Prerequisites
Before studying this topic, make sure you understand:
- Newton’s First Law - Understanding inertia and frames of reference
- Vectors - Force and momentum are vector quantities
- Motion in a Straight Line - Acceleration concepts
The Hook: Why Can’t Humans Punch Like Thor?
In the Thor movies, when Thor swings Mjolnir and strikes something, the impact is devastating. But when a regular human punches with all their might, the effect is much smaller. Why?
Both apply force. But Thor’s hammer has enormous mass. And according to Newton’s Second Law, for the same force, the acceleration depends on mass. But there’s more — the momentum transfer is what creates the destructive impact!
The Big Questions:
- How do force, mass, and acceleration relate?
- What exactly is momentum?
- Why do heavier objects hit harder?
The Core Concept: Force Changes Momentum
Newton’s Second Law - Original Form
$$\boxed{\vec{F} = \frac{d\vec{p}}{dt}}$$In words: The rate of change of momentum equals the applied force.
Where:
- $\vec{F}$ = Net external force
- $\vec{p}$ = Linear momentum = $m\vec{v}$
- $t$ = time
In simple terms: Force is what changes how much “motion” (momentum) an object has!
The Famous F = ma Form
For constant mass, the second law becomes:
$$\vec{F} = \frac{d(m\vec{v})}{dt} = m\frac{d\vec{v}}{dt} = m\vec{a}$$ $$\boxed{\vec{F} = m\vec{a}}$$This is the most commonly used form in JEE problems.
Interactive Demo: Visualize Newton’s Second Law
See how force, mass, and acceleration relate in real-time.
F = dp/dt is the general form (works even when mass changes, like rockets)
F = ma is the special case (only when mass is constant)
For JEE: Most problems use constant mass, so F = ma works. But for rocket/variable mass problems, use F = dp/dt!
Understanding Momentum
Definition
$$\boxed{\vec{p} = m\vec{v}}$$Momentum is the “quantity of motion” possessed by a body.
Units: kg⋅m/s or N⋅s
Properties of Momentum
- Vector quantity — has magnitude and direction
- Direction same as velocity
- Depends on both mass and velocity
- More momentum = harder to stop
Force Analysis: The Free Body Diagram (FBD)
Most important skill in mechanics: Drawing correct FBD!
Steps to Draw FBD
- Isolate the body — consider only ONE object at a time
- Identify all forces acting on that body:
- Weight ($mg$)
- Normal reaction ($N$)
- Tension ($T$)
- Friction ($f$)
- Applied force ($F$)
- Show force directions with arrows
- Choose coordinate system (usually: x-horizontal, y-vertical)
- Resolve forces into components if needed
Mistake 1: Including forces the body exerts on others
- Wrong: Including reaction force that the body exerts on the ground
- Correct: Only forces that other objects exert ON this body
Mistake 2: Forgetting to show all forces
- Always check: Weight? Normal? Friction? Tension? Applied force?
Mistake 3: Drawing pseudo forces in inertial frames
- Pseudo forces only in non-inertial (accelerating) frames!
Applying F = ma
One-Dimensional Motion
For motion along a line:
$$F_{\text{net}} = ma$$Sign convention matters:
- Choose positive direction (usually right/up)
- Forces in positive direction: positive
- Forces in negative direction: negative
Two-Dimensional Motion
Resolve into components:
$$F_x = ma_x$$ $$F_y = ma_y$$Key insight: Motion in x and y are independent (just like projectile motion!)
