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. 84Inelastic Collision 1D
8 MCQs1 revision card9-step worked example
Source: NCERT Work, Energy and PowerPYQ coverage: NEET 2024Official key: NTA-verifiedLast updated: 27 Sep 2026
Inelastic Collision 1D, explained for NEET
The trap here has one shape: the question says the bodies stick together, and the student reaches for the elastic formula anyway. (m₁−m₂)/(m₁+m₂) is burned in from practice, so it fires before the word "inelastic" has registered. Every mark lost on this topic is that one reflex.
An inelastic collision conserves momentum and nothing else. Kinetic energy is not conserved — some of it goes into deformation, heat and sound. The extreme case is the perfectly inelastic collision, where the bodies move off together as one object with a single common velocity. NCERT Class 11 Physics Chapter 5 states the momentum-conservation law on page 84 and gives the common-velocity result on page 85.
Momentum conservation for two bodies that coalesce gives
v = (m₁u₁ + m₂u₂)/(m₁ + m₂)
with u₁, u₂ signed along the line of motion. The kinetic energy lost follows:
ΔK = −½ · (m₁m₂)/(m₁+m₂) · (u₁ − u₂)²
Read that second expression carefully. The loss depends on the relative velocity (u₁ − u₂), not on either velocity alone — which is why two bodies moving at the same velocity, however fast, lose nothing when they touch. And because the bracket is squared, the loss is never negative: a perfectly inelastic collision always loses kinetic energy unless the bodies were already moving together.
For NEET, the equal-mass-one-at-rest case is worth having as an instant recall: m at u strikes m at rest, common velocity is u/2, and exactly half the kinetic energy is gone. That single fact answers most questions on this topic in under thirty seconds.
Watch-out: the standard formulas assume no external impulse during the brief contact. If the question adds a wall, a peg, or a held body, momentum of the pair is no longer conserved and the shortcut collapses.
Can you answer these Inelastic 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
In a perfectly inelastic collision between two bodies, which quantity is conserved?
Show answer and why every option is right or wrong
Answer: B. Linear momentum is conserved in all collisions where no net external force acts during impact; kinetic energy is conserved only in elastic collisions. NCERT Class 11 Physics Chapter 5, page 84, states the momentum-conservation law for collisions.
Why A is wrong: A is wrong because kinetic energy is precisely what an inelastic collision does not conserve — part of it becomes internal energy of the deformed bodies.
Why C is wrong: C is wrong because it describes an elastic collision; applying both conservation laws to an inelastic case is the defining error on this topic (trap: defaulting to elastic).
Why D is wrong: D is wrong because discarding momentum conservation as well leaves no usable law — momentum survives every collision type with no external impulse.
MCQ 2Easy RecallPractice
Two bodies undergo a collision and afterwards move off together with a single common velocity. This collision is described as:
Show answer and why every option is right or wrong
Answer: C. Bodies sticking together and sharing one final velocity is the definition of a perfectly inelastic collision, for which NCERT Class 11 Physics Chapter 5 gives the common-velocity formula on page 85.
Why A is wrong: A is wrong because a partially elastic collision leaves the bodies separated with some kinetic energy lost — they do not share one velocity.
Why B is wrong: B is wrong because in a perfectly elastic collision the bodies separate with distinct velocities and kinetic energy is conserved; coalescence is incompatible with that.
Why D is wrong: D is wrong because a super-elastic collision releases stored energy so that kinetic energy increases, which is not what coalescence describes.
MCQ 3Easy RecallPractice
In the expression ΔK = −½ · (m₁m₂)/(m₁+m₂) · (u₁ − u₂)² for a perfectly inelastic collision, the quantity (u₁ − u₂) is:
Show answer and why every option is right or wrong
Answer: C. The bracket is the difference of the two signed initial velocities, that is the relative velocity of approach. NCERT Class 11 Physics Chapter 5, page 85, sets out the kinetic-energy loss in this form.
Why A is wrong: A is wrong because the expression contains only initial (pre-collision) velocities; nothing after the collision appears in it.
Why B is wrong: B is wrong because the common velocity is the separate quantity (m₁u₁ + m₂u₂)/(m₁+m₂), a weighted mean rather than a difference.
Why D is wrong: D is wrong because both masses enter through the m₁m₂/(m₁+m₂) factor; the bracket does not single out either body's velocity.
MCQ 4Direct ApplicationPractice
A body of mass 3.0 kg moving at 4.0 m/s along the +x direction strikes a stationary body of mass 1.0 kg. The two stick together on impact. The common velocity of the pair immediately afterwards is:
Show answer and why every option is right or wrong
Answer: B. Momentum conservation gives v = (3.0 × 4.0 + 1.0 × 0)/(3.0 + 1.0) = 12/4.0 = 3.0 m/s. NCERT Class 11 Physics Chapter 5, page 85, gives this common-velocity result.
