If the net external force on a system is zero, the total linear momentum of the system remains constant. p_initial = p_final. This is a direct consequence of Newton's Second and Third laws applied to a closed system; basis of collision and recoil analysis.
-- NCERT Class 11 Physics, Ch. 4, p. 57Conservation Linear Momentum
8 MCQs4 revision cards9-step worked example
Source: NCERT Laws of MotionPYQ coverage: NEET 2022Official key: NTA-verifiedLast updated: 25 Sep 2026
Conservation Linear Momentum, explained for NEET
The trap that costs marks here: treating momentum as a number instead of a vector. A body at rest explodes into three fragments — you add their momenta as magnitudes and get a nonzero total. But momentum conservation is a vector equation: the x-components sum to zero and the y-components sum to zero, independently. Every NEET explosion or recoil problem tests whether you decompose or just add magnitudes.
The law. Newton's second law in the form F = dp/dt tells you that if the net external force on a system is zero, the total linear momentum does not change (NCERT Class 11 Physics Chapter 4, page 57). Internal forces — collision impacts, spring forces, explosion gases — always cancel by Newton's third law. They redistribute momentum among parts of the system but cannot change the total.
The condition is strict. "Net external force = 0" means zero over the time interval you care about. During a collision lasting milliseconds, gravity's impulse is negligible compared to the collision force, so momentum is conserved across the collision instant. But if you track a projectile for several seconds of free fall, gravity's impulse is not negligible — momentum is not conserved over that interval.
Component-wise application. Write Σ mᵢvᵢ (before) = Σ mᵢvᵢ (after) along each axis separately. In explosion problems, the initial momentum is zero (body at rest), so the vector sum of all fragment momenta must also be zero. Two fragments flying perpendicular do not cancel each other's momenta by simple subtraction — you need Pythagoras on the resultant.
Watch out: impulse (J = Δp) has the same units as momentum (kg·m/s), not the same units as force (N). Confusing impulse with force inflates or deflates your answer by a factor of seconds.
Can you answer these Conservation Linear Momentum MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Direct ApplicationPractice
A 10 kg body at rest explodes into two fragments of mass 4 kg and 6 kg. If the 4 kg fragment moves at 6 m/s, what is the speed of the 6 kg fragment?
Show answer and why every option is right or wrong
Answer: D. Initial momentum = 0. By conservation: 4 × 6 + 6 × v₂ = 0 (taking the 4 kg direction as positive, the 6 kg moves opposite). |v₂| = 24/6 = 4 m/s. (NCERT Class 11 Physics Chapter 4, page 57.)
Why A is wrong: A is wrong because it assumes both fragments move at the same speed, ignoring the mass difference. Conservation requires m₁v₁ = m₂v₂, not v₁ = v₂. (trap: momentum vector vs scalar confusion)
Why B is wrong: B is wrong because it divides 24 by 10 (total mass) instead of by 6 (the fragment's own mass). Only the 6 kg fragment's mass enters the denominator.
Why C is wrong: C is wrong because it adds the masses (4 + 6 = 10) and divides incorrectly. The product m₁v₁ = 24 kg·m/s must equal m₂v₂, giving v₂ = 4 m/s, not 10 m/s.
MCQ 2Easy RecallPractice
Linear momentum of a system is conserved when:
Show answer and why every option is right or wrong
Answer: A. Conservation of linear momentum requires the net external force on the system to be zero. Internal forces always cancel by Newton's third law and cannot change total momentum. (NCERT Class 11 Physics Chapter 4, page 57.)
Why B is wrong: B is wrong because internal forces always cancel in pairs by Newton's third law — they never affect total system momentum regardless of their magnitude. The condition is about external forces, not internal ones.
Why C is wrong: C is wrong because kinetic energy conservation defines elastic collisions specifically. Momentum is conserved in both elastic and inelastic collisions as long as net external force is zero.
Why D is wrong: D is wrong because collisions are internal interactions. Momentum is conserved precisely during collisions (when external impulse is negligible). The presence or absence of collisions is irrelevant to the conservation condition.
