The Momentum Formula gives the quantity of motion of a moving object as p = mv, where p is momentum, m is mass, and v is velocity. This formula is a cornerstone of Class 9 and Class 11 Physics under the NCERT curriculum. It forms the basis of Newton’s Second Law and the Law of Conservation of Momentum. JEE Main and NEET aspirants frequently encounter momentum-based problems in mechanics. This article covers the formula, its derivation, a complete physics formula sheet, three progressive solved examples, CBSE exam tips, and common mistakes to avoid.
Key Momentum Formulas at a Glance
Quick reference for the most important momentum formulas used in CBSE and competitive exams.
- 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: \( p_i = p_f \) i.e., \( m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \)
- Momentum from Kinetic Energy: \( p = \sqrt{2mKE} \)
- Angular Momentum: \( L = I\omega \)
- De Broglie Momentum (NEET): \( p = \frac{h}{\lambda} \)
What is Momentum Formula?
The Momentum Formula defines momentum as the product of an object’s mass and its velocity. In physics, momentum measures how difficult it is to stop a moving object. A heavier object moving at the same speed as a lighter object has greater momentum. Similarly, a faster object has more momentum than a slower one of equal mass.
Momentum is a vector quantity. This means it has both magnitude and direction. The direction of momentum is always the same as the direction of velocity. Its SI unit is kg·m/s or equivalently N·s (Newton-second).
In the NCERT curriculum, momentum is introduced in Class 9 Science, Chapter 9 — Force and Laws of Motion. It is revisited and expanded in Class 11 Physics, Chapter 5 — Laws of Motion. The concept underpins Newton’s Second and Third Laws. It is also the foundation of the Law of Conservation of Momentum, which is one of the most tested concepts in CBSE board exams and JEE/NEET.
The dimensional formula for momentum is [MLT¹], derived from mass [M] and velocity [LT¹].
Momentum Formula — Expression and Variables
The standard expression for linear momentum is:
\[ p = mv \]
In its Newton’s Second Law form, momentum relates to force and time as:
\[ F = \frac{\Delta p}{\Delta t} = \frac{m(v – u)}{t} \]
| 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) |
| F | Net Force applied | Newton (N) |
| Δp | Change in Momentum (Impulse) | kg·m/s (or N·s) |
| Δt | Time interval | second (s) |
| u | Initial velocity | m/s |
Derivation of the Momentum Formula
Newton’s Second Law states that the rate of change of momentum equals the net force acting on a body.
Step 1: An object of mass m moves with initial velocity u. A force F acts on it for time t. Its velocity changes to v.
Step 2: Initial momentum: \( p_i = mu \). Final momentum: \( p_f = mv \).
Step 3: Change in momentum: \( \Delta p = mv – mu = m(v – u) \).
Step 4: By Newton’s Second Law: \( F = \frac{\Delta p}{\Delta t} = \frac{m(v-u)}{t} \).
Step 5: Since acceleration \( a = \frac{v-u}{t} \), this simplifies to \( F = ma \), confirming the derivation. The momentum itself is simply \( p = mv \).
Complete Physics Formula Sheet — Momentum and Related Concepts
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| Linear Momentum | \( p = mv \) | m = mass, v = velocity | 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 | \( J = F \cdot \Delta t \) | F = force, Δt = time interval | N·s | Class 11, Ch 5 |
| Impulse-Momentum Theorem | \( J = \Delta p = m(v – u) \) | m = mass, v = final velocity, u = initial velocity | kg·m/s | Class 11, Ch 5 |
| Conservation of Momentum | \( m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \) | m = masses, u = initial velocities, v = final velocities | kg·m/s | Class 9, Ch 9 / Class 11, Ch 5 |
| Momentum from Kinetic Energy | \( p = \sqrt{2mKE} \) | m = mass, KE = kinetic energy | kg·m/s | Class 11, Ch 6 |
| Angular Momentum | \( L = I\omega \) | I = moment of inertia, ω = angular velocity | kg·m²/s | Class 11, Ch 7 |
| Angular Momentum (particle) | \( L = mvr \) | m = mass, v = velocity, r = radius | kg·m²/s | Class 11, Ch 7 |
| De Broglie Wavelength (NEET/JEE) | \( \lambda = \frac{h}{p} = \frac{h}{mv} \) | h = Planck’s constant, p = momentum | m | Class 12, Ch 11 |
| Recoil Velocity | \( v_{recoil} = -\frac{m_{bullet} \cdot v_{bullet}}{M_{gun}} \) | m = bullet mass, v = bullet velocity, M = gun mass | m/s | Class 9, Ch 9 |
Momentum Formula — Solved Examples
Example 1 (Class 9-10 Level) — Basic Momentum Calculation
Problem: A cricket ball of mass 0.15 kg is moving with a velocity of 40 m/s. Calculate its momentum.
