Edexcel IAL - Mechanics 1- 4.3 Momentum, Impulse and Conservation of Momentum- Study notes - New syllabus
Edexcel IAL – Mechanics 1- 4.3 Momentum, Impulse and Conservation of Momentum -Study notes- New syllabus
Edexcel IAL – Mechanics 1- 4.3 Momentum, Impulse and Conservation of Momentum -Study notes -Edexcel A level Maths- per latest Syllabus.
Key Concepts:
- 4.3 Momentum, Impulse and Conservation of Momentum
Momentum and Impulse
Momentum is a vector quantity that measures the quantity of motion of a particle. It depends on both the mass of the particle and its velocity.
If a particle of mass \( m \) is moving with velocity \( \mathbf{v} \), its momentum \( \mathbf{p} \) is given by
\( \mathbf{p} = m\mathbf{v} \)
The SI unit of momentum is \( \text{kg m s}^{-1} \).
Impulse
The impulse of a force is defined as the product of the force and the time for which it acts.
If a constant force \( \mathbf{F} \) acts for a time \( t \), the impulse is
\( \mathbf{I} = \mathbf{F}t \)
Impulse is also a vector quantity and has the same units as momentum.
Impulse–Momentum Principle
The impulse–momentum principle states that the impulse acting on a particle is equal to the change in momentum of the particle.
\( \mathbf{F}t = m\mathbf{v} – m\mathbf{u} \)
where \( \mathbf{u} \) is the initial velocity and \( \mathbf{v} \) is the final velocity.
This principle is particularly useful when a large force acts over a very short time, such as during a collision.
Example :
A particle of mass \( 2 \) kg is moving in a straight line with speed \( 5 \,\text{m s}^{-1} \). A constant force of \( 6 \) N acts on it for \( 3 \) s in the direction of motion. Find the final speed of the particle.
▶️ Answer/Explanation
Impulse:
\( Ft = 6 \times 3 = 18 \)
Change in momentum:
\( m(v – u) = 2(v – 5) \)
Using the impulse–momentum principle:
\( 18 = 2(v – 5) \Rightarrow v = 14 \)
Conclusion: The final speed of the particle is \( 14 \,\text{m s}^{-1} \).
Conservation of Momentum
The principle of conservation of momentum states that if no external resultant force acts on a system, the total momentum of the system remains constant.
This principle is commonly applied to collisions between particles.
Direct Collisions of Two Particles
In this syllabus, collisions are confined to one-dimensional motion, meaning all motion takes place along a straight line.
If two particles collide directly, momentum is conserved along the line of motion.
Let:
- Masses be \( m_1 \) and \( m_2 \)
- Initial velocities be \( u_1 \) and \( u_2 \)
- Final velocities be \( v_1 \) and \( v_2 \)
Then conservation of momentum gives
\( m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2 \)
No knowledge of Newton’s law of restitution is required. Questions will provide sufficient information to determine the final velocities.
Example :
A particle of mass \( 2 \) kg moving at \( 6 \,\text{m s}^{-1} \) collides directly with a particle of mass \( 3 \) kg moving at \( 2 \,\text{m s}^{-1} \) in the same direction. After the collision, the 2 kg particle moves at \( 1 \,\text{m s}^{-1} \). Find the velocity of the 3 kg particle after the collision.
▶️ Answer/Explanation
Initial momentum:
\( (2 \times 6) + (3 \times 2) = 12 + 6 = 18 \)
Final momentum:
\( (2 \times 1) + 3v = 2 + 3v \)
Using conservation of momentum:
\( 18 = 2 + 3v \Rightarrow v = \dfrac{16}{3} \)
Conclusion: The velocity of the 3 kg particle after the collision is \( \dfrac{16}{3} \,\text{m s}^{-1} \) in the original direction of motion.