MCQ 3Easy RecallPractice
A gun of mass 4.0 kg fires a bullet of mass 20 g horizontally with a speed of 400 m/s. The recoil speed of the gun is:
Show answer and why every option is right or wrong
Answer: B. B is correct. Gun and bullet start at rest, so the total momentum stays zero: m_bullet v_bullet + M_gun V_gun = 0. The speed of the gun is V = (0.020 kg × 400 m/s)/4.0 kg = 2.0 m/s, directed opposite to the bullet.
Why A is wrong: A is wrong because 2000 m/s uses the bullet's mass in grams, 20, instead of 0.020 kg.
Why C is wrong: C is wrong because 8.0 is the bullet's momentum, 0.020 × 400 = 8.0 kg·m/s, reported as a speed. It still has to be divided by the gun's mass.
Why D is wrong: D is wrong because 0.50 m/s is the ratio inverted: M/(mv) = 4.0/8.0. The heavier gun must move slower than the bullet by the ratio of the masses.
MCQ 4CalculationPractice
A ball of mass 0.2 kg is dropped from 5 m and rebounds to 1.25 m. Taking g = 10 m/s², the impulse exerted by the floor on the ball is:
Show answer and why every option is right or wrong
Answer: C. Speed before impact: v₁ = √(2 × 10 × 5) = 10 m/s (downward). Speed after rebound: v₂ = √(2 × 10 × 1.25) = 5 m/s (upward). Taking upward as positive: Δp = m(v₂ − (−v₁)) = 0.2 × (5 + 10) = 3 kg·m/s. (NCERT Class 11 Physics Chapter 4, page 55.)
Why A is wrong: A is wrong because it computes |v₁ − v₂| = |10 − 5| = 5 and then 0.2 × 5 = 1 kg·m/s. This ignores the velocity direction reversal at bounce — the velocities must be added, not subtracted. (trap: impulse rebound sign reversal)
Why B is wrong: B is wrong because it uses only the pre-impact speed: 0.2 × 10 = 2 kg·m/s. This treats the ball as stopping at the floor without rebounding, missing the upward momentum contribution.
Why D is wrong: D is wrong because it uses only the post-rebound speed: 0.2 × 5 × 0.5 = 0.5 kg·m/s. The impulse must account for the full momentum change including the direction reversal.
MCQ 5CalculationPractice
A 12 kg body at rest explodes into three equal fragments. Two fragments fly off perpendicular to each other, each at 8 m/s. The speed of the third fragment is:
Show answer and why every option is right or wrong
Answer: B. Each fragment has mass 4 kg. Fragment 1: momentum = 4 × 8 = 32 kg·m/s along x. Fragment 2: momentum = 4 × 8 = 32 kg·m/s along y. The third fragment's momentum must cancel the resultant: magnitude = √(32² + 32²) = 32√2 kg·m/s. Speed = 32√2 / 4 = 8√2 m/s. (NCERT Class 11 Physics Chapter 4, page 57.)
Why A is wrong: A is wrong because it assumes the third fragment moves at the same speed as the others. Since two perpendicular momenta combine by Pythagoras (not simple addition), the resultant is larger by a factor of √2. (trap: momentum vector vs scalar confusion — adding magnitudes instead of vectors)
Why C is wrong: C is wrong because it adds the momenta as scalars: 32 + 32 = 64 kg·m/s, giving v = 64/4 = 16 m/s. Perpendicular vectors combine by Pythagoras: √(32² + 32²) = 32√2, not 64. (trap: momentum vector vs scalar confusion)
Why D is wrong: D is wrong because it divides by the total mass (12 kg) instead of the third fragment's mass (4 kg): 32√2/12 ≈ 3.77 ≈ 4√2 m/s. Conservation applies to the third fragment alone.
