Explanation
Impulse–momentum theorem J = Δp links force applied over time to momentum change. Graphical method works for any F(t).
Problem Solving — L3 Laws of Motion
Six problems spanning this chapter’s NIOS syllabus. Every question is built from the notes and formula sheet: solve with equations first, then read the formal textbook-style write-up, the easy explanation, and the topic in depth (formulas, meaning, exam tips). Explanations open by default.
Draw a free-body diagram of a book resting on a table. Identify all forces on the book and explain, using Newton’s first law, why the book does not accelerate. If the table is suddenly jerked sideways with large acceleration, what happens to the book (qualitatively) and which property of matter is responsible?
Pencil sketch (labelled)
Solution — step by step with formulas
- Draw free-body diagram: weight mg downward; normal force N upward from table (see pencil sketch).
- Horizontal forces on a stationary book on a fixed table: none (or friction = 0 if no horizontal push).
- Vertical equilibrium: N = mg ⇒ ΣF = 0 ⇒ acceleration a = 0 (first law).
- When table jerks sideways, book tends to remain at rest relative to ground due to inertia until friction accelerates it.
Final answer: Book in equilibrium under N and mg (ΣF = 0). On sudden jerk, book lags due to inertia (mass).
Formulas used in this problem
Textbook formal language
Newton’s first law asserts that a body continues in its state of rest or of uniform motion in a straight line unless compelled by a net external force to change that state. The property of matter by virtue of which it resists change in its state of rest or uniform motion is inertia; mass is the quantitative measure of inertia. For the book on a stationary table, the vector sum of gravitational force and normal reaction vanishes; hence acceleration is zero. When the table is accelerated suddenly, the external horizontal force on the book is only friction, which is finite; for a sufficiently large acceleration of the table, the book does not instantly acquire the table’s velocity—manifestation of inertia.
Formulae applied: If ΣF = 0 ⇒ v = constant (including zero).
Easy language (same idea, plain words)
The book sits still because two forces cancel: gravity pulls down, the table pushes up the same amount. No leftover force means no speeding up or slowing down. If someone yanks the table sideways, the book “wants” to stay where it was for a moment—that laziness of matter is inertia. Friction may drag it along later, but it doesn’t jump instantly with the table.
Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.
Topic in depth — Inertia and Newton’s first law
Inertia appears in everyday life: passengers lean back when a bus starts and lurch forward when it stops. The first law defines an inertial frame as one in which free particles move with constant velocity. Always begin motion problems by asking whether net force is zero; if yes, velocity is constant even if non-zero (uniform motion).
Link to chapter notes (L3 — Inertia and Newton’s first law): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: If ΣF = 0 ⇒ v = constant (including zero). In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.
Exam tip
Quote the law in one line, then write If ΣF = 0 ⇒ v = constant (including zero) before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.
Common mistakes
- Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
- Using the wrong sign convention for work/heat/force direction.
- Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
- Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
Draw a sketch showing a ball hitting a wall and rebounding (label pᵢ and p_f). A ball of mass 0.20 kg moving east at 10 m·s⁻¹ hits a wall and rebounds west at 8.0 m·s⁻¹. Taking east as positive, calculate (i) initial momentum, (ii) final momentum, (iii) change in momentum. What does the large change in momentum imply about the wall’s force?
Pencil sketch (labelled)
Solution — step by step with formulas
- m = 0.20 kg; v_i = +10 m·s⁻¹; v_f = −8.0 m·s⁻¹.
- p_i = mv_i = 0.20 × 10 = +2.0 kg·m·s⁻¹.
- p_f = mv_f = 0.20 × (−8) = −1.6 kg·m·s⁻¹.
- Δp = p_f − p_i = −1.6 − 2.0 = −3.6 kg·m·s⁻¹.
- By Newton II, F_avg = Δp/Δt; large |Δp| in short contact time ⇒ large average force by the wall on the ball.
Final answer: p_i = +2.0 kg·m·s⁻¹; p_f = −1.6 kg·m·s⁻¹; Δp = −3.6 kg·m·s⁻¹ (large wall force).
Formulas used in this problem
Textbook formal language
Linear momentum of a particle is the product of its mass and velocity, a vector quantity collinear with velocity. The change in momentum is the vector difference of final and initial momenta. Newton’s second law in momentum form states that the net external force equals the time rate of change of momentum; hence a large momentum change in a small interval implies a large average force.
Formulae applied: p = mv; Δp = p_f − p_i.
