The Elastic Collision Formula gives the final velocities of two objects after a collision in which both momentum and kinetic energy are conserved. It is a core concept in Class 11 Physics (NCERT Chapter 6 — Work, Energy and Power) and appears regularly in CBSE board exams, JEE Main, JEE Advanced, and NEET. This article covers the complete formula, step-by-step derivation, a comprehensive formula sheet, three solved examples at progressive difficulty levels, CBSE exam tips, common mistakes, and JEE/NEET applications.
Key Elastic Collision Formulas at a Glance
Quick reference for the most important elastic collision formulas.
- Conservation of Momentum: \( m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \)
- Conservation of Kinetic Energy: \( \frac{1}{2}m_1 u_1^2 + \frac{1}{2}m_2 u_2^2 = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2 \)
- Final velocity of object 1: \( v_1 = \frac{(m_1 – m_2)u_1 + 2m_2 u_2}{m_1 + m_2} \)
- Final velocity of object 2: \( v_2 = \frac{(m_2 – m_1)u_2 + 2m_1 u_1}{m_1 + m_2} \)
- Coefficient of restitution: \( e = 1 \) (for perfectly elastic collision)
- Relative velocity condition: \( u_1 – u_2 = v_2 – v_1 \)
- Special case (equal masses): \( v_1 = u_2,\; v_2 = u_1 \)
What is the Elastic Collision Formula?
The Elastic Collision Formula describes the outcome of a collision between two objects in which the total kinetic energy of the system is conserved, along with the total linear momentum. In an elastic collision, no energy is lost to heat, sound, or deformation. This is an idealised model that closely approximates collisions between atomic and subatomic particles, as well as collisions between hard billiard balls.
According to NCERT Class 11 Physics, Chapter 6 (Work, Energy and Power), a collision is called elastic if the kinetic energy before and after the event remains unchanged. The two governing laws are Newton's Law of Conservation of Momentum and the Law of Conservation of Kinetic Energy. Together, these two conditions allow us to solve for the two unknown final velocities of the colliding bodies.
The Elastic Collision Formula is distinct from the inelastic collision formula, where kinetic energy is not conserved. Understanding this distinction is critical for scoring well in both CBSE Class 11 examinations and competitive entrance tests such as JEE and NEET.
Elastic Collision Formula — Expression and Variables
For two objects of masses \( m_1 \) and \( m_2 \) moving with initial velocities \( u_1 \) and \( u_2 \) along the same straight line, the final velocities after an elastic collision are:
\[ v_1 = \frac{(m_1 – m_2)\,u_1 + 2m_2\,u_2}{m_1 + m_2} \]
\[ v_2 = \frac{(m_2 – m_1)\,u_2 + 2m_1\,u_1}{m_1 + m_2} \]
These expressions come from solving the momentum and kinetic energy equations simultaneously.
| 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) |
| \( v_1 \) | Final velocity of object 1 | metre per second (m/s) |
| \( v_2 \) | Final velocity of object 2 | metre per second (m/s) |
| \( e \) | Coefficient of restitution | Dimensionless (= 1 for elastic) |
Derivation of the Elastic Collision Formula
Step 1 — Write the momentum conservation equation:
\[ m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \quad \cdots (1) \]
Step 2 — Write the kinetic energy conservation equation:
\[ \frac{1}{2}m_1 u_1^2 + \frac{1}{2}m_2 u_2^2 = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2 \quad \cdots (2) \]
Step 3 — Rearrange equation (1): \( m_1(u_1 – v_1) = m_2(v_2 – u_2) \)
Step 4 — Rearrange equation (2): \( m_1(u_1^2 – v_1^2) = m_2(v_2^2 – u_2^2) \), which factors to \( m_1(u_1 – v_1)(u_1 + v_1) = m_2(v_2 – u_2)(v_2 + u_2) \)
Step 5 — Divide Step 4 by Step 3: This gives \( u_1 + v_1 = v_2 + u_2 \), i.e., \( u_1 – u_2 = v_2 – v_1 \). This is the relative velocity condition.
