NEET Physics — Newton's Laws of Motion
Conservation of Linear Momentum – Complete Notes, Revision, Important Questions & Downloads
Conservation of Linear Momentum states that if no external force acts on a system (ΣF_ext = 0), the total momentum of the system remains constant. This topic covers three subtopics: Principle of Conservation (ΣF_ext = 0 → p_total = constant), Gun Recoil (initial momentum = 0 → m₁v₁ = −m₂v₂), and Rocket Propulsion (thrust F = −u × dm/dt, velocity equation v = u·ln(m₀/m) − gt). NEET tests all three as separate question types: statement-based questions on the law (independent of frame of reference, equivalent to Newton's Third Law), numerical problems on gun recoil and explosion problems, and conceptual questions on rocket thrust. The law is the most powerful conservation law in classical mechanics — valid for both elastic and inelastic collisions.
Theory + NumericalsNewton's Laws Ch.4ΣF=0 → p=const
Expected QuestionsQ
1–2
Conservation of momentum is tested every year in NEET — both as direct statement questions and as collision/explosion numericals. Gun recoil and rocket propulsion are standard NEET numerical scenarios.
Time Required⏱
90 min
20 min for the law statement and its properties (frame independence, equivalence to Third Law). 30 min for gun recoil and explosion problems. 40 min for rocket propulsion derivation, thrust formula, and numerical application.
Difficulty⚡
Medium
The law itself is simple. Gun recoil problems are easy (both initially at rest → m₁v₁ = m₂v₂). Rocket propulsion is harder — the thrust formula and log velocity equation require careful derivation. The logarithmic velocity equation v = u·ln(m₀/m) − gt is tested qualitatively (which variable increases thrust) and occasionally numerically.
NRI USA Curriculum GapUS
Low
AP Physics 1 covers conservation of momentum, collisions (elastic and inelastic), explosion problems, and rocket propulsion. The NEET-specific additions: the explicit equivalence between conservation of momentum and Newton's Third Law, the logarithmic rocket velocity equation v = u·ln(m₀/m) − gt as a named formula, and statement-based questions on the law's frame independence.
8Subtopics
8+Practice Questions
4Free Downloads
90 minPrep Time
NEET Weightage — Conservation of Linear Momentum
Newton's Laws of Motion (Chapter 4)| NEET Year | Questions from this Topic | Bar | Marks |
|---|---|---|---|
| 2024 | 1 | 4 | |
| 2023 | 1 | 4 | |
| 2022 | 1 | 4 | |
| 2021 | 1 | 4 | |
| 2020 | 1 | 4 | |
| 2019 | 1 | 4 | |
| 6-Year Total (2019–2024) | 3–6 | 12–24 |
Principle: If no external force acts on a system (called isolated) of constant mass, the total momentum of the system remains constant with time. ΣF_ext = 0 → p_total = constant → m₁v₁ + m₂v₂ + ... = constant. Applies to all types of collisions (elastic and inelastic) and to explosions.
Gun Recoil: Bullet (m) +gun (M) initially at rest → total initial p = 0. After firing: mv_bullet + MV_gun = 0 → MV_gun = −mv_bullet → V_gun = −(m/M)v_bullet. Recoil speed = m × v_bullet / M. The negative sign → recoil is opposite to bullet direction.
Rocket Propulsion: Thrust on rocket = −u × (dm/dt), where u = exhaust velocity relative to rocket; dm/dt = rate of fuel ejection (negative, as mass decreases). Net force = Thrust − mg = −u(dm/dt) − mg. Rocket velocity: v = u·ln(m₀/m) − gt, where m₀ = initial mass, m = current mass.
Gun Recoil: Bullet (m) +gun (M) initially at rest → total initial p = 0. After firing: mv_bullet + MV_gun = 0 → MV_gun = −mv_bullet → V_gun = −(m/M)v_bullet. Recoil speed = m × v_bullet / M. The negative sign → recoil is opposite to bullet direction.
Rocket Propulsion: Thrust on rocket = −u × (dm/dt), where u = exhaust velocity relative to rocket; dm/dt = rate of fuel ejection (negative, as mass decreases). Net force = Thrust − mg = −u(dm/dt) − mg. Rocket velocity: v = u·ln(m₀/m) − gt, where m₀ = initial mass, m = current mass.
