Further Mechanics: Impulse, Momentum and Circular Motion

Introduction to Further Mechanics

Welcome to Further Mechanics! In IAS Unit 1, you learned how forces cause linear acceleration and explored momentum along a single straight line. Here in IA2 Unit 4, we expand our physical toolkit into two crucial dimensions: changing momentum during impacts (impulse), analyzing collisions in two dimensions, and understanding how objects move along curved paths (circular motion).

Don't worry if these ideas seem a bit abstract at first. Mechanics is all around you—from seatbelts cushioning your body in a car stop to a roller coaster looping upside down without dropping its passengers. By breaking each topic down into clear, logical steps, you will quickly master the calculations and explanations needed for your exam.

1. Impulse and Newton's Second Law

In Unit 1, Newton's second law was often stated simply as \(\Sigma F = ma\). However, Sir Isaac Newton originally defined force in terms of the rate of change of momentum. When mass remains constant, this leads to one of the most powerful equations in mechanics.

Momentum and the Impulse Equation

Linear momentum \(p\) of an object of mass \(m\) moving with velocity \(v\) is defined as:
\(p = mv\)
Because velocity is a vector, momentum is also a vector quantity. Its SI unit is \(\text{kg m s}^{-1}\) (or equivalently, \(\text{N s}\)).

When an average resultant force \(F\) acts on an object over a time interval \(\Delta t\), the product is known as impulse:
\(\text{Impulse} = F \Delta t\)

According to Newton's second law, force is the rate of change of momentum:
\(F = \frac{\Delta p}{\Delta t}\)

Rearranging this gives the fundamental formula from your formula sheet:
\(F \Delta t = \Delta p\)

This tells us that impulse equals change of momentum.

Force-Time Graphs

Real-life forces—such as a bat hitting a cricket ball or a foot kicking a football—are rarely constant. They rise rapidly to a peak and then drop back to zero.
• On a graph of force against time (\(F\) on the vertical axis, \(t\) on the horizontal axis), the area under the graph represents the impulse, which equals the change in momentum \(\Delta p\).
• If the graph is a simple shape (like a triangle or rectangle), you can calculate the area geometrically: \(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}\).
• If the force is non-uniform, the mean force \(F_{\text{average}}\) acting over time \(\Delta t\) can be found using \(F_{\text{average}} = \frac{\Delta p}{\Delta t}\).

Real-World Application: Impact Safety

Why do cars have crumple zones and airbags? Why do gymnasts bend their knees when landing?
Consider a passenger coming to rest from a high speed: their change in momentum \(\Delta p\) is fixed by their initial speed and mass.
Since \(F = \frac{\Delta p}{\Delta t}\):
• By increasing the impact duration \(\Delta t\) (via crumple zones, airbags, or bending knees), the average impact force \(F\) acting on the human body is greatly reduced.
• A smaller force means a vastly lower risk of severe injury.

Core Practical 9 Link: Force and Momentum Change

In Core Practical 9, you investigate the force exerted on an object against its change of momentum. Typically, a trolley on a friction-compensated track is pulled by a falling mass or pushed against a force sensor. Light gates determine the initial and final velocities to calculate \(\Delta p = m(v - u)\), while a force sensor or data logger measures the force over the collision time \(\Delta t\). Plotting force against \(\frac{\Delta p}{\Delta t}\) confirms that the gradient is 1, verifying \(F = \frac{\Delta p}{\Delta t}\).

Key Takeaway for Section 1: Impulse is the product of force and time (\(F \Delta t\)), it equals the change in momentum (\(\Delta p\)), and it is given by the area under a force–time graph.

2. Collisions and Momentum in Two Dimensions

Momentum is always conserved in any closed system where no external resultant forces act. In Unit 4, we apply this law to collisions occurring on a plane (2D).

The Principle of Conservation of Linear Momentum

The principle states: For a system of interacting bodies, the total linear momentum remains constant, provided no external resultant force acts on the system.
Because momentum is a vector, it must be conserved independently along perpendicular directions, typically chosen as the \(x\)-axis (horizontal) and the \(y\)-axis (vertical):
• Total initial momentum in the \(x\)-direction = Total final momentum in the \(x\)-direction
• Total initial momentum in the \(y\)-direction = Total final momentum in the \(y\)-direction

Resolving 2D Collisions Step-by-Step

Suppose particle 1 (mass \(m_1\)) moves horizontally with speed \(u_1\) and strikes a stationary particle 2 (mass \(m_2\)). After collision, particle 1 moves at speed \(v_1\) at an angle \(\theta_1\) above the horizontal, and particle 2 moves at speed \(v_2\) at an angle \(\theta_2\) below the horizontal.

