Conservation of Linear Momentum β Recoil & Explosions (NCERT 4.7)
1. What Does Conservation Claim?
The law of conservation of linear momentum: for an isolated system β one feeling no net external force β the total linear momentum is constant, no matter how violently the members interact. Collisions, couplings, explosions, recoils: the books may balance in any crazy way internally, but the system's total p = mβvβ + mβvβ + β¦ reads exactly the same before and after. The law is not a new assumption; it is the second law and the third law audited together β and that derivation is itself the most-asked theory item from this chapter, so we run it line by line below.
The law's exam power is that it bypasses force entirely. During a bullet's flight down a barrel, forces spike and vanish in microseconds; nobody can integrate them honestly. But momentum needs only the endpoints: total p before = total p after. Recoil speeds, fragment velocities after an explosion, velocities after coupling β all are one-line bookkeeping problems once the external forces are ruled negligible.
2. Complete Theory: Derivation, Formulas, and the Energy Asymmetry
Derivation (learn it verbatim β it is a theory question). Consider a system of two interacting bodies with total momentum P = pβ + pβ. Differentiate: dP/dt = dpβ/dt + dpβ/dt. By the second law, dpβ/dt = Fββ (the force on body 1 from body 2) and dpβ/dt = Fββ. Adding: dP/dt = Fββ + Fββ. But the third law says Fββ = βFββ β an internal actionβreaction pair acting on different bodies β so dP/dt = 0 and P = constant. Generalise to any number of bodies: all internal forces come in cancelling pairs; only the external resultant Fext survives, giving dP/dt = Fext. Set Fext = 0 and conservation follows. Note the fine print: the law needs no net external force β frictionless tables, or a collision so brief that external impulses (like gravity's) are negligible compared to the enormous internal ones.
The 1-D workhorse equation. For two bodies along one axis: mβuβ + mβuβ = mβvβ + mβvβ. Sign convention decides everything: choose one direction positive and let negative numbers carry the opposite motion β a 10 m/s rightward approach against a 4 m/s leftward one enters as +10 and β4. The recoil special case (both start at rest, m fires v): 0 = mv + MV β V = β(m/M)v. The same equation, three costumes: gun-and-bullet, jumper-and-boat, nucleus-and-debris.
The energy asymmetry β the trap and the treasure. Momentum conservation never mentions kinetic energy, and that independence is the whole point: in inelastic interactions (bodies coupling, clay sticking, sand loading a trolley) kinetic energy visibly vanishes into heat, sound and deformation β yet momentum books stay balanced to the last kg m/s. In elastic interactions (idealised hard collisions, atoms) kinetic energy also survives, giving a second equation. Class 11 keeps the second equation for the collisions chapter of System of Particles; here the safe exam sentence is: momentum is conserved whenever the net external force is zero; kinetic energy is conserved only in elastic collisions. NTA statement-items pair these two sentences constantly.
Where external force is sneaky but forgivable. A projectile exploding mid-air: gravity acts on every fragment, so total momentum of the fragments is not conserved over long times β but during the microsecond explosion the impulse of gravity (mgΞt) is negligible, so "just before = just after" holds. Similarly a colliding pair on a rough floor: friction is external, but over the millisecond of impact its impulse is dwarfed by the contact impulse. NTA's fragmentation questions ("a 4 kg body moving at 5 m/s explodes into 3 kg and 1 kgβ¦") live exactly in this justification.
3. Visualising a Collision's Momentum Ledger
Figure 4.4 β Before/after: the ledger balances even when the trolleys couple
Top row: before β 4 kg at 5 m/s approaches a resting 2 kg. Bottom row: after β coupled, both move at 10/3 m/s. Compare total momentum in each panel.
Exam read-out: coupling costs 16.7 J of kinetic energy (heat + crunch) but zero momentum. Whenever an option claims "both momentum and KE are conserved" for sticking bodies, strike it β the ledger figure is the reasoning NTA wants in the explanation column.
4. Solved Examples
Example 1 β Bullet embeds in a block (perfectly inelastic)
A 10 g bullet moving horizontally at 500 m/s embeds itself in a 2 kg wooden block at rest on a frictionless table. Find the common velocity afterwards and the kinetic energy lost.
Solution (step by step): System = bullet + block; external horizontal force = 0 β momentum conserved. Initial p = 0.010 Γ 500 = 5 kg m/s. Final: (0.010 + 2)v = 2.010v. So v = 5/2.010 = 2.49 m/s in the bullet's direction. KE before = Β½ Γ 0.01 Γ 500Β² = 1250 J; KE after = Β½ Γ 2.01 Γ (2.49)Β² β 6.2 J β over 99% of the bullet's kinetic energy became heat and deformation, while momentum carried through untouched. That contrast is the page's whole message.
Example 2 β Jumping off a boat
A 50 kg girl jumps horizontally at 4 m/s from a 200 kg boat initially at rest (water resistance negligible). Find the boat's recoil velocity and the ratio of their kinetic energies.
