Momentum

— Physics · AS Level (7408) · Collisions & Conservation of Momentum —

★ Exam Tip: Momentum conservation applies EVERYWHERE in a closed system, not just in collisions (it even applies to a rock swinging past a star with no contact at all). If a question doesn't say "elastic," ALWAYS assume inelastic and just use ΣPᵢ=ΣP_f — never reach for the elastic-only equations unless told to.

1. Momentum & Impulse

Linear momentum: the product of an object's mass × velocity. Vector, measured in kg m s⁻¹. p = m×v

Recall: Rate of change of momentum = Force. F = dp/dt = Δp/Δt

Gradient of a p–t graph = Force. Area under an F–t graph = Δp.

Δp = F × Δt (Impulse of a force) | Δp = p_f − p_i

2. Principle of Conservation of Momentum

"For a closed system, the total momentum remains constant, given there is no resultant force on the system."

★ ΣP_i = ΣP_f

For 2D problems, resolve into components: ① ΣP_ix = ΣP_fx ② ΣP_iy = ΣP_fy

★ Trick — the "unifying factor": just like time links horizontal/vertical in projectile motion, when one direction's equation has too many unknowns, solve the OTHER direction first and substitute back. Break ΣP=ΣP into x and y and solve whichever has fewer unknowns first.

Trick: don't pre-assign a sign to an unknown velocity/force — write it as a bare symbol and let the algebra's result tell you the sign.

3. Collisions — Elastic vs Inelastic

Type	Momentum	Kinetic Energy
Inelastic	Conserved	NOT conserved
Elastic	Conserved	Conserved

★ Default rule: if the question doesn't say "elastic," ASSUME inelastic — use only ΣPᵢ=ΣP_f. Never assume elastic unless told.

Method order for ELASTIC collisions (try in this order, stop once one works):

u₁−u₂ = v₂−v₁ (fastest, when applicable)
ΣPᵢ = ΣP_f
ΣKEᵢ = ΣKE_f (most algebra — save for last)

Note: Energy is scalar — use unsigned speeds in the KE equation, no convention needed there.

4. Deriving u₁−u₂ = v₂−v₁

m₁ at u₁, m₂ at u₂ before; m₁ at v₁, m₂ at v₂ after (both rightwards, convention +)

① ΣPi=ΣPf: m₁u₁+m₂u₂=m₁v₁+m₂v₂ ⟹ m₁(u₁−v₁)=m₂(v₂−u₂) ⟹ m₁/m₂ = (v₂−u₂)/(u₁−v₁) ② ΣKEi=ΣKEf (×2 to clear the ½'s): m₁u₁²+m₂u₂²=m₁v₁²+m₂v₂² ⟹ m₁(u₁²−v₁²)=m₂(v₂²−u₂²) ⟹ m₁/m₂ = (v₂²−u₂²)/(u₁²−v₁²) Combine ① and ②: (v₂−u₂)/(u₁−v₁) = (v₂²−u₂²)/(u₁²−v₁²) Factor with a²−b²=(a−b)(a+b): 1 = (v₂−u₂)(v₂+u₂) / [(u₁−v₁)(u₁+v₁)] ÷ [(v₂−u₂)/(u₁−v₁)] — the (v₂−u₂) and (u₁−v₁) factors cancel ⟹ u₁+v₁ = v₂+u₂ ⟹ u₁−u₂ = v₂−v₁

★ u₁−u₂ = v₂−v₁ (Elastic collision condition)

In words: Relative speed of approach (u₁−u₂) = Relative speed of separation (v₂−v₁)

same speed, same direction

same direction, different speed

head-on approach

5. Coalescing Collisions

Coalescing: Always inelastic. Upon collision, the masses combine and move together as one unit.