Step 1: Write the horizontal (\(x\)) equation:
\(m_1 u_1 = m_1 v_1 \cos\theta_1 + m_2 v_2 \cos\theta_2\)

Step 2: Write the vertical (\(y\)) equation:
Since there was zero initial vertical motion:
\(0 = m_1 v_1 \sin\theta_1 - m_2 v_2 \sin\theta_2\)
Which means: \(m_1 v_1 \sin\theta_1 = m_2 v_2 \sin\theta_2\)

Vector Triangle Method: Because the total initial momentum vector equals the total final momentum vector, the vector sum of final momenta \(\vec{p}_1 + \vec{p}_2\) must close to form a triangle with initial momentum \(\vec{p}_{\text{initial}}\). For perpendicular final paths (common when two equal masses collide elastically with one at rest), this forms a right-angled vector triangle.

Core Practical 10 Link: ICT Analysis of Collisions

In Core Practical 10, you use digital video analysis or stroboscopic photography to study collisions between two small spheres (such as ball bearings or pucks on an air table). By tracking position coordinates over successive video frames, software calculates the \(x\)- and \(y\)-velocity components before and after impact, allowing you to test momentum conservation in two dimensions.

Elastic vs. Inelastic Collisions

While momentum is always conserved in all collisions (assuming no external resultant force), kinetic energy is not always conserved:
• Elastic collision: Total kinetic energy is conserved. (\(E_{k,\text{initial}} = E_{k,\text{final}}\)). Collisions between subatomic particles or idealized gas molecules are often elastic.
• Inelastic collision: Total kinetic energy is not conserved. Some kinetic energy is converted into thermal energy, sound, or the work done in permanently deforming the colliding objects. If the bodies stick together after colliding, it is described as completely inelastic.

Kinetic Energy Expressed in Terms of Momentum

For a non-relativistic particle of mass \(m\) moving with speed \(v\):
Kinetic energy is defined as:
\(E_k = \frac{1}{2}mv^2\)

We know that momentum is \(p = mv\), which means \(v = \frac{p}{m}\). Substituting this into the kinetic energy equation gives:
\(E_k = \frac{1}{2}m\left(\frac{p}{m}\right)^2 = \frac{1}{2}m\frac{p^2}{m^2}\)

Cancelling one factor of \(m\) produces the equation provided on the Unit 4 formula sheet:
\(E_k = \frac{p^2}{2m}\)

Why is this relationship useful? It allows you to find kinetic energy directly from momentum without calculating speed first. It also shows that if two particles have the same momentum, the lighter particle (smaller \(m\)) carries significantly more kinetic energy.

Key Takeaway for Section 2: Linear momentum is conserved independently in perpendicular directions. Always check whether kinetic energy is conserved to classify a collision as elastic or inelastic, and use \(E_k = \frac{p^2}{2m}\) for quick energy-momentum conversions.

3. Circular Motion: Kinematics and Angles

When an object travels along a circular path, its direction of travel is constantly changing. To describe this smoothly, physicists use rotational quantities rather than straight-line distances.

Radians vs. Degrees

The radian (\(\text{rad}\)) is the natural SI unit for measuring angles.
One radian is defined as the angle subtended at the centre of a circle by an arc whose length is equal to the radius of the circle.

Because the circumference of a complete circle of radius \(r\) is \(2\pi r\):
\(360^\circ = 2\pi \text{ radians}\)

From this relationship:
• To convert degrees to radians: multiply by \(\frac{2\pi}{360} = \frac{\pi}{180}\)
• To convert radians to degrees: multiply by \(\frac{180}{\pi}\)
• Common angles: \(90^\circ = \frac{\pi}{2}\text{ rad}\), \(180^\circ = \pi\text{ rad}\), \(270^\circ = \frac{3\pi}{2}\text{ rad}\), \(360^\circ = 2\pi\text{ rad}\).

For an angle \(\theta\) in radians, arc length \(s\) is given by:
\(s = r\theta\)

Angular Velocity and Period

Angular velocity (\(\omega\), Greek letter omega) is the rate of change of angular displacement with respect to time:
\(\omega = \frac{\Delta\theta}{\Delta t}\)
The SI unit of angular velocity is \(\text{rad s}^{-1}\).

