Chapter 4: Laws of Motion
Unit III – Laws of Motion
4.1 Aristotle’s FallacyVs. Newton’s Insight
Aristotle believed a force is needed to maintain motion. Galileo showed that on frictionless surfaces, a body continues moving without force. Newton formalized this insight into his three laws.
4.2 Newton’s First Law (Law of Inertia)
A body at rest or in uniform motion (constant velocity) continues in that state unless acted upon by a net external force.
- This defines inertia — the tendency of a body to resist change in its state.
- Inertia is measured by mass (greater mass = greater inertia).
- A frame in which Newton’s 1st law holds is called an inertial frame.
4.3 Newton’s Second Law
The rate of change of momentum of a body is proportional to the net external force and occurs in the direction of the force.
\[ \vec{F} = \frac{d\vec{p}}{dt} \quad \text{where} \quad \vec{p} = m\vec{v} \]
For constant mass:
\[ \boxed{\vec{F} = m\vec{a}} \]
- SI unit of force: Newton (N) = kg m/s²
- 1 N = force that gives 1 kg an acceleration of 1 m/s²
Impulse
\[ \vec{J} = \vec{F}\Delta t = \Delta\vec{p} = m\vec{v} - m\vec{u} \]
- Unit: N·s = kg·m/s
- Impulse = change in momentum
4.4 Newton’s Third Law
For every action, there is an equal and opposite reaction.
\[ \vec{F_{AB}} = -\vec{F_{BA}} \]
- Forces always occur in pairs.
- Action and reaction act on different bodies.
4.5 Conservation of Linear Momentum
If no external force acts on a system:
\[ \vec{p}_{total} = \text{constant} \]
\[ m_1\vec{v_1} + m_2\vec{v_2} = m_1\vec{u_1} + m_2\vec{u_2} \]
Application — Recoil of gun:
\[ 0 = m_b v_b + m_g v_g \Rightarrow v_g = -\frac{m_b v_b}{m_g} \]
4.6 Equilibrium of Concurrent Forces
A body is in equilibrium when the vector sum of all forces is zero:
\[ \sum \vec{F} = 0 \]
For three forces in equilibrium (Lami’s theorem):
\[ \frac{F_1}{\sin\alpha} = \frac{F_2}{\sin\beta} = \frac{F_3}{\sin\gamma} \]
4.7 Friction
Friction is the force that opposes relative motion between surfaces.
Types of Friction
| Type | Formula | Notes |
|---|---|---|
| Static friction | \(f_s \leq \mu_s N\) | Prevents motion; \(\mu_s\) = coefficient of static friction |
| Kinetic (sliding) friction | \(f_k = \mu_k N\) | During motion; \(\mu_k < \mu_s\) |
| Rolling friction | \(f_r = \mu_r N\) | Least; \(\mu_r \ll \mu_k\) |
\[ N = mg\cos\theta \quad \text{(on inclined plane)} \]
Angle of friction (\(\lambda\)): \(\tan\lambda = \mu_s\)
Angle of repose (\(\theta_r\)): maximum angle of incline before sliding: \(\tan\theta_r = \mu_s\)
4.8 Free Body Diagrams
Simple Block on Surface
4.9 Dynamics of Circular Motion
For a particle moving in a circle, the centripetal force provides the inward acceleration:
\[ F_c = \frac{mv^2}{r} \]
This is not a new type of force but the net resultant of existing forces directed toward the centre.