Why A is wrong: A is wrong because it is the equal-mass answer u/2, applied without checking that the masses here are 3.0 kg and 1.0 kg rather than equal.
Why C is wrong: C is wrong because it keeps the striking body's original speed, which would require the stationary mass to acquire momentum from nowhere.
Why D is wrong: D is wrong because it divides the initial momentum by the difference of the masses, 12/(3.0 − 1.0) = 6.0, instead of by their sum, 4.0 kg.
MCQ 5Direct ApplicationPractice
A body of mass m moving at speed u strikes a stationary body of the same mass m, and the two stick together. The fraction of the initial kinetic energy that is lost in the collision is:
Show answer and why every option is right or wrong
Answer: A. The common velocity is u/2, so the final kinetic energy is ½(2m)(u/2)² = ¼mu², against an initial ½mu² — exactly half is lost. NCERT Class 11 Physics Chapter 5, page 85, gives the loss expression that yields this result.
Why B is wrong: B is wrong because ¼mu² is the kinetic energy that remains, not the fraction lost; the fraction lost is that remainder measured against ½mu², which is one-half.
Why C is wrong: C is wrong because three-quarters would require the final speed to be u/4 rather than u/2.
Why D is wrong: D is wrong because it assumes kinetic energy is conserved, which is the elastic case; a perfectly inelastic collision between bodies of different velocities always loses energy (trap: defaulting to elastic).
MCQ 6Direct ApplicationPractice
Two bodies of masses 2.0 kg and 5.0 kg both move along the +x direction at 3.0 m/s. They collide and stick together. The kinetic energy lost in this collision is:
Show answer and why every option is right or wrong
Answer: A. The loss is proportional to (u₁ − u₂)², which vanishes when both bodies have the same velocity, so nothing is lost and the pair continues at 3.0 m/s. NCERT Class 11 Physics Chapter 5, page 85, gives the loss in terms of this velocity difference.
Why B is wrong: B is wrong because the half-loss result is specific to equal masses with one body at rest; it is not a general property of perfectly inelastic collisions.
Why C is wrong: C is wrong because a collision that removed all kinetic energy would leave the pair at rest, contradicting momentum conservation for two bodies already moving together.
Why D is wrong: D is wrong because the mass difference does not appear in the loss expression — the masses enter only through the product-over-sum factor, which multiplies a squared velocity difference of zero.
MCQ 7CalculationPractice
A body of mass 2.0 kg moving at 6.0 m/s along +x collides head-on with a body of mass 4.0 kg moving at 3.0 m/s along −x. The two stick together. The kinetic energy lost in the collision is:
Show answer and why every option is right or wrong
Answer: D. D is correct. Total momentum is 2.0(+6.0) + 4.0(−3.0) = 0, so the joined pair is at rest and every joule of kinetic energy is lost. Initial K = ½(2.0)(6.0)² + ½(4.0)(3.0)² = 36 + 18 = 54 J, and final K = 0, so the loss is 54 J. The standard expression agrees: ½ × (2.0 × 4.0)/(6.0) × (6.0 − (−3.0))² = ½ × (4/3) × 81 = 54 J. NCERT Class 11 Physics Chapter 5, page 85.
Why A is wrong: A is wrong because 12 J is the magnitude of one body's momentum contribution in kg·m/s, not an energy.
Why B is wrong: B is wrong because 27 J is half the initial kinetic energy, the equal-mass-one-at-rest result. Here the momenta cancel exactly, so the loss is the whole 54 J, not half of it.
Why C is wrong: C is wrong because 36 J is the kinetic energy of the 2.0 kg body alone; the 4.0 kg body's 18 J is lost as well.
MCQ 8CalculationPractice
A body of mass m moving at speed u strikes a stationary body of mass M and the two stick together. For which value of M is the fraction of kinetic energy lost the greatest?
Show answer and why every option is right or wrong
Answer: D. The fraction lost is M/(m + M), which rises towards 1 as M grows — a light body striking a very heavy stationary one loses almost all its kinetic energy. NCERT Class 11 Physics Chapter 5, page 85, gives the loss expression from which this fraction follows.
Why A is wrong: A is wrong because when M is very small the pair barely slows, M/(m+M) tends to zero and almost no kinetic energy is lost.
Why B is wrong: B is wrong because equal masses give M/(m+M) = ½, which is a loss but not the greatest available one.
Why C is wrong: C is wrong because the fraction M/(m+M) depends explicitly on the mass ratio; only the conservation of momentum, not the energy loss, is independent of it.
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Inelastic Collision 1D: quick recall before you leave
How do you solve a Inelastic Collision 1D question? A worked example
Pattern: perfectly inelastic collision, equal masses, one at rest (NEET 2024).
- 1
Given.
