Elastic Collision 2d

8 MCQs2 revision cards9-step worked example

Source: NCERT Work, Energy and PowerPYQ coverage: NEET 2021, 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026

Lesson

The trap: In 2D elastic collisions, students try to apply the 1D elastic-collision velocity formulas directly to each axis, forgetting that the 2D problem has additional unknowns (scattering angles) that the 1D formulas don't resolve. The 1D formulas assume head-on impact; a glancing collision in two dimensions requires you to decompose momentum along and perpendicular to the line of impact separately.

What NCERT says: NCERT Class 11 Physics Chapter 5, page 14 states that in an elastic collision, both momentum and kinetic energy are conserved. The text derives the 1D result explicitly but notes that 2D collisions require vector conservation of momentum — two independent component equations — plus the scalar KE conservation equation.

The 2D setup: Consider two bodies colliding on a frictionless surface. Choose the x-axis along the initial velocity of the projectile. Before collision: body 1 moves along x; body 2 is at rest. After collision: body 1 scatters at angle θ₁ above x; body 2 recoils at angle θ₂ below x.

Three conservation equations govern the outcome:

x-momentum: m₁u₁ = m₁v₁ cos θ₁ + m₂v₂ cos θ₂
y-momentum: 0 = m₁v₁ sin θ₁ − m₂v₂ sin θ₂
KE: ½m₁u₁² = ½m₁v₁² + ½m₂v₂²

Four unknowns (v₁, v₂, θ₁, θ₂), three equations — so one quantity must be given or measured experimentally.

Equal-mass special case: When m₁ = m₂ and body 2 is initially at rest, the KE equation combined with momentum conservation forces θ₁ + θ₂ = 90° — the two bodies always scatter at right angles. This is a high-frequency NEET fact.

Watch-out: Confusing elastic with inelastic changes the entire calculation. In an inelastic collision, KE is not conserved, so you lose one equation and gain a different constraint (coefficient of restitution or "stick together"). Always verify the collision type before setting up equations.

Practice 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 an elastic collision, which quantities are conserved?

MCQ 2Easy RecallPractice

Two balls of equal mass undergo a 2D elastic collision on a frictionless surface. Ball 1 is initially moving; ball 2 is at rest. After collision, ball 1 scatters at 30° to the original direction. What is the scattering angle of ball 2?

MCQ 3Direct ApplicationPractice

A body of mass 2m moving at velocity u collides elastically and head-on with a body of mass m at rest. What is the speed of the lighter body after collision?

MCQ 4Direct ApplicationPractice

Two identical billiard balls collide elastically on a level table. Ball A moves at 5.0 m/s; ball B is stationary. After the collision, ball A moves at 3.0 m/s. What is the speed of ball B after the collision?

MCQ 5Concept TrapPractice

In a 2D elastic collision between two unequal masses (target at rest), there are four unknowns (v₁, v₂, θ₁, θ₂) but only three conservation equations. What additional information is needed to solve the problem completely?

MCQ 6Direct ApplicationPractice

A particle of mass m collides elastically with an identical particle at rest. After collision, the first particle is observed moving at angle θ to its original direction. Which expression gives the speed of the first particle after collision?

MCQ 7Direct ApplicationPractice

Two bodies of equal mass undergo a perfectly inelastic collision (they stick together). One moves at velocity v; the other is at rest. A student claims the common velocity is v/3. What is the correct common velocity?

MCQ 8CalculationPractice

A ball of mass m₁ = 0.5 kg moving at 4.0 m/s collides elastically in 2D with a ball of mass m₂ = 0.5 kg at rest. After the collision, ball 1 moves at 2.0 m/s. What fraction of the initial kinetic energy is transferred to ball 2?

