How to Use This Page
1. Solve every exercise question independently first — then compare your method with the solutions below. 2. Focus on the reasoning behind each answer, especially in free-body diagrams and force resolution. 3. Note the "Common Mistakes" section — these are the most frequent exam errors for this chapter. 4. Use the Exam Checklist to confirm you have covered every key formula and concept before the exam. 5. Numerical examples in code blocks include SI units at every step — always do the same in your exam.
Key Concepts (Quick Recap)
Newton's Three Laws of Motion
| Law | Statement | Key Formula |
|---|
| First Law | A body continues in its state of rest or uniform motion unless acted upon by an external force. | Concept of inertia |
| Second Law | The rate of change of momentum is proportional to the applied force and in its direction. | F = ma (for constant mass) |
| Third Law | Every action has an equal and opposite reaction. | F_AB = −F_BA |
Momentum and Impulse
| Quantity | Formula | SI Unit |
|---|
| Linear momentum | p = mv | kg·m/s |
| Impulse | J = FΔt = Δp = m(v − u) | N·s |
| Force (from momentum) | F = dp/dt | N |
Friction
| Type | Formula | Notes |
|---|
| Static friction | fₛ ≤ μₛN | Self-adjusting, up to maximum μₛN |
| Kinetic friction | fₖ = μₖN | Constant, μₖ < μₛ |
| Rolling friction | fᵣ = μᵣN | μᵣ < μₖ < μₛ |
Circular Motion Dynamics
| Quantity | Formula |
|---|
| Centripetal force | F_c = mv²/r = mω²r |
| Centripetal acceleration | a_c = v²/r = ω²r |
| Angle of banking | tan θ = v²/(rg) |
Important Definitions
| Term | Definition |
|---|
| Inertia | Tendency of a body to resist any change in its state of rest or uniform motion |
| Impulse | Product of force and time interval; equals change in momentum |
| Force | Push or pull that changes or tends to change the state of motion of a body |
| Weight | W = mg (force of gravity on a body) |
| Normal reaction | Force exerted by a surface on a body, perpendicular to the surface |
I. Choose the Best Answer (MCQs)
1. A 5 kg object experiences a net force of 20 N. Its acceleration is:
Answer
- (a) 2 m/s²
- (b) 4 m/s²
- (c) 100 m/s²
- (d) 0.25 m/s²
Answer(b) 4 m/s²
ExplanationF = ma → a = F/m = 20/5 = 4 m/s².
2. When a bus suddenly starts, a passenger falls backward. This is explained by:
Answer
- (a) Newton's second law
- (b) Newton's third law
- (c) Newton's first law (inertia)
- (d) Law of conservation of momentum
Answer(c) Newton's first law (inertia)
ExplanationThe passenger's body tends to remain at rest (inertia of rest) while the bus moves forward.
3. The coefficient of static friction is always ________ the coefficient of kinetic friction.
Answer
- (a) Less than
- (b) Equal to
- (c) Greater than
- (d) Unrelated to
Answer(c) Greater than
ExplanationIt always requires more force to start motion than to maintain it, so μₛ > μₖ.
4. A body of mass 2 kg moving at 5 m/s is brought to rest in 0.5 s. The force acting on it is:
Answer
- (a) 10 N
- (b) 20 N
- (c) 5 N
- (d) 2.5 N
Answer(b) 20 N
ExplanationF = m(v − u)/t = 2(0 − 5)/0.5 = −20 N. Magnitude = 20 N (opposing motion).
5. In circular motion, the centripetal force acts:
Answer
- (a) Along the tangent
- (b) Along the radius toward the centre
- (c) Away from the centre
- (d) In the direction of motion
Answer(b) Along the radius toward the centre
ExplanationCentripetal force is always directed radially inward, providing the centripetal acceleration needed for circular motion.
6. Action and reaction forces act on:
Answer
- (a) The same body
- (b) Two different bodies
- (c) Either the same or different bodies
- (d) The heavier body only
Answer(b) Two different bodies
ExplanationNewton's third law states that action and reaction always act on two different bodies and hence never cancel each other.
7. A rocket works on the principle of conservation of:
Answer
- (a) Energy
- (b) Mass
- (c) Linear momentum
- (d) Angular momentum
Answer(c) Linear momentum
ExplanationThe rocket pushes exhaust gases backward; the gases push the rocket forward. Total momentum (rocket + exhaust) remains conserved.