1.0
Avg Questions / Year
24
Total Marks (6 yrs)
Direct
Pattern
Medium
Difficulty
How to Prepare Conservation of Linear Momentum for NEET
1
Master the law statement and its properties The law: 'If no external force acts on a system, the total momentum of the system remains constant.' Key properties tested in NEET: (1) The law is independent of frame of reference (even though linear momentum IS frame-dependent). (2) Conservation of linear momentum is equivalent to Newton's Third Law. (3) The law applies to isolated systems. All three properties appear in assertion-reason questions.
2
Drill explosion and gun recoil problems Template for all 'initially at rest → explode/fire' problems: initial p = 0 → final p₁ + p₂ = 0 → m₁v₁ = −m₂v₂. The vector equation: components must balance. For 2D explosions: ΣFx conservation AND ΣFy conservation separately. For straight-line problems: magnitudes are m₁v₁ = m₂v₂.
3
Learn rocket propulsion qualitatively and formula-wise Thrust F = −u·dm/dt (magnitude = u × mass flow rate). Larger exhaust velocity → larger thrust (for same rate of fuel ejection). Larger fuel flow rate → larger thrust. Net upward acceleration: a = F_thrust/m − g. Velocity equation v = u·ln(m₀/m) − gt — as fuel burns (m decreases), ln(m₀/m) increases → rocket accelerates. NEET also asks which factor increases thrust: answer is exhaust velocity, not just the fuel flow.
Study Materials — Conservation of Linear Momentum
PDF · Cheat Sheet · MCQ Set · PYQFull Notes
Law of Conservation of Linear Momentum. Properties (frame independence, Third Law equivalence). Gun recoil derivation. 2D gun/explosion examples. Rocket propulsion theory, thrust formula, and velocity equation derivation.
Download NotesFormula Sheet
ΣF_ext = 0 → p_total = constant; m₁u₁+m₂u₂ = m₁v₁+m₂v₂; Gun recoil: MV = −mv; Thrust: F = −u(dm/dt); Rocket velocity: v = u·ln(m₀/m) − gt.
Download SheetMCQ Practice
15 questions: law statement assertions, gun recoil numericals, 2D explosion problems, rocket thrust calculations, collision problems using momentum conservation, frame independence questions.
Download MCQsPYQ
Year-tagged NEET questions on momentum conservation — gun recoil, rocket propulsion, collision problems, and law statement questions.
Download PYQsSubtopics in Conservation of Linear Momentum
2-Column TableRapid Revision — Conservation of Linear Momentum
Concept → Trap → Example1) Principle of Conservation
CoreIf no external force acts on a system (called isolated) of constant mass, the total momentum of the system remains constant with time. ΣF_ext = 0 → d(p_total)/dt = 0 → p_total = constant. The law applies to both elastic and inelastic collisions, and to explosions.
- Key NEET property 1: Law of conservation of linear momentum is independent of frame of reference, though linear momentum depends on frame of reference. The conservation holds in ALL inertial frames even though the actual momentum values differ.
- Key NEET property 2: Conservation of linear momentum is equivalent to Newton's Third Law of motion. (Proof: from F_12 = −F_21 → dp₁/dt = −dp₂/dt → d(p₁+p₂)/dt = 0 → p_total = constant.)
- NEET: 'In a perfectly inelastic collision, is momentum conserved?' → Yes. Momentum is always conserved (provided ΣF_ext = 0). Kinetic energy is NOT conserved in inelastic collisions. These two are independent — learn which one is conserved in each collision type.
Example (NEET-style)A 5 kg cart moving at 4 m/s east collides with a stationary 3 kg cart, and they stick together. Final velocity: (5×4 + 3×0) = (5+3)v_f → v_f = 20/8 = 2.5 m/s east. Momentum: before = 20 kg·m/s east; after = 8 × 2.5 = 20 kg·m/s east ✓ (conserved). Kinetic energy: before = ½×5×16 = 40 J; after = ½×8×6.25 = 25 J. KE lost = 15 J (not conserved — perfectly inelastic). Momentum conserved, KE not — perfect example of the distinction.
2) Gun Recoil
CoreA gun and bullet are initially at rest. Total initial momentum = 0. After firing, momentum must still = 0: m_bullet × v_bullet + M_gun × V_gun = 0. Therefore: M_gun × V_gun = −m_bullet × v_bullet. Recoil speed: V_gun = (m_bullet × v_bullet) / M_gun. Recoil direction: exactly opposite to bullet direction.
- Derivation basis: both initially at rest → p_initial = 0. No external horizontal forces (assuming frictionless ground) → p_final = 0. The bullet goes forward; the gun recoils backward.