Banking of Roads
For a vehicle on a banked road (angle \(\theta\)):
\[ \tan\theta = \frac{v^2}{rg} \]
With friction:
\[ v_{max} = \sqrt{rg,\frac{\mu_s + \tan\theta}{1 - \mu_s\tan\theta}} \]
Key Formulas Summary
| Formula | Quantity |
|---|---|
| \(\vec{F} = m\vec{a}\) | Newton’s 2nd Law |
| \(J = F\Delta t = \Delta p\) | Impulse |
| \(f_k = \mu_k N\) | Kinetic friction |
| \(f_{s,max} = \mu_s N\) | Maximum static friction |
| \(\tan\theta = \mu_s\) | Angle of repose |
| \(\tan\theta = \dfrac{v^2}{rg}\) | Banking angle |
Practice Questions
Section A – MCQ (1 mark each)
Q1. A body of mass 2 kg is acted upon by a force that changes its velocity from 3 m/s to 7 m/s in 2 s. The force applied is:
(a) 2 N (b) 4 N (c) 6 N (d) 8 N
Answer
(b) 4 N
\(F = m \times a = 2 \times \dfrac{7-3}{2} = 2 \times 2 = 4 \text{ N}\)
Q2. A gun of mass 5 kg fires a bullet of mass 50 g with a velocity of 400 m/s. The recoil velocity of the gun is:
(a) 2 m/s (b) 4 m/s (c) 8 m/s (d) 10 m/s
Answer
(b) 4 m/s
By conservation of momentum: \(0 = 0.05 \times 400 + 5 \times v_g\)
\(v_g = -\dfrac{0.05 \times 400}{5} = -4 \text{ m/s}\)
Q3. The coefficient of static friction between a body (mass 10 kg) and a floor is 0.4. The force needed to just move it is (\(g = 10 \text{ m/s}^2\)):
(a) 20 N (b) 40 N (c) 60 N (d) 80 N
Answer
(b) 40 N — \(f_s = \mu_s mg = 0.4 \times 10 \times 10 = 40 \text{ N}\)
Section B – Short Answer (2–3 marks)
Q4. State and prove the law of conservation of linear momentum from Newton’s third law.
Answer
Consider two bodies A and B exerting forces on each other. By Newton’s 3rd law:
\[ \vec{F_{AB}} = -\vec{F_{BA}} \]
By Newton’s 2nd law:
\[ \frac{d\vec{p}_A}{dt} = -\frac{d\vec{p}_B}{dt} \]
\[ \frac{d(\vec{p}_A + \vec{p}_B)}{dt} = 0 \Rightarrow \vec{p}_A + \vec{p}_B = \text{constant} \]
Total momentum is conserved in the absence of external force.
Q5. A block of mass 5 kg rests on a rough horizontal surface. A horizontal force of 30 N just moves it. Find \(\mu_s\). If the same force is maintained, find the acceleration if \(\mu_k = 0.4\). (\(g = 10 \text{ m/s}^2\))
Answer
\[ \mu_s = \frac{F}{mg} = \frac{30}{5 \times 10} = 0.6 \]
\[ f_k = \mu_k mg = 0.4 \times 5 \times 10 = 20 \text{ N} \]
\[ a = \frac{F - f_k}{m} = \frac{30 - 20}{5} = \mathbf{2 \text{ m/s}^2} \]
Section D – Competency-Based Questions
Q6. (Case Study) A car of mass 1000 kg moves around a circular road of radius 50 m at a speed of 20 m/s. The road is banked.
(i) Calculate the banking angle required for safe turning without friction.
(ii) Calculate the centripetal force experienced by the car.
(iii) If friction is also present with \(\mu = 0.3\), what is the maximum speed for safe turning?
(\(g = 10 \text{ m/s}^2\))
Answer
(i) \(\tan\theta = \dfrac{v^2}{rg} = \dfrac{400}{500} = 0.8 \Rightarrow \theta = \tan^{-1}(0.8) \approx \mathbf{38.7°}\)
(ii) \(F_c = \dfrac{mv^2}{r} = \dfrac{1000 \times 400}{50} = \mathbf{8000 \text{ N}}\)
(iii)
\[ v_{max} = \sqrt{rg,\frac{\mu + \tan\theta}{1 - \mu\tan\theta}} = \sqrt{50 \times 10 \times \frac{0.3 + 0.8}{1 - 0.3 \times 0.8}} \]
\[ = \sqrt{500 \times \frac{1.1}{0.76}} = \sqrt{500 \times 1.447} = \sqrt{723.7} \approx \mathbf{26.9 \text{ m/s}} \]