Step 6 — Substitute back into equation (1) and solve simultaneously to obtain the final velocity expressions shown above.
Complete Physics Collision Formula Sheet
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| Momentum Conservation | \( m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \) | m = mass, u = initial velocity, v = final velocity | kg·m/s | Class 11, Ch 5 & 6 |
| Elastic Collision — Final velocity of body 1 | \( v_1 = \frac{(m_1 – m_2)u_1 + 2m_2 u_2}{m_1 + m_2} \) | m = mass, u = initial velocity | m/s | Class 11, Ch 6 |
| Elastic Collision — Final velocity of body 2 | \( v_2 = \frac{(m_2 – m_1)u_2 + 2m_1 u_1}{m_1 + m_2} \) | m = mass, u = initial velocity | m/s | Class 11, Ch 6 |
| Relative Velocity Condition | \( u_1 – u_2 = v_2 – v_1 \) | u = initial velocity, v = final velocity | m/s | Class 11, Ch 6 |
| Coefficient of Restitution | \( e = \frac{v_2 – v_1}{u_1 – u_2} \) | e = 1 (elastic), 0 < e < 1 (inelastic), e = 0 (perfectly inelastic) | Dimensionless | Class 11, Ch 6 |
| Equal Mass Elastic Collision | \( v_1 = u_2,\; v_2 = u_1 \) | Velocities exchange when \( m_1 = m_2 \) | m/s | Class 11, Ch 6 |
| Heavy body hits stationary light body | \( v_1 \approx u_1,\; v_2 \approx 2u_1 \) | When \( m_1 \gg m_2 \), \( u_2 = 0 \) | m/s | Class 11, Ch 6 |
| Light body hits stationary heavy body | \( v_1 \approx -u_1,\; v_2 \approx 0 \) | When \( m_1 \ll m_2 \), \( u_2 = 0 \) | m/s | Class 11, Ch 6 |
| Kinetic Energy (general) | \( KE = \frac{1}{2}mv^2 \) | m = mass, v = velocity | Joule (J) | Class 11, Ch 6 |
| Perfectly Inelastic Collision (common velocity) | \( v = \frac{m_1 u_1 + m_2 u_2}{m_1 + m_2} \) | Bodies stick together after collision | m/s | Class 11, Ch 6 |
Elastic Collision Formula — Solved Examples
Example 1 (Class 9-10 Level) — Equal Mass Elastic Collision
Problem: A ball of mass 2 kg moving at 6 m/s collides head-on elastically with another ball of mass 2 kg at rest. Find the final velocities of both balls.
Given: \( m_1 = 2\,\text{kg},\; m_2 = 2\,\text{kg},\; u_1 = 6\,\text{m/s},\; u_2 = 0 \)
Step 1: Since the masses are equal, apply the special case formula: \( v_1 = u_2 \) and \( v_2 = u_1 \).
Step 2: Substitute values: \( v_1 = 0\,\text{m/s} \) and \( v_2 = 6\,\text{m/s} \).
Step 3: Verify momentum: \( 2 \times 6 + 2 \times 0 = 2 \times 0 + 2 \times 6 = 12\,\text{kg·m/s} \). ✓
Step 4: Verify kinetic energy: \( \frac{1}{2}(2)(6^2) = 36\,\text{J} \) before; \( \frac{1}{2}(2)(6^2) = 36\,\text{J} \) after. ✓
Answer
The first ball comes to rest: \( v_1 = 0\,\text{m/s} \). The second ball moves forward at \( v_2 = 6\,\text{m/s} \). The velocities are exchanged — a hallmark of equal-mass elastic collisions.
Example 2 (Class 11-12 Level) — Unequal Masses
Problem: A 4 kg object moving at 8 m/s collides elastically with a 2 kg object moving at 2 m/s in the same direction. Calculate the final velocities of both objects.