★ Why coalescing is ALWAYS inelastic: after coalescing there is only ONE final velocity — no v₁ and v₂ anymore — so the relative speed of separation is necessarily ZERO. Since elastic requires (relative speed of approach) = (relative speed of separation), and separation speed is pinned at 0, this can only be satisfied if the approach speed was ALSO already zero (i.e. no real collision). So every real coalescing collision is inelastic by logical necessity.

Worked example — coalescingCalculation

Object① 2kg at +1 m/s, object② 5kg at −10 m/s. They coalesce. Find the combined velocity.

ΣPi=ΣPf: 2(1)+5(−10) = 7(v) ⟹ v=(2−50)/7 ⟹ v≈−7 m/s (≈7 m/s left)

Newton's Third Law in Collisions

① t_A = t_B = t (contact time is the same for both) | ② F_A = −F_B (N3L)

③ Δp=F×Δt ⟹ F_A×t = −F_B×t ⟹ Δp_B = −Δp_A

★ Alternative derivation (no N3L needed!): just rearrange ΣPi=ΣPf directly: P_Ai+P_Bi=P_Af+P_Bf ⟹ −(P_Bf−P_Bi) = (P_Af−P_Ai) ⟹ −Δp_B=Δp_A. This is pure algebra from momentum conservation — either derivation is valid and reaches the same true statement.

★ Exam-critical wording trap: "Every action has an equal and opposite reaction" scores ZERO for "state Newton's third law" — it doesn't name an actual pair. Required: "If object A applies a force on object B in one direction, object B will apply a force of the same magnitude on object A but in the opposite direction."

6. Worked MCQ Practice

Firework explodes into 3 piecesMCQ · 2D, resolve x & y

Stationary firework explodes: two 50g pieces at v₁, v₂ symmetric at 60° above horizontal; one 100g piece straight down at 8 m/s. Find v₁, v₂.

Horizontal: 0 = (50×10⁻³)v₁cos60 − (50×10⁻³)v₂cos60 ⟹ v₁=v₂ (cancels neatly — solve this direction first!) Vertical: 0 = (50×10⁻³)v₁sin60+(50×10⁻³)v₂sin60+(100×10⁻³)(−8) ⟹ 8=v₁sin60 ⟹ v₁=v₂=9.2 m/s

Answer: B (9.2, 9.2)

Golf ball dropped, reboundsMCQ · kinematics feeds into momentum

Golf ball mass m dropped from h₁, rebounds to h₂. What is the total change in momentum?

Trick: the momenta aren't given directly — build them from the heights via v²=u²+2as first, THEN combine with Δp=m(v_f−v_i).

Impact speed: v²=2gh₁ ⟹ v_i=−√(2gh₁) (down, convention up=+) Rebound speed: 0=v_f²−2gh₂ ⟹ v_f=+√(2gh₂) (up) Δp = m(v_f−v_i) = m(√(2gh₂)+√(2gh₁))

Answer: B — m√2gh₁ + m√2gh₂

Ball bounces off a wallMCQ · average force

Ball 2.0kg at 4.0 m/s hits a wall, rebounds at 2.8 m/s. Collision time 150ms. Average force on the wall?

F=Δp/Δt=[(2)(−2.8)−(2)(4)]/(150×10⁻³) = −13.6/0.15 ⟹ F=−91N (magnitude 91N)

Answer: D (91N)

Which collision is NOT elastic?MCQ · check u₁−u₂=v₂−v₁

Table of u₁, u₂, v₁, v₂ for 4 collisions. Which is NOT elastic?

A: u₁−u₂=2−(−5)=7 vs v₂−v₁=−2−(−5)=3 ⟹ 7≠3 NOT elastic B, C, D: all check out equal (elastic)

Answer: A

Slow + fast vehicle collide — which is NEVER possible?MCQ · conceptual

Slow and fast vehicle collide head-on. Which outcome is NEVER possible, regardless of masses?

A vehicle transferring momentum to another always costs it some of its own speed — a colliding body's speed can never INCREASE as a result of giving momentum away.

