Momentum Conservation Problems
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Direct answer
Fire a rifle and the butt kicks your shoulder with the same momentum the bullet carries forward — that symmetry is conservation of linear momentum: with no external force acting, total momentum stays constant along every axis, Σmv_before = Σmv_after. Momentum is a vector, so signs carry the physics; the method is always the same three lines — write momentum before, write it after, equate component by component. Recoil speed follows m_gun v_gun = m_bullet v_bullet; internal forces during the brief blast cancel in pairs (Newton's third law), which is why the law applies exactly during collisions, explosions and rocket thrust even though kinetic energy does not survive them.
What you must remember
- The law: the total linear momentum of an isolated system is conserved; apply it axis by axis in two-dimensional problems.
- Why collisions qualify: internal forces act in equal and opposite pairs, so impacts and explosions conserve momentum even when kinetic energy is lost to heat and deformation.
- Recoil template: v_gun = (m_bullet/m_gun) × v_bullet; equal and opposite momenta, but kinetic energies in inverse ratio of masses (KE = p²/2m).
- Explosion at rest: fragment momenta must vector-sum to zero; a body bursting mid-air keeps its centre of mass on the original trajectory — NCERT's centre-of-mass logic.
- Rocket equation: v − u = v_r ln(M_initial/M_final), with v_r the exhaust speed relative to the rocket; thrust comes from ejecting mass, not from pushing air.
- Man-on-boat template: a person of mass m walking length L along a boat of mass M (still water) shifts the boat back by mL/(M + m) — centre of mass stays fixed.
- Sign hygiene: fix one positive direction before the first line; every option list contains the correct magnitude with one wrong sign.
A recoil problem, fully reasoned
A 5 kg rifle fires a 50 g bullet at 400 m/s. Momentum before is zero, so 5 × v_gun + 0.05 × 400 = 0, giving v_gun = −4 m/s — four metres per second backward. Now compare energies: the bullet takes ½ × 0.05 × 400² = 4000 J, the rifle only ½ × 5 × 4² = 40 J. Equal momenta, energies standing in the inverse ratio of the masses — which is why the shoulder survives and the target does not, and why "ratio of kinetic energies" questions are really mass-ratio questions in disguise. Finish with the boat: a 60 kg person walks 4 m along a 120 kg boat; the boat drifts back 60 × 4/(180) ≈ 1.33 m, so the person actually covers only about 2.67 m relative to the water — the centre of mass never moved.
Where NEET sets the trap
The most common slip is scalar thinking: adding magnitudes when directions oppose, or dropping the minus sign on recoil. The second is the explosion-plus-projectile hybrid — a shell fired upward bursts at the top of its flight; where do the fragments land? Around the original impact point, because gravity keeps acting on the unchanged centre of mass while pieces scatter symmetrically about it. Assertion-reason items probe the boundary of the law: momentum is conserved during inelastic collisions (true) while kinetic energy is not (also true); a rocket accelerates in empty space (true, no medium needed); and during the flight of the projectile-fragment system, the centre of mass follows the original parabola regardless of the explosion. Numericals asking for "velocity of centre of mass" after collision are zero-computation gifts if this is understood.
Frequently asked questions
Why is momentum conserved in a collision even when kinetic energy is not?
Internal collision forces act in equal and opposite pairs (Newton's third law), so the total momentum cannot change; kinetic energy can convert into heat, sound and deformation.
What is the recoil velocity of a 5 kg gun firing a 50 g bullet at 400 m/s?
4 m/s backward, from 0 = 5v + 0.05 × 400; the gun's kinetic energy is a hundred times smaller than the bullet's.
A projectile explodes in mid-air. What happens to the fragments' centre of mass?
It continues on the original parabolic path — gravity is the only external force, so the centre of mass obeys the same trajectory as the unbroken shell.
How are two-dimensional explosion problems solved?
Conserve momentum separately along x and y; if the body was at rest, the vector sum of the fragments' momenta must be zero, which fixes magnitudes and directions together.
On what principle does rocket propulsion work?
Momentum conservation — ejecting exhaust backwards at high speed gives the rocket forward momentum; the ideal rocket equation is v − u = v_r ln(M_initial/M_final).