MCQ 6Direct ApplicationPractice
A bullet of mass 20 g moving at 500 m/s embeds itself in a 4.98 kg wooden block at rest. The velocity of the block-bullet system immediately after impact is:
Show answer and why every option is right or wrong
Answer: A. By conservation of momentum: 0.020 × 500 = (0.020 + 4.98) × v. So 10 = 5.00 × v, giving v = 2 m/s. This is a perfectly inelastic collision — momentum is conserved but kinetic energy is not. (NCERT Class 11 Physics Chapter 4, page 57.)
Why B is wrong: B is wrong because 5 m/s comes from sharing the bullet's momentum of 10 kg·m/s equally between the two bodies (10/2), as if momentum split by number of bodies rather than by mass. The combined 5.00 kg moves together, so v = 10/5.00 = 2 m/s.
Why C is wrong: C is wrong because 10 is the bullet's momentum, 0.020 × 500 = 10 kg·m/s, reported as a speed without dividing by the combined mass of 5.00 kg.
Why D is wrong: D is wrong because 0.5 m/s is the inverted ratio, total mass over momentum (5.00/10), instead of momentum over total mass.
MCQ 7Direct ApplicationPractice
A cannon of mass 1000 kg fires a shell of mass 1 kg at 500 m/s. The recoil speed of the cannon is:
Show answer and why every option is right or wrong
Answer: D. Initial momentum = 0 (both at rest). After firing: 1 × 500 + 1000 × (−v) = 0. So v = 500/1000 = 0.5 m/s. The cannon recoils in the direction opposite to the shell. (NCERT Class 11 Physics Chapter 4, page 57.)
Why A is wrong: A is wrong because it divides by 100 instead of 1000: 500/100 = 5 m/s. The cannon's mass is 1000 kg, not 100 kg.
Why B is wrong: B is wrong because it divides by 10000 instead of 1000: 500/10000 = 0.05 m/s. The denominator should be the cannon mass (1000 kg), not 10× that.
Why C is wrong: C is wrong because it divides by 10 instead of 1000: 500/10 = 50 m/s. This is an order-of-magnitude error in the cannon's mass.
MCQ 8Concept TrapPractice
A student applies conservation of momentum to a ball thrown vertically upward, claiming its momentum is the same at launch and at the highest point. This is incorrect because:
Show answer and why every option is right or wrong
Answer: C. Gravity is a net external force acting on the ball throughout the flight. Since the net external force is not zero, momentum is not conserved over this interval. At the top, the ball's momentum is zero — clearly different from launch. (NCERT Class 11 Physics Chapter 4, page 57.)
Why A is wrong: A is wrong because momentum is perfectly well-defined in any direction — it is a vector quantity (p = mv) with no restriction on the axis. The issue is the non-zero external force, not the definition of momentum.
Why B is wrong: B is wrong because there is no collision involved in vertical throw. Even if air resistance is present, the fundamental reason momentum is not conserved is the net external force (gravity), not 'inelasticity.'
Why D is wrong: D is wrong because the ball's mass does not change during flight (it is a rigid body, not a rocket). The non-conservation is due to the gravitational external force, not variable mass.
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Conservation Linear Momentum: quick recall before you leave
How do you solve a Conservation Linear Momentum question? A worked example
- 1
Given
A body of mass 5m, initially at rest, explodes into three fragments of mass m, 2m, and 2m. The two fragments of mass 2m each fly off at speed v, perpendicular to each other.
- 2
Required
Find the speed of the fragment of mass m.
- 3
Concept
Conservation of linear momentum. The body is initially at rest, so total initial momentum = 0. After explosion, the vector sum of all fragment momenta must equal zero.
