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 2D
8 MCQs2 revision cards9-step worked example
Source: NCERT Work, Energy and PowerOfficial key: NTA-verifiedLast updated: 25 Sep 2026
Inelastic Collision 2D, explained for NEET
The trap in two-dimensional inelastic collisions is treating momentum as one equation. It is two. Momentum is a vector, so conservation holds separately along x and along y — and the y-equation is the one aspirants skip when the incoming bodies happen to move along a single axis.
Take the commonest shape: a body moving along +x strikes a stationary body, and the two stick together. Before impact the total y-momentum is zero. That zero is a real constraint, not an absent one: it forces the combined body to move along +x, with no y-component. Many aspirants "solve" this by intuition and get the right answer, then fail the moment the second body is also moving — because they never wrote the y-equation down.
NCERT Class 11 Physics Chapter 5, page 84, states the conservation law that does the work here: in the absence of a net external force, the total momentum of the system is conserved during the collision. Resolve into components and you get
- m₁u₁ₓ + m₂u₂ₓ = (m₁ + m₂)vₓ
- m₁u₁ᵧ + m₂u₂ᵧ = (m₁ + m₂)vᵧ
Two equations, two unknowns — the perfectly inelastic 2D case is always fully determined. The final speed is √(vₓ² + vᵧ²) and the direction is arctan(vᵧ/vₓ).
Kinetic energy is not conserved, and its loss must be computed from the actual speeds: ΔK = ½(m₁+m₂)v² − (½m₁u₁² + ½m₂u₂²). Do not carry over a 1D shortcut formula for the loss; in 2D the relative-velocity term is a vector difference, and the algebraic 1D expression for ΔK does not transfer unchanged.
Watch out: having found vₓ and vᵧ, the answer to "find the velocity" is a magnitude and a direction. A stem that asks for speed alone still requires both components first — you cannot get √(vₓ² + vᵧ²) from the x-equation on its own.
Can you answer these Inelastic Collision 2D 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 moving in a plane, the conservation law stated in NCERT Class 11 Physics Chapter 5 applies to momentum in the following way:
Show answer and why every option is right or wrong
Answer: A. A is correct. Momentum is a vector, so conservation in the absence of a net external force holds component by component — the law on NCERT Class 11 Physics Chapter 5, page 84, applies to the vector, hence to each of its components.
Why B is wrong: B is wrong because no axis is privileged by the masses — the y-component equation holds exactly as the x-component one does, whichever body is heavier.
Why C is wrong: C is wrong because conserving a vector means conserving both magnitude and direction; a change of direction would require an external impulse.
Why D is wrong: D is wrong because momentum conservation during a collision depends on the absence of a net external force, not on any relation between the masses.
MCQ 2Easy RecallPractice
Two bodies collide obliquely in a plane and stick together. The number of independent scalar equations supplied by momentum conservation is:
Show answer and why every option is right or wrong
Answer: B. B is correct. Motion confined to a plane gives one scalar momentum equation per axis — x and y — so two independent equations, which exactly match the two unknowns (vₓ, vᵧ) of the single combined body.
Why A is wrong: A is wrong because it counts the momentum vector as a single scalar; a planar vector equation resolves into two scalar equations.
Why C is wrong: C is wrong because three component equations arise only for a genuinely three-dimensional collision, not for motion confined to a plane.
Why D is wrong: D is wrong because it presumably adds a kinetic-energy equation; kinetic energy is not conserved in an inelastic collision, so it supplies no equation here.
MCQ 3Easy RecallPractice
For a perfectly inelastic collision in a plane, the kinetic energy of the system after impact is:
Show answer and why every option is right or wrong
Answer: C. C is correct. In an inelastic collision only momentum is conserved; the lost kinetic energy is converted to internal energy such as heat and permanent deformation.
Why A is wrong: A is wrong because equality of kinetic energy before and after is the defining property of an elastic collision, not an inelastic one.
Why B is wrong: B is wrong because a collision cannot generate kinetic energy without an internal energy source, which is not present in a standard collision problem.
Why D is wrong: D is wrong because the combined body moves off with a non-zero common velocity whenever the initial total momentum is non-zero, and so retains kinetic energy.
