Educational

Conservation of Linear Momentum

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by

Saim Khalid
in

1. Introduction

When you watch a rocket launch into space, it almost feels like magic. A huge vehicle weighing thousands of tons rises against gravity, leaving Earth behind. But behind this spectacle lies a very simple yet powerful principle of physics: conservation of linear momentum.

Linear momentum is the “quantity of motion” of an object, and its conservation is one of the most universal laws of nature. From the recoil of a gun to the flight of rockets, the principle remains the same:

👉 “If no external force acts on a system, its total momentum remains constant.”

In this article, we’ll explore:

  • Basics of linear momentum
  • The law of conservation of momentum
  • Recoil examples (guns, cannons, etc.)
  • Rocket propulsion explained step by step
  • Derivation of the rocket equation
  • Real-world challenges in rocket launches
  • Applications in space science and daily life
  • Solved examples + practice problems

By the end, you’ll clearly understand how rockets work using nothing more than Newton’s laws and momentum conservation.

2. What is Linear Momentum?

Definition

Linear momentum (ppp) is defined as: p=mvp = m vp=mv

Where:

  • mmm = mass of object (kg)
  • vvv = velocity of object (m/s)

👉 Momentum is a vector (depends on direction as well as magnitude).

Examples

  • A heavy truck at slow speed can have the same momentum as a fast-moving motorcycle.
  • A cricket ball hit by a bat changes momentum drastically depending on velocity.

3. Newton’s Second Law in Terms of Momentum

Newton’s Second Law can be written as: F=dpdtF = \frac{dp}{dt}F=dtdp​

This means force equals the rate of change of momentum.

If mass is constant: F=maF = m aF=ma

If mass is variable (as in rockets losing fuel): F=vdmdtF = v \frac{dm}{dt}F=vdtdm​

👉 This is the key to understanding rocket propulsion.

4. Law of Conservation of Linear Momentum

Statement

“In an isolated system (no external force), the total momentum before an interaction equals the total momentum after the interaction.” pinitial=pfinalp_{initial} = p_{final}pinitial​=pfinal​

Examples

  • Recoil of a gun: Bullet moves forward, gun moves backward.
  • Jumping off a boat: Person moves one way, boat moves the other.
  • Explosions: Fragments fly in different directions, but total momentum remains zero if initial momentum was zero.

5. Recoil Motion – A Simple Analogy

Before understanding rockets, let’s see recoil examples.

Gun Example

  • A gun of mass MMM fires a bullet of mass mmm with velocity vvv.
  • Before firing: Total momentum = 0.
  • After firing:

mv+MV=0m v + M V = 0mv+MV=0 V=−mvMV = – \frac{m v}{M}V=−Mmv​

👉 Gun recoils backward with velocity VVV.

Rockets work on exactly the same principle, but continuously instead of a single shot.

6. Rocket Propulsion – The Core Idea

Rockets carry fuel. When the fuel burns, hot gases are expelled at high speed from the nozzle.

  • Gases move backward.
  • By momentum conservation, rocket moves forward.

This is an application of Newton’s Third Law:
“For every action, there is an equal and opposite reaction.”

👉 Action: gases ejected backward.
👉 Reaction: rocket pushed forward.

7. Derivation of Rocket Equation (Tsiolkovsky’s Equation)

Consider a rocket of mass MMM moving with velocity vvv.

  • In a short time dtdtdt, it expels a small mass dmdmdm of fuel with velocity uuu relative to the rocket.
  • By conservation of momentum:

(M)v=(M−dm)(v+dv)+(dm)(v−u)(M) v = (M – dm)(v + dv) + (dm)(v – u)(M)v=(M−dm)(v+dv)+(dm)(v−u)

Simplifying: Mdv=u dmM dv = u \, dmMdv=udm

Integrating from initial mass MiM_iMi​ to final mass MfM_fMf​: Δv=uln⁡MiMf\Delta v = u \ln \frac{M_i}{M_f}Δv=ulnMf​Mi​​

Where:

  • Δv\Delta vΔv = final velocity gain of rocket
  • uuu = exhaust velocity of gases
  • MiM_iMi​ = initial mass (rocket + fuel)
  • MfM_fMf​ = final mass (rocket after fuel burnt)

👉 This is the Tsiolkovsky Rocket Equation.

8. Factors Affecting Rocket Motion

  1. Exhaust Velocity (u): Higher exhaust speed → higher rocket speed.
  2. Mass Ratio (Mi/MfM_i/M_fMi​/Mf​): More fuel relative to rocket mass → greater velocity.
  3. Gravity & Air Resistance: External forces reduce efficiency.
  4. Multi-Stage Rockets: Used to discard extra weight and increase efficiency.

