CIE AS/A Level Physics Unit : 3.3 Linear momentum and its conservation Study Notes- 2025-2027 Syllabus
CIE AS/A Level Physics Unit : 3.3 Linear momentum and its conservation Study Notes
CIE AS/A Level Physics Unit : 3.3 Linear momentum and its conservation Study Notes at IITian Academy focus on specific topic and type of questions asked in actual exam. Study Notes focus on AS/A Level Physics Study Notes syllabus with Candidates should be able to:
- state the principle of conservation of momentum
- apply the principle of conservation of momentum to solve simple problems, including elastic and inelastic interactions between objects in both one and two dimensions (knowledge of the concept of coefficient of restitution is not required)
- recall that, for an elastic collision, total kinetic energy is conserved and the relative speed of approach is equal to the relative speed of separation
- understand that, while momentum of a system is always conserved in interactions between objects, some change in kinetic energy may take place
Principle of Conservation of Momentum
The principle of conservation of momentum states that the total momentum of a system remains constant, provided no external resultant force acts on the system.
\( \mathrm{Total\ Initial\ Momentum = Total\ Final\ Momentum} \)
or equivalently,
\( \mathrm{\sum \vec{p}_{\text{before}} = \sum \vec{p}_{\text{after}}} \)
Explanation:
- Momentum is conserved in all interactions and collisions (elastic or inelastic) if the net external force on the system is zero.
- For two or more interacting bodies, the total vector momentum before interaction equals the total vector momentum after interaction.
- Although individual momenta of bodies may change, their total combined momentum remains the same.
- This principle follows directly from Newton’s Third Law of Motion — forces between two interacting bodies are equal and opposite, so changes in momentum cancel out.
Mathematical Form:
For two bodies (1 and 2) before and after a collision:
\( \mathrm{m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2} \)
- \( \mathrm{m_1, m_2} \): masses of bodies 1 and 2
- \( \mathrm{u_1, u_2} \): initial velocities
- \( \mathrm{v_1, v_2} \): final velocities
Conditions for Momentum Conservation:
- No external resultant force acts on the system (only internal forces between interacting bodies).
- The system is isolated — e.g., no friction or external influence from the surroundings.
Applications:
- Collisions between moving objects (e.g. carts, balls, vehicles).
- Explosions (where total momentum before and after is equal, but in opposite directions).
- Recoil of a gun or rocket propulsion — the momentum of ejected material equals the opposite momentum of the system.
Example
A 2 kg cart moving at \( \mathrm{3\,m/s} \) collides with a stationary 1 kg cart. After the collision, the 2 kg cart slows to \( \mathrm{1\,m/s} \). Find the velocity of the 1 kg cart immediately after impact (assuming no external forces).
▶️ Answer / Explanation
By conservation of momentum:
\( \mathrm{(2)(3) + (1)(0) = (2)(1) + (1)(v)} \)
\( \mathrm{6 = 2 + v \Rightarrow v = 4\,m/s.} \)
The 1 kg cart moves forward at \( \mathrm{4\,m/s.} \)
Elastic and Inelastic Interactions in One and Two Dimensions
The principle of conservation of momentum states that the total momentum of an isolated system remains constant, provided no external resultant force acts on it.
\( \mathrm{\sum \vec{p}_{\text{before}} = \sum \vec{p}_{\text{after}}} \)
1. Elastic and Inelastic Interactions
Elastic Collision:
In an elastic collision, both momentum and total kinetic energy are conserved.
- \( \mathrm{Total\ momentum\ before = Total\ momentum\ after} \)
- \( \mathrm{Total\ kinetic\ energy\ before = Total\ kinetic\ energy\ after} \)
- Relative speed of approach = Relative speed of separation:
\( \mathrm{u_1 – u_2 = -(v_1 – v_2)} \)
Inelastic Collision:
In an inelastic collision, momentum is conserved, but kinetic energy is not (some energy is transformed into heat, sound, or deformation).
- \( \mathrm{Total\ momentum\ before = Total\ momentum\ after} \)
- \( \mathrm{Total\ kinetic\ energy\ before > Total\ kinetic\ energy\ after} \)
In a perfectly inelastic collision, the objects stick together and move with a common velocity after impact.
2. Momentum Conservation in One Dimension
For two colliding bodies moving along a straight line:
\( \mathrm{m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2} \)
- \( \mathrm{m_1, m_2} \): masses of the two bodies
- \( \mathrm{u_1, u_2} \): velocities before collision
- \( \mathrm{v_1, v_2} \): velocities after collision
Example
A 2.0 kg cart moving at \( \mathrm{3.0\,m/s} \) collides with a stationary 1.0 kg cart. The carts stick together after the collision. Find their common velocity immediately after impact.
▶️ Answer / Explanation
Using conservation of momentum:
\( \mathrm{(2.0)(3.0) + (1.0)(0) = (2.0 + 1.0)v} \)
\( \mathrm{6.0 = 3v \Rightarrow v = 2.0\,m/s.} \)
The carts move together at \( \mathrm{2.0\,m/s} \) after the collision.
3. Momentum Conservation in Two Dimensions
When two objects collide or explode in two dimensions (e.g., oblique impact), the principle of conservation of momentum applies independently in both the x and y directions:
\( \mathrm{\sum p_x^{\text{before}} = \sum p_x^{\text{after}}} \)
\( \mathrm{\sum p_y^{\text{before}} = \sum p_y^{\text{after}}} \)
Procedure:
- Resolve all velocities into horizontal and vertical components.
- Apply conservation of momentum separately in each direction.
- Use vector addition to determine the resultant velocities.
Example
A 0.5 kg ball moving at \( \mathrm{4.0\,m/s} \) east collides with a 0.5 kg ball moving at \( \mathrm{3.0\,m/s} \) north. After the collision, they stick together. Determine the magnitude and direction of their combined velocity.
