Chapter: Linear Momentum and Impulse
Welcome to one of the most exciting and practical topics in AS Physics! Have you ever wondered why catching a fast-moving cricket ball hurts your hands unless you pull them back, or why modern cars are built with crumple zones? The answers lie in linear momentum and impulse.
Don't worry if physics formulas sometimes feel overwhelming. In this chapter, we will break down each idea step-by-step using clear everyday examples. By the end of these notes, you will feel confident tackling any momentum question in your CCEA AS 1 exam!
---1. Understanding Linear Momentum
What is Momentum?
Informally, momentum is simply "mass on the move". It is a measure of how difficult it is to stop a moving object.
Think about this: A heavy lorry rolling very slowly at \(1\text{ m s}^{-1}\) is hard to stop because of its large mass. A tiny bullet flying at \(400\text{ m s}^{-1}\) is also hard to stop because of its high velocity. Both have a large amount of momentum.
The Definition and Formula
Linear momentum (\(p\)) is defined as the product of an object's mass (\(m\)) and its velocity (\(v\)).
\(p = m \times v\)
Where:
• \(p\) = linear momentum, measured in kilogram metres per second (\(\text{kg m s}^{-1}\)) or Newton-seconds (\(\text{N s}\))
• \(m\) = mass of the object, measured in kilograms (\(\text{kg}\))
• \(v\) = velocity of the object, measured in metres per second (\(\text{m s}^{-1}\))
Vector Nature of Momentum
Because velocity is a vector (it has both magnitude and direction), momentum is also a vector quantity. Its direction is always the same as the direction of the velocity.
Crucial Rule for Calculations: Always choose one direction as positive (e.g., to the right = \(+\)) and the opposite direction as negative (e.g., to the left = \(-\)). If an object bounces backwards, its new velocity must have a minus sign!
Did you know? Momentum has two equivalent units: \(\text{kg m s}^{-1}\) and \(\text{N s}\). Both are completely acceptable in your exam, but stick to standard SI units when calculating!
Key Takeaway: Momentum is \(p = mv\). It is a vector quantity, so direction (and plus/minus signs) matters enormously!
---2. Newton’s Second Law in Terms of Momentum
Connecting Force and Momentum
You probably already know Newton's Second Law as \(F = ma\). At AS Level, we look at Newton's original and more fundamental definition:
Newton's Second Law: The resultant force acting on an object is directly proportional to (and in the same direction as) the rate of change of momentum.
\(F = \frac{\Delta p}{\Delta t} = \frac{mv - mu}{t}\)
Where:
• \(F\) = resultant force applied (\(\text{N}\))
• \(\Delta p\) = change in momentum (\(\text{kg m s}^{-1}\) or \(\text{N s}\))
• \(\Delta t\) (or \(t\)) = time taken for the change (\(\text{s}\))
• \(u\) = initial velocity (\(\text{m s}^{-1}\))
• \(v\) = final velocity (\(\text{m s}^{-1}\))
How does this link to \(F = ma\)?
If the mass \(m\) stays constant, we can factor it out:
\(F = \frac{m(v - u)}{t}\)
Since acceleration is \(a = \frac{v - u}{t}\), substituting this gives:
\(F = ma\)
So \(F = ma\) is simply a special case of \(F = \frac{\Delta p}{\Delta t}\) when mass does not change!
Key Takeaway: Force is the rate of change of momentum (\(F = \frac{\Delta p}{\Delta t}\)). To create a big change in momentum quickly, you need a large resultant force.
---3. Impulse and Force-Time Graphs
What is Impulse?
When you rearrange the equation \(F = \frac{\Delta p}{\Delta t}\), you get:
\(\text{Impulse} = F \Delta t = \Delta p = mv - mu\)
Impulse is defined as the product of the average force and the time duration for which it acts. It is equal to the change in momentum of the object.
• Unit of Impulse: Newton-seconds (\(\text{N s}\)) or \(\text{kg m s}^{-1}\)
Force-Time (\(F\)-\(t\)) Graphs
In the real world, forces are rarely constant during collisions. A tennis racket hitting a ball exerts a force that starts small, peaks, and then drops back to zero.
• On a graph of Force (y-axis) against Time (x-axis), the area under the graph equals the Impulse (which is the total change in momentum, \(\Delta p\)).
• For a triangular force-time graph: \(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \text{Impulse}\)
Real-World Applications: Vehicle Safety and Sports
In any crash, an occupant's change in momentum (\(\Delta p\)) is fixed (they must be brought to rest from their driving speed). Since \(\Delta p = F \times \Delta t\):
• If you increase the collision time (\(\Delta t\)), you decrease the impact force (\(F\)) felt by the passenger.
