Edexcel IAL - Mechanics 2- 4.1 Momentum and Impulse- Study notes - New syllabus
Edexcel IAL – Mechanics 2- 4.1 Momentum and Impulse -Study notes- New syllabus
Edexcel IAL – Mechanics 2- 4.1 Momentum and Impulse -Study notes -Edexcel A level Maths- per latest Syllabus.
Key Concepts:
- 4.1 Momentum and Impulse
Momentum as a Vector, Impulse–Momentum Principle, and Conservation of Linear Momentum
Momentum plays a central role in mechanics, especially when dealing with interactions between particles. In this syllabus, momentum is treated as a vector quantity, and its principles are applied mainly to motion in one or two dimensions.
Momentum as a Vector
The momentum \( \mathbf{p} \) of a particle is defined as the product of its mass and velocity:
\( \mathbf{p} = m\mathbf{v} \)
Since velocity is a vector, momentum is also a vector and has both magnitude and direction.
If the velocity is written in component form, momentum can be written as
\( \mathbf{p} = m(v_x\mathbf{i} + v_y\mathbf{j}) \)
The SI unit of momentum is \( \text{kg m s}^{-1} \).
The Impulse–Momentum Principle (Vector Form)
The impulse of a force is defined as the product of the force and the time for which it acts.
In vector form, impulse is given by
\( \mathbf{I} = \mathbf{F}t \)
The impulse–momentum principle states that the impulse of the resultant force acting on a particle is equal to the change in momentum of the particle:
\( \mathbf{F}t = m\mathbf{v} – m\mathbf{u} \)
This principle is valid even when the force acts for a short time, such as during collisions.
Conservation of Linear Momentum
If no external resultant force acts on a system of particles, the total linear momentum of the system remains constant.
In vector form, for two particles:
Total momentum before interaction \( = \) total momentum after interaction
\( m_1\mathbf{u}_1 + m_2\mathbf{u}_2 = m_1\mathbf{v}_1 + m_2\mathbf{v}_2 \)
This principle is commonly applied to collisions and explosions. Newton’s law of restitution is not required for this syllabus.
Example :
A particle of mass \( 3 \) kg moves with velocity \( 4\mathbf{i} + 2\mathbf{j} \,\text{m s}^{-1} \). Find its momentum.
▶️ Answer/Explanation
Momentum is given by \( \mathbf{p} = m\mathbf{v} \).
\( \mathbf{p} = 3(4\mathbf{i} + 2\mathbf{j}) \)
\( \mathbf{p} = 12\mathbf{i} + 6\mathbf{j} \)
Conclusion: The momentum is \( 12\mathbf{i} + 6\mathbf{j} \,\text{kg m s}^{-1} \).
Example :
A force of \( 5\mathbf{i} \) N acts on a particle of mass \( 2 \) kg for \( 4 \) s. If the particle is initially at rest, find its velocity after \( 4 \) s.
▶️ Answer/Explanation
Using the impulse–momentum principle:
\( \mathbf{F}t = m\mathbf{v} – m\mathbf{u} \)
Since the particle starts from rest, \( \mathbf{u} = \mathbf{0} \).
\( 5\mathbf{i} \times 4 = 2\mathbf{v} \)
\( \mathbf{v} = 10\mathbf{i} \)
Conclusion: The velocity after \( 4 \) s is \( 10\mathbf{i} \,\text{m s}^{-1} \).
Example :
Two particles of masses \( 2 \) kg and \( 3 \) kg move along the same straight line. Their velocities before collision are \( 6 \,\text{m s}^{-1} \) and \( -2 \,\text{m s}^{-1} \) respectively. Find the common velocity after they stick together.
▶️ Answer/Explanation
Since the particles stick together, momentum is conserved.
Initial momentum:
\( (2)(6) + (3)(-2) = 12 – 6 = 6 \)
Total mass after collision:
\( 2 + 3 = 5 \)
Final velocity:
\( v = \dfrac{6}{5} \)
Conclusion: The common velocity after collision is \( \dfrac{6}{5} \,\text{m s}^{-1} \).