Solution: Initial p = 0. After: 50 Γ 4 + 200 Γ V = 0 β V = β200/200 = β1 m/s: the boat recoils at 1 m/s opposite her jump. KE ratio: Kgirl/Kboat = (Β½ Γ 50 Γ 16)/(Β½ Γ 200 Γ 1) = 400/100 = 4 β and in general, at equal momenta the KE splits inversely to mass (K1/K2 = mβ/mβ), the pΒ²/2m form from the first-law page doing quiet work. Sanity check: lighter jumper, larger speed, larger energy share β.
5. Practice Questions
Q1. A 60 kg astronaut floating in space throws a 2 kg wrench away at 6 m/s. Find her recoil velocity and explain which law guarantees it.
ANSWER: Initial total p = 0. After: 2 Γ 6 + 60 Γ V = 0 β V = β0.2 m/s β she drifts backward at 0.2 m/s. The third law makes the throw's forces an internal pair for the astronaut+wrench system, and with no external force in deep space the second law fixes dP/dt = 0. This is exactly how spacecraft manoeuvre with thrusters.
Q2 (MCQ). A 4 kg body moving at 5 m/s explodes into fragments of 3 kg and 1 kg. If the 3 kg fragment continues along the original direction at 4 m/s, the velocity of the 1 kg fragment is: (a) 8 m/s along the original direction (b) 3 m/s opposite (c) 8 m/s opposite (d) 5 m/s along the original direction
ANSWER: (a) β during the explosion gravity's impulse is negligible, so momentum books close: 4 Γ 5 = 3 Γ 4 + 1 Γ v β v = 8 m/s along the original direction. Method note: fix one positive axis, write both fragment momenta with signs, and solve β exactly one option closes the ledger.
Q3. A 20 g bullet is fired from a 4 kg gun. The bullet leaves at 300 m/s. Find the gun's recoil speed, and the impulse the gun received.
ANSWER: Books: bullet takes 0.02 Γ 300 = 6 kg m/s forward, so the gun takes 6 kg m/s backward β V = 6/4 = 1.5 m/s recoil. The gun's impulse is J = Ξp = 6 NΒ·s β equal and opposite to the bullet's impulse (third law), and equal to force Γ barrel-time from the impulseβmomentum theorem.
6. Key Formulas & Takeaways
| Relation | Condition / remark |
|---|---|
| mβuβ + mβuβ = mβvβ + mβvβ | 1-D, isolated system; signs carry directions |
| 0 = mv + MV (recoil) | Both at rest initially; V = β(m/M)v |
| dP/dt = Fext | Internal pairs cancel by the third law |
| Ξ£p constant β Fext = 0 | Or interaction so brief external impulse β 0 |
| K not necessarily conserved | Only elastic interactions keep KE; momentum never needs it |
| K1/K2 = mβ/mβ at equal p | Lighter fragment carries more energy share |
The full collision taxonomy (elastic, inelastic, coefficient of restitution) opens in System of Particles and Rotational Motion; this page's ledger discipline is the foundation it builds on.
7. Frequently Asked Questions
State and derive the law of conservation of linear momentum.
For an isolated system (no net external force), the total linear momentum remains constant. Derivation: the second law gives dP/dt = F_ext for the system's total momentum P. By the third law, internal forces occur in equal-and-opposite pairs acting on different bodies, so their contributions cancel in the sum and only the external force remains. Setting F_ext = 0 gives dP/dt = 0, hence P = constant. Component-wise, each direction conserves momentum independently.
Why is momentum conserved even in collisions where kinetic energy is lost?
Momentum conservation requires only that the net external force on the system be zero β it says nothing about energy. In an inelastic collision, kinetic energy converts to heat, sound and deformation, but every such internal process transfers momentum between the bodies in equal-and-opposite pairs (third law), leaving the total unchanged. That is why a lump of clay sticking to a wall-and-trolley system, or two trolleys coupling together, still obeys m1u1 + m2u2 = (m1 + m2)v exactly.
What is the recoil velocity of a gun firing a bullet?
For a gun of mass M firing a bullet of mass m with velocity v, both initially at rest, conservation gives 0 = mv + MV, so the recoil velocity V = -(m/M)v β the gun moves backward with speed (m/M)v. The negative sign records the opposite direction. The same formula covers a person jumping off a boat, a nucleus emitting a particle, and an astronaut throwing a wrench β in every case the lighter product carries more speed, and the momenta are equal and opposite.
A body explodes in mid-air. Is momentum conserved despite the explosion forces?
Yes, for the system of fragments. The explosive forces are internal β each fragment is pushed by its neighbours with equal and opposite third-law partners, so they cancel in the sum. Gravity is external but during the very short explosion the impulse of gravity (mg Ξt) is negligible, so momentum just before equals momentum just after the explosion. Afterwards, of course, gravity changes the total momentum of the fragments continuously. Typical NTA questions give the pre-explosion momentum and ask for one fragment's velocity, solved by straight bookkeeping.
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