The time period (\(T\)) is the time taken for the object to complete one full revolution (\(2\pi\) radians). Therefore:
\(\omega = \frac{2\pi}{T}\)
Rearranging for time period gives the equation on your formula sheet:
\(T = \frac{2\pi}{\omega}\)

Since frequency \(f = \frac{1}{T}\), we can also relate angular velocity to frequency:
\(\omega = 2\pi f\)

Connecting Linear Speed and Angular Velocity

Consider an object moving along a circle of radius \(r\). In one period \(T\), it covers a distance equal to the circumference \(2\pi r\).
Its linear speed \(v\) is:
\(v = \frac{\text{distance}}{\text{time}} = \frac{2\pi r}{T}\)

Since \(\frac{2\pi}{T} = \omega\), we obtain:
\(v = \omega r\)

Memory tip: Think of a spinning record or carousel. Every point on the disc has the exact same angular velocity \(\omega\) (they all turn through the same angle in one second), but points further from the centre (larger \(r\)) must travel faster along a larger circle, so they have a higher linear speed \(v\).

Key Takeaway for Section 3: Angles must be expressed in radians for circular motion calculations (\(2\pi\text{ rad} = 360^\circ\)). Angular velocity is \(\omega = \frac{2\pi}{T}\), and linear speed along the circumference is \(v = \omega r\).

4. Centripetal Acceleration and Centripetal Force

This is one of the most critical conceptual areas in A Level physics: Can an object accelerate if its speed is constant? Yes! Velocity is a vector with both magnitude (speed) and direction. When an object follows a curved path at constant speed, its direction changes continuously. Because velocity is changing, the object must be accelerating.

Deriving Centripetal Acceleration from Vector Diagrams

Understanding this derivation is an explicit requirement of the Edexcel specification.

Consider a particle moving at constant speed \(v\) along an arc of a circle of radius \(r\) from point A to point B in a short time interval \(\Delta t\).
• At point A, the velocity vector is \(\vec{v}_A\), tangential to the circle.
• At point B, the velocity vector is \(\vec{v}_B\), also of magnitude \(v\), but turned through an angle \(\Delta\theta\).
• The angle subtended at the centre of the circle between radius vectors to A and B is \(\Delta\theta\).

The change in velocity is given by vector subtraction:
\(\Delta\vec{v} = \vec{v}_B - \vec{v}_A\)

Now, draw the vector triangle formed by \(-\vec{v}_A\), \(\vec{v}_B\), and \(\Delta\vec{v}\):
1. The two velocity vectors \(\vec{v}_A\) and \(\vec{v}_B\) both have the same length \(v\).
2. The angle between \(\vec{v}_A\) and \(\vec{v}_B\) is \(\Delta\theta\).
3. For very small angles \(\Delta\theta\) (as \(\Delta t \to 0\)), the vector triangle is similar to the spatial position triangle formed by the two radii and the arc length \(\Delta s\).
4. Using the ratio of side lengths from these similar triangles:
\(\frac{|\Delta\vec{v}|}{v} \approx \frac{\Delta s}{r}\)
Therefore:
\(|\Delta\vec{v}| \approx \frac{v \Delta s}{r}\)

Divide both sides by the time interval \(\Delta t\):
\(\frac{|\Delta\vec{v}|}{\Delta t} \approx \frac{v}{r} \left(\frac{\Delta s}{\Delta t}\right)\)

As \(\Delta t\) approaches zero:
• \(\frac{|\Delta\vec{v}|}{\Delta t}\) becomes the magnitude of acceleration \(a\).
• \(\frac{\Delta s}{\Delta t}\) is the linear speed \(v\).

Substituting \(v\) gives the formula for centripetal acceleration:
\(a = \frac{v^2}{r}\)

Because \(v = \omega r\), we can substitute \(v\) to find the alternative form:
\(a = \frac{(\omega r)^2}{r} = \frac{\omega^2 r^2}{r} = r\omega^2\)

Both forms are supplied on your exam data sheet:
\(a = \frac{v^2}{r} = r\omega^2\)

Direction of acceleration: Looking at the vector triangle, as \(\Delta\theta \to 0\), the vector \(\Delta\vec{v}\) points directly towards the centre of the circle. Thus, the acceleration is directed radially inwards towards the centre of the circular path. This is why it is called centripetal (meaning "centre-seeking").