▶️ Answer / Explanation
Apply conservation of momentum in both directions.
Horizontal component: \( \mathrm{p_x = (0.5)(4.0) = 2.0\,kg\,m/s} \)
Vertical component: \( \mathrm{p_y = (0.5)(3.0) = 1.5\,kg\,m/s} \)
Total momentum: \( \mathrm{p = \sqrt{(2.0)^2 + (1.5)^2} = 2.5\,kg\,m/s} \)
Combined mass: \( \mathrm{1.0\,kg} \)
Velocity: \( \mathrm{v = p/m = 2.5/1.0 = 2.5\,m/s} \)
Direction: \( \mathrm{\tan\theta = \dfrac{1.5}{2.0} \Rightarrow \theta = 36.9^\circ} \) north of east.
Elastic Collisions — Kinetic Energy Conservation
In an elastic collision:
- Total momentum is conserved.
- Total kinetic energy is conserved.
Hence, for a two-body system:
\( \mathrm{\tfrac{1}{2}m_1 u_1^2 + \tfrac{1}{2}m_2 u_2^2 = \tfrac{1}{2}m_1 v_1^2 + \tfrac{1}{2}m_2 v_2^2} \)
and
\( \mathrm{u_1 – u_2 = -(v_1 – v_2)} \)
This means the relative speed of approach before collision equals the relative speed of separation after collision.
Elastic vs Inelastic Collisions:
| Property | Elastic Collision | Inelastic Collision |
|---|---|---|
| Momentum | Conserved | Conserved |
| Kinetic Energy | Conserved | Not conserved (partly converted to heat, sound, etc.) |
| Relative Speed | \( \mathrm{u_1 – u_2 = -(v_1 – v_2)} \) | No fixed relation |
| Example | Elastic ball collision | Car crash, clay impact |
Example
A 2.0 kg cart moving at \( \mathrm{3.0\,m/s} \) collides elastically with a stationary 1.0 kg cart. Determine their velocities after collision.
▶️ Answer / Explanation
Using conservation of momentum:
\( \mathrm{(2)(3) + (1)(0) = (2)v_1 + (1)v_2} \Rightarrow \mathrm{6 = 2v_1 + v_2.} \)
Using conservation of kinetic energy:
\( \mathrm{(2)(3^2) = (2)v_1^2 + (1)v_2^2 \Rightarrow 18 = 2v_1^2 + v_2^2.} \)
Solving the two equations gives \( \mathrm{v_1 = 1.0\,m/s} \) and \( \mathrm{v_2 = 4.0\,m/s.} \)
Check: Relative speed of approach = \( \mathrm{3.0\,m/s} \); relative speed of separation = \( \mathrm{4.0 – 1.0 = 3.0\,m/s.} \) Hence, kinetic energy and relative speed are both conserved — collision is elastic.
Conservation of Momentum and Change in Kinetic Energy
In all interactions between objects such as collisions or explosions the total momentum of the system is always conserved, provided no external resultant force acts on it. However, the total kinetic energy of the system may change depending on the nature of the interaction.
\( \mathrm{\sum \vec{p}_{\text{before}} = \sum \vec{p}_{\text{after}}} \)
but
\( \mathrm{\sum E_{k,\text{before}} \neq \sum E_{k,\text{after}}} \)
Key Idea:
- Momentum is always conserved in an isolated system (no external forces).
- Kinetic energy may not be conserved because some of it can be transformed into other forms of energy (such as heat, sound, deformation, or internal energy).
- This means an interaction can conserve momentum but not kinetic energy.
1. Elastic Interactions
In an elastic collision:
- Both momentum and kinetic energy are conserved.
- No mechanical energy is lost to heat or deformation.
- Relative speed of approach = Relative speed of separation.
\( \mathrm{u_1 – u_2 = -(v_1 – v_2)} \)
2. Inelastic Interactions
In an inelastic collision:
- Momentum is conserved, but kinetic energy is not.
- Some kinetic energy is transformed into other forms (e.g., sound, heat, or deformation).
- In a perfectly inelastic collision, the objects stick together and move with a common velocity after impact.
Energy Transformations in Inelastic Collisions:
- Loss of kinetic energy = gain in other forms (e.g., internal energy or heat).
- This energy transformation ensures total energy (mechanical + internal) remains conserved — consistent with the law of conservation of energy.
Summary of Conservation Rules:
| Type of Interaction | Momentum Conserved? | Kinetic Energy Conserved? | Example |
|---|---|---|---|
| Elastic Collision | Yes | Yes | Billiard balls, gas molecules |
| Inelastic Collision | Yes | No (some energy → heat/sound) | Car crash, clay impact |
| Explosion | Yes | No (chemical energy → kinetic energy) | Gun recoil, fireworks |
Example
A 3.0 kg trolley moving at \( \mathrm{4.0\,m/s} \) collides with a stationary 1.0 kg trolley. After the collision, the trolleys stick together. Calculate
(a) their common velocity after collision and
(b) the loss in kinetic energy.
▶️ Answer / Explanation
(a) Common velocity:
Using momentum conservation:
\( \mathrm{(3)(4) + (1)(0) = (3+1)v} \Rightarrow \mathrm{12 = 4v \Rightarrow v = 3.0\,m/s.} \)
(b) Kinetic energy loss:
Initial \( \mathrm{E_k = \tfrac{1}{2}(3)(4^2) = 24\,J.} \)
Final \( \mathrm{E_k = \tfrac{1}{2}(4)(3^2) = 18\,J.} \)
Loss in kinetic energy = \( \mathrm{24 – 18 = 6\,J.} \)
The 6 J loss in kinetic energy is converted into sound, heat, and deformation energy during the impact.