Given: \( m_1 = 4\,\text{kg},\; m_2 = 2\,\text{kg},\; u_1 = 8\,\text{m/s},\; u_2 = 2\,\text{m/s} \)
Step 1: Write the formula for \( v_1 \):
\[ v_1 = \frac{(m_1 – m_2)u_1 + 2m_2 u_2}{m_1 + m_2} \]
Step 2: Substitute values into \( v_1 \):
\[ v_1 = \frac{(4 – 2)(8) + 2(2)(2)}{4 + 2} = \frac{16 + 8}{6} = \frac{24}{6} = 4\,\text{m/s} \]
Step 3: Write the formula for \( v_2 \):
\[ v_2 = \frac{(m_2 – m_1)u_2 + 2m_1 u_1}{m_1 + m_2} \]
Step 4: Substitute values into \( v_2 \):
\[ v_2 = \frac{(2 – 4)(2) + 2(4)(8)}{4 + 2} = \frac{-4 + 64}{6} = \frac{60}{6} = 10\,\text{m/s} \]
Step 5: Verify momentum: \( 4(8) + 2(2) = 36\,\text{kg·m/s} \); \( 4(4) + 2(10) = 36\,\text{kg·m/s} \). ✓
Step 6: Verify kinetic energy: Before = \( \frac{1}{2}(4)(64) + \frac{1}{2}(2)(4) = 128 + 4 = 132\,\text{J} \); After = \( \frac{1}{2}(4)(16) + \frac{1}{2}(2)(100) = 32 + 100 = 132\,\text{J} \). ✓
Answer
Final velocity of object 1: \( v_1 = 4\,\text{m/s} \). Final velocity of object 2: \( v_2 = 10\,\text{m/s} \). Both momentum and kinetic energy are conserved, confirming the collision is elastic.
Example 3 (JEE/NEET Level) — Head-On Elastic Collision with Opposite Velocities
Problem: A 3 kg ball moving at 10 m/s to the right collides head-on elastically with a 6 kg ball moving at 4 m/s to the left. Find the final velocities and the kinetic energy transferred to the heavier ball.
Given: \( m_1 = 3\,\text{kg},\; m_2 = 6\,\text{kg},\; u_1 = +10\,\text{m/s},\; u_2 = -4\,\text{m/s} \) (taking right as positive)
Step 1: Apply the elastic collision formula for \( v_1 \):
\[ v_1 = \frac{(3 – 6)(10) + 2(6)(-4)}{3 + 6} = \frac{-30 – 48}{9} = \frac{-78}{9} = -8.67\,\text{m/s} \]
Step 2: Apply the formula for \( v_2 \):
\[ v_2 = \frac{(6 – 3)(-4) + 2(3)(10)}{3 + 6} = \frac{-12 + 60}{9} = \frac{48}{9} = 5.33\,\text{m/s} \]
Step 3: Calculate initial KE of the heavier ball: \( KE_{2i} = \frac{1}{2}(6)(4^2) = 48\,\text{J} \)
Step 4: Calculate final KE of the heavier ball: \( KE_{2f} = \frac{1}{2}(6)(5.33^2) = \frac{1}{2}(6)(28.41) \approx 85.2\,\text{J} \)
Step 5: Kinetic energy transferred to the heavier ball: \( \Delta KE_2 = 85.2 – 48 = 37.2\,\text{J} \)
Step 6: Verify total momentum: Before = \( 3(10) + 6(-4) = 30 – 24 = 6\,\text{kg·m/s} \); After = \( 3(-8.67) + 6(5.33) = -26 + 32 = 6\,\text{kg·m/s} \). ✓
Answer
Final velocity of the 3 kg ball: \( v_1 \approx -8.67\,\text{m/s} \) (bounces back to the left). Final velocity of the 6 kg ball: \( v_2 \approx 5.33\,\text{m/s} \) (moves to the right). Approximately 37.2 J of kinetic energy is transferred to the heavier ball during the collision.
CBSE Exam Tips 2025-26
- Always assign signs to velocities. Choose a positive direction before substituting. Failing to do so is the single most common reason for wrong answers in collision problems.
