The Inelastic Collision Formula describes how two objects collide when kinetic energy is not conserved, expressed as \ ( m_1 u_1 + m_2 u_2 = (m_1 + m_2)v \) for a perfectly inelastic collision. This concept is covered in NCERT Class 11 Physics, Chapter 6 (Work, Energy and Power). It is also a high-weightage topic in JEE Main, JEE Advanced, and NEET. This article covers the formula, its derivation, a complete formula sheet, three solved examples, CBSE exam tips, and JEE/NEET applications.
Key Inelastic Collision Formulas at a Glance
Quick reference for the most important inelastic collision formulas.
- Conservation of Momentum: \( m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \)
- Perfectly Inelastic Collision: \( m_1 u_1 + m_2 u_2 = (m_1 + m_2)v \)
- Final velocity: \( v = \frac{m_1 u_1 + m_2 u_2}{m_1 + m_2} \)
- Kinetic Energy Lost: \( \Delta KE = \frac{1}{2}m_1 u_1^2 + \frac{1}{2}m_2 u_2^2 – \frac{1}{2}(m_1+m_2)v^2 \)
- Coefficient of Restitution: \( e = \frac{v_2 – v_1}{u_1 – u_2} \) (0 < e < 1 for inelastic)
- Energy Lost (compact form): \( \Delta KE = \frac{m_1 m_2 (u_1 – u_2)^2}{2(m_1 + m_2)} \)
What is Inelastic Collision Formula?
The Inelastic Collision Formula is a mathematical expression derived from the Law of Conservation of Momentum. It applies when two objects collide and kinetic energy is not fully conserved. Some energy converts into heat, sound, or deformation during the impact.
In an inelastic collision, total momentum before the collision equals total momentum after the collision. However, total kinetic energy decreases. This is the key difference from an elastic collision, where both momentum and kinetic energy are conserved.
A perfectly inelastic collision is a special case. Here, the two objects stick together after impact. They move as one combined mass. This gives the simplest form of the inelastic collision formula.
This topic is covered in NCERT Class 11 Physics, Chapter 6 — Work, Energy and Power. Students encounter it again in Class 12 revision and in competitive exam preparation. Understanding this formula is essential for scoring well in CBSE board exams and entrance tests like JEE and NEET.
Inelastic Collision Formula — Expression and Variables
For a perfectly inelastic collision, the two objects move together after impact. The formula is:
\[ m_1 u_1 + m_2 u_2 = (m_1 + m_2)v \]
Rearranging to find the final velocity:
\[ v = \frac{m_1 u_1 + m_2 u_2}{m_1 + m_2} \]
For a general inelastic collision (objects do not stick together), momentum conservation gives:
\[ m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \]
The kinetic energy lost in a perfectly inelastic collision is:
\[ \Delta KE = \frac{m_1 m_2 (u_1 – u_2)^2}{2(m_1 + m_2)} \]
| Symbol | Quantity | SI Unit |
|---|---|---|
| \( m_1 \) | Mass of object 1 | kilogram (kg) |
| \( m_2 \) | Mass of object 2 | kilogram (kg) |
| \( u_1 \) | Initial velocity of object 1 | metre per second (m/s) |
| \( u_2 \) | Initial velocity of object 2 | metre per second (m/s) |
| \( v \) | Final common velocity (perfectly inelastic) | metre per second (m/s) |
| \( v_1 \) | Final velocity of object 1 (general case) | metre per second (m/s) |
| \( v_2 \) | Final velocity of object 2 (general case) | metre per second (m/s) |
| \( \Delta KE \) | Kinetic energy lost | Joule (J) |
| \( e \) | Coefficient of restitution | Dimensionless |
Derivation of the Inelastic Collision Formula
The derivation uses Newton’s Third Law and the Law of Conservation of Momentum.
Step 1: Consider two objects of masses \( m_1 \) and \( m_2 \) moving with velocities \( u_1 \) and \( u_2 \) before collision.
Step 2: By Newton’s Third Law, the force on object 1 from object 2 is equal and opposite to the force on object 2 from object 1. The collision time is the same for both objects.
Step 3: Total momentum before collision: \( p_i = m_1 u_1 + m_2 u_2 \)
Step 4: For a perfectly inelastic collision, both objects move with the same final velocity \( v \). Total momentum after: \( p_f = (m_1 + m_2)v \)
Step 5: Setting \( p_i = p_f \): \( m_1 u_1 + m_2 u_2 = (m_1 + m_2)v \). Solving for \( v \) gives the inelastic collision formula.
