Linear momentum and its conservation

1. Overview

Linear momentum is a vector quantity defined as the product of an object's mass and its velocity. In physics, the study of interactions—whether they are collisions between subatomic particles or the macroscopic impact of vehicles—relies on the Principle of Conservation of Momentum. This principle dictates that in a closed system, the total momentum remains constant over time, regardless of the internal forces acting between objects. While momentum is universally conserved in these closed systems, kinetic energy may be transformed into other forms, such as thermal energy or sound, leading to the distinction between elastic and inelastic interactions. Mastering this topic requires a rigorous approach to vector addition and a clear understanding of the conditions under which energy is conserved.

---

Key Definitions

• Linear Momentum ($p$): The product of an object's mass and its velocity. It is a vector quantity, meaning it has both magnitude and direction. The SI unit is $\text{kg m s}^{-1}$ or $\text{N s}$.
• Principle of Conservation of Momentum: The total momentum of a closed system remains constant, provided no external resultant force acts on the system.
• Closed System: An isolated system where no matter or energy enters or leaves, and the only forces acting are internal forces between the objects within the system.
• Elastic Collision: An interaction in which total momentum is conserved and total kinetic energy is also conserved.
• Inelastic Collision: An interaction in which total momentum is conserved, but total kinetic energy is not conserved. Some kinetic energy is transferred into internal (thermal) energy, sound, or work done in deforming the objects.
• Perfectly Inelastic Collision: A specific case of an inelastic collision where the objects stick together after impact and move with a common velocity. This results in the maximum possible loss of kinetic energy consistent with the conservation of momentum.
• Relative Speed of Approach: The magnitude of the difference in velocities of two objects as they move toward each other.
• Relative Speed of Separation: The magnitude of the difference in velocities of two objects as they move away from each other after a collision.

---

Content

3.3.1 The Principle of Conservation of Momentum

The conservation of momentum is a direct consequence of Newton’s Third Law. When two objects interact, the force exerted by object A on object B is equal and opposite to the force exerted by object B on object A ($F_{AB} = -F_{BA}$). Since the time of contact ($\Delta t$) is the same for both, the impulse ($F\Delta t$) is equal and opposite, leading to equal and opposite changes in momentum ($\Delta p_A = -\Delta p_B$). Consequently, the total change in momentum of the system is zero.

The Mathematical Expression: For a system of two colliding objects (1 and 2): $$\mathbf{m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2}$$

Where:

• $m_1, m_2$ = masses of the objects (kg)
• $u_1, u_2$ = initial velocities (m s⁻¹)
• $v_1, v_2$ = final velocities (m s⁻¹)

Vector Nature and Sign Conventions: Because momentum is a vector, you must define a coordinate system before starting a calculation.

• Choose one direction (e.g., to the right) as positive (+).
• The opposite direction (e.g., to the left) is negative (-).
• If an object is moving left at $5 \text{ m s}^{-1}$, its velocity in the equation must be entered as $-5 \text{ m s}^{-1}$.

3.3.2 Elastic and Inelastic Collisions

In any interaction within a closed system, the total energy is always conserved. However, the total kinetic energy ($E_k$) is only conserved in specific circumstances.

1. Elastic Collisions:

• Momentum: Conserved ($\sum p_{initial} = \sum p_{final}$).
• Kinetic Energy: Conserved ($\sum E_{k, initial} = \sum E_{k, final}$).
• Macroscopic Examples: Collisions between hard steel spheres or billiard balls (though these are only "near-elastic").
• Microscopic Examples: Collisions between gas molecules (ideal gas assumption).

2. Inelastic Collisions:

• Momentum: Conserved.
• Kinetic Energy: Not conserved. $E_{k, final} < E_{k, initial}$.
• The "lost" kinetic energy is typically converted into thermal energy (due to internal friction/deformation) or sound energy.
• Perfectly Inelastic: The objects coalesce (stick together). The final state is represented by $(m_1 + m_2)v$.

3.3.3 Relative Speed in Elastic Collisions

For any perfectly elastic collision, a unique relationship exists between the velocities of the two objects. The speed at which they approach each other is exactly equal to the speed at which they move away from each other.

The Relative Speed Equation: $$\mathbf{u_1 - u_2 = v_2 - v_1}$$

• $(u_1 - u_2)$ is the relative speed of approach.
• $(v_2 - v_1)$ is the relative speed of separation.

