Edexcel A Level (IAL) Physics-4.1 Impulse- Study Notes- New Syllabus
Edexcel A Level (IAL) Physics -4.1 Impulse- Study Notes- New syllabus
Edexcel A Level (IAL) Physics -4.1 Impulse- Study Notes -Edexcel A level Physics – per latest Syllabus.
Key Concepts:
- understand how to use the equation impulse = F∆t =∆p (Newton’s second law of motion)
Impulse and Change in Momentum: \( \text{Impulse} = F\Delta t = \Delta p \)
Impulse describes the effect of a force acting over a period of time. It is directly related to the change in momentum of an object and follows from Newton’s second law of motion.
Definition of Momentum
Momentum \( p \) is defined as:
\( p = mv \)
- \( p \) = momentum (kg m s⁻¹)
- \( m \) = mass (kg)
- \( v \) = velocity (m s⁻¹)
Definition of Impulse
Impulse is defined as the product of force and the time for which it acts:
\( \text{Impulse} = F\Delta t \)
- \( F \) = force (N)
- \( \Delta t \) = time interval (s)
Unit of impulse: N s, which is equivalent to kg m s⁻¹.
Connection to Newton’s Second Law
Newton’s second law states:
\( F = \dfrac{\Delta p}{\Delta t} \)
Rearranging:
\( F\Delta t = \Delta p \)
Thus:
\( \text{Impulse} = \Delta p \)
This shows that the impulse applied to an object equals its change in momentum.
Understanding the Equation Physically
- A large force acting for a short time can produce the same impulse as a small force acting for a long time.
- Increasing the stopping time reduces the force for the same change in momentum.
- This principle is used in safety features (airbags, crumple zones).
Direction and Vectors
- Momentum is a vector quantity.
- Impulse has the same direction as the change in momentum.
- A change in direction of motion implies a change in momentum even if speed is constant.
Force–Time Graph Interpretation
For a varying force, impulse is given by the area under the force–time graph:
\( \text{Impulse} = \text{area under } F\text{–}t \text{ graph} \)
This still equals the change in momentum.
Example (Easy)
A constant force of \( 10\ \mathrm{N} \) acts on a ball for \( 0.20\ \mathrm{s} \). Find the impulse.
▶️ Answer / Explanation
\( \text{Impulse} = F\Delta t = 10 \times 0.20 = 2.0\ \mathrm{N\,s} \)
Example (Medium)
A \( 0.50\ \mathrm{kg} \) ball initially at rest is struck and moves off at \( 8.0\ \mathrm{m\,s^{-1}} \). Calculate the impulse delivered to the ball.
▶️ Answer / Explanation
Initial momentum:
\( p_i = 0 \)
Final momentum:
\( p_f = mv = 0.50 \times 8.0 = 4.0\ \mathrm{kg\,m\,s^{-1}} \)
Impulse:
\( \Delta p = 4.0\ \mathrm{kg\,m\,s^{-1}} = 4.0\ \mathrm{N\,s} \)
Example (Hard)
A car of mass \( 1200\ \mathrm{kg} \) moving at \( 20\ \mathrm{m\,s^{-1}} \) is brought to rest uniformly in \( 4.0\ \mathrm{s} \). Find the average force acting on the car.
▶️ Answer / Explanation
Change in momentum:
\( \Delta p = mv_f – mv_i = 0 – (1200 \times 20) = -24000\ \mathrm{kg\,m\,s^{-1}} \)
Use impulse equation:
\( F = \dfrac{\Delta p}{\Delta t} = \dfrac{-24000}{4.0} = -6000\ \mathrm{N} \)
Magnitude of force: \( 6000\ \mathrm{N} \) (opposite to motion).