Prerequisites
Before studying this topic, make sure you understand:
- Newton’s Second Law - Force, momentum, and F = dp/dt
- Newton’s Third Law - Action-reaction pairs
- Vectors - Vector addition and components
The Hook: Why Do Cricket Players Pull Their Hands Back While Catching?
Imagine catching a cricket ball hit hard by a batsman (like the intense scenes in Indian cricket matches). If you keep your hands rigid, it hurts! But if you pull your hands back while catching, it feels much softer. Why?
Both cases bring the ball to rest (same change in momentum), but pulling back increases the time → reduces the force on your hands!
This is the Impulse-Momentum Theorem in action.
The Big Questions:
- What’s the connection between force and momentum change?
- Why does increasing time reduce impact force?
- How is momentum conserved in collisions?
The Core Concept: Impulse
Definition of Impulse
Impulse is the product of force and the time interval for which it acts.
$$\boxed{\vec{J} = \int_{t_1}^{t_2} \vec{F} \, dt}$$For constant force:
$$\boxed{\vec{J} = \vec{F} \cdot \Delta t}$$Units: N·s or kg·m/s (same as momentum!)
Physical meaning: Impulse measures the “total effect” of a force acting over time.
Impulse-Momentum Theorem
$$\boxed{\vec{J} = \Delta \vec{p} = \vec{p}_f - \vec{p}_i}$$In words: Impulse equals the change in momentum.
Derivation: From Newton’s second law:
$$\vec{F} = \frac{d\vec{p}}{dt}$$ $$\vec{F} \, dt = d\vec{p}$$Integrating:
$$\int \vec{F} \, dt = \int d\vec{p}$$ $$\boxed{\vec{J} = \Delta \vec{p}}$$Understanding Impulse Graphically
Force-Time Graph
Impulse = Area under F-t curve
For constant force:
- Graph is a rectangle
- Area = $F \times \Delta t = J$
For variable force:
- Area under curve = $\int F \, dt = J$
For variable force, we can define average force:
$$\vec{F}_{\text{avg}} = \frac{\vec{J}}{\Delta t} = \frac{\Delta \vec{p}}{\Delta t}$$This is the constant force that would produce the same impulse in the same time.
Interactive Demo: Visualize Force and Momentum
Explore how impulse relates to momentum change.
Applications of Impulse
1. Catching a Ball (Cricket/Baseball)
Scenario: Ball of mass $m$ moving at velocity $v$ is brought to rest.
Change in momentum: $\Delta p = 0 - mv = -mv$
Impulse needed: $J = mv$
Using $J = F \cdot \Delta t$:
$$F = \frac{mv}{\Delta t}$$Key insight:
- Rigid catch (small $\Delta t$) → Large $F$ → Hurts!
- Soft catch (large $\Delta t$) → Small $F$ → Doesn’t hurt
In Pathaan, when Shah Rukh Khan lands from a jump onto a car, the car roof crumples. This increases collision time $\Delta t$, reducing impact force $F = \frac{\Delta p}{\Delta t}$.
Same principle used in:
- Airbags (increase collision time)
- Crash barriers (crumple zones)
- Bending knees when landing
2. Hammer and Nail
Scenario: Hammer strikes nail, drives it into wood.
Large force for short time creates impulse that changes nail’s momentum from 0 to some value.
3. Follow-Through in Sports
Golf, cricket, tennis: Players follow through after hitting the ball.
Why? Follow-through increases contact time $\Delta t$, allowing same impulse with reduced peak force → better control and reduced injury risk.
Conservation of Momentum
Law of Conservation of Momentum
$$\boxed{\text{If } \vec{F}_{\text{ext}} = 0, \text{ then } \vec{p}_{\text{total}} = \text{constant}}$$In words: If no external force acts on a system, the total momentum of the system remains constant.
Derivation from Newton’s Laws
Consider two objects A and B interacting (no external forces).
