Core Theory I – Classical Mechanics
B.Sc. Physics – Core Science
Scope:
Kinematics, Newton’s Laws, Work–Energy Theorem, Conservation of Energy & Momentum – the foundational framework for all classical mechanics problems in the B.Sc. Physics curriculum.
1. Simple Explanation
Classical mechanics is the branch of physics that describes how objects move under the action of forces.
It is built on three pillars:
| Pillar | What it gives you |
|---|---|
| Kinematics | Describes motion (position, velocity, acceleration) without reference to why it happens. |
| Dynamics | Uses Newton’s Laws to link forces to the change in motion. |
| Energetics | Relates work done by forces to energy (kinetic, potential). The Work‑Energy Theorem and conservation laws provide powerful tools to solve problems. |
2. Key Points
| Concept | Formula / Relation | Notes |
|---|---|---|
| Displacement | (\vec{s} = \vec{r}_f - \vec{r}_i) | Vector quantity |
| Velocity | (\vec{v} = \frac{d\vec{r}}{dt}) | Instantaneous or average |
| Acceleration | (\vec{a} = \frac{d\vec{v}}{dt}) | Constant (a) → simple equations |
| Newton’s First Law | Inertia: (\sum \vec{F} = 0 \Rightarrow \vec{a}=0) | Reference to inertial frames |
| Newton’s Second Law | (\sum \vec{F} = m\vec{a}) | Vector equation |
| Newton’s Third Law | (\vec{F}{12} = -\vec{F}{21}) | Action–reaction pairs |
| Work | (W = \int \vec{F}\cdot d\vec{s}) | Scalar, unit: J |
| Kinetic Energy | (K = \frac12 m v^2) | Energy of motion |
| Potential Energy (gravitational) | (U = m g h) | Reference level arbitrary |
| Work‑Energy Theorem | (W_{\text{total}} = \Delta K) | Total (net) work = change in kinetic energy |
| Conservation of Energy | (E_{\text{total}} = K + U = \text{const}) | No non‑conservative forces |
| Conservation of Linear Momentum | (\sum \vec{p} = \text{const}) | (\vec{p}=m\vec{v}) |
| Impulse | (\vec{J} = \int \vec{F},dt = \Delta \vec{p}) | Useful for collision problems |
3. Worked Example
Problem:
A (5;\text{kg}) block starts from rest at the top of a (30^{\circ}) frictional incline that is (4;\text{m}) high.
Coefficient of kinetic friction ( \mu_k = 0.20 ).
Find the block’s speed at the bottom.
Step‑by‑Step Solution
-
Geometry & Height
The vertical height (h=4;\text{m}).
-
Gravitational Potential Energy at the top
[ U_i = m g h = (5)(9.81)(4) = 196.2;\text{J} ]
-
Kinetic Energy at the top
(K_i = 0) (starts from rest)
-
Work done by friction
(F_{\text{fric}} = \mu_k N)
Normal force (N = m g \cos\theta)
[ N = (5)(9.81)\cos30^{\circ} = 5 \times 9.81 \times 0.866 = 42.5;\text{N} ]
[ F_{\text{fric}} = 0.20 \times 42.5 = 8.5;\text{N} ]
Distance along the incline (s):
[ s = \frac{h}{\sin\theta} = \frac{4}{\sin30^{\circ}} = \frac{4}{0.5} = 8;\text{m} ]
Work by friction (negative because it opposes motion):
[ W_{\text{fric}} = -F_{\text{fric}}, s = -(8.5)(8) = -68;\text