- Generalisation for any explosion (object at rest disintegrating into two parts): m₁v₁ + m₂v₂ = 0 → m₁v₁ = −m₂v₂ (in vector form). The two fragments move in opposite directions; their momenta are equal and opposite.
- NEET: 'A shell at rest explodes into two fragments of masses 4 kg and 6 kg. If the 4 kg fragment moves at 6 m/s, what is the speed of the 6 kg fragment?' → 4×6 = 6×v → v = 4 m/s (opposite direction).
Example (NEET-style)Gun mass M = 2 kg, bullet mass m = 0.02 kg, bullet velocity v = 500 m/s forward. Recoil velocity: V = (m × v) / M = (0.02 × 500) / 2 = 10/2 = 5 m/s backward. Check: p_after = 0.02×500 + 2×(−5) = 10 − 10 = 0 = p_before ✓. If the person holds the gun, the 'gun' system includes part of the person's arm — the effective M is larger, making V smaller. This is why holding a gun firmly against the shoulder reduces apparent recoil.
3) Rocket Propulsion
CoreRocket ejects exhaust gas at velocity u (relative to rocket) to move forward. Thrust: F_thrust = −u × (dm/dt), where dm/dt < 0 (mass decreasing) so thrust is positive (forward). Net force on rocket: F_net = −u(dm/dt) − mg. Rocket velocity equation: v = u·log_e(m₀/m) − gt, where m₀ = initial mass, m = current mass.
- Derivation: at time t, rocket mass = m, velocity = v. Ejects gas dm at velocity (v−u) relative to ground in time dt. Momentum at t: mv. Momentum at t+dt: (m−dm)(v+dv) + dm(v−u). Conservation: mv = (m−dm)(v+dv) + dm(v−u). Simplifying: m·dv = u·dm (where dm < 0). Integrating: v = u·ln(m₀/m) − gt.
- Thrust F = u × |dm/dt|. To maximise thrust: use large exhaust velocity u (use high-energy fuels) OR increase mass ejection rate |dm/dt| (use large engines). NEET asks which parameter: answer — exhaust velocity u is more efficient (multiplied directly into thrust).
- NEET: 'A rocket consumes 60 kg/s of fuel, exhaust velocity = 2000 m/s. Calculate thrust.' → F = u × |dm/dt| = 2000 × 60 = 120,000 N = 120 kN.
Example (NEET-style)Rocket: initial mass m₀ = 1000 kg, burns fuel at 10 kg/s, exhaust velocity u = 500 m/s. Thrust = 500 × 10 = 5000 N. After 20 s: remaining mass = 1000 − 10×20 = 800 kg. Velocity gained (ignoring gravity): v = u·ln(m₀/m) = 500 × ln(1000/800) = 500 × ln(1.25) = 500 × 0.2231 = 111.6 m/s. With gravity correction: v = 500 × ln(1.25) − 10 × 20 = 111.6 − 196 = negative. This shows gravity dominates at low thrust — real rockets need thrust > weight for lift-off: F_thrust > m₀g. Here 5000 N < 1000×10 = 10000 N, so this rocket cannot lift off. Increasing |dm/dt| or u would be required.
US Curriculum Gaps — Conservation of Linear Momentum
Topics in this section are tested in NEET but organised differently in standard US physics courses.Momentum Conservation Equivalent to Newton's Third Law (AP Physics 1 Gap)
AP Physics 1 covers momentum conservation and Newton's Third Law separately. NEET tests the explicit statement 'Conservation of linear momentum is equivalent to Newton's Third Law of motion' as a standalone assertion. The derivation (Third Law → equal and opposite forces → equal and opposite impulses → total momentum conserved) is a NEET-specific connection. Additionally, NEET tests 'The law of conservation of linear momentum is independent of frame of reference, though linear momentum depends on frame' — another assertion-reason item not explicitly in AP Physics 1.
- NEET: 'Conservation of linear momentum is equivalent to Newton's Third Law' → True (assertion-reason question)
- NEET: 'Law is independent of frame of reference' → True — tested verbatim
- AP Physics 1: these equivalences are not tested as standalone assertions
Rocket Velocity Equation with Natural Logarithm (AP Physics C Gap)
AP Physics C: Mechanics covers the rocket thrust equation (F = v_e × dm/dt) and may derive the velocity equation using calculus. However, the specific NCERT form v = u·log_e(m₀/m) − gt with the natural logarithm is tested as a named formula in NEET. NEET numerical problems using this equation (find velocity after burning given fuel) appear. AP Physics 1 does not cover the logarithmic rocket velocity equation at all — only AP Physics C does, and even there it is less emphasised as a formula to memorise with a specific form.