By Newton’s Third Law:
$$\vec{F}_{AB} = -\vec{F}_{BA}$$For object A:
$$\vec{F}_{BA} = \frac{d\vec{p}_A}{dt}$$For object B:
$$\vec{F}_{AB} = \frac{d\vec{p}_B}{dt}$$Adding:
$$\frac{d\vec{p}_A}{dt} + \frac{d\vec{p}_B}{dt} = \vec{F}_{BA} + \vec{F}_{AB} = 0$$ $$\frac{d(\vec{p}_A + \vec{p}_B)}{dt} = 0$$ $$\boxed{\vec{p}_A + \vec{p}_B = \text{constant}}$$Conservation of momentum is a consequence of Newton’s Third Law!
Internal forces (action-reaction pairs) don’t change total momentum because they cancel out.
Collisions
Types of Collisions
1. Elastic Collision
Properties:
- Momentum is conserved
- Kinetic energy is conserved
- Objects bounce off each other
- Coefficient of restitution $e = 1$
Examples: Collision between hard steel balls, atomic particles
2. Inelastic Collision
Properties:
- Momentum is conserved
- Kinetic energy is NOT conserved (some lost as heat, sound, deformation)
- Objects may stick together or bounce with reduced speed
- Coefficient of restitution $0 < e < 1$
Examples: Car crashes, ball hitting ground and bouncing
3. Perfectly Inelastic Collision
Properties:
- Momentum is conserved
- Maximum kinetic energy loss
- Objects stick together after collision
- Coefficient of restitution $e = 0$
Examples: Bullet embedding in wood, clay balls colliding
Head-On Elastic Collision
Two Objects with Different Masses
Before collision:
- Object 1: mass $m_1$, velocity $u_1$
- Object 2: mass $m_2$, velocity $u_2$
After collision:
- Object 1: velocity $v_1$
- Object 2: velocity $v_2$
Conservation of momentum:
$$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \quad \text{...(1)}$$Conservation of kinetic energy (elastic):
$$\frac{1}{2}m_1 u_1^2 + \frac{1}{2}m_2 u_2^2 = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2 \quad \text{...(2)}$$Solving these equations:
$$\boxed{v_1 = \frac{(m_1 - m_2)u_1 + 2m_2 u_2}{m_1 + m_2}}$$ $$\boxed{v_2 = \frac{(m_2 - m_1)u_2 + 2m_1 u_1}{m_1 + m_2}}$$Special Cases
Case 1: Equal masses ($m_1 = m_2 = m$)
$$v_1 = u_2$$ $$v_2 = u_1$$Result: Velocities are exchanged!
Case 2: Second object at rest ($u_2 = 0$)
$$v_1 = \frac{(m_1 - m_2)u_1}{m_1 + m_2}$$ $$v_2 = \frac{2m_1 u_1}{m_1 + m_2}$$Sub-case 2a: $m_1 = m_2$ (equal masses, one at rest)
$$v_1 = 0, \quad v_2 = u_1$$First object stops, second moves with first’s initial velocity!
Sub-case 2b: $m_1 \ll m_2$ (light object hits heavy)
$$v_1 \approx -u_1, \quad v_2 \approx 0$$Light object bounces back, heavy barely moves. (Ball hitting wall)
Sub-case 2c: $m_1 \gg m_2$ (heavy object hits light)
$$v_1 \approx u_1, \quad v_2 \approx 2u_1$$Heavy object continues almost unchanged, light object shoots forward at twice the speed!
“Equal masses EXCHANGE velocities”
$$m_1 = m_2 \implies v_1 = u_2, \, v_2 = u_1$$“Light bounces BACK, heavy stays PUT”
$$m_1 \ll m_2 \implies v_1 \approx -u_1, \, v_2 \approx 0$$Head-On Perfectly Inelastic Collision
Objects stick together after collision.
Conservation of momentum:
$$m_1 u_1 + m_2 u_2 = (m_1 + m_2) v$$ $$\boxed{v = \frac{m_1 u_1 + m_2 u_2}{m_1 + m_2}}$$Loss in kinetic energy:
$$\Delta KE = KE_i - KE_f$$ $$\Delta KE = \frac{1}{2}m_1 u_1^2 + \frac{1}{2}m_2 u_2^2 - \frac{1}{2}(m_1+m_2)v^2$$After substituting and simplifying:
$$\boxed{\Delta KE = \frac{1}{2}\frac{m_1 m_2}{m_1 + m_2}(u_1 - u_2)^2}$$This is the reduced mass formula for energy loss!
In perfectly inelastic collision, kinetic energy loss is maximum (for given initial conditions).