Answer: C (the fast vehicle's speed increases) — impossible; A, B, D are all achievable with the right mass ratio

Rosetta probe — orbiter & lander separateMCQ · isolated system, no collision

Orbiter 170kg + lander 100kg, isolated system. Lander pushed towards comet, Δv=3.0 m/s. Change in speed of the orbiter?

ΣPi=ΣPf: (170+100)(0) = 100(3)+170(v) ⟹ v=−300/170 ⟹ ≈−1.8 m/s

Answer: A (1.8 m/s)

Two identical spheres, head-on elastic collisionMCQ · statement check

Two spheres, mass m, speed v, moving towards each other, HEAD-ON ELASTIC collision. Which statement is correct?

ΣKEi = ½mv²+½mv² = mv² (total, before) ΣPi = mv+m(−v) = 0 (total, before)

Answer: B — total KE after impact is mv² (elastic ⟹ KE conserved ⟹ KE after = KE before = mv²)

Rock R orbits star S — "gravitational slingshot"MCQ · momentum without contact

Rock R (1.0×10²⁷kg) at 1.0×10⁴ m/s reverses direction around star S (1.0×10³⁰kg), leaving at the same speed. Which statement is correct?

Trick: no physical collision occurs, but it's still a closed system — ΣPi=ΣPf applies regardless.

ΣPi=ΣPf: (1×10²⁷)(1×10⁴)+0 = (1×10²⁷)(−1×10⁴)+(1×10³⁰)v v = 2(1×10²⁷)(1×10⁴)/(1×10³⁰) ⟹ v=20 m/s

Answer: B (change in velocity of S ≈ 20 m/s). Note: option C is wrong — by N3L/momentum conservation, R and S's momentum CHANGES are always EQUAL in magnitude (not different by 10³) — only S's velocity change is tiny because its mass is huge.

Nitrogen molecule P collides with stationary Q (2D, angled)Calculation · resolve x & y

P at 320 m/s collides with stationary Q (same mass, no external forces). After: P at 180 m/s at 55° to its original path; Q at 34° to P's initial path. Find v (Q's speed).

y: 0 = m(180sin55) − m(vsin34) ⟹ v = 180sin55/sin34 x (cross-check): 320m = m(180cos55)+m(vcos34) ⟹ v=(320−180cos55)/cos34 Both give v≈260 m/s

Answer: D (260 m/s)

Elastic collision X (mass m) & Y (mass 4m)Calculation · u₁−u₂=v₂−v₁

X at speed v before, 3v/5 in the OPPOSITE direction after. Y (4m) stationary before. Find KE of Y after.

u₁−u₂=v₂−v₁: v−0 = v_Y−(−3v/5) ⟹ v_Y=2v/5 KE_Y=½(4m)(2v/5)² = 8mv²/25 = 16/50 mv²

Answer: C (16/50 mv²)

Two balls, masses m and 2m, head-onCalculation · loss of KE

Ball m at 2v, ball 2m at v (opposite directions), collide head-on. After, ball m rebounds LEFT at v. Find the loss of KE.

ΣPi=ΣPf: m(2v)+2m(−v) = m(−v)+2m(v₂) ⟹ v₂=v/2 ΔKE = [½mv²+½(2m)(v/2)²] − [½m(2v)²+½(2m)v²] = (¾mv²) − (3mv²) = −9/4 mv² (a loss)

Answer: C (9/4 mv²)

7. Structured Problems

Block X (5v) collides with stationary Y, coalesceStructured Q — ratio derivation

Block X (mass mₓ, speed 5v) collides head-on with stationary block Y (mass m_Y), sticks, moves at common speed v.

(a) State Newton's 3rd law: "If object A applies a force on object B in one direction, object B will apply a force of the same magnitude on object A but in the opposite direction."