- NEET: v = u·ln(m₀/m) − gt is a named formula, tested in numericals
- AP Physics 1: rocket propulsion is conceptual only (thrust = force from gas ejection)
- AP Physics C: calculus derivation possible but specific form may differ
NEET-Style Practice Questions — Conservation of Linear Momentum
4 Questions1A shell of mass 10 kg is at rest. It explodes into two parts of masses 6 kg and 4 kg. The 6 kg fragment moves at 5 m/s. What is the speed of the 4 kg fragment?Explosion/Recoil
5 m/s
7.5 m/s
3 m/s
12 m/s
Initial momentum = 0 (at rest). After explosion: p_total = 0. 6 × 5 + 4 × v₂ = 0 → 30 + 4v₂ = 0 → v₂ = −7.5 m/s. Speed of 4 kg fragment = 7.5 m/s (in the opposite direction to the 6 kg fragment). Check: 6×5 = 30 kg·m/s forward; 4×7.5 = 30 kg·m/s backward. Sum = 0 ✓. Trap: do not add masses — the two fragments have opposite velocities; use the vector equation m₁v₁ + m₂v₂ = 0.
2Assertion: Law of conservation of linear momentum is independent of frame of reference. Reason: Linear momentum itself depends on the frame of reference.Assertion-Reason
Both assertion and reason are correct, and reason is the correct explanation
Both assertion and reason are correct, but reason is NOT a correct explanation of assertion
Assertion is correct but reason is incorrect
Both are incorrect
Both statements are true. Assertion: The LAW (Δp_total = 0 when ΣF_ext = 0) is valid in all inertial frames — TRUE. Reason: Linear momentum values DO change with frame (a ball at rest in one frame has non-zero momentum in another) — TRUE. However, the reason does NOT explain the assertion. The reason why the conservation LAW is frame-independent is that Newton's Second Law (from which conservation is derived) is valid in all inertial frames — not because momentum itself is frame-dependent. The reason and assertion are both true but are independent facts. Answer: B.
3A rocket of initial mass 1000 kg has an exhaust velocity of 500 m/s. What is the thrust when the fuel consumption rate is 20 kg/s?Rocket Thrust
20,000 N
10,000 N
500 N
2,500 N
Thrust = u × |dm/dt| = 500 × 20 = 10,000 N. The rocket weight at t=0: W = m₀g = 1000×10 = 10,000 N. In this case, thrust = weight → the rocket is on the verge of lift-off (net force = 0, a = 0). For the rocket to accelerate upward, thrust > weight, so |dm/dt| must be increased OR u must be increased. This threshold scenario (thrust = weight) is a standard NEET comprehension question.
4A man of mass 60 kg stands in a stationary boat of mass 200 kg. He walks 2 m toward the bow. How far does the boat move (assuming no external horizontal forces)?Conservation in Extended Problem
0.46 m
2 m
0.6 m
No displacement
Initial momentum of system (man + boat) = 0. No external horizontal force → centre of mass doesn't move. Man moves 2 m forward relative to boat. Let boat move d backward (relative to ground). Man moves forward relative to ground = (2 − d). Conservation of centre of mass position: m_man × displacement_man + m_boat × displacement_boat = 0. 60×(2−d) + 200×(−d) = 0 → 120 − 60d − 200d = 0 → 260d = 120 → d = 120/260 = 0.46 m. Boat moves 0.46 m in the direction opposite to man's walking. Man moves 2 − 0.46 = 1.54 m relative to ground.
Practice Problems — Conservation of Linear Momentum
Click "Reveal Answer" after attempting1A rifle of mass 3 kg fires a bullet of mass 10 g at a speed of 600 m/s. Find (a) the recoil velocity of the rifle, and (b) the ratio of the kinetic energy of the bullet to that of the rifle.
