Welcome to Dynamics!
Welcome to one of the most exciting areas of your CCEA AS 1 Physics course: Dynamics! While kinematics describes how objects move, dynamics explains why they move by investigating the forces responsible for motion. Whether you are analyzing a car braking, a skydiver reaching terminal velocity, or a football arcing through the air, dynamics provides the mathematical toolkit to understand it all.
Don't worry if this chapter feels challenging at first! By breaking down forces into simple components and taking a step-by-step approach to calculations, you will master these concepts in no time.
1. Resistive Forces and Free-Body Force Analysis
Before calculating how an object accelerates, we must first identify all the forces acting upon it. In the real world, motion is almost always opposed by resistive forces.
Types of Resistive Forces
1. Friction: A contact force that opposes the relative motion between two solid surfaces in direct contact.
2. Drag (Fluid Resistance / Air Resistance): A resistive force exerted on an object moving through a fluid (either a liquid or a gas). Drag always acts in the direction directly opposite to the instantaneous direction of motion.
Free-Body Diagrams (FBDs)
A free-body diagram is a diagrammatic representation showing all the forces acting on a single isolated body. Follow these core rules when drawing FBDs for CCEA examinations:
• Represent the object simply (such as a box or dot) and draw all force vectors as arrows pointing away from the centre of mass.
• Make the length of each arrow proportional to the magnitude of the force.
• Clearly label every force (e.g., Weight \(W\), Normal Reaction \(R\), Friction \(F\), Drag \(D\)).
• Examiner Warning: Never include the "resultant force" as an extra arrow on a free-body diagram! An FBD must only display actual physical forces acting on the body.
Resolving Forces on an Inclined Plane
When an object sits on a slope tilted at an angle \(\theta\) to the horizontal, its weight \(W = mg\) acts vertically downwards. To analyse the motion, we resolve the weight into two perpendicular components:
• Component parallel to the slope (pulling the object down the ramp):
\(W_\parallel = mg \sin\theta\)
• Component perpendicular to the slope (pressing into the ramp):
\(W_\perp = mg \cos\theta\)
• Normal reaction force (\(R\)):
When there are no extra forces acting perpendicular to the ramp, the surface pushes back equally: \(R = mg \cos\theta\).
Helpful Memory Trick: Remember that \(\cos\) is "close" to the angle \(\theta\) between the vertical and the perpendicular axis, so the perpendicular component gets \(\cos\theta\), while the parallel component gets \(\sin\theta\).
Key Takeaway: Always resolve forces along the plane of motion and perpendicular to it. On a slope of angle \(\theta\), the downhill component of gravity is always \(mg \sin\theta\).
2. Resultant Force and Newton's Second Law
When the forces acting on an object are unbalanced, a net force exists, causing the object to accelerate.
Newton's Second Law
The net resultant force \(F_{\text{net}}\) acting on a body of constant inertial mass \(m\) produces an acceleration \(a\) in the direction of the resultant force:
\(F_{\text{net}} = \sum F = ma\)
Where:
• \(F_{\text{net}}\) is the resultant force measured in newtons (\(\text{N}\))
• \(m\) is the mass in kilograms (\(\text{kg}\))
• \(a\) is the acceleration in metres per second squared (\(\text{m s}^{-2}\))
Understanding Terminal Velocity
Consider an object falling vertically from rest through a fluid (such as a skydiver falling through air):
Step 1: Release at \(t = 0\)
The instantaneous velocity is zero (\(v = 0\text{ m s}^{-1}\)), meaning drag is zero (\(F_{\text{drag}} = 0\text{ N}\)). The only force acting is weight (\(W = mg\)). The downward acceleration is at its maximum: \(a = g = 9.81\text{ m s}^{-2}\).
Step 2: Gaining Speed
As the object falls faster, the upward drag force increases. The net downward force decreases:
\(F_{\text{net}} = mg - F_{\text{drag}} = ma\)
Because \(F_{\text{net}}\) decreases, the acceleration \(a\) decreases, but the object is still speeding up.