(i) Show m_Y/mₓ=4: ΣPi=ΣPf: mₓ(5v)+m_Y(0) = (mₓ+m_Y)v ⟹ 5mₓ=mₓ+m_Y ⟹ m_Y/mₓ=4 (ii) Ratio (ΣKE_f)/(ΣKEi) = [½(mₓ+m_Y)v²]/[½mₓ(5v)²] = (mₓ+m_Y)/(25mₓ) = (mₓ+4mₓ)/(25mₓ) = 5/25 ⟹ 1:5

(iii) Ratio for a PERFECTLY ELASTIC collision → 1:1 (KE always conserved in an elastic collision, by definition)

(c) Momentum-time graph for block X given (contact at t=20ms). (i)1. t=0→20ms: "Zero resultant force" (constant momentum, no contact yet). 2. t=20→40ms: "Leftwards constant resultant force" (linear decrease = constant force).

★ Sketching trick for block Y's momentum: count the grid boxes X's momentum line moves through during contact, then draw Y's line changing by the SAME number of boxes over the SAME time interval, but in the OPPOSITE direction (Δp_Y=−Δp_X exactly). Flat at 0 before contact; flat at its new value after.

Toy car propeller — jet of air, N3L, frictionStructured Q — momentum chains into F=ma

Propeller pushes air left at 1.8 m/s in a uniform cylinder (radius 0.045m), air density 1.3 kg/m³.

(i) Mass of air in 2.0s: l=1.8×2=3.6m; m=ρ×πr²×l=1.3×π(0.045)²×3.6 ⟹ m=0.030 kg (ii)1. Δp=p_f−p_i=0.030(1.8)−0 ⟹ Δp=0.054 Ns (ii)2. F=Δp/Δt=0.054/2 ⟹ F=0.027 N

(iii) N3L: "The force exerted by the air on the propeller is equal and opposite to the force exerted by the propeller on the air."

Trick: reuse the 0.027N directly as the car's forward driving force (by N3L) — don't recompute it — then apply ΣF=ma for the final part.

(iv) ΣF=0.027−F_r; (0.20)(0.075)=0.027−F_r ⟹ F_r=0.012 N

Two blocks — relative speed of approach/separation checkStructured Q — elastic or inelastic?

Block A (3M) at 0.40 m/s right, block B (M) at 0.25 m/s left collide. After: A at 0.20 m/s right, B at v.

(i) N3L explains Δp_A=−Δp_B: "We know that F_A=−F_B. We also know their contact times are the same and that Δp=F×Δt ⟹ F_A×t=−F_B×t ⟹ Δp_A=−Δp_B."

(ii) ΣPi=ΣPf: 3M(0.40)+M(−0.25) = 3M(0.20)+M(v) ⟹ v=0.35 m/s (iii)1. Relative speed of approach: u₁−u₂=0.40−(−0.25)=0.65 m/s (iii)2. Relative speed of separation: v₂−v₁=0.35−0.20=0.15 m/s

(iv) Elastic or inelastic? "Inelastic as relative speed of approach is not equal to relative speed of separation."

✓ Quick Recap — Full Chapter Cheat-Sheet

p = mv (kg m s⁻¹) | F = dp/dt | Δp = FΔt (impulse)
ΣPᵢ = ΣP_f in any closed system — with or without an actual collision.
★ No "elastic" stated ⟹ ASSUME inelastic, use only ΣP conservation.
Elastic collision method order: ① u₁−u₂=v₂−v₁ ② ΣPᵢ=ΣP_f ③ ΣKEᵢ=ΣKE_f
Coalescing is ALWAYS inelastic (relative speed of separation is necessarily 0).
Δp_B=−Δp_A — from N3L (equal/opposite force, same contact time) OR by simply rearranging ΣP conservation.
State N3L properly: "if A applies a force on B, B applies an equal force on A in the opposite direction" — NOT the vague "every action has an equal and opposite reaction."
2D problems: resolve into x and y; solve whichever direction has fewer unknowns first, substitute into the other (like the "time" trick in projectile motion).
Never pre-assign a sign to an unknown — let the algebra's result tell you the direction.