Why use this? In exam problems involving elastic collisions, you are often given the masses and initial velocities and asked to find both final velocities ($v_1$ and $v_2$). Using the conservation of kinetic energy ($\frac{1}{2}mv^2$) leads to difficult quadratic equations. The relative speed equation is linear and much easier to solve simultaneously with the momentum equation.

3.3.4 Collisions in Two Dimensions

When objects collide at angles (oblique collisions), momentum must be conserved in two independent dimensions (usually horizontal $x$ and vertical $y$).

Step-by-Step Approach:

1. Resolve all initial and final velocity vectors into $x$ and $y$ components using trigonometry:
   • $v_x = v \cos \theta$
   • $v_y = v \sin \theta$
2. Apply Conservation of Momentum for the x-axis: $$\sum p_{x, before} = \sum p_{x, after}$$
3. Apply Conservation of Momentum for the y-axis: $$\sum p_{y, before} = \sum p_{y, after}$$
4. Kinetic Energy: Since $E_k$ is a scalar, do not resolve it into components. Total $E_k$ is simply $\frac{1}{2}mv^2$ using the actual magnitude of the velocity.

---

Worked Example 1 — 1D Elastic Collision

A sphere of mass $0.40 \text{ kg}$ moving at $6.0 \text{ m s}^{-1}$ hits a stationary sphere of mass $0.20 \text{ kg}$ head-on. The collision is perfectly elastic. Calculate the final velocity of each sphere.

Step 1: Conservation of Momentum Let $v_1$ be the final velocity of the $0.40 \text{ kg}$ mass and $v_2$ be the final velocity of the $0.20 \text{ kg}$ mass. $$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$$ $$(0.40 \times 6.0) + (0.20 \times 0) = 0.40 v_1 + 0.20 v_2$$ $$2.4 = 0.40 v_1 + 0.20 v_2$$ Divide by 0.20 to simplify: $$12 = 2v_1 + v_2 \quad \text{--- (Equation 1)}$$

Step 2: Relative Speed Equation (since collision is elastic) $$u_1 - u_2 = v_2 - v_1$$ $$6.0 - 0 = v_2 - v_1$$ $$v_2 = 6.0 + v_1 \quad \text{--- (Equation 2)}$$

Step 3: Solve Simultaneously Substitute (2) into (1): $$12 = 2v_1 + (6.0 + v_1)$$ $$12 = 3v_1 + 6.0$$ $$6.0 = 3v_1 \implies v_1 = 2.0 \text{ m s}^{-1}$$ Substitute $v_1$ back into (2): $$v_2 = 6.0 + 2.0 = 8.0 \text{ m s}^{-1}$$

Answer: $v_1 = 2.0 \text{ m s}^{-1}$ and $v_2 = 8.0 \text{ m s}^{-1}$ (both in the original direction).

---

Worked Example 2 — 2D Inelastic Collision

A block of mass $M = 2.0 \text{ kg}$ is sliding at $5.0 \text{ m s}^{-1}$ along the positive x-axis. It collides with a second block of mass $m = 3.0 \text{ kg}$ moving at $4.0 \text{ m s}^{-1}$ along the positive y-axis. After the collision, the two blocks stick together. Calculate the magnitude and direction of their common velocity.

Step 1: Momentum in the x-direction $$p_{x, before} = M u_x + m(0) = 2.0 \times 5.0 = 10.0 \text{ kg m s}^{-1}$$ $$p_{x, after} = (M + m) V_x = 5.0 V_x$$ $$10.0 = 5.0 V_x \implies V_x = 2.0 \text{ m s}^{-1}$$

Step 2: Momentum in the y-direction $$p_{y, before} = M(0) + m u_y = 3.0 \times 4.0 = 12.0 \text{ kg m s}^{-1}$$ $$p_{y, after} = (M + m) V_y = 5.0 V_y$$ $$12.0 = 5.0 V_y \implies V_y = 2.4 \text{ m s}^{-1}$$

Step 3: Resultant Velocity Magnitude $$V = \sqrt{V_x^2 + V_y^2} = \sqrt{2.0^2 + 2.4^2} = \sqrt{4.0 + 5.76} = \sqrt{9.76}$$ $$V \approx 3.12 \text{ m s}^{-1}$$

Step 4: Direction (Angle $\theta$ from the x-axis) $$\theta = \tan^{-1}\left(\frac{V_y}{V_x}\right) = \tan^{-1}\left(\frac{2.4}{2.0}\right) = \tan^{-1}(1.2)$$ $$\theta \approx 50.2^\circ$$

Answer: $3.1 \text{ m s}^{-1}$ at an angle of $50^\circ$ to the x-axis.

