The Impulse Formula is expressed as \( J = F imes t \), where J is impulse, F is the applied force, and t is the time interval over which the force acts. This formula is a core concept in Class 11 Physics (NCERT Chapter 5 — Laws of Motion) and appears regularly in CBSE board exams, JEE Main, and NEET. In this article, we cover the complete derivation, a full physics formula sheet, three progressive solved examples, CBSE exam tips for 2025-26, and JEE/NEET application patterns.
Key Impulse Formulas at a Glance
Quick reference for the most important impulse and momentum formulas.
- Basic Impulse: \( J = F imes t \)
- Impulse-Momentum Theorem: \( J = \Delta p = p_f – p_i \)
- Impulse with variable force: \( J = \int F \, dt \)
- Momentum: \( p = mv \)
- Change in momentum: \( \Delta p = m(v – u) \)
- Newton’s Second Law (impulse form): \( F = rac{\Delta p}{\Delta t} \)
- SI unit of Impulse: Newton-second (N·s) or kg·m/s
What is the Impulse Formula?
The Impulse Formula defines impulse as the product of the net force acting on an object and the time duration for which that force acts. In simpler terms, impulse measures the “push” delivered to an object over a specific time interval. A large force applied for a short time can produce the same impulse as a small force applied for a longer time.
Impulse is a vector quantity. It has both magnitude and direction. Its direction is always the same as the direction of the applied force.
In NCERT Class 11 Physics, Chapter 5 (Laws of Motion), impulse is introduced as a direct consequence of Newton’s Second Law. The concept bridges force, time, and the change in momentum of an object. This connection is formally stated in the Impulse-Momentum Theorem: the impulse acting on a body equals the change in its linear momentum.
Everyday examples include a cricket bat striking a ball, a car airbag slowing a passenger, and a karate chop breaking a wooden board. Each of these involves a force acting over a time interval, which is exactly what the Impulse Formula captures.
Impulse Formula — Expression and Variables
The standard expression for the Impulse Formula is:
\[ J = F \times t \]
The Impulse-Momentum Theorem gives an equivalent and equally important expression:
\[ J = \Delta p = m(v – u) \]
For a variable force, the impulse is computed using integration:
\[ J = \int_{t_1}^{t_2} F \, dt \]
| Symbol | Quantity | SI Unit |
|---|---|---|
| J | Impulse | Newton-second (N·s) or kg·m/s |
| F | Net applied force | Newton (N) |
| t | Time interval | Second (s) |
| Δp | Change in linear momentum | kg·m/s |
| m | Mass of the object | Kilogram (kg) |
| v | Final velocity | m/s |
| u | Initial velocity | m/s |
Derivation of the Impulse Formula
The derivation starts from Newton’s Second Law of Motion. According to Newton’s Second Law, the net force on an object equals the rate of change of its momentum:
\[ F = \frac{\Delta p}{\Delta t} \]
Rearranging this equation by multiplying both sides by \( \Delta t \):
\[ F \times \Delta t = \Delta p \]
The left-hand side is defined as impulse \( J \). Therefore:
\[ J = F \times t = \Delta p = m(v – u) \]
This result is the Impulse-Momentum Theorem. It shows that impulse and change in momentum are physically equivalent quantities with the same SI units.
Complete Physics Formula Sheet — Impulse, Momentum and Related Laws
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| Impulse (constant force) | \( J = F \times t \) | J = Impulse, F = Force, t = time | N·s | Class 11, Ch 5 |
| Impulse-Momentum Theorem | \( J = \Delta p = m(v – u) \) | m = mass, v = final vel., u = initial vel. | kg·m/s | Class 11, Ch 5 |
| Linear Momentum | \( p = mv \) | p = momentum, m = mass, v = velocity | kg·m/s | Class 11, Ch 5 |
| Newton’s Second Law | \( F = ma \) | F = force, m = mass, a = acceleration | N | 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 |
| Conservation of Momentum | \( m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \) | m = mass, u = initial vel., v = final vel. | kg·m/s | Class 11, Ch 5 |
| Impulse (variable force) | \( J = \int_{t_1}^{t_2} F \, dt \) | F = force as a function of time | N·s | Class 11, Ch 5 |
| Average Force from Impulse | \( F_{avg} = \frac{J}{\Delta t} \) | J = impulse, Δt = time interval | N | Class 11, Ch 5 |
| Kinetic Energy | \( KE = \frac{1}{2}mv^2 \) | m = mass, v = velocity | Joule (J) | Class 11, Ch 6 |
| Work-Energy Theorem | \( W = \Delta KE \) | W = work done, KE = kinetic energy | Joule (J) | Class 11, Ch 6 |
Impulse Formula — Solved Examples
Example 1 (Class 9-10 Level — Direct Application)
Problem: A force of 20 N acts on a football for 0.5 seconds. Calculate the impulse delivered to the ball.
Given:
- Force, F = 20 N
- Time, t = 0.5 s
Step 1: Write the Impulse Formula: \( J = F \times t \)
Step 2: Substitute the given values: \( J = 20 \times 0.5 \)
Step 3: Calculate: \( J = 10 \) N·s
Answer
Impulse J = 10 N·s
Example 2 (Class 11-12 Level — Using Impulse-Momentum Theorem)
Problem: A cricket ball of mass 0.15 kg is moving at 30 m/s towards a batsman. The batsman hits it back in the opposite direction at 40 m/s. The ball is in contact with the bat for 0.02 seconds. Find (a) the impulse on the ball and (b) the average force exerted by the bat.