Complete Physics Collision Formula Sheet
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| Perfectly Inelastic Collision (Final Velocity) | \( v = \frac{m_1 u_1 + m_2 u_2}{m_1 + m_2} \) | m = mass, u = initial velocity, v = final velocity | m/s | Class 11, Ch 6 |
| Conservation of Momentum (General) | \( m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \) | m = mass, u = initial velocity, v = final velocity | kg·m/s | Class 11, Ch 6 |
| Kinetic Energy Lost (Perfectly Inelastic) | \( \Delta KE = \frac{m_1 m_2 (u_1-u_2)^2}{2(m_1+m_2)} \) | m = mass, u = initial velocities | Joule (J) | Class 11, Ch 6 |
| Coefficient of Restitution | \( e = \frac{v_2 – v_1}{u_1 – u_2} \) | u = initial velocity, v = final velocity | Dimensionless | Class 11, Ch 6 |
| Elastic Collision — Final Velocity of Object 1 | \( v_1 = \frac{(m_1-m_2)u_1 + 2m_2 u_2}{m_1+m_2} \) | m = mass, u = initial velocity | m/s | Class 11, Ch 6 |
| Elastic Collision — Final Velocity of Object 2 | \( v_2 = \frac{(m_2-m_1)u_2 + 2m_1 u_1}{m_1+m_2} \) | m = mass, u = initial velocity | m/s | Class 11, Ch 6 |
| Linear Momentum | \( p = mv \) | m = mass, v = velocity | kg·m/s | Class 11, Ch 5 |
| Kinetic Energy | \( KE = \frac{1}{2}mv^2 \) | m = mass, v = velocity | Joule (J) | Class 11, Ch 6 |
| Impulse | \( J = F \cdot \Delta t = \Delta p \) | F = force, t = time, p = momentum | N·s | Class 11, Ch 5 |
| Energy Lost as Fraction | \( \frac{\Delta KE}{KE_i} = \frac{m_2}{m_1+m_2}\left(1 – \frac{u_2}{u_1}\right)^2 \) (when \( u_2=0 \): \( \frac{m_2}{m_1+m_2} \)) | m = mass, u = initial velocity | Dimensionless | Class 11, Ch 6 |
Inelastic Collision Formula — Solved Examples
Example 1 (Class 9-10 Level — Perfectly Inelastic Collision)
Problem: A ball of mass 3 kg moving at 4 m/s collides with a stationary ball of mass 1 kg. They stick together after the collision. Find the final velocity of the combined system.
Given: \( m_1 = 3 \) kg, \( u_1 = 4 \) m/s, \( m_2 = 1 \) kg, \( u_2 = 0 \) m/s
Step 1: Write the perfectly inelastic collision formula:
\[ v = \frac{m_1 u_1 + m_2 u_2}{m_1 + m_2} \]
Step 2: Substitute the given values:
\[ v = \frac{(3)(4) + (1)(0)}{3 + 1} = \frac{12}{4} = 3 \text{ m/s} \]
Answer
The final velocity of the combined system is 3 m/s in the original direction of motion.
Example 2 (Class 11-12 Level — Energy Loss Calculation)
Problem: A truck of mass 2000 kg moving at 10 m/s collides with a car of mass 500 kg moving at 2 m/s in the same direction. They stick together after the collision. Calculate (a) the final velocity and (b) the kinetic energy lost in the collision.
Given: \( m_1 = 2000 \) kg, \( u_1 = 10 \) m/s, \( m_2 = 500 \) kg, \( u_2 = 2 \) m/s
Step 1: Apply the perfectly inelastic collision formula to find \( v \):
\[ v = \frac{m_1 u_1 + m_2 u_2}{m_1 + m_2} = \frac{(2000)(10) + (500)(2)}{2000 + 500} \]
\[ v = \frac{20000 + 1000}{2500} = \frac{21000}{2500} = 8.4 \text{ m/s} \]
Step 2: Calculate initial kinetic energy:
\[ KE_i = \frac{1}{2}(2000)(10)^2 + \frac{1}{2}(500)(2)^2 = 100000 + 1000 = 101000 \text{ J} \]
Step 3: Calculate final kinetic energy:
\[ KE_f = \frac{1}{2}(2500)(8.4)^2 = \frac{1}{2}(2500)(70.56) = 88200 \text{ J} \]
Step 4: Calculate kinetic energy lost:
\[ \Delta KE = KE_i – KE_f = 101000 – 88200 = 12800 \text{ J} \]
Answer
(a) Final velocity = 8.4 m/s. (b) Kinetic energy lost = 12800 J = 12.8 kJ.
Example 3 (JEE/NEET Level — Ballistic Pendulum Concept)
Problem: A bullet of mass 20 g is fired horizontally at 300 m/s into a wooden block of mass 980 g suspended by a string. The bullet embeds itself in the block. Find (a) the velocity of the block-bullet system just after impact, and (b) the height to which the system rises. (Take \( g = 10 \) m/s²)
Given: \( m_1 = 0.02 \) kg, \( u_1 = 300 \) m/s, \( m_2 = 0.98 \) kg, \( u_2 = 0 \) m/s
Step 1: Apply the inelastic collision formula for the combined velocity:
\[ v = \frac{m_1 u_1 + m_2 u_2}{m_1 + m_2} = \frac{(0.02)(300) + 0}{0.02 + 0.98} = \frac{6}{1} = 6 \text{ m/s} \]
Step 2: After the collision, use conservation of energy to find the height \( h \). The kinetic energy of the system converts to potential energy:
\[ \frac{1}{2}(m_1 + m_2)v^2 = (m_1 + m_2)gh \]
Step 3: Simplify and solve for \( h \):
\[ h = \frac{v^2}{2g} = \frac{(6)^2}{2 \times 10} = \frac{36}{20} = 1.8 \text{ m} \]
Step 4: Verify energy loss. Initial KE of bullet: \( \frac{1}{2}(0.02)(300)^2 = 900 \) J. Final KE: \( \frac{1}{2}(1)(6)^2 = 18 \) J. Energy lost = 882 J. This confirms the collision is highly inelastic.