Energy is converted to heat, sound, deformation, etc.
Coefficient of Restitution (e)
Definition
The coefficient of restitution measures how bouncy a collision is.
$$\boxed{e = \frac{\text{Relative velocity of separation}}{\text{Relative velocity of approach}}}$$ $$e = \frac{v_2 - v_1}{u_1 - u_2}$$Range: $0 \leq e \leq 1$
Types:
- $e = 1$: Perfectly elastic
- $0 < e < 1$: Inelastic
- $e = 0$: Perfectly inelastic
Using e to Find Final Velocities
For collision with e given:
Momentum conservation:
$$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$$Restitution equation:
$$v_2 - v_1 = e(u_1 - u_2)$$Solve these two equations simultaneously to find $v_1$ and $v_2$.
Collision with Wall/Ground
Object hits a fixed wall/ground elastically.
Since wall is infinitely massive ($m_2 \to \infty$):
- Wall doesn’t move ($v_2 = 0$)
- Object bounces back
For elastic collision ($e = 1$):
$$v = -u$$Object rebounds with same speed, opposite direction.
For inelastic collision (coefficient $e$):
$$v = -eu$$Object rebounds with reduced speed.
Example: Ball dropped from height $h$ on ground with restitution coefficient $e$.
After first bounce, reaches height:
$$h_1 = e^2 h$$After second bounce:
$$h_2 = e^4 h$$After $n$ bounces:
$$h_n = e^{2n} h$$In action movies like Jawan, when objects (or people!) fall and bounce, the bounce height is always less than the drop height (unless it’s super slow-mo CGI!).
This is because $e < 1$ for real collisions → energy is lost → can’t bounce back to original height.
Interactive Animation
Explore different collision scenarios:
Adjust masses, velocities, and coefficient of restitution to see how momentum and energy change!
Multi-Dimensional Collisions
Oblique Collisions
When collision is not head-on, we must consider vector momentum.
Key principle: Momentum is conserved in both x and y directions separately.
$$m_1 u_{1x} + m_2 u_{2x} = m_1 v_{1x} + m_2 v_{2x}$$ $$m_1 u_{1y} + m_2 u_{2y} = m_1 v_{1y} + m_2 v_{2y}$$For elastic oblique collision, additional equation from energy conservation.
Center of Mass and Momentum
Center of Mass Velocity
$$\vec{v}_{\text{CM}} = \frac{m_1 \vec{v}_1 + m_2 \vec{v}_2}{m_1 + m_2}$$In terms of total momentum:
$$\vec{p}_{\text{total}} = (m_1 + m_2) \vec{v}_{\text{CM}}$$Key insight: If no external force, $\vec{v}_{\text{CM}}$ remains constant!
The center of mass moves with constant velocity even if objects are colliding, exploding, or interacting internally.
Explosions and Recoil
Explosion
Initially: Single object at rest (or moving) Finally: Breaks into fragments
Momentum conservation:
$$\vec{p}_{\text{initial}} = \sum \vec{p}_{\text{fragments}}$$For object initially at rest:
$$\sum m_i \vec{v}_i = 0$$Recoil (Revisited)
Gun-bullet system:
Initially at rest: $p_{\text{total}} = 0$
After firing:
$$m_{\text{bullet}} v_{\text{bullet}} + m_{\text{gun}} v_{\text{gun}} = 0$$ $$v_{\text{gun}} = -\frac{m_{\text{bullet}}}{m_{\text{gun}}} v_{\text{bullet}}$$Memory Tricks & Patterns
Mnemonic for Impulse
“Just Forget Delta-P” → $J = F \Delta t = \Delta p$
Mnemonic for Collision Types
“E-PIE”
- Elastic: Both momentum and Energy conserved
- Perfectly Inelastic: Stick together, maximum Energy loss
- (General) Inelastic: Momentum conserved, Energy lost
Pattern Recognition
| Collision Type | $e$ | Momentum | KE | After collision |
|---|---|---|---|---|
| Elastic | 1 | Conserved | Conserved | Bounce off |
| Inelastic | 0 < e < 1 | Conserved | Lost | Bounce with less speed |
| Perfectly Inelastic | 0 | Conserved | Max loss | Stick together |
Common Mistakes to Avoid
Wrong: In all collisions, both momentum and kinetic energy are conserved.