II. Fill in the Blanks
1. The SI unit of force is ________. Newton (N) 2. According to Newton's second law, F = ________. ma (or dp/dt) 3. The maximum value of static friction is called ________ friction. Limiting 4. The tendency of a body to resist change in its state of motion is called ________. Inertia 5. In a free-body diagram, ________ forces are replaced by arrows. All external 6. The centripetal force required for a body of mass m moving at speed v in a circle of radius r is ________. mv²/r 7. Impulse equals change in ________. Momentum 8. The normal reaction on a body of mass m placed on a horizontal surface is ________. mg 9. On an inclined plane at angle θ, the component of weight along the plane is ________. mg sin θ 10. The banking angle θ for a road of radius r at speed v satisfies tan θ = ________. v²/(rg) 11. When two bodies are in contact, the force between them perpendicular to the surface is called ________ reaction. Normal 12. Momentum is a ________ quantity (scalar/vector). Vector
III. True or False (with corrections)
1. A body can have acceleration without any external force acting on it.
- False. By Newton's second law, F = ma. If F = 0, then a = 0. A body cannot accelerate without an external force.
2. Static friction is a self-adjusting force.
- True. Static friction adjusts its magnitude up to μₛN to exactly balance the applied force and prevent motion.
3. Action and reaction forces cancel each other because they are equal and opposite.
- False. Action and reaction act on two different bodies, so they cannot cancel each other. They produce effects on different bodies.
4. A heavier body has more inertia than a lighter body.
- True. Inertia is measured by mass. Greater mass means greater resistance to change in motion.
5. Centripetal force does work on a body in uniform circular motion.
- False. Centripetal force is always perpendicular to the displacement (which is tangential), so W = F·d·cos90° = 0. No work is done.
IV. Match the Following
| Column A | Column B |
|---|
| 1. Newton's first law | (d) Law of inertia |
| 2. Newton's second law | (f) F = ma |
| 3. Newton's third law | (a) Action-reaction on different bodies |
| 4. Static friction | (e) Self-adjusting up to μₛN |
| 5. Kinetic friction | (b) Constant value μₖN |
| 6. Centripetal force | (c) mv²/r, directed toward centre |
V. Answer Briefly (2-mark Questions)
1. State Newton's first law of motion and give one example.
Answer
Newton's first law states: A body continues in its state of rest or uniform motion in a straight line unless acted upon by an external force. Example: A book on a table remains at rest until a force pushes it. When a moving car stops suddenly, passengers lurch forward because their bodies tend to continue moving.
2. What is a free-body diagram? Why is it useful?
Answer
A free-body diagram (FBD) is a sketch showing a body isolated from its surroundings with all external forces acting on it represented by arrows. It is useful because it helps identify all forces, resolve them into components, and correctly apply Newton's second law to each axis. Errors in FBDs are the single biggest cause of wrong answers in mechanics.
3. Distinguish between static friction and kinetic friction.
Answer
Static friction acts on a body at rest, adjusting up to a maximum value μₛN. Kinetic friction acts on a body in relative motion and has a fixed value μₖN. The coefficient of static friction μₛ is greater than μₖ, which is why it is harder to start an object sliding than to keep it sliding.
4. Why is a road banked at curves?
Answer
A road is banked so that a component of the normal reaction provides the necessary centripetal force for a vehicle negotiating the curve, reducing reliance on friction. At the ideal banking angle θ, tan θ = v²/(rg), and a vehicle can negotiate the curve at design speed even with zero friction.
VI. Answer in Detail (5-mark Questions with Numerical Examples)
1. State and prove the law of conservation of linear momentum. Solve: A 10 g bullet is fired from a 5 kg gun with a speed of 400 m/s. Find the recoil velocity of the gun.
Answer
Statement: The total linear momentum of an isolated system remains constant if no external force acts on it.
Proof: Consider two bodies A and B with momenta p_A = m_A v_A and p_B = m_B v_B. They collide and exert forces on each other. By Newton's third law, F_AB = −F_BA. By Newton's second law, F_AB = dp_A/dt and F_BA = dp_B/dt. So dp_A/dt + dp_B/dt = 0, hence d(p_A + p_B)/dt = 0, meaning p_A + p_B = constant.
Numerical:
Given: m_bullet = 10 g = 0.01 kg, v_bullet = 400 m/s
m_gun = 5 kg
Before firing, total momentum = 0 (both at rest).
By conservation of momentum:
0 = m_bullet × v_bullet + m_gun × v_gun
0 = 0.01 × 400 + 5 × v_gun
0 = 4 + 5 × v_gun
v_gun = −4/5 = −0.8 m/s
The negative sign indicates the gun recoils opposite to
the bullet's direction.
