The Momentum And Its Conservation Formula is one of the most fundamental concepts in physics, expressing the quantity of motion as p = mv, where p is momentum, m is mass, and v is velocity. Covered in NCERT Class 9 (Chapter 9) and revisited in Class 11 (Chapter 5), this formula is essential for CBSE board exams and forms the backbone of collision problems in JEE Main, JEE Advanced, and NEET. This article covers the definition, derivation, complete formula sheet, solved examples at three difficulty levels, CBSE exam tips, common mistakes, and JEE/NEET applications.
Key Momentum Formulas at a Glance
Quick reference for the most important momentum and conservation formulas.
- Linear momentum: \( p = mv \)
- Newton’s second law (momentum form): \( F = \frac{\Delta p}{\Delta t} \)
- Impulse: \( J = F \cdot \Delta t = \Delta p \)
- Conservation of momentum: \( m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \)
- Elastic collision (velocity exchange): \( v_1 = \frac{(m_1 – m_2)u_1 + 2m_2 u_2}{m_1 + m_2} \)
- Perfectly inelastic collision: \( v = \frac{m_1 u_1 + m_2 u_2}{m_1 + m_2} \)
- Restitution coefficient: \( e = \frac{v_2 – v_1}{u_1 – u_2} \)
What is Momentum And Its Conservation Formula?
The Momentum And Its Conservation Formula describes two closely related ideas. First, momentum (p) is the product of an object’s mass and its velocity. It tells us how much motion an object carries. Second, the law of conservation of momentum states that the total momentum of an isolated system remains constant when no external force acts on it.
This concept is introduced in NCERT Class 9, Chapter 9 — Force and Laws of Motion. It is revisited and extended in NCERT Class 11, Chapter 5 — Laws of Motion. The conservation principle follows directly from Newton’s third law. When two objects interact, the forces they exert on each other are equal and opposite. Therefore, the change in momentum of one object is equal and opposite to the change in momentum of the other. The total change is zero, so total momentum is conserved.
Momentum is a vector quantity. Its direction is the same as the direction of velocity. The SI unit of momentum is kg·m/s (kilogram metre per second), which is also written as N·s (newton second).
Momentum And Its Conservation Formula — Expression and Variables
Linear Momentum
\[ p = mv \]
Law of Conservation of Momentum (Two-Body Collision)
\[ m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \]
Momentum-Force Relationship (Newton’s Second Law)
\[ F_{net} = \frac{\Delta p}{\Delta t} = \frac{p_f – p_i}{\Delta t} \]
| Symbol | Quantity | SI Unit |
|---|---|---|
| \( p \) | Linear momentum | kg·m/s (or N·s) |
| \( m \) | Mass of the object | kilogram (kg) |
| \( v \) | Velocity of the object | metre per second (m/s) |
| \( m_1, m_2 \) | Masses of two colliding bodies | kg |
| \( u_1, u_2 \) | Initial velocities before collision | m/s |
| \( v_1, v_2 \) | Final velocities after collision | m/s |
| \( F_{net} \) | Net external force on system | Newton (N) |
| \( \Delta t \) | Time interval | second (s) |
| \( J \) | Impulse | N·s (or kg·m/s) |
Derivation of Conservation of Momentum
Consider two objects A (mass \( m_1 \)) and B (mass \( m_2 \)) colliding. Let their initial velocities be \( u_1 \) and \( u_2 \), and final velocities be \( v_1 \) and \( v_2 \).
Step 1: By Newton’s third law, the force exerted by A on B is \( F_{AB} \), and the force by B on A is \( F_{BA} = -F_{AB} \).
Step 2: Using Newton’s second law for each body over time \( \Delta t \):
\[ F_{AB} \cdot \Delta t = m_2(v_2 – u_2) \quad \text{and} \quad F_{BA} \cdot \Delta t = m_1(v_1 – u_1) \]
Step 3: Since \( F_{AB} = -F_{BA} \), adding both equations gives:
\[ m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \]
This proves that total momentum before collision equals total momentum after collision, provided no external force acts on the system.