- 4
Formula
Σ mᵢvᵢ = 0 (since p_initial = 0)
- 5
Substitution
Let fragment 1 (mass 2m) move along +x at speed v, and fragment 2 (mass 2m) move along +y at speed v. Fragment 3 (mass m) moves at speed u in some direction.
x-component: 2m·v + 0 + m·u_x = 0 → u_x = −2v
y-component: 0 + 2m·v + m·u_y = 0 → u_y = −2v - 6
Calculation
Speed of fragment 3: u = √(u_x² + u_y²) = √((2v)² + (2v)²) = √(4v² + 4v²) = √(8v²) = 2v√2
Note: The mass ratio 5m = m + 2m + 2m is an exact counting relationship. The factor 2 and √2 are mathematical constants. Neither contributes to significant-figure considerations. - 7
Final answer
The fragment of mass m flies off at speed 2√2 v in the direction diagonally opposite to the resultant momentum of the other two fragments (at 225° from the +x axis, or equivalently at 45° below the −x axis).
- 8
Common trap
Adding momenta as scalars: 2mv + 2mv = 4mv, then setting m·u = 4mv → u = 4v. This is wrong because the two 2m fragments move perpendicular to each other, not in the same direction. Their combined momentum magnitude is √((2mv)² + (2mv)²) = 2mv√2, not 4mv. The correct answer is 2√2 v, not 4v.
- 9
Similar NEET-style question
A body of mass 3m at rest breaks into three equal fragments. Two fragments move with speed v at 60° to each other. Find the speed of the third fragment.
Approach: Resolve the two momenta along and perpendicular to their bisector. The resultant of the two equal fragments along the bisector = 2 × mv × cos 30° = mv√3. The perpendicular components cancel. The third fragment moves opposite to the bisector at speed √3 v.
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What to remember before solving Conservation Linear Momentum questions
Which Conservation Linear Momentum formulas do you need for NEET?
1 formula — click to collapse
Conservation of linear momentum
If the net external force on a system of particles is zero, the total linear momentum of the system is conserved (vector equality of total p before and after any internal interaction). Basis of all collision and recoil analysis.
| Symbol | Quantity | SI Unit |
|---|---|---|
| F_ext | Net external force on system | N |
| p_total | Sum of m_i*v_i over all particles | kg*m/s |
Valid when
- Net EXTERNAL force is zero (internal forces always cancel by Newton's 3rd law)
- Conservation is vector — apply componentwise (x and y separately)
- Holds independent of whether collisions are elastic or inelastic
Do NOT use when
- External impulses present (gravity over a long time, friction)
Where do students lose marks on Conservation Linear Momentum?
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: Sign Convention
Student adds magnitudes of momenta of fragments instead of vectors; ignores cancellation when fragments fly perpendicular or in opposite directions.
When it triggers
Question describes a body at rest exploding into multiple fragments with given mass ratios and partial velocity info.
How to avoid
Total momentum is a VECTOR. Initial p = 0; therefore Σ m_i v_i = 0 as a vector equation. Decompose along chosen axes (often natural symmetry axes); sum = 0 in each.
Root cause: concept gap
Correction
Linear momentum is conserved only when the net EXTERNAL force is zero. Gravity over a finite time changes momentum. For collision problems we use conservation because the collision happens over a brief time interval where external impulses are negligible compared to internal collision forces.
Wrong option pattern
Distractor sets initial momentum = final momentum for a free-fall problem where gravity has acted for several seconds.
More in Laws of Motion: 14 exam traps and mistakes · 8 formulas · 9 question patterns from its other lessons.
Conservation Linear Momentum questions from past NEET papers
1 question from NEET 2022. Answers verified against NTA official keys. — click to collapse
NEET 2022
132v
2v
32v
422v
NTA Answer: Option 4(final)
How does NEET ask about Conservation Linear Momentum?
1 recurring pattern from past papers — click to collapse
A body at rest explodes into 3 fragments; mass ratios given; some fragments' velocities given; find velocity of remaining fragment. Apply vector momentum conservation: Σ m_i v_i = 0 (since initial p = 0). Common shape: 2:2:1 mass ratio, two equal-mass fragments fly perpendicular with given speed; third fragment recoils.
Multi StepMedium
Common distractors
scalar sum of momenta
Treating momentum as scalar; ignores vector cancellation
Sources
NCERT refs: Class 11 Physics Chapter 4, p.57
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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