Correct: Momentum is ALWAYS conserved (no external force). Kinetic energy is conserved only in elastic collisions.
JEE trap: Problem asks about collision but doesn’t specify elastic → Don’t assume KE conservation!
Wrong: Treating momentum as scalar in 2D collisions.
Correct: Momentum is a VECTOR. In 2D/3D collisions, apply conservation separately to each component.
$$p_x \text{ conserved}, \quad p_y \text{ conserved}, \quad p_z \text{ conserved}$$Common mistake: Not choosing a consistent positive direction.
Correct approach:
- Choose positive direction (say, right)
- Velocities to the right: positive
- Velocities to the left: negative
- Apply momentum conservation with correct signs
Wrong: Using momentum conservation when external forces act.
Example: Two blocks colliding on a rough surface (friction acts).
Correct: Momentum is conserved only if NO external force (or external forces cancel). With friction, momentum is NOT conserved!
Practice Problems
Level 1: Foundation (NCERT)
A force of 50 N acts on a body for 0.2 s. Calculate the impulse.
Solution:
$$J = F \cdot \Delta t = 50 \times 0.2 = 10 \text{ N·s}$$Answer: 10 N·s
A cricket ball of mass 150 g moving at 30 m/s is caught by a player in 0.05 s. Find the average force exerted by the player.
Solution:
Change in momentum:
$$\Delta p = m(v - u) = 0.15(0 - 30) = -4.5 \text{ kg·m/s}$$(Negative because ball is brought to rest)
Average force:
$$F = \frac{\Delta p}{\Delta t} = \frac{-4.5}{0.05} = -90 \text{ N}$$Magnitude: 90 N (opposite to ball’s motion)
Answer: 90 N
A 2 kg object moving at 5 m/s collides with a 3 kg object at rest. They stick together. Find their common velocity.
Solution:
Momentum conservation:
$$m_1 u_1 + m_2 u_2 = (m_1 + m_2)v$$ $$2 \times 5 + 3 \times 0 = (2 + 3)v$$ $$10 = 5v$$ $$v = 2 \text{ m/s}$$Answer: 2 m/s
Level 2: JEE Main
A ball of mass 0.5 kg moving at 10 m/s collides elastically with another identical ball at rest. Find the velocities after collision.
Solution:
For elastic collision with equal masses, velocities exchange:
$$v_1 = u_2 = 0 \text{ m/s}$$ $$v_2 = u_1 = 10 \text{ m/s}$$Physical interpretation: First ball stops, second ball moves with first’s initial velocity.
Answers:
- First ball: 0 m/s
- Second ball: 10 m/s
A 5 kg gun fires a 20 g bullet at 300 m/s. Find the recoil velocity of the gun.
Solution:
Momentum conservation (initially at rest):
$$0 = m_b v_b + m_g v_g$$ $$v_g = -\frac{m_b v_b}{m_g} = -\frac{0.02 \times 300}{5} = -1.2 \text{ m/s}$$Negative sign: opposite to bullet’s direction.
Answer: 1.2 m/s (backward)
A 4 kg object moving at 6 m/s collides head-on with a 2 kg object moving at 3 m/s in opposite direction. They stick together. Find: (a) Final velocity (b) Loss in kinetic energy
Solution:
Taking 4 kg’s direction as positive:
$$u_1 = 6 \text{ m/s}, \quad u_2 = -3 \text{ m/s}$$(a) Momentum conservation:
$$4 \times 6 + 2 \times (-3) = (4 + 2)v$$ $$24 - 6 = 6v$$ $$v = 3 \text{ m/s}$$(b) Energy loss:
$$KE_i = \frac{1}{2} \times 4 \times 6^2 + \frac{1}{2} \times 2 \times 3^2 = 72 + 9 = 81 \text{ J}$$ $$KE_f = \frac{1}{2} \times 6 \times 3^2 = 27 \text{ J}$$ $$\Delta KE = 81 - 27 = 54 \text{ J}$$Answers: (a) 3 m/s (b) 54 J
Level 3: JEE Advanced
A ball of mass 1 kg moving at 5 m/s collides with another ball of mass 2 kg at rest. If coefficient of restitution is 0.5, find the velocities after collision.