Quick recall before you leave

Worked Example

1
Given
A ball of mass 0.20 kg moves at 6.0 m/s along the x-axis and strikes an identical ball (0.20 kg) at rest on a frictionless surface. After the elastic collision, the first ball is observed moving at 30° above the x-axis.
2
Required
(a) Speed of each ball after collision. (b) Direction of ball 2 after collision.
3
Concept
In a 2D elastic collision between equal masses with the target at rest, both momentum and kinetic energy are conserved. The equal-mass constraint forces the two balls to scatter at right angles: θ₁ + θ₂ = 90°.
4
Formula
- KE conservation: u₁² = v₁² + v₂² (equal masses cancel) - Equal-mass right-angle rule: θ₂ = 90° − θ₁ - From momentum + KE: v₁ = u₁ cos θ₁, v₂ = u₁ sin θ₁
5
Substitution
- θ₁ = 30°, so θ₂ = 90° − 30° = 60° (below x-axis) - v₁ = 6.0 × cos 30° = 6.0 × (√3/2) - v₂ = 6.0 × sin 30° = 6.0 × (1/2)
6
Calculation
- v₁ = 6.0 × 0.866 = 5.196 ≈ 5.2 m/s - v₂ = 6.0 × 0.500 = 3.0 m/s Note on exact constants: cos 30° = √3/2 and sin 30° = 1/2 are exact trigonometric values and do not limit significant figures. The given speed 6.0 m/s (2 significant figures) controls the precision of the final answer.
7
Final answer
- Ball 1: 5.2 m/s at 30° above the x-axis - Ball 2: 3.0 m/s at 60° below the x-axis Verification: v₁² + v₂² = (5.196)² + (3.0)² = 27.0 + 9.0 = 36.0 = (6.0)² = u₁². KE conserved. ✓
8
Common trap
A common confusion is applying the 1D velocity-exchange result ("equal masses swap velocities") directly to 2D. In 1D head-on collisions, ball 1 stops and ball 2 moves at u₁. In 2D glancing collisions, ball 1 does NOT stop — it continues at reduced speed at angle θ₁. The 90° scattering rule replaces the "velocity exchange" special case when the collision is not head-on.
9
Similar NEET-style question
A proton moving at 4.0 × 10⁵ m/s collides elastically with a stationary proton. After the collision, the incident proton is deflected by 45°. Find the speeds of both protons after the collision and verify that KE is conserved. ---

Before solving, remember these

Law

Elastic and inelastic collisions

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. 14

Formulas

10 formulas — click to collapse

Elastic collision — 1D final velocities

v_{1} = \frac{( m _{1} - m _{2} ) u _{1} + 2 m _{2} u _{2}}{m _{1} + m _{2}}, v_{2} = \frac{( m _{2} - m _{1} ) u _{2} + 2 m _{1} u _{1}}{m _{1} + m _{2}}

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)

Gravitational potential energy (near Earth)

U = m g h

Potential energy of a body of mass m at height h above the chosen reference level, in a region where g is approximately constant.

Symbol	Quantity	SI Unit
U	Gravitational PE	J
m	Mass	kg
g	Gravitational acceleration	m/s^2
h	Height above reference level	m

Valid when

Region small enough that g is uniform (typically near Earth's surface)
Reference level is freely chosen — only DIFFERENCES in U have physical meaning

Do NOT use when

Large altitude changes (use U = -GMm/r general form)
Below the reference level — h is signed

Kinetic energy

K = \frac{1}{2} m v^{2}

Energy a body possesses by virtue of its motion. Always non-negative. Frame-dependent through v.

Symbol	Quantity	SI Unit
K	Kinetic energy	J
m	Mass	kg
v	Speed	m/s

Valid when

Non-relativistic speeds (v << c)
Translational motion only (rotational KE = (1/2) I omega^2 separately)

Conservation of mechanical energy

K_{i} + U_{i} = K_{f} + U_{f}

If only conservative forces do work on a system, the total mechanical energy E = K + U is constant in time.