Body A: mass m, speed u along +x. Body B: mass m, at rest. The bodies stick together on impact. No external horizontal force acts during the collision.
- 2
Required.
The common velocity after impact, and the kinetic energy lost as a fraction of the initial kinetic energy.
- 3
Concept.
The bodies coalesce, so this is a perfectly inelastic collision: linear momentum is conserved, kinetic energy is not. The elastic-collision velocity formulas do not apply here, because they are derived by imposing kinetic-energy conservation as a second condition — a condition this collision violates.
- 4
Formula.
v = (m₁u₁ + m₂u₂)/(m₁ + m₂), with the kinetic energies computed from K = ½mv² before and after.
- 5
Substitution.
m₁ = m, u₁ = u, m₂ = m, u₂ = 0:
v = (m·u + m·0)/(m + m) - 6
Calculation.
v = mu/(2m) = u/2
Initial kinetic energy: K_i = ½mu² + 0 = ½mu².
Final kinetic energy: K_f = ½(2m)(u/2)² = ½ · 2m · u²/4 = ¼mu².
Loss: ΔK = K_f − K_i = ¼mu² − ½mu² = −¼mu².
The2in(m + m) = 2m, the½in the kinetic-energy definition and the4from squaringu/2are exact counting factors from the algebra. They carry no measurement uncertainty and so place no limit on the significant figures of a numerical answer built from this result. - 7
Final answer.
The common velocity is u/2 in the original direction of motion. The kinetic energy lost is ¼mu², which is one-half of the initial ½mu².
- 8
Common trap.
The standing error is to reach for v₁ = (m₁−m₂)u₁/(m₁+m₂) because the elastic formula is the better-practised one. With equal masses that formula gives v₁ = 0 and v₂ = u — the bodies exchange velocities and separate, which directly contradicts the statement that they stick together. Read for the words "stick together", "move with a common velocity", or "completely inelastic" before choosing any collision formula; they rule the elastic route out. The second error is to state the common velocity correctly and then assert no kinetic energy was lost, which is the same elastic assumption arriving one step later.
- 9
Similar NEET-style question.
A body of mass 1.0 kg moving at 8.0 m/s along +x strikes a stationary body of mass 3.0 kg and the two move off together. Find the common velocity and the fraction of the initial kinetic energy lost. (Answer: 2.0 m/s along +x; three-quarters lost, since the fraction lost is M/(m+M) = 3.0/4.0.)
What to remember before solving Inelastic Collision 1D questions
If two bodies of masses m₁ and m₂ with velocities u₁ and u₂ stick together after collision, the common final velocity is v = (m₁ u₁ + m₂ u₂) / (m₁ + m₂). KE loss = ½ [m₁ m₂ / (m₁+m₂)] (u₁ - u₂)².
-- NCERT Class 11 Physics, Ch. 5, p. 84Which Inelastic Collision 1D formulas do you need for NEET?
1 formula — click to collapse
Perfectly inelastic 1D collision — common final velocity and KE loss
When two bodies stick together after a 1D collision, the common velocity is given by momentum conservation. The KE lost is converted to internal energy (heat, deformation).
| Symbol | Quantity | SI Unit |
|---|---|---|
| v | Common final velocity | m/s |
| m1, m2 | Masses | kg |
| u1, u2 | Initial velocities | m/s |
| Delta_K | Change in kinetic energy | J |
Valid when
- Bodies stick together immediately after collision (perfectly inelastic)
- Net external force = 0 during the brief collision (momentum conservation)
Where do students lose marks on Inelastic Collision 1D?
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
2 items — click to collapse
Category: Similar Terms
Student uses the elastic-collision velocity formulas (m1-m2)/(m1+m2) when the question explicitly says 'completely inelastic' (bodies stick).
When it triggers
Question says 'inelastic', 'stick together', 'after collision moves with common velocity'.
How to avoid
Perfectly inelastic 1D: v_common = (m₁ u₁ + m₂ u₂)/(m₁ + m₂). KE is NOT conserved; loss = ½ (m₁ m₂)/(m₁+m₂) × (u₁ - u₂)². Don't use elastic formulas.
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: 10 exam traps and mistakes · 9 formulas · 6 question patterns from its other lessons.
Inelastic Collision 1D questions from past NEET papers
1 question from NEET 2024. Answers verified against NTA official keys. — click to collapse
NEET 2024Revised key
11 : 2
22 : 1
34 : 1
41 : 4
NTA Answer: Option 2(revised_final)
How does NEET ask about Inelastic Collision 1D?
1 recurring pattern from past papers — click to collapse
Two bodies of equal mass m, one moving at v₁ and one at rest, undergo completely inelastic 1D collision. Common velocity = v₁/2 (momentum conservation). KE loss = ½ m v₁² × ½ = ¼ m v₁².
Direct ApplicationEasy
Common distractors
no KE loss stated
Treats collision as elastic by default
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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