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 2D
8 MCQs1 revision card9-step worked example
Source: NCERT Work, Energy and PowerOfficial key: NTA-verifiedLast updated: 25 Sep 2026
Elastic Collision 2D, explained for NEET
A two-dimensional elastic collision is the one place in this chapter where counting equations matters more than recalling a formula. Two bodies collide on a plane and separate along different lines. You have three conservation statements: momentum along x, momentum along y, and kinetic energy. The unknowns are four — two final speeds and two final directions. Three equations, four unknowns. The system is underdetermined. One extra piece of information — an impact parameter, a scattering angle, or a stated symmetry — must come from the question. If it isn't there, the final velocities are genuinely not fixed, and no amount of algebra will produce them.
NCERT Class 11 Physics Chapter 5 states the conservation law on page 84: in an elastic collision both linear momentum and kinetic energy are conserved. Momentum is the vector statement — it holds component by component, so m₁u₁ₓ + m₂u₂ₓ = m₁v₁ₓ + m₂v₂ₓ and the same again for y. Kinetic energy is scalar; it involves full speeds, so ½m₁u₁² + ½m₂u₂² = ½m₁v₁² + ½m₂v₂², with v² = vₓ² + v_y². There is no separate "kinetic energy along x."
Resolve along the line of impact and perpendicular to it. The one-dimensional elastic result applies only along the line of impact; perpendicular to it, each body's velocity component is unchanged when the surfaces are smooth.
One result worth holding: when a moving body collides elastically with an equal, stationary mass and the collision is oblique, the two bodies separate at 90° to each other. Check it against the algebra — it drops out of the two conservation statements when m₁ = m₂ and u₂ = 0, and it is not a general rule for unequal masses.
Watch-out: writing an energy equation per axis. Energy has no components.
Can you answer these Elastic 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
For a two-dimensional elastic collision between two bodies moving in the x–y plane, the complete set of independent conservation equations available is:
Show answer and why every option is right or wrong
Answer: B. B is correct. Momentum is a vector and is conserved separately along each of the two axes of the plane; kinetic energy is a scalar and supplies exactly one equation. NCERT Class 11 Physics Chapter 5, page 84, states both conservations for an elastic collision.
Why A is wrong: A is wrong because it treats momentum conservation as a single scalar statement. Momentum is a vector; in a plane it yields one independent equation per axis, so two.
Why C is wrong: C is wrong because it invents a per-axis kinetic-energy equation. Kinetic energy is a scalar built from the full speed v² = vₓ² + v_y²; it cannot be split into an x-energy and a y-energy.
Why D is wrong: D is wrong because the motion is confined to a plane, which has two axes, not three. A third momentum equation exists only if the motion is genuinely three-dimensional.
MCQ 2Easy RecallPractice
In a two-dimensional elastic collision between two bodies, the number of unknown quantities describing the motion after the collision (two final speeds and two final directions) is four. Given the conservation equations available, the system is:
Show answer and why every option is right or wrong
Answer: A. A is correct. Three equations (two momentum components plus kinetic energy, per NCERT Class 11 Physics Chapter 5, page 84) cannot fix four unknowns; the question must supply an impact parameter, a scattering angle, or a symmetry condition.
Why B is wrong: B is wrong because it assumes three equations settle four unknowns. Counting the unknowns — v₁, v₂ and the two final direction angles — shows one short.
Why C is wrong: C is wrong because there are fewer equations than unknowns, not more. An overdetermined system would have redundant or conflicting constraints; here the constraints are simply insufficient.
Why D is wrong: D is wrong because kinetic-energy conservation is precisely what defines an elastic collision, in two dimensions as in one. Dimensionality does not repeal it.
MCQ 3Easy RecallPractice
The standard one-dimensional elastic-collision velocity formulas quoted in NCERT Class 11 Physics Chapter 5 may be applied to a two-dimensional collision:
Show answer and why every option is right or wrong
Answer: C. C is correct. The conditions of use attached to the one-dimensional elastic result specify a head-on collision; for a two-dimensional case, the velocities are decomposed along and perpendicular to the line of impact, and the formula applies to the along-impact components. NCERT Class 11 Physics Chapter 5, page 85.
Why A is wrong: A is wrong because applying the elastic velocity formulas along both axes independently would impose kinetic-energy conservation twice, once per axis. Energy is a scalar and is conserved once, for the full speeds.
Why B is wrong: B is wrong because the incident body's original direction and the line of impact coincide only in a head-on collision. In an oblique collision the exchange of momentum acts along the line joining the centres at contact, not along the incident velocity.
Why D is wrong: D is wrong because the formulas do apply, restricted to the along-impact direction. The perpendicular components are handled separately — they are simply unchanged for smooth bodies.