Answer: Recoil velocity of the gun = 0.8 m/s (opposite to bullet)
2. Explain the types of friction with examples. Solve: A block of mass 5 kg rests on a rough horizontal surface. The coefficient of static friction is 0.4 and kinetic friction is 0.3. Find (a) the minimum force to start the block moving, and (b) the acceleration once it starts moving.
Answer
Types of friction:
- Static friction (fₛ): Acts when a body tends to slide but is still at rest. It is self-adjusting up to a maximum value fₛ(max) = μₛN. Example: A book on a tilted table that does not slide.
- Kinetic friction (fₖ): Acts when a body slides over a surface. fₖ = μₖN, always less than limiting friction. Example: A sliding block on a floor.
- Rolling friction (fᵣ): Acts when a body rolls over a surface. fᵣ = μᵣN, and μᵣ < μₖ < μₛ. Example: A ball rolling on a floor.
Numerical:
Given: m = 5 kg, μₛ = 0.4, μₖ = 0.3, g = 10 m/s²
(a) Minimum force to start moving (must overcome static friction):
N = mg = 5 × 10 = 50 N
fₛ(max) = μₛ × N = 0.4 × 50 = 20 N
Minimum applied force = 20 N
(b) Acceleration once moving (kinetic friction opposes motion):
Kinetic friction: fₖ = μₖ × N = 0.3 × 50 = 15 N
Net force once it moves: F_net = Applied force − fₖ
= 20 − 15 = 5 N
a = F_net / m = 5 / 5 = 1 m/s²
Answer: (a) Minimum force to start = 20 N
(b) Acceleration = 1 m/s²
3. Derive the expression for centripetal force and explain banking of roads. Solve: A car of mass 1000 kg moves on a circular track of radius 50 m at 20 m/s. Find the centripetal force and the ideal banking angle (g = 10 m/s²).
Answer
Derivation of centripetal force: In uniform circular motion, the velocity direction changes continuously. The acceleration toward the centre is a_c = v²/r. By Newton's second law, the force required to produce this acceleration is F_c = ma_c = mv²/r. This centripetal force can be provided by tension, friction, normal reaction component, or gravity component.
Banking of roads: On a banked road at angle θ, the normal reaction N has a horizontal component N sin θ that provides centripetal force, and a vertical component N cos θ that balances weight. At ideal speed (no friction needed): N sin θ = mv²/r and N cos θ = mg. Dividing: tan θ = v²/(rg).
Numerical:
Given: m = 1000 kg, r = 50 m, v = 20 m/s, g = 10 m/s²
Centripetal force:
F_c = mv²/r = 1000 × (20)² / 50
F_c = 1000 × 400 / 50 = 400000/50 = 8000 N
Banking angle:
tan θ = v²/(rg) = (20)² / (50 × 10)
tan θ = 400 / 500 = 0.8
θ = tan⁻¹(0.8) ≈ 38.66°
Answer: Centripetal force = 8000 N
Banking angle θ ≈ 38.66°
Common Mistakes to Avoid
1. Confusing mass and weight — mass (kg) is constant; weight W = mg (N) varies with g. Use F = ma with m in kg, not weight.
2. Forgetting to draw a free-body diagram — always sketch the FBD with all forces before writing equations. Missing one force leads to wrong answers.
3. Not resolving forces on inclined planes — always resolve weight into mg sin θ (along the plane) and mg cos θ (perpendicular to the plane). Never use the full weight directly in the motion equation.
4. Assuming friction always opposes gravity — friction opposes relative motion or tendency of motion. On an inclined plane, if a block is about to slide down, friction acts up the plane.
5. Mixing up static and kinetic friction — static friction is self-adjusting (f ≤ μₛN); only use μₛ when the body is at rest. Once the body moves, switch to kinetic friction μₖN.
6. Applying Newton's third law incorrectly — action and reaction act on different bodies. They never cancel each other on the same body.
7. Sign errors in connected-body problems — always define a positive direction and stick to it for all bodies in the system.
8. Forgetting centripetal force is not an additional force — it is the name for the net force toward the centre provided by existing forces (tension, friction, normal, gravity). Do not add it separately.
Exam Checklist
- [ ] Can state Newton's three laws with examples
- [ ] Can prove conservation of linear momentum using Newton's third law
- [ ] Know F = ma and can solve problems with multiple forces
- [ ] Can draw correct free-body diagrams for any situation
- [ ] Understand static vs kinetic friction and when to use which
- [ ] Can solve inclined plane problems (resolve weight into components)
- [ ] Know centripetal force formula F_c = mv²/r and its derivation
- [ ] Understand banking of roads (tan θ = v²/rg)
- [ ] Can solve impulse-momentum problems (J = Δp)
- [ ] Remember: action-reaction act on different bodies, never cancel