Given: Mass, m = 0.15 kg; Velocity, v = 40 m/s
Step 1: Write the momentum formula: \( p = mv \)
Step 2: Substitute the given values: \( p = 0.15 \times 40 \)
Step 3: Calculate: \( p = 6 \) kg·m/s
Answer
The momentum of the cricket ball is 6 kg·m/s in the direction of motion.
Example 2 (Class 11-12 Level) — Conservation of Momentum
Problem: A gun of mass 3 kg fires a bullet of mass 0.03 kg with a muzzle velocity of 500 m/s. Both the gun and bullet are initially at rest. Find the recoil velocity of the gun.
Given: Mass of gun, M = 3 kg; Mass of bullet, m = 0.03 kg; Initial velocity of both = 0; Velocity of bullet after firing, v = 500 m/s
Step 1: Apply the Law of Conservation of Momentum. Total initial momentum = 0 (system at rest).
Step 2: Write the conservation equation: \( M \cdot V + m \cdot v = 0 \)
Step 3: Solve for recoil velocity V of the gun: \( V = -\frac{m \cdot v}{M} \)
Step 4: Substitute values: \( V = -\frac{0.03 \times 500}{3} = -\frac{15}{3} = -5 \) m/s
Step 5: The negative sign indicates the gun recoils in the direction opposite to the bullet’s motion.
Answer
The recoil velocity of the gun is 5 m/s in the direction opposite to the bullet’s motion.
Example 3 (JEE/NEET Level) — Momentum and Kinetic Energy Relationship
Problem: Two objects A and B have masses 4 kg and 1 kg respectively. They have equal kinetic energies. Find the ratio of their momenta \( p_A : p_B \).
Given: \( m_A = 4 \) kg; \( m_B = 1 \) kg; \( KE_A = KE_B = KE \)
Step 1: Use the relation between momentum and kinetic energy: \( p = \sqrt{2mKE} \)
Step 2: Write momentum for object A: \( p_A = \sqrt{2 m_A KE} = \sqrt{2 \times 4 \times KE} = \sqrt{8 \cdot KE} \)
Step 3: Write momentum for object B: \( p_B = \sqrt{2 m_B KE} = \sqrt{2 \times 1 \times KE} = \sqrt{2 \cdot KE} \)
Step 4: Find the ratio: \( \frac{p_A}{p_B} = \frac{\sqrt{8 \cdot KE}}{\sqrt{2 \cdot KE}} = \sqrt{\frac{8}{2}} = \sqrt{4} = 2 \)
Step 5: Therefore, \( p_A : p_B = 2 : 1 \).
Answer
The ratio of momenta is \( p_A : p_B = \mathbf{2 : 1} \). Objects with equal kinetic energy do NOT have equal momenta. The heavier object has greater momentum.
CBSE Exam Tips 2025-26
- State the formula first: Always write \( p = mv \) before substituting values. CBSE examiners award one mark for the correct formula statement alone.
- Mention direction: Momentum is a vector. Always state the direction of momentum in your answer (e.g., “in the direction of motion”). Missing direction can cost half a mark.
- Use correct SI units: Write kg·m/s or N·s. Both are acceptable. Avoid writing just “units” without specifying them.