MCQ 4Direct ApplicationPractice
Two smooth spheres collide obliquely and elastically. Taking the line of impact as the x-axis, the component of each sphere's velocity perpendicular to the line of impact (the y-component):
Show answer and why every option is right or wrong
Answer: D. D is correct. For smooth surfaces the contact force acts only along the line of impact, so there is no impulse perpendicular to it and each sphere's perpendicular velocity component is unaltered. This is the decomposition prescribed by the conditions of use of the elastic-collision result, NCERT Class 11 Physics Chapter 5, page 85.
Why A is wrong: A is wrong because reversal requires an impulse in that direction. Smooth surfaces exert no tangential force at contact, so nothing acts perpendicular to the line of impact.
Why B is wrong: B is wrong because a velocity component cannot vanish without an impulse to remove it. The spheres continue to drift perpendicular to the line of impact exactly as before.
Why C is wrong: C is wrong because exchange of velocity is the equal-mass result along the line of impact, not a perpendicular effect. The perpendicular direction carries no impulse at all.
MCQ 5Direct ApplicationPractice
A body of mass m moving in the +x direction collides elastically and obliquely with a stationary body of the same mass m. After the collision the two bodies move along directions separated by an angle of:
Show answer and why every option is right or wrong
Answer: A. A is correct. For equal masses with one body initially at rest, momentum conservation gives u = v₁ + v₂ and energy conservation gives u² = v₁² + v₂²; squaring the first and comparing forces v₁·v₂ = 0, so the separation angle is 90°. NCERT Class 11 Physics Chapter 5, page 84.
Why B is wrong: B is wrong because 60° would require v₁·v₂ to be positive, contradicting the result that squaring the momentum equation and subtracting the energy equation leaves 2v₁·v₂ = 0.
Why C is wrong: C is wrong because 45° is the separation only if the collision is symmetric AND each body leaves at 22.5°, which the algebra does not give. The dot product v₁·v₂ vanishes, forcing a right angle regardless of how the speed is shared.
Why D is wrong: D is wrong because 180° describes the two bodies moving apart along one line, which is the head-on case, not an oblique one. It would also leave no net x-momentum unless one speed exceeded the other in a way energy conservation forbids.
MCQ 6Direct ApplicationPractice
A body of mass 0.30 kg travelling at 5.0 m/s in the +x direction collides elastically with a stationary body of mass 0.30 kg. After the collision the first body moves at 4.0 m/s in a direction making 37° with the +x axis, and cos 37° = 0.80 (exact for this question). The x-component of the second body's velocity after the collision is:
Show answer and why every option is right or wrong
Answer: C. C is correct. Momentum along x: (0.30)(5.0) = (0.30)(4.0)(0.80) + (0.30)v₂ₓ, so 5.0 = 3.2 + v₂ₓ and v₂ₓ = 1.8 m/s. The equal masses cancel throughout. The data fit an elastic collision: for equal masses with the target at rest, the first body leaves at u cos 37° = 4.0 m/s. NCERT Class 11 Physics Chapter 5, page 84.
Why A is wrong: A is wrong because 4.0 m/s is the first body's full speed after the collision. Its x-component is smaller by the factor cos 37°, and the second body's share is smaller still.
Why B is wrong: B is wrong because 3.2 m/s is the first body's own x-component, (4.0)(0.80), not the second body's. It is the term subtracted from the initial momentum, not the remainder.
Why D is wrong: D is wrong because 5.0 m/s is the total initial x-velocity, which would leave the first body with no x-momentum at all. The first body plainly retains an x-component of 3.2 m/s.
MCQ 7Concept TrapPractice
A question describes an oblique elastic collision between two bodies of unequal masses, giving both initial velocities and stating that the collision is elastic, but supplying no impact parameter, scattering angle, or symmetry condition. A student writes down conservation of momentum along x, along y, and of kinetic energy, and then solves for all four post-collision quantities. The flaw in this approach is that:
Show answer and why every option is right or wrong
Answer: B. B is correct. The student's three equations are all valid, but the post-collision state needs four numbers; without a fourth datum the final velocities are not uniquely fixed. NCERT Class 11 Physics Chapter 5, page 84.
Why A is wrong: A is wrong because elastic means kinetic energy is conserved, by definition, whether the collision is head-on or oblique. The equation is sound — there are simply too few of them.
Why C is wrong: C is wrong because momentum along y is conserved whatever its value, and its value is zero only if neither body has an initial y-velocity. Even then the equation constrains the final y-components to be equal and opposite, which is real information.
Why D is wrong: D is wrong because momentum conservation holds component by component for any masses. Equal masses give the special 90°-separation result, but they are not a precondition for resolving momentum along axes.
MCQ 8CalculationPractice
A body of mass m moving at speed u along +x collides elastically with a stationary body of mass m. After the collision the incident body moves at 60° to the +x axis. Taking cos 60° = 0.50 and sin 60° = 0.866 (both exact for this question), the speed of the incident body after the collision is:
Show answer and why every option is right or wrong
Answer: D. D is correct. For equal masses with one at rest, the two bodies separate at 90°, so the second body leaves at −30° to the +x axis. Momentum along y gives v₁ sin 60° = v₂ sin 30°, and momentum along x gives v₁ cos 60° + v₂ cos 30° = u; solving, v₁ = u cos 60° = 0.50 u. NCERT Class 11 Physics Chapter 5, page 84.
