In an elastic collision, both linear momentum AND kinetic energy are conserved. In an inelastic collision, momentum is conserved but kinetic energy is NOT (some K converts to internal energy / heat / deformation). In a perfectly inelastic collision the bodies stick together after collision.
-- NCERT Class 11 Physics, Ch. 5, p. 84Elastic Collision 1D
8 MCQs1 revision card9-step worked example
Source: NCERT Work, Energy and PowerPYQ coverage: NEET 2026Official key: NTA-verifiedLast updated: 21 Sep 2026
Elastic Collision 1D, explained for NEET
Two conditions define an elastic collision, and dropping either one is where marks go. Momentum is conserved in every collision — elastic, inelastic, sticking, exploding. What makes a collision elastic is the second condition: total kinetic energy before equals total kinetic energy after. A question that gives you two masses and two initial velocities and says "elastic" is handing you two equations, not one.
NCERT Class 11 Physics Chapter 5 states the law on page 84: when a collision conserves both momentum and kinetic energy, it is elastic. Solving those two simultaneous equations for a head-on (one-dimensional) collision gives the standard result on page 85:
v₁ = [(m₁ − m₂)u₁ + 2m₂u₂] / (m₁ + m₂) v₂ = [(m₂ − m₁)u₂ + 2m₁u₁] / (m₁ + m₂)
Velocities here are signed. A body moving left in a rightward-positive frame enters as a negative number. Substituting a magnitude where a signed value belongs produces a clean-looking wrong answer, which is exactly what a distractor is built from.
Three limiting cases are worth memorising, because NEET prefers them to brute algebra:
- Equal masses (m₁ = m₂), target at rest: the first body stops dead, the second leaves with u₁. The velocities are exchanged.
- Very heavy target at rest (m₂ ≫ m₁): v₁ → −u₁. The light body rebounds with almost its original speed; the heavy one barely moves.
- Very light target at rest (m₂ ≪ m₁): v₁ → u₁ and v₂ → 2u₁. The struck body leaves at roughly twice the incident speed.
Note what the first case means physically: an elastic collision between equal masses transfers all the kinetic energy. Nothing is lost — that is the definition — but nothing is retained by the striker either.
Watch out: the formulas above hold only when kinetic energy is conserved. If a problem says the bodies stick, or move off with a common velocity, or quotes a coefficient of restitution below 1, these equations do not apply.
Can you answer these Elastic Collision 1D MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
Which pair of quantities is conserved in an elastic collision, according to NCERT Class 11 Physics Chapter 5?
Show answer and why every option is right or wrong
Answer: A. A is correct. Page 84 of NCERT Class 11 Physics Chapter 5 defines an elastic collision as one in which both linear momentum and total kinetic energy are conserved.
Why B is wrong: B is wrong because momentum conservation is not optional — it follows from the absence of a net external force during the collision, and holds regardless of whether the collision is elastic.
Why C is wrong: C is wrong because no potential-energy change is involved in a contact collision between free bodies; the second conserved quantity is kinetic energy.
Why D is wrong: D is wrong because momentum alone is conserved in every collision, elastic or not. Momentum-only conservation is the defining feature of an inelastic collision, not an elastic one.
MCQ 2Easy RecallPractice
In the standard one-dimensional elastic-collision formula on page 85 of NCERT Class 11 Physics Chapter 5, the quantities u₁, u₂, v₁ and v₂ are:
Show answer and why every option is right or wrong
Answer: D. D is correct. The formula on page 85 of NCERT Class 11 Physics Chapter 5 uses signed velocities along the line of motion — a body travelling opposite to the chosen positive direction enters as a negative value.
Why A is wrong: A is wrong because substituting bare magnitudes loses the direction information the formula needs; the signs are carried inside the algebra, not applied afterwards.
Why B is wrong: B is wrong because the formula is written in the laboratory frame with velocities measured there, not in the centre-of-mass frame.
Why C is wrong: C is wrong because each symbol is a velocity in m/s, not a momentum-derived quantity.