Solution:
Given: $m_1 = 1$ kg, $u_1 = 5$ m/s, $m_2 = 2$ kg, $u_2 = 0$, $e = 0.5$
Momentum conservation:
$$1 \times 5 + 2 \times 0 = 1 \times v_1 + 2 \times v_2$$ $$v_1 + 2v_2 = 5 \quad \text{...(1)}$$Restitution equation:
$$v_2 - v_1 = e(u_1 - u_2) = 0.5 \times 5 = 2.5$$ $$v_2 - v_1 = 2.5 \quad \text{...(2)}$$From (2): $v_2 = v_1 + 2.5$
Substituting in (1):
$$v_1 + 2(v_1 + 2.5) = 5$$ $$v_1 + 2v_1 + 5 = 5$$ $$3v_1 = 0$$ $$v_1 = 0 \text{ m/s}$$ $$v_2 = 0 + 2.5 = 2.5 \text{ m/s}$$Answers:
- First ball: 0 m/s (stops)
- Second ball: 2.5 m/s
A ball is dropped from height $h = 10$ m onto the ground. Coefficient of restitution is $e = 0.6$. Find: (a) Height after first bounce (b) Total distance traveled before coming to rest (g = 10 m/s²)
Solution:
(a) Velocity just before hitting ground:
$$v = \sqrt{2gh} = \sqrt{2 \times 10 \times 10} = \sqrt{200} = 10\sqrt{2} \text{ m/s}$$Velocity just after bounce:
$$v' = ev = 0.6 \times 10\sqrt{2} = 6\sqrt{2} \text{ m/s}$$Height reached:
$$h_1 = \frac{v'^2}{2g} = \frac{(6\sqrt{2})^2}{20} = \frac{72}{20} = 3.6 \text{ m}$$Alternative: $h_1 = e^2 h = 0.36 \times 10 = 3.6$ m
(b) Total distance:
Ball travels down from $h$, bounces to $h_1 = e^2 h$, down again, bounces to $h_2 = e^4 h$, and so on.
$$\text{Total} = h + 2h_1 + 2h_2 + 2h_3 + \ldots$$ $$= h + 2(h_1 + h_2 + h_3 + \ldots)$$ $$= h + 2(e^2h + e^4h + e^6h + \ldots)$$ $$= h + 2e^2 h(1 + e^2 + e^4 + \ldots)$$Geometric series: $\sum_{n=0}^{\infty} r^n = \frac{1}{1-r}$ for $|r| < 1$
$$= h + 2e^2 h \cdot \frac{1}{1 - e^2}$$ $$= h \left(1 + \frac{2e^2}{1-e^2}\right)$$ $$= h \cdot \frac{1 - e^2 + 2e^2}{1 - e^2}$$ $$= h \cdot \frac{1 + e^2}{1 - e^2}$$ $$= 10 \cdot \frac{1 + 0.36}{1 - 0.36} = 10 \cdot \frac{1.36}{0.64} = 10 \times 2.125 = 21.25 \text{ m}$$Answers: (a) 3.6 m (b) 21.25 m
A bomb of mass 3 kg initially at rest explodes into three fragments of masses 1 kg, 1 kg, and 1 kg. Two fragments fly off at right angles to each other with speeds 30 m/s and 40 m/s. Find the velocity of the third fragment.
Solution:
Let the two known fragments be along x and y axes.
Fragment 1: $m_1 = 1$ kg, $\vec{v}_1 = 30\hat{i}$ m/s Fragment 2: $m_2 = 1$ kg, $\vec{v}_2 = 40\hat{j}$ m/s Fragment 3: $m_3 = 1$ kg, $\vec{v}_3 = ?$
Momentum conservation (initially at rest):
$$\vec{0} = m_1 \vec{v}_1 + m_2 \vec{v}_2 + m_3 \vec{v}_3$$ $$0 = 1 \times 30\hat{i} + 1 \times 40\hat{j} + 1 \times \vec{v}_3$$ $$\vec{v}_3 = -30\hat{i} - 40\hat{j}$$Magnitude:
$$v_3 = \sqrt{30^2 + 40^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50 \text{ m/s}$$Direction:
$$\tan\theta = \frac{40}{30} = \frac{4}{3}$$ $$\theta = \tan^{-1}(4/3) \approx 53.1°$$Direction is opposite to the first quadrant, so third quadrant (or 180° + 53.1° = 233.1° from +x axis).