---

Worked Example 3 — Explosion (Recoil)

A stationary radioactive nucleus of mass $210 \text{ u}$ decays by emitting an alpha particle (mass $4 \text{ u}$) with a velocity of $1.6 \times 10^7 \text{ m s}^{-1}$. Calculate the recoil velocity of the daughter nucleus.

Step 1: Conservation of Momentum Initial momentum = 0 (stationary). Final momentum = $m_{\alpha} v_{\alpha} + m_{nucleus} v_{nucleus}$ $$0 = (4 \text{ u} \times 1.6 \times 10^7 \text{ m s}^{-1}) + (206 \text{ u} \times v_{nucleus})$$ (Note: $210 - 4 = 206 \text{ u}$ for the daughter nucleus. Units of 'u' cancel out.)

Step 2: Solve for $v_{nucleus}$ $$- (4 \times 1.6 \times 10^7) = 206 \times v_{nucleus}$$ $$- 6.4 \times 10^7 = 206 \times v_{nucleus}$$ $$v_{nucleus} = \frac{-6.4 \times 10^7}{206} \approx -3.1 \times 10^5 \text{ m s}^{-1}$$

Answer: $3.1 \times 10^5 \text{ m s}^{-1}$ in the opposite direction to the alpha particle.

---

Key Equations

Quantity / Principle	Equation	Formula Sheet?
Linear Momentum	$\mathbf{p = mv}$	Yes
Conservation of Momentum	$\mathbf{\sum p_{initial} = \sum p_{final}}$	No
Kinetic Energy (Momentum form)	$\mathbf{E_k = \frac{p^2}{2m}}$	No
Relative Speed (Elastic only)	$\mathbf{u_1 - u_2 = v_2 - v_1}$	No
Perfectly Inelastic Collision	$\mathbf{m_1 u_1 + m_2 u_2 = (m_1 + m_2)v}$	No
2D Momentum (x-component)	$\mathbf{m_1 u_{1x} + m_2 u_{2x} = m_1 v_{1x} + m_2 v_{2x}}$	No

---

Common Mistakes to Avoid

• ❌ Ignoring Direction: Adding $2 \text{ kg m s}^{-1}$ and $3 \text{ kg m s}^{-1}$ to get $5 \text{ kg m s}^{-1}$ when the objects are moving toward each other.
  • ✅ Right: Assign signs. If right is positive, the sum is $(+2) + (-3) = -1 \text{ kg m s}^{-1}$.
• ❌ Assuming $E_k$ Conservation: Assuming a collision is elastic because no energy loss was mentioned.
  • ✅ Right: Only assume $E_k$ is conserved if the question explicitly uses the word "elastic". Otherwise, assume it is inelastic.
• ❌ Mass Unit Errors: Using mass in grams (g).
  • ✅ Right: Convert all masses to kilograms (kg) immediately ($1 \text{ g} = 10^{-3} \text{ kg}$).
• ❌ Resolving Kinetic Energy: Trying to find the "x-component of kinetic energy."
  • ✅ Right: Kinetic energy is a scalar. Calculate it using the total magnitude of velocity: $E_k = \frac{1}{2}m(v_x^2 + v_y^2)$.
• ❌ Confusing "Total Energy" and "Kinetic Energy": Stating that total energy is not conserved in an inelastic collision.
  • ✅ Right: Total energy is always conserved (Principle of Conservation of Energy). Only Kinetic Energy is not conserved in inelastic collisions.

---

Exam Tips

1. The "Show that" Elasticity Question: If asked to prove a collision is elastic, you must calculate:
   • Total $E_k$ before ($\frac{1}{2}m_1u_1^2 + \frac{1}{2}m_2u_2^2$)
   • Total $E_k$ after ($\frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2$)
   • Compare the two values. If they are equal (within rounding), it is elastic.
2. Vector Diagrams: For 2D momentum, the initial momentum vector must equal the vector sum of the final momentum vectors. Drawing a closed vector triangle can often solve 2D problems faster than resolving components.
3. Internal vs. External Forces: If a question asks why momentum isn't conserved for a falling ball, the answer is that the Earth's gravitational pull is an external force. If you include the Earth in the system, momentum is conserved.
4. Significant Figures: Always provide your final answer to the same number of significant figures as the least precise data point given in the question (usually 2 or 3 s.f.).
5. The $E_k = \frac{p^2}{2m}$ Shortcut: This is extremely useful for multiple-choice questions. If two objects have the same momentum, the lighter one has more kinetic energy. If they have the same kinetic energy, the heavier one has more momentum.