9. Real-Life Applications

  1. Space Exploration: Satellites, space stations, interplanetary missions all rely on rocket propulsion.
  2. Missiles & Defense: Momentum principles apply to guided missiles.
  3. Jet Engines: Air-breathing engines also work on reaction principle.
  4. Fireworks & Explosions: Fragments move in opposite directions conserving momentum.
  5. Everyday Example: A person jumping off a skateboard pushes backward on the board.

10. Solved Numerical Problems

Example 1: Recoil of Gun

A 5 kg gun fires a 50 g bullet at 200 m/s. Find recoil velocity. MV+mv=0M V + m v = 0MV+mv=0 5V+0.05×200=05 V + 0.05 \times 200 = 05V+0.05×200=0 5V+10=0⇒V=−2 m/s5 V + 10 = 0 \quad \Rightarrow V = -2 \, m/s5V+10=0⇒V=−2m/s

👉 Gun recoils at 2 m/s backward.

Example 2: Rocket Velocity

A rocket has initial mass 2000 kg (including fuel). Final mass after burning fuel is 500 kg. Exhaust velocity = 1000 m/s. Δv=uln⁡MiMf\Delta v = u \ln \frac{M_i}{M_f}Δv=ulnMf​Mi​​ =1000ln⁡2000500=1000ln⁡4= 1000 \ln \frac{2000}{500} = 1000 \ln 4=1000ln5002000​=1000ln4 =1000×1.386=1386 m/s= 1000 \times 1.386 = 1386 \, m/s=1000×1.386=1386m/s

👉 Rocket speed = 1386 m/s.

Example 3: Two-Stage Rocket

A two-stage rocket has:

  • Stage 1: Mi=1000 kgM_i = 1000 \, kgMi​=1000kg, Mf=400 kgM_f = 400 \, kgMf​=400kg.
  • Stage 2: Mi=400 kgM_i = 400 \, kgMi​=400kg, Mf=200 kgM_f = 200 \, kgMf​=200kg.
  • Exhaust velocity = 2000 m/s.

Velocity after stage 1: Δv1=2000ln⁡1000400=2000ln⁡2.5\Delta v_1 = 2000 \ln \frac{1000}{400} = 2000 \ln 2.5Δv1​=2000ln4001000​=2000ln2.5 =2000×0.916=1832 m/s= 2000 \times 0.916 = 1832 \, m/s=2000×0.916=1832m/s

Velocity after stage 2: Δv2=2000ln⁡400200=2000ln⁡2\Delta v_2 = 2000 \ln \frac{400}{200} = 2000 \ln 2Δv2​=2000ln200400​=2000ln2 =2000×0.693=1386 m/s= 2000 \times 0.693 = 1386 \, m/s=2000×0.693=1386m/s

Total velocity = 1832+1386=3218 m/s1832 + 1386 = 3218 \, m/s1832+1386=3218m/s.

👉 Multi-staging greatly improves rocket performance.

11. Challenges in Rocket Propulsion

  1. Weight of Fuel: Rockets need huge amounts of fuel compared to payload.
  2. Atmospheric Drag: Reduces efficiency in initial stages.
  3. Gravity Losses: Constant pull of Earth requires extra energy.
  4. Heat & Pressure: Extreme conditions in combustion chamber.
  5. Engineering Complexity: Requires advanced materials, guidance systems, and cooling.

12. Misconceptions

  • Rockets don’t need air to push against. They work in space because gases are expelled backward.
  • Momentum is always conserved—but external forces (gravity, drag) must be considered.
  • Heavier rockets are not always better. Efficiency depends on mass ratio.

13. Key Formulas Recap

  1. Momentum:

p=mvp = m vp=mv

  1. Conservation of Momentum:

pinitial=pfinalp_{initial} = p_{final}pinitial​=pfinal​

  1. Recoil Velocity:

V=−mvMV = – \frac{m v}{M}V=−Mmv​

  1. Rocket Equation:

Δv=uln⁡MiMf\Delta v = u \ln \frac{M_i}{M_f}Δv=ulnMf​Mi​​

  1. Force (variable mass):

F=vdmdtF = v \frac{dm}{dt}F=vdtdm​

14. Practice Problems

  1. A 10 kg cannon fires a 0.2 kg shell at 100 m/s. Find recoil velocity of cannon.
  2. A rocket expels gases at 2000 m/s. If initial mass is 1000 kg and final mass is 500 kg, find rocket velocity.
  3. A 60 kg person jumps from a stationary boat of 240 kg with speed 5 m/s. Find boat’s recoil velocity.
  4. A two-stage rocket has exhaust velocity 1500 m/s. Stage 1: Mi=2000M_i = 2000Mi​=2000, Mf=1000M_f = 1000Mf​=1000. Stage 2: Mi=1000M_i = 1000Mi​=1000, Mf=400M_f = 400Mf​=400. Find total velocity.
  5. Why does a rocket perform better in space than in atmosphere? Explain using momentum conservation.

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