Given:
- Mass, m = 0.15 kg
- Initial velocity, u = −30 m/s (towards batsman, taken as negative)
- Final velocity, v = +40 m/s (away from batsman)
- Contact time, t = 0.02 s
Step 1: Apply the Impulse-Momentum Theorem: \( J = m(v – u) \)
Step 2: Substitute values: \( J = 0.15 imes (40 – (-30)) \)
Step 3: Simplify: \( J = 0.15 imes 70 = 10.5 \) N·s
Step 4: Find average force using \( F_{avg} = rac{J}{\Delta t} \):
\( F_{avg} = rac{10.5}{0.02} = 525 \) N
Answer
(a) Impulse J = 10.5 N·s
(b) Average Force = 525 N
Example 3 (JEE/NEET Level — Variable Force & Concept Application)
Problem: A force acts on a body of mass 2 kg such that it varies with time as \( F = (3t^2 + 2t) \) N, where t is in seconds. The body starts from rest. Find the velocity of the body at t = 3 s using the impulse-momentum approach.
Given:
- Mass, m = 2 kg
- Initial velocity, u = 0 (starts from rest)
- Force, F(t) = \( 3t^2 + 2t \) N
- Time interval: t = 0 to t = 3 s
Step 1: Use the variable force impulse formula: \( J = \int_0^3 F \, dt \)
Step 2: Substitute F(t): \( J = \int_0^3 (3t^2 + 2t) \, dt \)
Step 3: Integrate term by term:
\( J = \left[ t^3 + t^2 \right]_0^3 = (27 + 9) – (0) = 36 \) N·s
Step 4: Apply the Impulse-Momentum Theorem: \( J = m(v – u) \)
\( 36 = 2 \times (v – 0) \)
\( v = \frac{36}{2} = 18 \) m/s
Answer
Velocity at t = 3 s is 18 m/s
CBSE Exam Tips 2025-26
- Always state the theorem: In CBSE 2025-26 exams, writing the Impulse-Momentum Theorem (\( J = \Delta p \)) before solving a problem earns method marks. We recommend writing it explicitly in every answer.
- Sign convention is critical: Always define a positive direction before substituting velocity values. A common error is ignoring the sign when a ball reverses direction. Assign positive and negative directions clearly at the start.
- Units must match: Convert all values to SI units (kg, m/s, s, N) before substituting. Mixed units are a frequent cause of mark deductions in board exams.
- Distinguish impulse from force: Impulse is \( F imes t \), not just F. Many students write force when the question asks for impulse. Read the question carefully.
- Graphical questions: In CBSE Class 11, the area under a Force-Time graph equals the impulse. Our experts suggest practising at least 3 graphical problems before the board exam.
- Link to momentum conservation: Questions on collisions often require impulse as an intermediate step. Practice combining the impulse formula with the law of conservation of momentum for 5-mark problems.
Common Mistakes to Avoid with the Impulse Formula
- Ignoring direction (sign): Impulse is a vector. Students often treat it as a scalar and ignore the direction of the initial and final velocities. Always use a consistent sign convention.
- Confusing impulse with work: Work equals force times displacement (\( W = F \cdot d \)). Impulse equals force times time (\( J = F \cdot t \)). These are completely different physical quantities. Do not mix them up.
- Using wrong time interval: The time ‘t’ in the impulse formula is the duration of contact or the duration for which the force acts. It is not the total time of motion. Identify the correct time interval from the problem.
- Forgetting variable force integration: When the force changes with time, you cannot simply multiply F by t. You must integrate \( F(t) \) over the given time interval. This is a common JEE-level mistake.
- Incorrect unit conversion: N·s and kg·m/s are equivalent units for impulse and momentum. Students sometimes treat them as different quantities. Always verify: 1 N·s = 1 kg·m/s.
JEE/NEET Application of the Impulse Formula
In our experience, JEE aspirants encounter the Impulse Formula in at least 1-2 questions per year, either directly or as part of a collision problem. NEET also tests it in the context of biomechanics and force-time graphs. Here are the three most common application patterns.
Pattern 1: Force-Time Graph (Area = Impulse)
JEE Main frequently presents a Force vs. Time graph and asks for the change in momentum. The key insight is that the area under the F-t curve equals the impulse. For a rectangular graph, \( J = F imes t \). For a triangular graph, \( J = \frac{1}{2} \times base \times height \). Practise calculating areas for different graph shapes.
Pattern 2: Collision Problems (Average Force Calculation)
Both JEE and NEET present collision scenarios where a ball strikes a wall or a bat. The question gives initial and final velocities, mass, and contact time. You must use \( J = m(v – u) \) to find impulse, then divide by contact time to get the average force. This is a standard 4-mark JEE Main question type.
Pattern 3: Variable Force Integration
JEE Advanced tests impulse with a time-dependent force such as \( F(t) = at^2 + bt \). Students must integrate this expression over the given time limits. The result equals the change in momentum. If mass is given, the final velocity can be found. Our experts suggest mastering basic integration of polynomial and trigonometric functions for this pattern.
For NEET, the Impulse Formula appears in the context of injury prevention (airbags, helmets) and sports biomechanics. Understanding why increasing contact time reduces average force (at constant impulse) is a conceptual question that NEET tests directly.
FAQs on Impulse Formula
Explore more related physics formulas on ncertbooks.net. Visit our complete Physics Formulas hub for a full list of Class 11 and Class 12 formulas. You may also find these articles useful: Angular Speed Formula, Average Acceleration Formula, and Electric Flux Formula. For official NCERT syllabus details, refer to the NCERT official website.