Common Force Scenarios
1. Block on Horizontal Surface
FBD:
- Weight: $mg$ (downward)
- Normal: $N$ (upward)
- Applied force: $F$ (horizontal)
- Friction: $f$ (opposite to motion/tendency)
Vertical equilibrium: $N = mg$ (no vertical motion)
Horizontal motion: $F - f = ma$
2. Block on Inclined Plane
Choose tilted axes (x along plane, y perpendicular)
Weight components:
- Along plane: $mg\sin\theta$
- Perpendicular: $mg\cos\theta$
Equations:
- Perpendicular: $N = mg\cos\theta$ (no motion perpendicular to plane)
- Along plane: $mg\sin\theta - f = ma$
“SIN goes DOWN the plane”
- Component along slope (down) = $mg\sin\theta$
- Component into slope (normal) = $mg\cos\theta$
Why? When $\theta = 90°$ (vertical cliff), $\sin 90° = 1$, so full weight acts downward ✓
3. Pulley Systems (Two Blocks)
Key points:
- Tension is same throughout a massless, frictionless pulley
- Constraint: If string is inextensible, $a_1 = a_2$ (magnitudes)
For block 1 (mass $m_1$): $T - m_1g = m_1a$
For block 2 (mass $m_2$): $m_2g - T = m_2a$
Solve simultaneously:
$$a = \frac{(m_2 - m_1)g}{m_1 + m_2}$$ $$T = \frac{2m_1m_2g}{m_1 + m_2}$$Impulse: Force Over Time
Definition
$$\boxed{\vec{J} = \int \vec{F} \, dt = \Delta \vec{p}}$$Impulse is the change in momentum.
For constant force:
$$\boxed{J = F \cdot \Delta t = m(v - u)}$$Units: N⋅s (same as momentum: kg⋅m/s)
Impulse-Momentum Theorem
$$\boxed{F \cdot \Delta t = m \Delta v}$$Key insight: The same change in momentum can result from:
- Large force for short time (hammer hitting nail)
- Small force for long time (pushing a car)
In Pathaan (2023), when Shah Rukh Khan lands on a car roof after jumping from a building, why doesn’t he get severely injured?
The car roof crumples — increasing the collision time $\Delta t$. By impulse-momentum theorem:
$$F = \frac{m\Delta v}{\Delta t}$$Larger $\Delta t$ → smaller $F$ → reduced injury!
This is why:
- Airbags work (increase collision time)
- Athletes bend knees when landing (increase stopping time)
- Catching a cricket ball with moving hands (increase contact time)
Variable Mass Systems
Rocket Equation
For systems where mass changes (rocket burning fuel):
$$\boxed{F_{\text{thrust}} = v_{\text{rel}} \frac{dm}{dt}}$$Where:
- $v_{\text{rel}}$ = velocity of ejected gases relative to rocket
- $dm/dt$ = rate of mass ejection
For a rocket in free space (no gravity, no air resistance):
$$m\frac{dv}{dt} = v_{\text{rel}}\frac{dm}{dt}$$Tsiolkovsky Rocket Equation:
$$\boxed{v - u = v_{\text{rel}} \ln\left(\frac{m_i}{m_f}\right)}$$Where:
- $m_i$ = initial mass
- $m_f$ = final mass
- $\ln$ = natural logarithm
Memory Tricks & Patterns
Mnemonic for F = ma
“Fantastic Men Accelerate”
- Force = mass × acceleration
Mnemonic for Impulse
“Just Forget Delta-P” → J = F⋅Δt = Δp
- Just = Impulse (J is symbol)
- Forget = Force
- Delta-P = change in momentum
Pattern Recognition
| Given | Find | Use |
|---|---|---|
| Force, mass | Acceleration | $a = F/m$ |
| Force, time | Change in momentum | $\Delta p = F \cdot \Delta t$ |
| Force-time graph | Impulse | Area under curve |
| Mass, velocity | Momentum | $p = mv$ |
| Two masses on pulley | Acceleration | $a = \frac{(m_2-m_1)g}{m_1+m_2}$ |
Common Mistakes to Avoid
Wrong: $F = ma = 5 \times 2 = 10$ N
Problem: Direction matters! If forces are in different directions, you must add them vectorially.
Correct approach:
- Resolve all forces into components
- Find net force in each direction
- Apply F = ma separately for each component
Scenario: A rocket burns fuel and accelerates.
Wrong: $F = ma$ (mass is changing!)
Correct: $F = \frac{dp}{dt}$
Wrong: “The weight of the object is 10 kg”
Correct:
- Mass = 10 kg (scalar, doesn’t change)
- Weight = $mg$ = 10 × 10 = 100 N (force, vector, changes with g)
In pulley problems with different masses:
Wrong: Tension = weight of hanging mass = $mg$
Correct: Tension is ALWAYS between the two weights. Calculate using:
$$T = \frac{2m_1m_2g}{m_1+m_2}$$If $m_1 = m_2$, then $T = mg$ (special case)
Practice Problems
Level 1: Foundation (NCERT)
A force of 20 N acts on a body of mass 4 kg. Find the acceleration produced.