Centripetal Force

By Newton's second law, whenever there is an acceleration, there must be a resultant force in the same direction: \(\Sigma F = ma\).
Therefore, to keep an object of mass \(m\) moving in a circle, there must be a resultant force directed towards the centre:
\(F = \frac{mv^2}{r} = mr\omega^2\)

CRITICAL EXAM WARNING: Centripetal force is NOT a new, separate kind of force! It is simply the name given to the net resultant force acting towards the centre. It is always provided by physical, identifiable forces such as tension, friction, gravity, or normal reaction force.
• In free-body diagrams, never label a force as "centripetal force". Label the actual physical force (e.g., "Friction" or "Tension").

Common Circular Motion Scenarios

1. Car Rounding a Flat Horizontal Bend:
The centripetal force is provided entirely by the sideways friction between the tyres and the road surface:
\(F_{\text{friction}} = \frac{mv^2}{r}\)
If the road is icy or the car travels too fast, the required centripetal force exceeds the maximum available friction, and the car skids outward along a tangential path.

2. The Conical Pendulum / Banked Track:
When an aircraft banks or a pendulum swings in a horizontal circle, the tension \(T\) (or normal reaction \(R\)) acts at an angle \(\theta\) to the vertical.
• Resolving vertically (equilibrium): \(T \cos\theta = mg\)
• Resolving horizontally (towards centre): \(T \sin\theta = \frac{mv^2}{r}\)
Dividing the horizontal equation by the vertical equation eliminates mass and tension:
\(\tan\theta = \frac{v^2}{rg}\)

3. Vertical Circles (e.g., Ferris Wheel or Roller Coaster Loop):
Here, gravity and normal contact force act along the same line, but their directions change relative to the centre:
• At the bottom of the loop: Upward normal reaction \(R\) points towards the centre; weight \(mg\) points downwards away from the centre.
Resultant force towards centre: \(R - mg = \frac{mv^2}{r} \implies R = mg + \frac{mv^2}{r}\)
This is why you feel heavier at the bottom of a roller coaster drop!
• At the top of the loop (inside track): Both normal reaction \(R\) and weight \(mg\) point downwards towards the centre.
Resultant force towards centre: \(R + mg = \frac{mv^2}{r} \implies R = \frac{mv^2}{r} - mg\)
• Minimum speed to complete the loop: The car stays on the track as long as it maintains contact, meaning \(R \ge 0\). The minimum speed occurs just as \(R = 0\):
\(mg = \frac{mv^2}{r} \implies v_{\text{min}} = \sqrt{gr}\)

Key Takeaway for Section 4: Centripetal acceleration is directed towards the centre and given by \(a = \frac{v^2}{r} = r\omega^2\). The resultant force towards the centre is \(F = \frac{mv^2}{r} = mr\omega^2\), provided by real physical forces like tension, friction, or normal reaction.

Summary of Essential Equations

All of the following equations are provided in the IA2 Unit 4 exam formula sheet. You do not need to memorize them, but you must know how and when to apply them:
• Impulse: \(F \Delta t = \Delta p\)
• Kinetic Energy in terms of Momentum: \(E_k = \frac{p^2}{2m}\)
• Linear Speed in Circular Motion: \(v = \omega r\)
• Time Period of Rotation: \(T = \frac{2\pi}{\omega}\)
• Centripetal Acceleration: \(a = \frac{v^2}{r} = r\omega^2\)
• Centripetal Force: \(F = \frac{mv^2}{r} = mr\omega^2\)

Exam Technique & Common Pitfalls

• Forgetting Radian Mode: Always check your calculator mode! When computing angles or trigonometric functions in circular motion equations, ensure your calculator is set to radians.
• Confusing Impulse with Force: Force is measured in newtons (\(\text{N}\)), while impulse is force multiplied by time (\(\text{N s}\)). Check whether a question asks for the force or the impulse.
• Centrifugal Force Trap: "Centrifugal force" does not exist in an inertial frame of reference. Never write that a body is pushed outwards by centrifugal force. An object wants to continue in a straight line at a tangent because of Newton's first law (inertia); it requires an inward centripetal force to keep curving.
• Resolving Vectors in 2D Collisions: Ensure you assign consistent positive and negative signs to opposite directions (e.g., upward velocities positive, downward negative) so momentum components don't cancel out incorrectly.