MCQ 3Easy RecallPractice
A body of mass m moving with speed u strikes a stationary body of identical mass m head-on and elastically. After the collision:
Show answer and why every option is right or wrong
Answer: D. D is correct. Setting m₁ = m₂ and u₂ = 0 in the formula on page 85 of NCERT Class 11 Physics Chapter 5 gives v₁ = 0 and v₂ = u — equal masses exchange velocities.
Why A is wrong: A is wrong because a common final speed of u/2 for both bodies is the perfectly inelastic result, where kinetic energy is lost. Here total kinetic energy must be unchanged, and ½m(u/2)² + ½m(u/2)² = ¼mu² is only half the initial ½mu².
Why B is wrong: B is wrong because rebounding with the full incident speed off an equal mass would leave initial momentum mu becoming −mu, violating momentum conservation. That outcome belongs to a collision with a very heavy target.
Why C is wrong: C is wrong because it violates momentum conservation: initial momentum is mu, this option gives 2mu.
MCQ 4Direct ApplicationPractice
A ball of mass 0.20 kg moving at 4.0 m/s along the +x direction collides elastically head-on with a stationary ball of mass 0.60 kg. The final velocity of the 0.20 kg ball is:
Show answer and why every option is right or wrong
Answer: C. C is correct. Applying v₁ = [(m₁ − m₂)u₁ + 2m₂u₂]/(m₁ + m₂) from page 85 of NCERT Class 11 Physics Chapter 5 with u₂ = 0 gives v₁ = (0.20 − 0.60)(4.0)/0.80 = −2.0 m/s: the lighter ball rebounds.
Why A is wrong: A is wrong because it drops the sign of (m₁ − m₂). The mass difference is negative here, so v₁ must be negative — the lighter ball comes back.
Why B is wrong: B is wrong because a dead stop occurs only for equal masses; these masses differ by a factor of three.
Why D is wrong: D is wrong because +1.0 m/s is the common velocity if the two balls stuck together, (0.20 × 4.0)/0.80. The problem states the collision is elastic, so kinetic energy is conserved and the bodies separate.
MCQ 5Direct ApplicationPractice
In the same collision — 0.20 kg at 4.0 m/s striking a stationary 0.60 kg elastically — the final velocity of the 0.60 kg ball is:
Show answer and why every option is right or wrong
Answer: B. B is correct. Using v₂ = [(m₂ − m₁)u₂ + 2m₁u₁]/(m₁ + m₂) from page 85 of NCERT Class 11 Physics Chapter 5 with u₂ = 0: v₂ = 2(0.20)(4.0)/0.80 = +2.0 m/s.
Why A is wrong: A is wrong because a velocity exchange of the full 4.0 m/s happens only for equal masses; the target here is three times heavier.
Why C is wrong: C is wrong because +1.0 m/s is the common velocity of a perfectly inelastic collision between these bodies, not the elastic result.
Why D is wrong: D is wrong because 2u₁ = 8.0 m/s is the limiting result for a very light target, and 6.0 m/s corresponds to no correct substitution; check the momentum sum: 0.20(−2.0) + 0.60(2.0) = 0.80 kg·m/s equals the initial 0.20(4.0), while 0.60(6.0) alone already exceeds it.
MCQ 6Direct ApplicationPractice
A light ball moving at speed u strikes a stationary wall of effectively infinite mass in a head-on elastic collision. The final velocity of the ball, taking the incident direction as positive, is closest to:
Show answer and why every option is right or wrong
Answer: B. B is correct. Letting m₂ ≫ m₁ in v₁ = [(m₁ − m₂)u₁ + 2m₂u₂]/(m₁ + m₂) from page 85 of NCERT Class 11 Physics Chapter 5, the ratio (m₁ − m₂)/(m₁ + m₂) → −1, so v₁ → −u.
Why A is wrong: A is wrong because a ball stopping dead would lose all its kinetic energy, which an elastic collision forbids.
Why C is wrong: C is wrong because it has the ball passing through unchanged; the collision must reverse the light body's direction when the target is far heavier.