(a) 2 m/s, (b) 300:1
(a) 2 m/s, (b) 150:1
(a) 20 m/s, (b) 300:1
(a) 0.2 m/s, (b) 30:1
👁 Reveal Answer
(a) Initial momentum = 0. After firing: 0.010×600 + 3×V_rifle = 0 → V_rifle = −6/3 = −2 m/s. Recoil speed = 2 m/s (backward). (b) KE_bullet = ½×0.010×600² = ½×0.010×360000 = 1800 J. KE_rifle = ½×3×2² = ½×3×4 = 6 J. Ratio = 1800/6 = 300. KE of bullet is 300 times KE of rifle, even though their momenta are equal. This is because KE = p²/(2m) → KE ∝ 1/m for equal momenta. Lighter bullet (smaller m) has higher KE. The ratio equals M/m = 3/0.010 = 300. This is the standard result: for equal momenta, KE ratio = inverse of mass ratio.
2Two ice skaters (masses 50 kg and 80 kg) are initially at rest and push each other apart. The 50 kg skater moves at 4 m/s east. Find the velocity of the 80 kg skater.
2.5 m/s west
6.4 m/s west
4 m/s west
5 m/s west
👁 Reveal Answer
Initial momentum = 0. After push: 50×4 + 80×v₂ = 0 → 200 + 80v₂ = 0 → v₂ = −200/80 = −2.5 m/s. The 80 kg skater moves at 2.5 m/s west. This is a classic conservation momentum problem where the system starts at rest and internal forces (push) cannot change the total momentum. The lighter skater gets a proportionally larger speed. Apply KE check: KE₅₀ = ½×50×16 = 400 J; KE₈₀ = ½×80×6.25 = 250 J. Total KE = 650 J (came from chemical energy stored in skaters' muscles). Momentum: 50×4 − 80×2.5 = 200 − 200 = 0 ✓.
3A rocket has mass 10,000 kg (including 7,000 kg of fuel). Exhaust velocity = 1000 m/s. (a) What is the thrust when dm/dt = 100 kg/s? (b) Can the rocket lift off Earth? (c) What is the velocity when half the fuel is burned? (g = 10 m/s²)
(a) 100,000 N, (b) Yes (thrust > weight), (c) 485 m/s upward (no gravity correction)
(a) 100,000 N, (b) No, (c) 350 m/s
(a) 1,000,000 N, (b) Yes, (c) 693 m/s
(a) 100,000 N, (b) Yes, (c) 250 m/s
👁 Reveal Answer
(a) Thrust = u × |dm/dt| = 1000 × 100 = 100,000 N = 100 kN. (b) Weight at lift-off: W = m₀g = 10,000 × 10 = 100,000 N. Thrust = Weight exactly at t=0 → net force = 0. As fuel burns, mass decreases, weight decreases, thrust stays constant → rocket begins accelerating upward. So YES, the rocket lifts off (marginal at t=0, accelerates as fuel burns). (c) Half fuel burned: mass burned = 3,500 kg. Remaining mass m = 10,000 − 3,500 = 6,500 kg. Time to burn 3500 kg at 100 kg/s: t = 35 s. Velocity: v = u·ln(m₀/m) − gt = 1000·ln(10000/6500) − 10×35 = 1000×ln(1.538) − 350 = 1000×0.430 − 350 = 430 − 350 = 80 m/s. (Gravity penalty is large — 350 m/s lost to gravity versus 430 m/s gained from thrust.)
4Prove using momentum conservation that the centre of mass of an isolated system (ΣF_ext = 0) moves at constant velocity.
Centre of mass moves faster as energy increases
Centre of mass velocity = total momentum / total mass = constant when total momentum is constant
Centre of mass does not move for isolated systems
Centre of mass accelerates if internal forces are large
👁 Reveal Answer
Centre of mass velocity: v_cm = p_total / m_total (total momentum divided by total mass). For an isolated system: ΣF_ext = 0 → dp_total/dt = 0 → p_total = constant. Therefore v_cm = p_total / m_total = constant / constant = constant. The centre of mass of an isolated system moves at constant velocity (Newton's First Law for the centre of mass). Internal forces (Newton's Third Law pairs) cancel in the total momentum — only external forces change v_cm. Application: in an explosion, the centre of mass continues its original path; the fragments scatter around the trajectory the original object would have taken.
Physics — Newton's Laws of Motion Revision Checklist
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FAQ — Conservation of Linear Momentum
Notes · Downloads · Revision · Important QuestionsWhat is the law of conservation of linear momentum?
If no external force acts on a system (called isolated) of constant mass, the total momentum of the system remains constant with time. Mathematically: ΣF_ext = 0 → dp_total/dt = 0 → m₁v₁ + m₂v₂ + ... = constant. The law applies to all types of collisions (elastic and inelastic), to explosions, and to any isolated multi-body system. It is one of the most fundamental laws of classical mechanics.