Complete Physics Formula Sheet: Momentum and Collisions
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| Linear Momentum | \( p = mv \) | m = mass, v = velocity | kg·m/s | Class 9, Ch 9 |
| Conservation of Momentum | \( m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \) | m = masses, u = initial vel., v = final vel. | kg·m/s | Class 9, Ch 9; Class 11, Ch 5 |
| Newton’s Second Law (momentum form) | \( F = \frac{\Delta p}{\Delta t} \) | F = force, Δp = change in momentum, Δt = time | N | Class 11, Ch 5 |
| Impulse | \( J = F \cdot \Delta t = \Delta p \) | J = impulse, F = force, Δt = time interval | N·s | Class 11, Ch 5 |
| Perfectly Inelastic Collision | \( v = \frac{m_1 u_1 + m_2 u_2}{m_1 + m_2} \) | v = common final velocity | m/s | Class 11, Ch 6 |
| Elastic Collision — Final Velocity of Body 1 | \( v_1 = \frac{(m_1 – m_2)u_1 + 2m_2 u_2}{m_1 + m_2} \) | Standard elastic collision variables | m/s | Class 11, Ch 6 |
| Elastic Collision — Final Velocity of Body 2 | \( v_2 = \frac{(m_2 – m_1)u_2 + 2m_1 u_1}{m_1 + m_2} \) | Standard elastic collision variables | m/s | Class 11, Ch 6 |
| Coefficient of Restitution | \( e = \frac{v_2 – v_1}{u_1 – u_2} \) | e = 1 (elastic), e = 0 (perfectly inelastic) | Dimensionless | Class 11, Ch 6 |
| Kinetic Energy (for momentum context) | \( KE = \frac{p^2}{2m} \) | p = momentum, m = mass | Joule (J) | Class 11, Ch 6 |
| Rocket Propulsion (thrust) | \( F = -v_{rel} \frac{dm}{dt} \) | v_rel = exhaust velocity, dm/dt = mass flow rate | N | Class 11, Ch 5 |
Momentum And Its Conservation Formula — Solved Examples
Example 1 (Class 9-10 Level): Finding Momentum of a Moving Object
Problem: A cricket ball of mass 0.15 kg is bowled at a speed of 30 m/s. Calculate the momentum of the ball.
Given: Mass, m = 0.15 kg; Velocity, v = 30 m/s
Step 1: Write the momentum formula: \( p = mv \)
Step 2: Substitute the values: \( p = 0.15 \times 30 \)
Step 3: Calculate: \( p = 4.5 \) kg·m/s
Answer
The momentum of the cricket ball is 4.5 kg·m/s in the direction of motion.
Example 2 (Class 11-12 Level): Conservation of Momentum in a Two-Body Collision
Problem: A trolley of mass 4 kg moving at 6 m/s collides with a stationary trolley of mass 2 kg. After the perfectly inelastic collision, both trolleys move together. Find their common velocity after the collision.
Given: \( m_1 = 4 \) kg, \( u_1 = 6 \) m/s; \( m_2 = 2 \) kg, \( u_2 = 0 \) m/s
Step 1: Use the perfectly inelastic collision formula: \( v = \frac{m_1 u_1 + m_2 u_2}{m_1 + m_2} \)
Step 2: Substitute values: \( v = \frac{(4 \times 6) + (2 \times 0)}{4 + 2} \)
Step 3: Simplify numerator: \( v = \frac{24 + 0}{6} \)
Step 4: Calculate: \( v = \frac{24}{6} = 4 \) m/s
Verification: Initial momentum = \( 4 \times 6 + 2 \ times 0 = 24 \) kg·m/s. Final momentum = \( (4 + 2) \times 4 = 24 \) kg·m/s. Momentum is conserved. ✓
Answer
The common velocity after the collision is 4 m/s in the original direction of motion of the first trolley.
Example 3 (JEE/NEET Level): Elastic Collision with Velocity Exchange
Problem: A ball of mass 3 kg moving at 8 m/s collides elastically head-on with a ball of mass 1 kg moving at 2 m/s in the opposite direction. Find the velocities of both balls after the collision.
Given: \( m_1 = 3 \) kg, \( u_1 = +8 \) m/s; \( m_2 = 1 \) kg, \( u_2 = -2 \) m/s (negative because opposite direction)
Step 1: Apply the elastic collision formula for \( v_1 \):
\( v_1 = \frac{(m_1 – m_2)u_1 + 2m_2 u_2}{m_1 + m_2} \)
\( v_1 = \frac{(3 – 1)(8) + 2(1)(-2)}{3 + 1} = \frac{16 – 4}{4} = \frac{12}{4} = 3 \) m/s
Step 2: Apply the elastic collision formula for \( v_2 \):
\( v_2 = \frac{(m_2 – m_1)u_2 + 2m_1 u_1}{m_1 + m_2} \)
\( v_2 = \frac{(1 – 3)(-2) + 2(3)(8)}{3 + 1} = \frac{4 + 48}{4} = \frac{52}{4} = 13 \) m/s
Step 3: Verify momentum conservation: Initial = \( 3(8) + 1(-2) = 24 – 2 = 22 \) kg·m/s. Final = \( 3(3) + 1(13) = 9 + 13 = 22 \) kg·m/s. ✓
Step 4: Verify kinetic energy conservation: Initial KE = \( \frac{1}{2}(3)(64) + \frac{1}{2}(1)(4) = 96 + 2 = 98 \) J. Final KE = \( \frac{1}{2}(3)(9) + \frac{1}{2}(1)(169) = 13.5 + 84.5 = 98 \) J. ✓
Answer
After the elastic collision: \( v_1 = \) 3 m/s (in original direction) and \( v_2 = \) 13 m/s (in original direction of ball 1). Both momentum and kinetic energy are conserved.