Safety Examples:
• Crumple Zones: The front of a car collapses progressively during a collision, extending the impact time (\(\Delta t\)) and reducing the average force (\(F\)) on passengers.
• Airbags & Seatbelts: They stretch and compress, increasing stopping time and spreading the force over a larger area of the body.
• Catching a ball: Wicketkeepers pull their hands back with the ball to increase contact time (\(\Delta t\)), reducing the painful impact force.
Common Mistake to Avoid: Never say "crumple zones reduce the change in momentum". The change in momentum (\(\Delta p\)) is identical whether the crash is soft or hard! Crumple zones increase the time taken for that change to happen, which reduces the force.
Key Takeaway: Impulse \(= F \Delta t = \Delta p = \text{Area under } F\text{-}t \text{ graph}\). Increasing time reduces force!
---4. Principle of Conservation of Linear Momentum
The Law
The Principle of Conservation of Linear Momentum states:
For a closed system (with no external resultant forces acting), the total linear momentum before an interaction equals the total linear momentum after the interaction.
\(\text{Total Momentum Before} = \text{Total Momentum After}\)
\(m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2\)
Where:
• \(m_1, m_2\) = masses of bodies 1 and 2
• \(u_1, u_2\) = initial velocities of bodies 1 and 2
• \(v_1, v_2\) = final velocities of bodies 1 and 2
Step-by-Step Guide to Solving Momentum Problems
Step 1: Draw a quick sketch showing the objects before and after the collision.
Step 2: Choose a positive direction (e.g., right = \(+\)).
Step 3: Write down the mass and velocity of each object, assigning a minus sign to any velocity pointing in the negative direction.
Step 4: Set up the equation: \(\text{Total initial momentum} = \text{Total final momentum}\).
Step 5: Solve for the unknown variable.
Special Cases:
1. Objects sticking together: If two objects collide and coalesce (stick together), they move off with a shared final velocity \(v\):
\(m_1 u_1 + m_2 u_2 = (m_1 + m_2)v\)
2. Explosions: If a stationary object splits into two parts (like a cannon firing a cannonball), the initial momentum is zero (\(0\)):
\(0 = m_1 v_1 + m_2 v_2 \implies m_1 v_1 = - m_2 v_2\)
The two pieces fly apart in opposite directions with equal and opposite momenta!
Key Takeaway: Momentum is always conserved in any collision or explosion, provided no external resultant forces act on the system.
---5. Elastic and Inelastic Collisions
While total momentum is always conserved in every collision, total kinetic energy is not always conserved.
Elastic Collisions
• Definition: A collision in which total kinetic energy is conserved.
• \(\text{Total } E_k \text{ before} = \text{Total } E_k \text{ after}\)
• No energy is lost as heat, sound, or permanent deformation.
• Example: Collisions between gas molecules or subatomic particles (and idealised billiard ball collisions).
Inelastic Collisions
• Definition: A collision in which kinetic energy is not conserved.
• \(\text{Total } E_k \text{ after} < \text{Total } E_k \text{ before}\)
• Some kinetic energy is transformed into thermal energy (heat), sound energy, or used to permanently deform the objects.
• Example: Most everyday collisions, like two cars crashing or two clay balls sticking together.
How to Test for Elasticity in an Exam:
If an exam question asks: "Determine whether this collision is elastic or inelastic", follow these two steps:
1. Calculate total initial kinetic energy: \(E_{k(\text{initial})} = \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2\)
2. Calculate total final kinetic energy: \(E_{k(\text{final})} = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2\)
3. Compare the two values:
• If \(E_{k(\text{initial})} = E_{k(\text{final})}\), state that the collision is elastic.
• If \(E_{k(\text{initial})} \neq E_{k(\text{final})}\), state that the collision is inelastic.
Key Takeaway: Momentum is ALWAYS conserved in all collisions. Kinetic energy is ONLY conserved in elastic collisions.
---6. Summary & Quick Review Checklist
Before sitting your exam, make sure you can:
• State the definition and formula for momentum: \(p = mv\) (\(\text{kg m s}^{-1}\) or \(\text{N s}\)).
• State Newton's Second Law in terms of momentum: \(F = \frac{\Delta p}{\Delta t}\).
• Define impulse as \(\text{Impulse} = F \Delta t = \Delta p\).
• Find impulse by calculating the area under a force-time graph.
• Explain vehicle safety features using \(F = \frac{\Delta p}{\Delta t}\) (increasing \(\Delta t\) reduces \(F\)).
• State and apply the Principle of Conservation of Linear Momentum: \(m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2\).
• Remember to assign negative signs to velocities in the opposite direction.
• Distinguish between elastic (KE conserved) and inelastic (KE not conserved) collisions by calculating \(\frac{1}{2}mv^2\).