MCQ 4Direct ApplicationPractice
A body of mass 2.0 kg moving at 3.0 m/s along +x strikes a body of mass 3.0 kg moving at 2.0 m/s along +y. The two stick together. The x-component of their common velocity is:
Show answer and why every option is right or wrong
Answer: D. D is correct. Along x only the first body carries momentum: (2.0)(3.0) = 6.0 kg·m/s, shared by the combined mass 5.0 kg, giving vₓ = 6.0/5.0 = 1.2 m/s. This is the x-component equation of the law on NCERT Class 11 Physics Chapter 5, page 84.
Why A is wrong: A is wrong because 0.60 m/s is 3.0/5.0, the first body's speed divided by the combined mass; its momentum is (2.0)(3.0) = 6.0 kg·m/s, and that is what is shared.
Why B is wrong: B is wrong because 2.5 m/s averages the two given speeds; speeds along perpendicular axes cannot be averaged, as each axis has its own momentum equation.
Why C is wrong: C is wrong because it divides the first body's momentum by the wrong mass (6.0/4.0); the combined mass after sticking is m₁ + m₂ = 5.0 kg.
MCQ 5Direct ApplicationPractice
A body of mass 4.0 kg moving at 5.0 m/s along +x collides with a stationary body of mass 6.0 kg and the two stick together. The direction of motion of the combined body is:
Show answer and why every option is right or wrong
Answer: D. D is correct. The total y-momentum before impact is zero, so the y-equation forces vᵧ = 0 and the combined body must move along +x. The zero is a constraint, not a missing equation.
Why A is wrong: A is wrong because the two component equations are sufficient to fix both vₓ and vᵧ; the perfectly inelastic case is always fully determined, so no extra geometric input is needed.
Why B is wrong: B is wrong because a 45° deflection requires vₓ = vᵧ, whereas the y-equation gives vᵧ = 0 here.
Why C is wrong: C is wrong because it would demand y-momentum appearing from nothing, which the conservation law forbids in the absence of an external impulse.
MCQ 6Direct ApplicationPractice
Two bodies stick together after an oblique collision, and their common velocity has components vₓ = 3.0 m/s and vᵧ = 4.0 m/s. The speed of the combined body is:
Show answer and why every option is right or wrong
Answer: B. B is correct. Speed is the magnitude of the velocity vector, √(vₓ² + vᵧ²) = √(9.0 + 16) = 5.0 m/s. The integers 3.0 and 4.0 here are measured values, not exact ones.
Why A is wrong: A is wrong because it subtracts the components; perpendicular components combine by Pythagoras, never by subtraction.
Why C is wrong: C is wrong because it averages the two components, which is not how a vector magnitude is formed.
Why D is wrong: D is wrong because it adds the components arithmetically — valid only if they were along the same axis, which perpendicular components are not.
MCQ 7Concept TrapPractice
A student is given an oblique collision in which two bodies of known masses and known initial velocities stick together, and is asked for the final velocity. The student writes down both component momentum equations and also sets the total kinetic energy after impact equal to the total kinetic energy before impact. The student's method is:
Show answer and why every option is right or wrong
Answer: A. A is correct. Writing the kinetic-energy equation imports a property of elastic collisions into an inelastic one; kinetic energy is lost to internal energy here, so the extra equation is not merely redundant but wrong and will contradict the momentum result.
Why B is wrong: B is wrong because an extra equation helps only if it is true; a false constraint added to two correct ones makes the system inconsistent.
Why C is wrong: C is wrong because momentum conservation holds along both axes in a perfectly inelastic collision — the sticking affects the energy bookkeeping, not the momentum law.
Why D is wrong: D is wrong because it treats the kinetic-energy step as a harmless detour; it is a false statement about this collision, not surplus correct work.
MCQ 8CalculationPractice
A body of mass 1.0 kg moving at 4.0 m/s along +x collides with a body of mass 1.0 kg moving at 4.0 m/s along +y, and the two stick together. The kinetic energy lost in the collision is:
Show answer and why every option is right or wrong
Answer: C. C is correct. Components: vₓ = 4.0/2.0 = 2.0 m/s, vᵧ = 4.0/2.0 = 2.0 m/s, so v² = 8.0 m²/s². Final K = ½(2.0)(8.0) = 8.0 J against an initial K of ½(1.0)(16) + ½(1.0)(16) = 16 J, a loss of 8.0 J.