Why D is wrong: D is wrong because losing half the speed means losing three-quarters of the kinetic energy. Speed magnitude is preserved in the heavy-target limit.
MCQ 7Concept TrapPractice
Two trolleys approach each other on a frictionless track and collide head-on. A student is told only that momentum before equals momentum after. From this alone the student can conclude:
Show answer and why every option is right or wrong
Answer: A. A is correct. Momentum is conserved in every collision with no net external force, so it cannot distinguish elastic from inelastic. Page 84 of NCERT Class 11 Physics Chapter 5 requires kinetic energy to be checked as the second condition.
Why B is wrong: B is wrong because the quoted information is consistent with any degree of elasticity; nothing indicates the trolleys stick.
Why C is wrong: C is wrong because momentum conservation carries no information about elasticity — it holds equally for a collision in which the trolleys stick together and lose kinetic energy to heat.
Why D is wrong: D is wrong because equal masses do not make a collision elastic. Two equal-mass trolleys can couple together on impact, conserving momentum while losing kinetic energy.
MCQ 8CalculationPractice
A body of mass m moving at speed u collides head-on with a stationary body of mass 3m. The collision is elastic. The fraction of the incident kinetic energy retained by the incident body after the collision is:
Show answer and why every option is right or wrong
Answer: C. C is correct. From page 85 of NCERT Class 11 Physics Chapter 5, v₁ = (m − 3m)u/(4m) = −u/2. The retained kinetic-energy fraction is (v₁/u)² = (1/2)² = 1/4.
Why A is wrong: A is wrong because 3/4 is the fraction transferred to the heavier body, not the fraction retained. Check: the two fractions must sum to 1 for an elastic collision, and retained + transferred = 1/4 + 3/4.
Why B is wrong: B is wrong because it takes the speed ratio |v₁/u| = 1/2 as the energy ratio. Kinetic energy scales as the square of speed, so halving the speed leaves one quarter of the energy.
Why D is wrong: D is wrong because the incident body stops dead only for equal masses; here the target is three times heavier, so the striker rebounds and keeps some kinetic energy.
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Elastic Collision 1D: quick recall before you leave
How do you solve a Elastic Collision 1D question? A worked example
- 1
Given
• Mass of the moving body, m₁ = 1.0 kg• Initial velocity of m₁, u₁ = +6.0 m/s (taking the direction of its motion as positive)• Mass of the stationary body, m₂ = 2.0 kg• Initial velocity of m₂, u₂ = 0 m/s (at rest — exact)• The collision is head-on and elastic.
- 2
Required
The final velocity of each body, and the kinetic energy of the system after the collision.
- 3
Concept
An elastic head-on collision conserves both linear momentum and total kinetic energy (NCERT Class 11 Physics Chapter 5, page 84). Those two conditions together determine both final velocities uniquely — no extra information is needed. The kinetic-energy total after the collision must therefore equal the total before it; computing it separately is a check, not a new calculation.
- 4
Formula
v₁ = [(m₁ − m₂)u₁ + 2m₂u₂] / (m₁ + m₂)
v₂ = [(m₂ − m₁)u₂ + 2m₁u₁] / (m₁ + m₂)
(NCERT Class 11 Physics Chapter 5, page 85.) - 5
Substitution
v₁ = [(1.0 − 2.0)(6.0) + 2(2.0)(0)] / (1.0 + 2.0)
v₂ = [(2.0 − 1.0)(0) + 2(1.0)(6.0)] / (1.0 + 2.0) - 6
Calculation
v₁ = (−1.0 × 6.0) / 3.0 = −6.0 / 3.0 = −2.0 m/s
v₂ = (2 × 1.0 × 6.0) / 3.0 = 12 / 3.0 = +4.0 m/s
The factor 2 in the numerator of each expression is a counting number from the algebra, and the 3.0 kg total mass is a sum of the given masses; neither the counting 2 nor the exact u₂ = 0 contributes to the significant-figure count. The given data carry two significant figures, so the answers are quoted to two.
