CIE AS/A Level Maths-4.3 Momentum - Study Notes- New Syllabus - 2026-2027
CIE AS/A Level Maths-4.3 Momentum- Study Notes- New Syllabus
Ace AS/A Level Maths Exam with CIE AS/A Level Maths-4.3 Momentum- Study Notes
Key Concepts:
- Linear Momentum
- Conservation of Linear Momentum
Linear Momentum
Linear Momentum
Linear momentum of a particle is defined as the product of its mass and velocity:
\( p = mv \)
- It is a vector quantity because velocity is a vector.
- In one-dimensional motion, the direction of momentum is the same as the direction of velocity.
- Units: \( \text{kg m s}^{-1} \).
Key Points:
- If mass is constant, a change in momentum comes only from a change in velocity.
- A large momentum means either a large mass or high velocity.
- In 1D, we can take one direction as positive (usually right/forward), so momentum may be positive or negative.
Example:
A ball of mass \(2 \, \text{kg}\) is moving to the right with velocity \(3 \, \text{m/s}\). Find its momentum. Then find its momentum if it moves to the left with the same speed.
▶️ Answer/Explanation
Case 1: Motion to the right
\( p = mv = 2 \times 3 = 6 \, \text{kg m/s} \) (positive, since right is taken as positive).
Case 2: Motion to the left
\( p = mv = 2 \times (-3) = -6 \, \text{kg m/s} \).
Final Answer:
Momentum is \(\boxed{6 \, \text{kg m/s}}\) to the right, and \(\boxed{-6 \, \text{kg m/s}}\) to the left.
Conservation of Linear Momentum
Conservation of Linear Momentum
Principle:
The total linear momentum of a system of particles remains constant, provided that no external resultant force acts on the system.
In symbols (1D motion):
\( m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \)
where:
- \(m_1, m_2\) = masses of the two bodies
- \(u_1, u_2\) = initial velocities before impact
- \(v_1, v_2\) = final velocities after impact
Special Case – Coalescence (bodies stick together):
If the two bodies stick together after impact, then:
\( m_1 u_1 + m_2 u_2 = (m_1 + m_2) v \)
- This is often called an inelastic collision in one dimension.
- Kinetic energy is not conserved in this case, but momentum is always conserved.
Example:
A ball of mass \(3 \, \text{kg}\) moving with velocity \(4 \, \text{m/s}\) collides head-on with another ball of mass \(2 \, \text{kg}\) moving in the opposite direction with velocity \(1 \, \text{m/s}\). After collision, the \(3 \, \text{kg}\) ball moves at \(2 \, \text{m/s}\) to the right. Find the final velocity of the \(2 \, \text{kg}\) ball.
▶️ Answer/Explanation
Step 1: Apply momentum conservation
\( m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \)
Step 2: Substitute values
\( (3)(4) + (2)(-1) = (3)(2) + (2)(v_2) \)
\( 12 – 2 = 6 + 2v_2 \)
Step 3: Solve for \(v_2\)
\( 10 = 6 + 2v_2 \)
\( v_2 = 2 \, \text{m/s} \)
Final Answer:
The \(2 \, \text{kg}\) ball moves with \(\boxed{2 \, \text{m/s}}\) to the right.
Example:
A truck of mass \(1000 \, \text{kg}\) moving at \(6 \, \text{m/s}\) collides with a stationary car of mass \(500 \, \text{kg}\). After the impact, the two vehicles stick together. Find their common velocity after the collision.
▶️ Answer/Explanation
Step 1: Apply conservation of momentum
\( m_1 u_1 + m_2 u_2 = (m_1 + m_2) v \)
Step 2: Substitute values
\( (1000)(6) + (500)(0) = (1000 + 500)v \)
\( 6000 = 1500v \)
Step 3: Solve for \(v\)
\( v = \dfrac{6000}{1500} = 4 \, \text{m/s} \)
Final Answer:
The combined vehicles move with velocity \(\boxed{4 \, \text{m/s}}\) in the original direction of the truck.