Why A is wrong: A is wrong because it assumes kinetic energy is conserved; in a perfectly inelastic collision it is not, and the momentum result already fixes a smaller final kinetic energy.
Why B is wrong: B is wrong because it takes v = 2.0 m/s as the whole speed rather than one component, giving a final kinetic energy of 4.0 J and hence the wrong loss.
Why D is wrong: D is wrong because it reports the entire initial kinetic energy as lost, which would require the combined body to end at rest; its momentum is non-zero, so it cannot.
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Inelastic Collision 2D: quick recall before you leave
How do you solve a Inelastic Collision 2D question? A worked example
- 1
Given.
Body 1: m₁ = 2.0 kg, moving at u₁ = 6.0 m/s along +x. Body 2: m₂ = 2.0 kg, moving at u₂ = 6.0 m/s along +y. The bodies stick together on impact. No external force acts during the collision.
- 2
Required.
(a) The common velocity after impact — magnitude and direction. (b) The kinetic energy lost.
- 3
Concept.
Momentum is a vector. With no net external force, it is conserved during the collision, and that vector statement resolves into one scalar equation per axis. Kinetic energy is not conserved: the bodies stick, so some of it becomes internal energy.
- 4
Formula.
• x: m₁u₁ₓ + m₂u₂ₓ = (m₁ + m₂)vₓ• y: m₁u₁ᵧ + m₂u₂ᵧ = (m₁ + m₂)vᵧ• K = ½mv², with v² = vₓ² + vᵧ²
- 5
Substitution.
• x: (2.0)(6.0) + (2.0)(0) = (4.0)vₓ• y: (2.0)(0) + (2.0)(6.0) = (4.0)vᵧ
- 6
Calculation.
vₓ = 12/4.0 = 3.0 m/s; vᵧ = 12/4.0 = 3.0 m/s.
v = √(3.0² + 3.0²) = √18 = 4.2 m/s (2 s.f.).
Direction: arctan(3.0/3.0) = 45° from the +x axis, into the first quadrant.
Initial K = ½(2.0)(6.0²) + ½(2.0)(6.0²) = 36 + 36 = 72 J.
Final K = ½(4.0)(18) = 36 J.
Loss = 72 − 36 = 36 J.
The factor ½ in K = ½mv² and the 2 in the squares are exact mathematical constants, and the 45° here is an exact angle arising from equal components; none of them contributes to the significant-figure count. The count is governed by the measured 2.0 kg and 6.0 m/s, each 2 significant figures. - 7
Final answer.
The combined body moves at 4.2 m/s at 45° to the +x axis. Kinetic energy lost = 36 J, exactly half the initial kinetic energy.
- 8
Common trap.
Solving only the x-equation and reporting 3.0 m/s as "the speed." The y-equation is not optional here: body 2 carries y-momentum, so vᵧ ≠ 0, and the speed is the magnitude √(vₓ² + vᵧ²), not either component alone. The same slip inverted appears when the second body is at rest — then vᵧ = 0 is a result of the y-equation, and a student who never writes that equation gets the right answer for no reason and fails the case above.
- 9
Similar NEET-style question.
A body of mass 3.0 kg moving at 4.0 m/s along +x collides with a body of mass 1.0 kg moving at 4.0 m/s along −y. The two stick together. Find the magnitude of the common velocity and the fraction of kinetic energy lost. *(Route: vₓ = 12/4.0 = 3.0 m/s, vᵧ = −4.0/4.0 = −1.0 m/s; speed = √10 ≈ 3.2 m/s; initial K = 24 + 8.0 = 32 J, final K = ½(4.0)(10) = 20 J, fraction lost = 12/32 = 0.38.)*
What to remember before solving Inelastic Collision 2D questions
More in Work, Energy and Power: 12 exam traps and mistakes · 10 formulas · 7 question patterns from its other lessons.
Inelastic Collision 2D questions from past NEET papers
No question in our NEET 2020–2025 set targets this topic directly.
Sources
NCERT refs: Class 11 Physics Chapter 5, p.84
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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