Momentum check: before, (1.0)(6.0) + 0 = 6.0 kg·m/s; after, (1.0)(−2.0) + (2.0)(4.0) = −2.0 + 8.0 = 6.0 kg·m/s. ✓
Kinetic-energy check: before, ½(1.0)(6.0)² = 18 J; after, ½(1.0)(2.0)² + ½(2.0)(4.0)² = 2.0 + 16 = 18 J. ✓ - 7
Final answer
The 1.0 kg body rebounds at 2.0 m/s (opposite to its original direction); the 2.0 kg body moves forward at 4.0 m/s. The total kinetic energy after the collision is 18 J, unchanged from before — as the elastic condition requires.
- 8
Common trap
The negative sign on v₁. Substituting masses as bare magnitudes, or reading (m₁ − m₂) as a size rather than a signed difference, returns +2.0 m/s and a momentum total of 10 kg·m/s that fails the check. Whenever the struck body is heavier, expect the striker to come back.
A second trap sits in the problem statement rather than the algebra: if this question had said the bodies move off together, these formulas would not apply at all — that is a perfectly inelastic collision with a single common velocity of 2.0 m/s and kinetic energy lost to deformation. Read for the word "elastic" before choosing the formula. - 9
Similar NEET-style question
A 2.0 kg block moving at 5.0 m/s collides head-on and elastically with a stationary 8.0 kg block. Find the velocity of each block after the collision and verify that kinetic energy is conserved.
*(Answer: v₁ = −3.0 m/s, v₂ = +2.0 m/s; KE = 25 J before and after.)*
What to remember before solving Elastic Collision 1D questions
Two bodies of masses m₁, m₂ with initial velocities u₁, u₂ undergoing a head-on elastic collision: v₁ = [(m₁-m₂)u₁ + 2 m₂ u₂] / (m₁+m₂); v₂ = [(m₂-m₁)u₂ + 2 m₁ u₁] / (m₁+m₂). Special cases: equal masses exchange velocities; m₂ ≫ m₁ at rest reflects m₁ with reversed velocity.
-- NCERT Class 11 Physics, Ch. 5, p. 85Which Elastic Collision 1D formulas do you need for NEET?
1 formula — click to collapse
Elastic collision — 1D final velocities
Final velocities of two bodies after an elastic head-on (1D) collision — momentum AND kinetic energy are both conserved. Special cases: equal masses exchange velocities; very heavy m2 at rest reflects m1 with reversed velocity.
| Symbol | Quantity | SI Unit |
|---|---|---|
| m1, m2 | Masses of the two bodies | kg |
| u1, u2 | Initial velocities (signed, before collision) | m/s |
| v1, v2 | Final velocities (signed, after collision) | m/s |
Valid when
- Collision is ELASTIC (kinetic energy conserved)
- Head-on (1D) — for 2D collisions decompose along/perpendicular to line of impact
Do NOT use when
- Inelastic collision (KE not conserved; use momentum-only + restitution)
- Bodies stick together (perfectly inelastic case has its own formula)
Where do students lose marks on Elastic Collision 1D?
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
1 item — click to collapse
Mistake
Student applies elastic-collision velocity formulas to an inelastic collision (or vice versa).
Root cause: concept gap
Correction
Elastic: BOTH momentum and kinetic energy conserved -> use the (m1-m2)/(m1+m2) form. Inelastic: ONLY momentum conserved; KE generally lost to heat. Perfectly inelastic: bodies stick together -> common velocity = (m1*u1 + m2*u2)/(m1+m2). Identify which type from the problem before choosing a formula.
Wrong option pattern
Distractor uses elastic-collision formulas for two bodies that the problem says stick together after impact.
More in Work, Energy and Power: 11 exam traps and mistakes · 9 formulas · 7 question patterns from its other lessons.
Elastic Collision 1D questions from past NEET papers
1 question from NEET 2026. Answers verified against NTA official keys. — click to collapse
NEET 2026
18
27
35
42.5
NTA Answer: Option 3(final)
Sources
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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