Answer
(a) Velocity of block-bullet system = 6 m/s. (b) Height risen = 1.8 m. This is the classic ballistic pendulum problem frequently asked in JEE and NEET.
CBSE Exam Tips 2025-26
- Always state momentum conservation first. CBSE examiners award marks for writing the principle before applying the formula. Write: “By the Law of Conservation of Momentum…”
- Distinguish between types of collisions. In 2025-26 exams, one-mark questions often ask: “Is kinetic energy conserved in an inelastic collision?” The answer is always No.
- Use the compact energy-loss formula. We recommend memorising \( \Delta KE = \frac{m_1 m_2 (u_1-u_2)^2}{2(m_1+m_2)} \). It saves calculation time in board exams.
- Check direction of velocities. Always assign a positive direction before substituting. Assign “rightward” as positive. Objects moving left get negative velocity values.
- The coefficient of restitution \( e = 0 \) for perfectly inelastic collisions. This fact appears in short-answer questions. For elastic collisions, \( e = 1 \).
- Draw a before-and-after diagram. Our experts suggest sketching the collision scenario. It helps avoid sign errors and earns presentation marks in CBSE.
Common Mistakes to Avoid
Students frequently lose marks due to these errors. Study them carefully before your exam.
- Mistake 1: Confusing momentum and kinetic energy conservation. Momentum is always conserved in any collision (elastic or inelastic). Kinetic energy is only conserved in elastic collisions. Never write that kinetic energy is conserved in an inelastic collision.
- Mistake 2: Forgetting to use the combined mass. In a perfectly inelastic collision, the final mass is \( m_1 + m_2 \). Students sometimes use only \( m_1 \) or \( m_2 \) in the denominator. Always add both masses.
- Mistake 3: Ignoring the sign of velocity. If two objects move in opposite directions before collision, one velocity must be negative. Failing to assign signs leads to wrong final velocity values.
- Mistake 4: Assuming all collisions in real life are elastic. Real-world collisions (car crashes, clay balls sticking together) are inelastic. Elastic collisions are an idealisation, mostly seen in atomic or subatomic interactions.
- Mistake 5: Applying the perfectly inelastic formula to a general inelastic collision. The formula \( v = \frac{m_1 u_1 + m_2 u_2}{m_1+m_2} \) only applies when objects stick together. For other inelastic collisions, use the general momentum conservation equation along with the coefficient of restitution.
JEE/NEET Application of Inelastic Collision Formula
In our experience, JEE aspirants encounter the inelastic collision formula in at least one question per paper. NEET also tests this concept in the context of biomechanics and particle physics. Here are the three most common application patterns.
Pattern 1: Ballistic Pendulum
A bullet embeds in a suspended block. The block swings upward. This is a two-stage problem. Stage 1 uses the inelastic collision formula to find the velocity after impact. Stage 2 uses energy conservation to find the height. JEE Main 2019, 2021, and 2023 all featured this pattern. Always apply momentum conservation only during the collision phase. Apply energy conservation only after the collision.
Pattern 2: Fraction of Energy Lost
JEE questions often ask for the fraction of kinetic energy lost. The formula is:
\[ \text{Fraction lost} = \frac{\Delta KE}{KE_i} = \frac{m_2}{m_1 + m_2} \quad \text{(when } u_2 = 0\text{)} \]
This result is elegant. When a moving object hits a stationary one and they stick together, the fraction of energy lost depends only on the mass ratio. In our experience, this formula appears as a one-step MCQ in JEE Main. Memorising it directly saves 2-3 minutes per paper.
Pattern 3: Two-Dimensional Inelastic Collision
JEE Advanced sometimes presents collisions in 2D. Here, momentum is conserved separately along the x-axis and y-axis. Write two equations:
\[ m_1 u_{1x} + m_2 u_{2x} = (m_1 + m_2) v_x \]
\[ m_1 u_{1y} + m_2 u_{2y} = (m_1 + m_2) v_y \]
Then find the magnitude: \( v = \sqrt{v_x^2 + v_y^2} \). Practice resolving velocities into components before attempting 2D collision problems. NEET does not test 2D collisions, but JEE Advanced does.
FAQs on Inelastic Collision Formula
For more Physics formulas, explore our Complete Physics Formulas Hub. You may also find these related articles useful: Angular Speed Formula, Average Acceleration Formula, and Electric Flux Formula. For the official NCERT syllabus, refer to ncert.nic.in.