CBSE Exam Tips 2025-26
- Define before calculating: In CBSE 2025-26 board exams, always state the law of conservation of momentum in words before applying the formula. Examiners award marks for the statement itself.
- Direction matters: Momentum is a vector. Always assign positive and negative signs to velocities before substituting. We recommend drawing a direction arrow in your answer.
- Identify collision type: In problems, first determine whether the collision is elastic (KE conserved), inelastic (KE not conserved), or perfectly inelastic (bodies stick together). The formula changes accordingly.
- Impulse-momentum theorem: For problems involving force and time, use \( J = \Delta p = F \cdot \Delta t \). This is a frequent 2-mark question in Class 11 boards.
- Unit consistency: Always convert mass to kg and velocity to m/s before substituting. Express final momentum in kg·m/s or N·s (both are accepted).
- Verification step: In 5-mark problems, verify your answer by checking that total momentum before equals total momentum after. This earns the final mark.
Common Mistakes to Avoid
- Ignoring direction (sign error): Many students treat momentum as a scalar. If two objects move in opposite directions, one velocity must be negative. Forgetting this leads to completely wrong answers in collision problems.
- Applying conservation when external forces act: The law of conservation of momentum applies only to isolated systems. If friction or gravity has a component along the direction of motion, momentum is NOT conserved. Always check the condition first.
- Confusing impulse with momentum: Impulse (\( J = F \cdot \Delta t \)) is the change in momentum, not the momentum itself. The two are equal in magnitude only when the initial momentum is zero.
- Using the wrong collision formula: Students often apply the perfectly inelastic formula (bodies stick together) to elastic collisions. Identify the collision type before choosing the formula.
- Forgetting that KE is not conserved in inelastic collisions: Only momentum is conserved in all collisions (when no external force acts). Kinetic energy is conserved only in perfectly elastic collisions. Do not use energy conservation to find final velocities in inelastic collisions unless specifically given.
JEE/NEET Application of Momentum And Its Conservation Formula
In our experience, JEE aspirants encounter the Momentum And Its Conservation Formula in 2-3 questions every year, spanning both JEE Main and JEE Advanced. NEET also tests this concept in the context of nuclear reactions and ballistic problems. Here are the key application patterns to master:
Pattern 1: Oblique Collisions (JEE Advanced)
In two-dimensional collisions, momentum is conserved separately along the x-axis and the y-axis. Apply \( m_1 u_{1x} + m_2 u_{2x} = m_1 v_{1x} + m_2 v_{2x} \) and the same for y-components. This is a common JEE Advanced multi-concept problem combining vectors and momentum.
Pattern 2: Recoil Problems (JEE Main / NEET)
When a gun fires a bullet, the initial momentum of the system is zero. By conservation: \( 0 = m_{bullet} v_{bullet} + m_{gun} v_{gun} \). This gives \( v_{gun} = -\frac{m_{bullet} v_{bullet}}{m_{gun}} \). The negative sign confirms the gun recoils backward. NEET frequently tests this with rockets and radioactive decay (alpha emission).
Pattern 3: Variable Mass / Rocket Equation (JEE Advanced)
The thrust force on a rocket is \( F = -v_{rel} \frac{dm}{dt} \), derived directly from the momentum principle. JEE Advanced problems combine this with kinematics equations. Our experts suggest practising at least five rocket-propulsion problems before the exam, as this is a high-difficulty, high-reward topic.
Pattern 4: Kinetic Energy in Terms of Momentum (JEE/NEET)
The relation \( KE = \frac{p^2}{2m} \) is extremely useful. If two objects have the same momentum, the lighter one has more kinetic energy. If two objects have the same kinetic energy, the heavier one has more momentum. JEE Main frequently tests this comparison-type question.
FAQs on Momentum And Its Conservation Formula
We hope this comprehensive guide on the Momentum And Its Conservation Formula has helped you build a strong foundation. For more related physics formulas, explore our Physics Formulas hub. You may also find it helpful to study the Spring Constant Formula and the Flow Rate Formula, both of which involve similar principles of force and motion. For official NCERT resources, visit ncert.nic.in.