Symbol	Quantity	SI Unit
K_i, K_f	Initial, final kinetic energy	J
U_i, U_f	Initial, final potential energy	J

Valid when

Only CONSERVATIVE forces do work (gravity, spring, electrostatic — not friction or drag)
Closed system; no energy exchange with surroundings

Do NOT use when

Friction, drag, or other non-conservative forces do work
External forces add energy to the system

Perfectly inelastic 1D collision — common final velocity and KE loss

v = \frac{m _{1} u _{1} + m _{2} u _{2}}{m _{1} + m _{2}}, Δ K = - \frac{1}{2} \frac{m _{1} m _{2}}{m _{1} + m _{2}} (u_{1} - u_{2})^{2}

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)

Instantaneous power

P = \frac{d W}{d t} = F \cdot v

Instantaneous power is the time rate of doing work; equivalently, the dot product of the force and velocity. Average power over an interval is W/t.

Symbol	Quantity	SI Unit
P	Power (instantaneous)	W
F	Force	N
v	Velocity	m/s

Valid when

F.v form when force and velocity are both known at the instant

Spring potential energy (Hooke's law regime)

U = \frac{1}{2} k x^{2}

Elastic potential energy stored in an ideal spring of force constant k that has been displaced by x from its natural length. Independent of the sign of x (compression or extension).

Symbol	Quantity	SI Unit
U	Elastic PE	J
k	Spring constant (force per unit displacement)	N/m
x	Displacement from natural length	m

Valid when

Spring obeys Hooke's law (F = -k*x) over the displacement range
No internal damping / hysteresis assumed

Do NOT use when

Beyond elastic limit (Hooke's law fails)
Real springs with finite damping (some elastic PE goes to internal energy)

Work done by a constant force

W = F \cdot s = F s cos θ

The work done by a constant force F on an object that undergoes a displacement s is the dot product F.s. Equivalently, W = (magnitude of F) * (magnitude of s) * cos(angle between them). Work is a scalar but has a sign.

Symbol	Quantity	SI Unit
W	Work (scalar, signed)	J
F	Constant force (vector)	N
s	Displacement (vector)	m
theta	Angle between F and s	rad/deg

Valid when

Force is CONSTANT in magnitude and direction over the displacement
Use the COMPONENT of force along the displacement, not magnitude alone

Do NOT use when

Force varies with position (use the variable-force integral)
Multiple forces act — apply this to each one separately or use net force

Work-energy theorem

W_{net} = Δ K = \frac{1}{2} m (v^{2} - v_{0}^{2})

The net work done by all forces on a particle equals the change in its kinetic energy. Holds for both constant and variable forces, in 1D and higher dimensions.

Symbol	Quantity	SI Unit
W_net	Net work done by all forces	J
Delta_K	Change in kinetic energy	J
m	Mass of particle	kg
v, v0	Final, initial speed	m/s

Valid when

W_net is the NET (vector-sum) work, not work of any one force
Particle (point-mass) idealisation; for extended bodies handle internal energy separately

Work done by a variable force

W = \int_{x_{i}}^{x_{f}} F (x) d x

When a force varies along the path, the work done is the line integral of force over displacement. In one dimension this is the area under the F-vs-x curve between the start and end positions.

Symbol	Quantity	SI Unit
W	Work done	J
F(x)	Force as a function of position	N
dx	Infinitesimal displacement	m
x_i, x_f	Initial and final positions	m

Valid when

Force may depend on position (e.g. spring force F = -k*x)
Generalises to W = integral F.dr in higher dimensions

Exam Traps & Common Mistakes

These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.

12 items — click to collapse

Trap

Linear speed-distance ratio under uniform deceleration

Category: Overthinking

Student assumes proportionality of speed to remaining distance under uniform deceleration. In fact, KE drops linearly with distance (v² is the linear quantity, not v): v² = u² - 2as. Speed-vs-distance is a sqrt-curve, not a line.

When it triggers

Question describes a body decelerating through stages with given speed at one stage; asks for distance to stop or speed at another stage.

How to avoid

Always work with v², not v, when uniform deceleration is in play. The work-energy theorem gives the same answer faster: ½ m v² = work done against constant force over distance.

Trap

Incline-with-friction: missing μ cos θ term

Category: Sign Convention

Student writes a = g sin θ for a rough incline (which is the smooth-incline answer); forgets to subtract μ g cos θ.

When it triggers

Question contrasts rough vs smooth inclines, or asks for acceleration on a rough incline.