- Conservation problems — write both sides: For conservation of momentum problems, explicitly write the initial momentum equals final momentum equation. This earns full method marks even if arithmetic is wrong.
- Dimensional analysis: We recommend verifying your answer using dimensional analysis. The dimension of momentum is [MLT¹]. This quick check can prevent silly errors.
- Negative sign in recoil: In gun-bullet or explosion problems, always explain what the negative sign means. Write: “The negative sign indicates the direction is opposite to the direction of the bullet/fragment.”
Common Mistakes to Avoid
- Mistake 1 — Confusing momentum with velocity: Momentum depends on both mass AND velocity. A slow-moving truck can have far greater momentum than a fast-moving tennis ball. Always multiply both quantities.
- Mistake 2 — Ignoring the vector nature: Students often add momenta algebraically without considering direction. In head-on collisions, one momentum must be taken as negative. Use a sign convention and stick to it.
- Mistake 3 — Applying conservation of momentum when external forces act: The Law of Conservation of Momentum holds ONLY when the net external force on the system is zero. Do not apply it blindly when friction or gravity acts externally.
- Mistake 4 — Confusing impulse with force: Impulse equals \( F \times \Delta t \), not just F. A large force acting for a very short time can produce the same impulse (and same change in momentum) as a small force acting for a longer time.
- Mistake 5 — Wrong formula for momentum from KE: Many students write \( p = \frac{KE}{v} \) instead of using \( p = \sqrt{2mKE} \). The second form is far more useful when velocity is not directly given.
JEE/NEET Application of Momentum Formula
In our experience, JEE aspirants encounter the Momentum Formula in at least 2–3 questions per paper, spanning mechanics, modern physics, and wave-particle duality. Here are the key application patterns to master.
Pattern 1 — Collision Problems (JEE Main & NEET)
Both elastic and inelastic collision problems use conservation of momentum as the primary equation. In an elastic collision, both momentum and kinetic energy are conserved. In a perfectly inelastic collision, only momentum is conserved. The combined formula for elastic collisions gives:
\[ v_1 = \frac{(m_1 – m_2)u_1 + 2m_2 u_2}{m_1 + m_2}, \quad v_2 = \frac{(m_2 – m_1)u_2 + 2m_1 u_1}{m_1 + m_2} \]
JEE Advanced frequently tests the coefficient of restitution alongside momentum conservation. Mastering both together is essential.
Pattern 2 — Impulse and Variable Force (JEE Advanced)
When force varies with time, impulse is found by integrating: \( J = \int F \, dt \). This equals the change in momentum. JEE Advanced questions often provide a force-time graph and ask for the change in momentum. The area under the F-t graph equals the impulse.
Pattern 3 — De Broglie Wavelength (NEET & JEE)
In modern physics, the Momentum Formula connects to quantum mechanics through the de Broglie relation: \( \lambda = \frac{h}{p} = \frac{h}{mv} \). NEET regularly asks students to calculate the de Broglie wavelength of electrons, protons, or alpha particles. A particle accelerated through a potential difference V gains kinetic energy \( eV = \frac{p^2}{2m} \), giving \( p = \sqrt{2meV} \). This is a high-frequency NEET question type. Our experts suggest practising at least 10 such problems before the exam.
Pattern 4 — Rocket Propulsion (JEE Advanced)
Rocket propulsion is a direct application of conservation of momentum. As exhaust gases are ejected backward, the rocket accelerates forward. The thrust force on a rocket is given by \( F = v_{rel} \frac{dm}{dt} \), where \( v_{rel} \) is the exhaust velocity relative to the rocket and \( \frac{dm}{dt} \) is the rate of mass ejection. This is a classic JEE Advanced topic derived entirely from the Momentum Formula.
FAQs on Momentum Formula
We hope this comprehensive guide on the Momentum Formula has helped you build a strong foundation. For related physics formulas, explore our detailed articles on the Spring Constant Formula, the Angular Displacement Formula, and the Flow Rate Formula. For the complete list of physics formulas, visit our Physics Formulas hub. You can also refer to the official NCERT website to download the Class 9 and Class 11 Physics textbooks for further reading.