§9.3 — Conservation of Momentum

Conservation of momentum is one of the most fundamental laws in physics and follows directly from Newton's second law for the center of mass (§9.2). This section shows that in isolated systems (no external forces), total momentum never changes and how to use this principle to solve complex problems.

Mathematical statement

From Newton's second law for the center of mass (§9.2):

\[ \mathbf{F}_{\text{ext}} = \frac{d\mathbf{P}}{dt} \]

If the system is isolated or external forces are zero:

\[ \mathbf{F}_{\text{ext}} = 0 \implies \frac{d\mathbf{P}}{dt} = 0 \]

Therefore:

\[ \mathbf{P} = \text{constant} \]

Or between initial and final states:

\[ \mathbf{P}_i = \mathbf{P}_f \]

\[ \sum_{i} m_i \mathbf{v}_{i,\text{initial}} = \sum_{i} m_i \mathbf{v}_{i,\text{final}} \]

For a two-particle system:

\[ m_1 \mathbf{v}_{1,i} + m_2 \mathbf{v}_{2,i} = m_1 \mathbf{v}_{1,f} + m_2 \mathbf{v}_{2,f} \]

Conditions for momentum conservation

1. System is isolated: No net external force acts (or external forces balance).

2. Only internal forces: Only forces between components act. These redistribute momentum among particles but do not change total momentum.

3. Before and after: The system must be isolated before and after an event (e.g., collision). During collision, internal forces are huge, but they are internal, not external.

Example 1: Two carts on frictionless track

Two carts on a horizontal frictionless track:

• Cart A: mass \( m_A = 4 \) kg, initial velocity \( v_{A,i} = 3 \) m/s
• Cart B: mass \( m_B = 6 \) kg, initial velocity \( v_{B,i} = 0 \) m/s

They collide and stick together. After collision, cart A moves at \( v_{A,f} = 0.6 \) m/s.

Question: What is the final velocity of cart B?

Solution:

Initial momentum: \[ \mathbf{P}_i = 4(3) + 6(0) = 12 \text{ kg·m/s} \]

Final momentum: \[ \mathbf{P}_f = 4(0.6) + 6 v_{B,f} = 2.4 + 6 v_{B,f} \]

From conservation: \[ \mathbf{P}_i = \mathbf{P}_f \]

\[ 12 = 2.4 + 6 v_{B,f} \]

\[ v_{B,f} = \frac{12 - 2.4}{6} = \frac{9.6}{6} = 1.6 \text{ m/s} \]

Cart B moves at 1.6 m/s after collision.

Example 2: Gun and bullet

A rifle sits on a frictionless table:

• Rifle: mass \( M = 2 \) kg
• Bullet: mass \( m = 0.02 \) kg

The gun is fired. Bullet velocity (relative to ground): \( v_{\text{bullet}} = 200 \) m/s.

Question: What is the rifle's recoil velocity?

Solution:

Initially, everything is at rest: \[ \mathbf{P}_i = 0 \]

After firing: \[ \mathbf{P}_f = M v_{\text{rifle}} + m v_{\text{bullet}} \]

From conservation: \[ 0 = 2 \cdot v_{\text{rifle}} + 0.02 \cdot 200 \]

\[ 2 v_{\text{rifle}} = -4 \]

\[ v_{\text{rifle}} = -2 \text{ m/s} \]

The rifle recoils backward (negative direction) at 2 m/s.

Example 3: Bomb exploding in a plane

An airplane (at rest) on a runway:

• Airplane: mass \( M = 1000 \) kg
• Bomb inside: mass \( m = 50 \) kg

The bomb explodes. Fragments:

• Fragment 1: mass \( m_1 = 30 \) kg, velocity \( v_1 = 100 \) m/s forward
• Fragment 2: mass \( m_2 = 20 \) kg, velocity \( v_2 = ? \) forward

Question: How does the airplane move?

Solution:

The situation seems complex. Use momentum conservation.

Initial momentum is zero. After explosion:

\[ \mathbf{P}_i = 0 \]

If we consider just the bomb fragments (not the airplane fuselage):

\[ m_1 v_1 + m_2 v_2 + (M + m) v_{\text{rest}} = 0 \]

If the airplane and most material remains moving together (the fuselage doesn't fragment), and only the bomb explodes internally:

\[ 30(100) + 20 v_2 = -(950) v_{\text{rest}} \]

For simplicity, if the airplane stays at rest (\( v_{\text{rest}} = 0 \)):

\[ 3000 + 20 v_2 = 0 \]

\[ v_2 = -150 \text{ m/s} \]

Fragment 2 shoots backward at 150 m/s.

Important notes

1. Momentum is a vector: Direction matters. In Example 2 (gun), bullet and gun move in opposite directions to keep total momentum zero.

2. Choice of system: The system must be chosen correctly. An isolated system has conserved momentum.

3. Not the same as energy conservation: Momentum conservation does not imply energy conservation. In inelastic collisions, kinetic energy becomes heat and deformation (Chapters 7–8).

4. Before and after: To apply momentum conservation, use before and after states. The during state (e.g., during collision) is usually unknown.

Contrast: Momentum vs mechanical energy

Property	Momentum	Mechanical energy
Conserved?	No external force	Only conservative forces
Inelastic collision	Conserved	Lost
Elastic collision	Conserved	Conserved
Explosion	Conserved	Increases (from chemical energy)

What you should be able to do

• State momentum conservation for isolated systems
• Distinguish external and internal forces
• Apply momentum conservation to solve collision and explosion problems
• Recognize that momentum and energy conservation are independent

Preview of §9.4

Now we can solve collision problems. But different collisions lose different amounts of energy. This leads to elastic and inelastic collisions — where conservation of energy and momentum work together.

📚 See also: Halliday Vol 1, Ch 9, §9.3.

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