Is momentum conservation always valid, or only in elastic collisions?
Momentum conservation is valid in ALL collisions — elastic, inelastic, and perfectly inelastic — as long as no external force acts on the system. It is not restricted to elastic collisions. In elastic collisions: BOTH momentum AND kinetic energy are conserved. In inelastic collisions: momentum IS conserved but kinetic energy is NOT (some KE converts to heat, sound, deformation). This distinction is a classic NEET trap.
Why is conservation of linear momentum equivalent to Newton's Third Law?
Consider two bodies with no external forces. Newton's Third Law: F₁₂ = −F₂₁ (force on body 1 by body 2 = negative of force on body 2 by body 1). So dp₁/dt = −dp₂/dt → d(p₁+p₂)/dt = 0 → p₁+p₂ = constant. Thus, Newton's Third Law directly implies conservation of total momentum. Conversely, if p_total is conserved for any isolated two-body system, Newton's Third Law must hold. The two are mathematically equivalent for systems with no external forces.
Is the law of conservation of momentum frame-dependent?
The LAW (the statement 'total momentum is conserved when ΣF_ext = 0') is INDEPENDENT of frame of reference. The ACTUAL MOMENTUM VALUES (the numerical magnitude and direction of p) ARE frame-dependent — the same body has different momentum in different frames. But whether total momentum is conserved (stays constant over time) is true in all inertial frames. This distinction is precisely what NEET tests in assertion-reason questions.
How does a rocket work without any external medium to push against?
A rocket works entirely through momentum conservation. The rocket ejects exhaust gas backward (action). By Newton's Third Law, the gas pushes the rocket forward (reaction). As more gas is ejected, the rocket gains forward momentum equal and opposite to the momentum of the ejected gas. No external medium (water, air, ground) is required — the rocket carries its own reaction mass (propellant). This is why rockets work in the vacuum of space. Thrust = exhaust velocity × mass flow rate = u × |dm/dt|.
Why does the kinetic energy of the bullet greatly exceed the kinetic energy of the recoiling gun, even though their momenta are equal?
After firing, |p_bullet| = |p_gun| (equal momenta by conservation). Kinetic energy: KE = p²/(2m). Therefore KE ∝ 1/m for equal momenta. Since m_bullet ≪ M_gun, KE_bullet ≫ KE_gun. Specifically: KE_bullet/KE_gun = M_gun/m_bullet. For a rifle (M = 3 kg, m = 10 g): ratio = 3/0.010 = 300. The bullet has 300 times the kinetic energy. The unequal energy comes from the chemical energy of the gunpowder — most of the energy goes to the light bullet, a small fraction to the heavy gun. Equal momentum does NOT mean equal kinetic energy.
A bomb at rest explodes into two unequal fragments. Do the fragments necessarily move in opposite directions?
Yes, for a simple two-fragment explosion where the initial momentum = 0. By conservation: m₁v₁ + m₂v₂ = 0 → m₁v₁ = −m₂v₂. The vectors v₁ and v₂ must be antiparallel (equal and opposite in direction). The fragments move in exactly opposite directions. If the bomb had initial velocity v₀ (non-zero), then p_total = (m₁+m₂)v₀ ≠ 0, and the fragments would NOT necessarily be in opposite directions — their momenta must sum to p_total, not zero.
What is the condition for a rocket to lift off the ground?
For lift-off: thrust > weight. Thrust = u × |dm/dt|. Weight at launch = m₀g. Condition: u × |dm/dt| > m₀g. As fuel burns, mass m decreases, so weight mg decreases, while thrust stays roughly constant (u and |dm/dt| are approximately constant) → the rocket accelerates more as it climbs. This is why rockets appear to speed up as they gain altitude (in addition to decreasing air drag). The first stage of any rocket has the highest fuel consumption rate because it must overcome the heaviest weight.
Can momentum be conserved in a system with friction?
No, not exactly. Friction is an external force (from the ground or another surface) on the system. If friction acts, ΣF_ext ≠ 0, and total momentum changes. However, if we include the ground as part of the system (Earth included), then friction becomes an internal force, and for the Earth + objects system, momentum IS conserved (Earth recoils). For practical NEET problems: if friction acts on the objects of interest, momentum is NOT conserved for that system. NEET explosion and gun problems usually specify 'on a smooth/frictionless surface' or 'in mid-air' to ensure external forces are zero and momentum is conserved.
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