Solution: Given: $F = 20$ N, $m = 4$ kg
Using Newton’s second law:
$$a = \frac{F}{m} = \frac{20}{4} = 5 \text{ m/s}^2$$Answer: 5 m/s²
A cricket ball of mass 150 g is moving at 20 m/s. Calculate its momentum.
Solution: Given: $m = 150$ g $= 0.15$ kg, $v = 20$ m/s
$$p = mv = 0.15 \times 20 = 3 \text{ kg⋅m/s}$$Answer: 3 kg⋅m/s
A force of 100 N acts on a body for 0.5 s. Calculate the impulse.
Solution: Given: $F = 100$ N, $\Delta t = 0.5$ s
$$J = F \cdot \Delta t = 100 \times 0.5 = 50 \text{ N⋅s}$$Answer: 50 N⋅s
Level 2: JEE Main
Two masses 3 kg and 5 kg are connected by a string over a frictionless pulley. Find: (a) Acceleration of the system (b) Tension in the string (Take g = 10 m/s²)
Solution:
For 5 kg mass (moving down):
$$5g - T = 5a \quad \text{...(1)}$$For 3 kg mass (moving up):
$$T - 3g = 3a \quad \text{...(2)}$$Adding equations (1) and (2):
$$5g - 3g = 5a + 3a$$ $$2g = 8a$$ $$a = \frac{g}{4} = \frac{10}{4} = 2.5 \text{ m/s}^2$$Finding tension from equation (2):
$$T = 3g + 3a = 3(10) + 3(2.5) = 30 + 7.5 = 37.5 \text{ N}$$Verification from equation (1):
$$T = 5g - 5a = 5(10) - 5(2.5) = 50 - 12.5 = 37.5 \text{ N}$$✓
Answers: (a) Acceleration = 2.5 m/s² (b) Tension = 37.5 N
A block of mass 10 kg rests on a smooth inclined plane of angle 30°. Find the acceleration of the block when released. (g = 10 m/s²)
Solution:
Since the plane is smooth (frictionless), only two forces act on the block:
- Weight component along plane: $mg\sin\theta$ (down the plane)
- Normal reaction $N$ (perpendicular to plane, no motion in this direction)
Using F = ma along the plane:
$$mg\sin\theta = ma$$ $$a = g\sin\theta = 10 \times \sin 30° = 10 \times 0.5 = 5 \text{ m/s}^2$$Note: Acceleration is independent of mass!
Answer: 5 m/s² down the plane
A ball of mass 200 g moving at 10 m/s strikes a wall and rebounds at 8 m/s. If the contact time is 0.02 s, find the average force exerted by the wall.
Solution:
Taking direction away from wall as positive:
- Initial velocity: $u = -10$ m/s (toward wall)
- Final velocity: $v = +8$ m/s (away from wall)
- Mass: $m = 0.2$ kg
- Time: $\Delta t = 0.02$ s
Change in momentum:
$$\Delta p = m(v - u) = 0.2[8 - (-10)] = 0.2 \times 18 = 3.6 \text{ kg⋅m/s}$$Using impulse-momentum theorem:
$$F \cdot \Delta t = \Delta p$$ $$F = \frac{\Delta p}{\Delta t} = \frac{3.6}{0.02} = 180 \text{ N}$$Answer: 180 N (away from wall)
A force acts on a 2 kg mass according to the graph shown (not drawn here). The force is 10 N for 3 seconds, then 0 N for 2 seconds, then 5 N for 2 seconds. If the body starts from rest, find its final velocity.