Answer: 50 m/s at 233.1° from positive x-axis (or $-30\hat{i} - 40\hat{j}$ m/s)
Quick Revision Box
| Concept | Formula | Key Point |
|---|---|---|
| Impulse | $J = F \cdot \Delta t = \Delta p$ | Change in momentum |
| Momentum conservation | $\vec{p}_{\text{total}} = \text{const}$ | No external force |
| Elastic collision | Momentum + KE conserved | $e = 1$ |
| Inelastic | Momentum conserved, KE lost | $0 < e < 1$ |
| Perfectly inelastic | Stick together | $e = 0$, max KE loss |
| Restitution | $e = \frac{v_2-v_1}{u_1-u_2}$ | Bounciness measure |
| Equal mass elastic | $v_1 = u_2$, $v_2 = u_1$ | Velocities exchange |
| Bounce height | $h_n = e^{2n}h$ | After $n$ bounces |
When to Use These Concepts
Use Impulse ($J = F\Delta t$) when:
- Force varies with time (F-t graph given)
- Finding average force during collision
- Time of contact is important
- Relating force to momentum change
Use Momentum Conservation when:
- No external force (or external forces cancel)
- Collision/explosion problems
- Multi-body systems
- Recoil problems
Check collision type:
- Elastic: Use momentum + energy conservation
- Coefficient given: Use momentum + restitution equation
- Perfectly inelastic: Use momentum, objects stick together
Don’t use momentum conservation if:
- External forces present (friction, applied force)
- Single object changing velocity due to external force
JEE Exam Strategy
Weightage
- JEE Main: 4-6 questions per year (very high yield!)
- JEE Advanced: 3-4 problems, often complex scenarios (4-6 marks each)
Common Question Types
- Elastic collision equal masses (easy, 2 marks)
- Recoil problems (easy-moderate, 2-3 marks)
- Coefficient of restitution (moderate, 3-4 marks)
- Ball bouncing (moderate, 3 marks)
- Explosion into fragments (moderate-advanced, 4 marks)
- Oblique collision (advanced, 5+ marks)
- Variable mass (rocket) (advanced, 4-5 marks)
Time-Saving Tricks
Trick 1: Equal masses elastic collision, one at rest → velocities exchange (no calculation needed!)
Trick 2: Perfectly inelastic → use $v = \frac{m_1u_1 + m_2u_2}{m_1+m_2}$ directly
Trick 3: Ball bouncing with restitution $e$ → height after $n$ bounces: $h_n = e^{2n}h_0$
Trick 4: For 2D collisions, always resolve into components and apply conservation separately
Trick 5: If problem mentions “smooth surface” or “in space” → no external force → use momentum conservation!
Teacher’s Summary
- Impulse = Change in momentum — $J = F\Delta t = \Delta p$ (connects force and momentum)
- Momentum is ALWAYS conserved — if no external force (internal forces cancel by Third Law)
- Three collision types:
- Elastic: Both momentum & KE conserved ($e=1$)
- Inelastic: Only momentum conserved ($0 < e < 1$)
- Perfectly inelastic: Stick together, max KE loss ($e=0$)
- Equal masses elastic — velocities exchange (super useful shortcut!)
- Coefficient of restitution — measures bounciness: $e = \frac{\text{separation}}{\text{approach}}$
- Impulse applications — larger time → smaller force (catching, airbags, crumple zones)
- Vector nature — in 2D/3D, conserve momentum in each direction separately
“Momentum is stubborn — it doesn’t change unless pushed by an external force!”
Related Topics
Within Laws of Motion
- Newton’s Second Law — F = dp/dt leads to impulse-momentum
- Newton’s Third Law — Explains momentum conservation
- Friction — External force that violates momentum conservation
Connected Chapters
- Work-Energy-Power — Energy approach to collisions
- Collisions (detailed) — Extended treatment
- Center of Mass — CM motion and momentum
Math Connections
- Vectors — 2D/3D momentum conservation
- Integration — Variable force impulse
- Sequences and Series — Bouncing ball total distance