- State both conservation laws explicitly. In CBSE board exams, you earn marks for writing the momentum equation and the kinetic energy equation separately before solving. Never skip these steps.
- Memorise the special cases. Equal masses exchanging velocities, and the limiting cases for \( m_1 \gg m_2 \) and \( m_1 \ll m_2 \), are asked directly in short-answer questions. We recommend making a small flashcard for these.
- Verify your answer. After finding \( v_1 \) and \( v_2 \), always substitute back and check that momentum and kinetic energy are conserved. This takes 30 seconds and can save 2–3 marks.
- Use the relative velocity shortcut. The condition \( u_1 – u_2 = v_2 – v_1 \) is faster than solving two simultaneous equations. Use it whenever you already know one final velocity.
- Watch units. Ensure masses are in kg and velocities in m/s. If the problem gives velocities in km/h, convert to m/s first. This is a common source of unit errors in 2025-26 exams.
Common Mistakes to Avoid
- Ignoring the sign of velocity. Many students treat all velocities as positive magnitudes. If two objects move in opposite directions, one velocity must be negative. Always define a positive direction at the start of your solution.
- Confusing elastic with inelastic collisions. In an elastic collision, kinetic energy is conserved. In an inelastic collision, it is not. Students sometimes apply the elastic formula to a perfectly inelastic scenario (where bodies stick together), leading to completely wrong results.
- Forgetting the coefficient of restitution. The coefficient of restitution \( e = 1 \) for elastic collisions. If a problem states \( e \neq 1 \), the collision is not perfectly elastic. Do not apply the elastic formula blindly.
- Swapping \( m_1 \) and \( m_2 \) in the formula. The formula for \( v_1 \) contains \( (m_1 – m_2) \) in the numerator. Students sometimes write \( (m_2 – m_1) \) instead. Double-check which mass belongs to which object before substituting.
- Not verifying conservation laws. Skipping the verification step means you cannot catch arithmetic errors. A 5-minute verification can prevent losing full marks on a multi-step question.
JEE/NEET Application of the Elastic Collision Formula
In our experience, JEE aspirants encounter the Elastic Collision Formula in at least one question per year in JEE Main, and it appears in JEE Advanced problems involving multi-body systems, oblique collisions, and energy transfer. NEET also tests this concept in the context of nuclear physics (alpha particle scattering) and biomechanics.
Application Pattern 1 — Maximum Energy Transfer
JEE frequently asks: “For what mass ratio is the kinetic energy transferred to the target body maximum?” The answer is \( m_1 = m_2 \). The fraction of kinetic energy transferred is:
\[ \frac{\Delta KE}{KE_i} = \frac{4m_1 m_2}{(m_1 + m_2)^2} \]
This fraction equals 1 (100% transfer) only when \( m_1 = m_2 \). This result is tested both as a derivation and as a direct MCQ.
Application Pattern 2 — Oblique Elastic Collisions
In JEE Advanced, elastic collisions are sometimes extended to two dimensions. When a moving ball strikes an identical stationary ball obliquely, the two balls move at 90° to each other after the collision. This is a direct consequence of momentum conservation and kinetic energy conservation in 2D. Our experts suggest practising this result with vector diagrams.
Application Pattern 3 — Nuclear Scattering (NEET)
In NEET, elastic collision concepts appear in Rutherford's alpha scattering experiment and in nuclear reactions. When an alpha particle (mass \( m \)) collides elastically with a stationary nucleus of mass \( M \), the energy transferred to the nucleus is:
\[ E_{\text{transferred}} = \frac{4mM}{(m+M)^2} \cdot E_0 \]
where \( E_0 \) is the initial kinetic energy of the alpha particle. Recognising this as the elastic collision energy transfer formula is the key to solving such NEET questions quickly.
FAQs on Elastic Collision Formula
For more related Physics formulas, explore our comprehensive guide on the Uniform Circular Motion Formula, the Electric Field Formula, and the Refractive Index Formula. You can also browse our complete Physics Formulas hub for Class 11 and Class 12 reference sheets. For official NCERT textbook content, visit the NCERT official website.