How to avoid

On a rough incline (block sliding down): a = g(sin θ - μ_k cos θ). On a rough incline (block sliding up): a = -g(sin θ + μ_k cos θ). Smooth case (μ = 0): just g sin θ.

Trap

Inelastic collision: defaulting to elastic

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.

Trap

Lift-power: friction term overlooked

Category: Overthinking

Student computes P = Mgv (just lifting against gravity) and ignores the friction-opposing-motion term.

When it triggers

Question describes a lift moving at constant speed with explicit friction force on cable or guides.

How to avoid

At constant speed, net force = 0, so cable tension T = Mg + f_friction. Power = T × v = (Mg + f) × v. Always add friction when stated.

Trap

Power problems: efficiency / loss factor dropped

Category: Overthinking

Student computes ideal power and forgets the (1 - loss_fraction) or efficiency multiplier.

When it triggers

Question gives turbine, motor, or transformer with stated efficiency or loss percentage.

How to avoid

Always read the question for efficiency η or loss%. Useful power P_useful = η × P_input or P_input × (1 - loss). Don't drop the factor even if the rest of the calc is in the unrelated parts of the problem.

Trap

Instantaneous power: x vs v

Category: Similar Terms

Student plugs displacement x into P = F·v formula instead of velocity v.

When it triggers

Question gives displacement x(t) explicitly and a constant force; asks for instantaneous power.

How to avoid

P = F · v where v = dx/dt. Compute v first (differentiate x(t) once), THEN plug into F·v. P is NOT F·x.

Trap

Spring PE linear vs quadratic scaling

Category: Overthinking

Student treats spring PE as proportional to displacement (linear) instead of displacement-squared (quadratic). Common error: 'doubling the stretch doubles the PE'. Actual: doubling the stretch gives 4× the PE.

When it triggers

Question gives U at one stretch and asks for U at another. Distractors include linear-scaling answer (×2 instead of ×4 for double stretch).

How to avoid

U = ½ k x² is QUADRATIC. The PE-to-stretch ratio is the SQUARE of the stretch ratio: if stretch goes from x₁ to x₂, U_new / U_old = (x₂/x₁)².

Trap

Vertical circle: energy conservation alone is insufficient

Category: Overthinking

Student uses ½ m v₀² = m g (2L) (energy to reach top) and forgets the additional v_top² ≥ gL constraint for tension.

When it triggers

Question asks for minimum v₀ at lowest point so the bob can complete a full vertical circle.

How to avoid

TWO constraints: (1) energy: v_top² = v₀² - 4gL; (2) tension at top ≥ 0: v_top² ≥ gL. Combined: v₀² ≥ 5gL. Energy alone gives only v₀² ≥ 4gL which is insufficient.

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.

Mistake

Student applies K_i + U_i = K_f + U_f in a problem where friction or other non-conservative forces clearly do work.

Root cause: formula misuse

Correction

Mechanical-energy conservation requires that ONLY conservative forces do work. When friction or drag is present, use the work-energy theorem directly: K_f - K_i = W_conservative + W_non-conservative, where W_non-conservative is typically negative (energy goes to heat).

Wrong option pattern

Distractor sets m*g*h = (1/2)*m*v^2 for a block sliding down a rough incline.

Mistake

Student treats gravitational potential energy as having an absolute value, instead of being defined relative to a chosen reference level.

Root cause: concept gap

Correction

PE is defined up to an additive constant. Only DIFFERENCES in PE have physical meaning (they equal the negative work done by the conservative force between two points). Choosing the ground as zero is a convention, not a derivation.

Wrong option pattern

Distractor offers a numerical PE value where two different reference levels would give different correct numbers.

Mistake

Student claims the normal reaction does positive (or negative) work on a block sliding on a horizontal surface.

Root cause: direction ignored

Correction

Work = F . s = F * s * cos(theta). For a block sliding horizontally, the normal force is perpendicular to displacement (theta = 90 deg, cos = 0), so the work done by N is ZERO. Always check the angle between force and displacement before computing work.