Solution:
Method: Use impulse-momentum theorem
Total impulse = Area under F-t graph
$$J_{\text{total}} = (10 \times 3) + (0 \times 2) + (5 \times 2) = 30 + 0 + 10 = 40 \text{ N⋅s}$$Since impulse = change in momentum:
$$J = \Delta p = m(v - u)$$ $$40 = 2(v - 0)$$ $$v = 20 \text{ m/s}$$Answer: 20 m/s
Level 3: JEE Advanced
Two blocks A (5 kg) and B (3 kg) are connected by a light inextensible string. A is on a smooth horizontal table and B hangs vertically over a frictionless pulley at the edge. Find the acceleration of the system and tension in the string. (g = 10 m/s²)
Solution:
For block A (horizontal motion):
$$T = 5a \quad \text{...(1)}$$For block B (vertical motion):
$$3g - T = 3a \quad \text{...(2)}$$From (1): $T = 5a$
Substituting in (2):
$$30 - 5a = 3a$$ $$30 = 8a$$ $$a = 3.75 \text{ m/s}^2$$Tension:
$$T = 5a = 5 \times 3.75 = 18.75 \text{ N}$$Answers:
- Acceleration = 3.75 m/s²
- Tension = 18.75 N
A block of mass $m$ is placed on a smooth wedge of mass $M$ and angle $\theta$. The wedge is on a smooth horizontal surface. When the system is released, find the acceleration of the wedge.
Solution:
Key insight: Since both surfaces are smooth, there’s no friction. The only forces are weights and normal reactions.
For block (mass m): Let wedge accelerate to the right with acceleration $a_M$. Let block accelerate down the wedge with $a_m$ (relative to wedge).
In ground frame, block has:
- Horizontal component: $a_M - a_m\cos\theta$ (to the right)
- Vertical component: $a_m\sin\theta$ (downward)
Vertical equilibrium of block:
$$mg - N\cos\theta = ma_m\sin\theta \quad \text{...(1)}$$Horizontal motion of block:
$$N\sin\theta = m(a_M - a_m\cos\theta) \quad \text{...(2)}$$For wedge (mass M): Horizontal force on wedge = horizontal component of N (to the left)
$$N\sin\theta = Ma_M \quad \text{...(3)}$$From (2) and (3):
$$m(a_M - a_m\cos\theta) = Ma_M$$ $$ma_M - ma_m\cos\theta = Ma_M$$ $$a_m\cos\theta = \frac{m - M}{m}a_M \quad \text{...(4)}$$From (1):
$$N = \frac{mg - ma_m\sin\theta}{\cos\theta}$$Substituting in (3):
$$\frac{mg - ma_m\sin\theta}{\cos\theta} \cdot \sin\theta = Ma_M$$ $$mg\sin\theta - ma_m\sin^2\theta = Ma_M\cos\theta$$Using (4) to eliminate $a_m$:
After algebraic manipulation:
$$\boxed{a_M = \frac{mg\sin\theta\cos\theta}{M + m\sin^2\theta}}$$Answer: $a_M = \frac{mg\sin\theta\cos\theta}{M + m\sin^2\theta}$
A uniform chain of length $L$ and mass $M$ is placed on a smooth table with length $b$ hanging over the edge. Find the time taken for the chain to slip off the table.
Solution:
Let $x$ = length hanging at time $t$
Mass hanging = $\frac{M}{L}x$
Equation of motion:
$$F = \frac{M}{L}x \cdot g = M \cdot a$$ $$a = \frac{gx}{L}$$But $a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = v\frac{dv}{dx}$
$$v\frac{dv}{dx} = \frac{gx}{L}$$ $$v \, dv = \frac{g}{L}x \, dx$$Integrating from $x = b$ (initially) to $x = L$ (finally):
$$\int_0^v v \, dv = \int_b^L \frac{g}{L}x \, dx$$ $$\frac{v^2}{2} = \frac{g}{L} \cdot \frac{x^2}{2} \Big|_b^L = \frac{g}{2L}(L^2 - b^2)$$ $$v^2 = \frac{g}{L}(L^2 - b^2)$$ $$v = \sqrt{\frac{g(L^2 - b^2)}{L}}$$But we need time, not velocity at the end. So we use:
$$v = \frac{dx}{dt}$$From earlier: $v^2 = \frac{g}{L}(x^2 - b^2)$ (generalizing)
$$v = \sqrt{\frac{g(x^2 - b^2)}{L}}$$ $$\frac{dx}{dt} = \sqrt{\frac{g}{L}} \sqrt{x^2 - b^2}$$ $$\frac{dx}{\sqrt{x^2 - b^2}} = \sqrt{\frac{g}{L}} \, dt$$Integrating:
$$\int_b^L \frac{dx}{\sqrt{x^2 - b^2}} = \sqrt{\frac{g}{L}} \int_0^T dt$$ $$\cosh^{-1}\left(\frac{x}{b}\right) \Big|_b^L = \sqrt{\frac{g}{L}} \cdot T$$ $$\cosh^{-1}\left(\frac{L}{b}\right) = \sqrt{\frac{g}{L}} \cdot T$$ $$\boxed{T = \sqrt{\frac{L}{g}} \cosh^{-1}\left(\frac{L}{b}\right)}$$Or in logarithmic form:
$$T = \sqrt{\frac{L}{g}} \ln\left(\frac{L}{b} + \sqrt{\frac{L^2}{b^2} - 1}\right)$$Answer: $T = \sqrt{\frac{L}{g}} \cosh^{-1}(L/b)$
Quick Revision Box
| Concept | Formula | Key Point |
|---|---|---|
| Second Law | $\vec{F} = \frac{d\vec{p}}{dt}$ | General form |
| F = ma | $\vec{F} = m\vec{a}$ | Constant mass only |
| Momentum | $\vec{p} = m\vec{v}$ | Quantity of motion |
| Impulse | $J = F \cdot \Delta t = \Delta p$ | Change in momentum |
| Atwood Machine | $a = \frac{(m_2-m_1)g}{m_1+m_2}$ | Two masses, pulley |
| Incline | $a = g\sin\theta$ | Smooth plane |
| FBD | Draw all forces | Most critical step! |
When to Use This Law
Use Newton’s Second Law when:
- Finding acceleration given forces
- Finding force given acceleration
- Analyzing motion with known/unknown forces
- Pulley, incline, constraint problems
Use Impulse-Momentum when:
- Force varies with time
- Given force-time graph
- Collision problems
- Finding average force during impact
Use Work-Energy when:
- Displacement is involved
- No time information given
- Energy conservation problems
JEE Exam Strategy
Weightage
- JEE Main: 4-6 questions per year (high yield!)
- JEE Advanced: 2-3 multi-concept problems (complex, 4+ marks each)
Common Question Types
- Atwood machine variations (very common, 3 marks)
- Block on incline (with/without friction, 2-4 marks)
- Wedge-block systems (advanced, 4 marks)
- Impulse-momentum (moderate, 2-3 marks)
- Force-time graphs (area = impulse, 2 marks)
Time-Saving Tricks
Trick 1: For Atwood machine, memorize:
$$a = \frac{|m_2 - m_1|}{m_1 + m_2}g, \quad T = \frac{2m_1m_2}{m_1+m_2}g$$Trick 2: For smooth incline, $a = g\sin\theta$ (independent of mass!)
Trick 3: If F-t graph is given, area under curve = impulse = change in momentum
Trick 4: For two-body problems, always write equations for BOTH bodies, then solve simultaneously
Teacher’s Summary
- F = dp/dt is fundamental — F = ma is just a special case for constant mass
- Momentum = mass × velocity — measures “quantity of motion”
- Impulse = F⋅Δt = Δp — same momentum change from large F (short time) or small F (long time)
- Free Body Diagram is CRITICAL — draw FBD for EVERY problem, no exceptions!
- Vector nature matters — always resolve into components for 2D problems
- For systems: Write F = ma for EACH body, use constraints, solve simultaneously
“Force doesn’t cause motion — it causes CHANGE in motion (acceleration)!”
Related Topics
Within Laws of Motion
- Newton’s First Law — What happens when F = 0
- Newton’s Third Law — How forces come in pairs
- Friction — A key force in most problems
- Impulse and Momentum — Detailed treatment of collisions
Connected Chapters
- Kinematics — Motion description (now we know WHY objects accelerate!)
- Work-Energy-Power — Alternative approach to dynamics
- Circular Motion — Centripetal force applications
- Rotational Motion — Angular analog: $\tau = I\alpha$
- Gravitation — Gravitational force applications
Math Connections
- Vectors — Force resolution and addition
- Differentiation — dp/dt concept
- Integration — For variable forces