Easy language (same idea, plain words)
Momentum is “mass times how fast and which way.” East was +, so start with +2.0. After bounce it goes west, so momentum is negative (−1.6). The change is final minus initial: −1.6 − 2.0 = −3.6. That big flip in a short hit means the wall pushed hard on the ball.
Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.
Topic in depth — Linear momentum
Momentum is conserved only for isolated systems; a single ball hitting a wall is not isolated—the Earth–wall system supplies external force. Always fix a positive direction before assigning signs to velocity and momentum.
Link to chapter notes (L3 — Linear momentum): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: p = mv; Δp = p_f − p_i. In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.
Exam tip
Quote the law in one line, then write p = mv; Δp = p_f − p_i before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.
Common mistakes
- Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
- Using the wrong sign convention for work/heat/force direction.
- Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
- Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
A constant net force of 12 N acts on a 3.0 kg body initially at rest. Find (i) acceleration, (ii) velocity after 4.0 s, (iii) momentum after 4.0 s. Verify that F = Δp/Δt gives the same force.
Solution — step by step with formulas
- a = F/m = 12/3.0 = 4.0 m·s⁻².
- From rest: v = u + at = 0 + 4.0×4.0 = 16 m·s⁻¹.
- p = mv = 3.0×16 = 48 kg·m·s⁻¹; p_i = 0 ⇒ Δp = 48.
- Δp/Δt = 48/4.0 = 12 N = F (consistent).
Final answer: a = 4.0 m·s⁻²; v = 16 m·s⁻¹; p = 48 kg·m·s⁻¹; F = Δp/Δt checks.
Formulas used in this problem
Textbook formal language
For constant mass, Newton’s second law reduces to F = ma, where F is the net external force and a the acceleration of the centre of mass. Equivalently F = dp/dt. With constant F and u = 0, v = at and p = mat, so Δp/Δt = ma = F.
Working formula set for this problem: F = ma; F = Δp/Δt. In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.
Easy language (same idea, plain words)
Force 12 N on 3 kg means it speeds up by 4 m/s every second. After 4 s it moves at 16 m/s, momentum 48. Force is also “how fast momentum changes”: 48 in 4 s is 12 N—same number, two views of one law.
Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.
Topic in depth — Newton’s second law
Use F = ma only for constant mass. For rockets (mass varying) retain F_ext + v_rel(dm/dt) forms. Net force means vector sum after free-body diagram; forgotten friction or components cause most exam errors.
Link to chapter notes (L3 — Newton’s second law): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: F = ma; F = Δp/Δt. In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.
Exam tip
Quote the law in one line, then write F = ma; F = Δp/Δt before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.
Common mistakes
- Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
- Using the wrong sign convention for work/heat/force direction.
- Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
- Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
A 0.15 kg cricket ball arrives at 20 m·s⁻¹ and is stopped by a fielder in 0.040 s. Find the magnitude of average force exerted by the fielder. Why does drawing the hands backward while catching reduce injury risk?
Solution — step by step with formulas
- Δp = m(v_f − v_i) = 0.15(0 − 20) = −3.0 kg·m·s⁻¹; |Δp| = 3.0.
- F_avg = |Δp|/Δt = 3.0/0.040 = 75 N.
- If hands move back, Δt increases; same |Δp| ⇒ smaller |F_avg|.
Final answer: F_avg = 75 N; larger Δt (drawing hands back) reduces average force.
Formulas used in this problem
Textbook formal language
Impulse delivered by a force in time Δt is J = ∫F dt = F_avg Δt and equals the change in momentum of the body. For fixed Δp, increasing interaction time decreases the average force. Fielders increase Δt by drawing the hands backward, thereby reducing peak force on hands and ball.
Working formula set for this problem: J = F_avg Δt = Δp. In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.
Easy language (same idea, plain words)
Stopping the ball means removing all its forward momentum. That change spread over 0.04 s needs about 75 N on average. If you pull hands back, you take longer to stop the ball, so the push on your hands is gentler—same “momentum cancel,” longer time.
Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.
Topic in depth — Impulse–momentum theorem
Airbags, crumple zones, and soft landings all trade longer collision time for smaller peak force. Graphically, impulse is the area under the F–t curve.
Link to chapter notes (L3 — Impulse–momentum theorem): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: J = F_avg Δt = Δp. In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.
Exam tip
Impulse–momentum is the integrated form of Newton’s second law. Same Δp with larger Δt means smaller average force — classic catching / airbag idea in NIOS notes.
Common mistakes
- Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
- Using the wrong sign convention for work/heat/force direction.
- Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
- Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
Draw a labelled diagram of gun recoil (bullet forward, gun backward). A gun of mass 4.0 kg fires a 20 g bullet at 300 m·s⁻¹. If the system is initially at rest and recoils freely, find the recoil speed of the gun. State the principle used and one condition for its validity.
Pencil sketch (labelled)
Solution — step by step with formulas
- m_b = 0.020 kg; m_g = 4.0 kg; v_b = +300 m·s⁻¹ (forward).
- Initial total p = 0.
- m_g v_g + m_b v_b = 0 ⇒ v_g = −(m_b/m_g)v_b = −(0.020/4.0)×300 = −1.5 m·s⁻¹.
- Recoil speed = 1.5 m·s⁻¹ opposite to bullet.
Final answer: Recoil speed 1.5 m·s⁻¹ backward; conservation of momentum (isolated system).
Formulas used in this problem
Textbook formal language
In the absence of external force in a given direction, the component of total linear momentum of an isolated system along that direction remains constant. For gun + bullet initially at rest, total momentum is zero; after firing, momenta are equal in magnitude and opposite in direction: m_g v_g = −m_b v_b.
Working formula set for this problem: m_g v_g + m_b v_b = 0 (initially at rest); v_g = −(m_b/m_g) v_b. In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.
Easy language (same idea, plain words)
Before the shot, nothing is moving, so total momentum is zero. The bullet goes one way with small mass but high speed; the gun must go the other way so the two momenta cancel. Light bullet × big speed = heavy gun × small speed → about 1.5 m/s kick.
Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.
Topic in depth — Conservation of linear momentum
External forces (shoulder, ground friction) can reduce observed recoil; the ideal free-gun model assumes no external horizontal force. Rockets use the same principle with continuous exhaust of mass.
Link to chapter notes (L3 — Conservation of linear momentum): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: m_g v_g + m_b v_b = 0 (initially at rest); v_g = −(m_b/m_g) v_b. In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.
Exam tip
Quote the law in one line, then write m_g v_g + m_b v_b = 0 (initially at rest); v_g = −(m_b/m_g) v_b before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.
Common mistakes
- Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
- Using the wrong sign convention for work/heat/force direction.
- Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
- Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
Draw the free-body diagram of a block on a rough horizontal surface (label F, f, N, mg). A 5.0 kg block rests on a horizontal surface with μ_s = 0.40 and μ_k = 0.30. Take g = 10 m·s⁻². (i) Find maximum static friction. (ii) What happens if a horizontal force of 15 N is applied? (iii) If 25 N is applied and the block moves, find kinetic friction and acceleration.
Pencil sketch (labelled)
Solution — step by step with formulas
- N = mg = 5.0×10 = 50 N.
- f_s(max) = μ_s N = 0.40×50 = 20 N.
- 15 N < 20 N ⇒ block does not start; f_s = 15 N (balances applied force).
- 25 N > 20 N ⇒ motion begins; f_k = μ_k N = 0.30×50 = 15 N.
- a = (F − f_k)/m = (25 − 15)/5.0 = 2.0 m·s⁻².
Final answer: (i) 20 N (ii) remains at rest, f_s=15 N (iii) f_k=15 N, a=2.0 m·s⁻²
Formulas used in this problem
Textbook formal language
Limiting static friction is μ_s N; static friction adjusts up to this limit to prevent relative motion. Once sliding occurs, kinetic friction is approximately μ_k N, usually with μ_k < μ_s. Net force along the surface equals ma.
Working formula set for this problem: f_s(max) = μ_s N; f_k = μ_k N; N = mg (horizontal). In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.
Easy language (same idea, plain words)
Max “grip” before sliding is 20 N. A 15 N push is weaker than grip, so the block stays put and friction equals 15 N. A 25 N push breaks the grip; sliding friction drops to 15 N, leaving 10 N net force, so it accelerates at 2 m/s².
Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.
Topic in depth — Static and kinetic friction
μ depends on surface pair, not area (ideal Amontons–Coulomb model). On an incline, N = mg cos θ and the component mg sin θ competes with friction—standard banking and ladder problems use the same laws.
Link to chapter notes (L3 — Static and kinetic friction): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: f_s(max) = μ_s N; f_k = μ_k N; N = mg (horizontal). In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.
Exam tip
Quote the law in one line, then write f_s(max) = μ_s N; f_k = μ_k N; N = mg (horizontal) before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.
Common mistakes
- Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
- Using the wrong sign convention for work/heat/force direction.
- Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
- Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).