Why A is wrong: A is wrong because no body can leave an elastic collision faster than the total energy allows; 2.0 u carries four times the initial kinetic energy. A speed greater than u is impossible here since the target was at rest.
Why B is wrong: B is wrong because 0.866 u is the second body's speed, u cos 30°. The body scattered at the larger angle from the original direction keeps the smaller speed, not the larger.
Why C is wrong: C is wrong because retaining the full speed u would leave the second body motionless, which contradicts it having received x-momentum. Checking energy: u² would already be used up by the first body alone.
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Elastic Collision 2D: quick recall before you leave
How do you solve a Elastic Collision 2D question? A worked example
- 1
Given
• Mass of incident body: m₁ = 0.20 kg• Mass of target body: m₂ = 0.20 kg, initially at rest• Initial speed of incident body: u₁ = 5.0 m/s along +x• Collision is elastic and oblique• After the collision, the incident body moves at 30.0° above the +x axis• cos 30.0° = 0.866, sin 30.0° = 0.500 (exact trigonometric values for the stated angle)
- 2
Required
The speed of each body after the collision, and the direction of the target body.
- 3
Concept
An elastic collision conserves both linear momentum and kinetic energy (NCERT Class 11 Physics Chapter 5, page 84). Momentum conservation is a vector statement and holds separately along x and along y. Kinetic energy is scalar and involves the full speeds. Three equations, four unknowns — but here the question supplies the fourth datum: the incident body's scattering direction. With that, the system closes.
Because the masses are equal and the target is at rest, a shortcut is available: the two bodies separate at 90°. That is derived, not assumed — squaring u₁ = v₁ + v₂ gives u₁² = v₁² + v₂² + 2v₁·v₂, and energy conservation (with equal masses cancelling) gives u₁² = v₁² + v₂², so v₁·v₂ = 0. - 4
Formula
Momentum along x: m₁u₁ = m₁v₁cos θ₁ + m₂v₂cos θ₂
Momentum along y: 0 = m₁v₁sin θ₁ − m₂v₂sin θ₂
Kinetic energy: ½m₁u₁² = ½m₁v₁² + ½m₂v₂² - 5
Substitution
Equal masses cancel from every term. With θ₁ = 30.0° and the separation angle 90.0°, the target leaves at θ₂ = 60.0° below the +x axis.
x: 5.0 = v₁(0.866) + v₂(0.500)
y: 0 = v₁(0.500) − v₂(0.866) - 6
Calculation
From the y-equation: v₂ = v₁(0.500/0.866) = 0.5774 v₁.
Substituting into the x-equation:
5.0 = 0.866 v₁ + 0.500(0.5774 v₁) = 0.866 v₁ + 0.2887 v₁ = 1.1547 v₁
v₁ = 5.0 / 1.1547 = 4.330 m/s
v₂ = 0.5774 × 4.330 = 2.500 m/s
Check against the energy equation: v₁² + v₂² = 18.75 + 6.25 = 25.0 = u₁². ✓
The masses (0.20 kg each), the trigonometric values, and the separation angle of 90.0° are exact quantities here — the masses cancel algebraically, and exact constants do not limit significant figures. Only u₁ = 5.0 m/s is a measured value, carrying two significant figures, so the answers are rounded to two. - 7
Final answer
v₁ = 4.3 m/s at 30.0° above the +x axis; v₂ = 2.5 m/s at 60.0° below the +x axis.
- 8
Common trap
Writing an energy equation for each axis — "½m u₁ₓ² = ½m v₁ₓ² + ½m v₂ₓ²" alongside a y-version. That is a fourth and fifth equation that do not exist. Momentum splits into components; kinetic energy does not, because v² = vₓ² + v_y² couples the axes. A student who does this will find an apparently determined system that gives a wrong answer and silently violates the real energy equation. The check in step 6 — verifying v₁² + v₂² = u₁² with full speeds — catches it.
- 9
Similar NEET-style question
A ball of mass 0.40 kg moving at 6.0 m/s strikes a stationary ball of the same mass elastically and obliquely. If the struck ball moves off at 45.0° to the original direction of motion, find the speed of each ball after the collision and the direction of the first ball. (Use the equal-mass 90°-separation result; expect 4.2 m/s each.)
What to remember before solving Elastic Collision 2D questions
Which Elastic Collision 2D 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)
More in Work, Energy and Power: 12 exam traps and mistakes · 9 formulas · 7 question patterns from its other lessons.
Elastic Collision 2D questions from past NEET papers
No question in our NEET 2020–2025 set targets this topic directly.
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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