Step 3: Reaching Terminal Velocity (\(v_t\))
Eventually, the velocity increases until the upward drag force equals the downward weight:
\(F_{\text{drag}} = mg \implies F_{\text{net}} = 0\text{ N}\)
With zero net force, the acceleration becomes zero (\(a = 0\text{ m s}^{-2}\)). The object now falls at a constant, maximum velocity known as terminal velocity.
Common Exam Mistake: Never say "gravity stops acting" or "weight becomes zero" at terminal velocity! The weight remains completely unchanged at \(mg\); it is the resultant force that becomes zero because drag balances weight.
Key Takeaway: When resistive forces match the driving force, \(F_{\text{net}} = 0\text{ N}\) and acceleration ceases, leading to steady terminal velocity.
3. Projectile Motion (2D Dynamics under Gravity)
A projectile is any object launched into the air that moves under the influence of gravity alone. The fundamental principle of projectile motion is the independence of orthogonal components: horizontal motion and vertical motion do not affect each other and must be analysed completely separately.
Core Assumptions for Ideal Projectiles
• Horizontal acceleration is zero: \(a_x = 0\text{ m s}^{-2}\) (air resistance is neglected).
• Vertical acceleration is constant and directed downwards: \(a_y = -g = -9.81\text{ m s}^{-2}\) (taking upwards as positive).
Resolving Initial Velocity
If an object is launched with an initial velocity \(u\) at an angle \(\theta\) above the horizontal:
• Initial horizontal velocity: \(u_x = u \cos\theta\)
• Initial vertical velocity: \(u_y = u \sin\theta\)
Horizontal Motion Analysis (\(a_x = 0\))
Because there is no horizontal acceleration, horizontal velocity remains constant throughout the entire flight:
\(v_x = u_x = u \cos\theta\)
Horizontal distance (displacement \(x\)) covered in time \(t\):
\(x = u_x t = (u \cos\theta) t\)
Vertical Motion Analysis (\(a_y = -g\))
Vertical motion behaves exactly like linear motion with constant acceleration \(g = 9.81\text{ m s}^{-2}\) acting downwards:
• Velocity at time \(t\):
\(v_y = u_y - gt = u \sin\theta - gt\)
• Vertical displacement \(y\) at time \(t\):
\(y = u_y t - \frac{1}{2}gt^2 = (u \sin\theta)t - \frac{1}{2}gt^2\)
• Velocity-displacement relationship:
\(v_y^2 = u_y^2 - 2gy\)
Key Projectile Formulae (Level Launch and Landing)
For a projectile launched from ground level and returning to the same ground level:
• Time to reach maximum height (\(t_{\text{apex}}\)): At the top of the trajectory, vertical velocity momentarily drops to zero (\(v_y = 0\)):
\(0 = u \sin\theta - gt_{\text{apex}} \implies t_{\text{apex}} = \frac{u \sin\theta}{g}\)
• Total time of flight (\(T\)): Because the trajectory is symmetrical under uniform gravity:
\(T = 2 t_{\text{apex}} = \frac{2u \sin\theta}{g}\)
• Maximum vertical height (\(H\)): Using \(v_y^2 = u_y^2 - 2gH\) with \(v_y = 0\):
\(H = \frac{u^2 \sin^2\theta}{2g}\)
• Horizontal range (\(R\)): The total horizontal distance traveled over time \(T\):
\(R = u_x T = (u \cos\theta)\left(\frac{2u \sin\theta}{g}\right) = \frac{u^2 \sin(2\theta)}{g}\)
Finding the Resultant Instantaneous Velocity
At any point during the flight, the actual velocity vector \(v\) is found by recombining the horizontal and vertical velocity components using Pythagoras' theorem and trigonometry:
• Speed (magnitude):
\(v = \sqrt{v_x^2 + v_y^2}\)
• Direction (angle \(\alpha\) relative to the horizontal):
\(\alpha = \arctan\left(\frac{v_y}{v_x}\right)\)
Key Takeaway: Time \(t\) is the single variable shared by both horizontal and vertical equations. Solve for time using one direction, then substitute it into the other!
4. CCEA Exam Pitfalls & Top Tips
Make sure you avoid these common traps reported by CCEA examiners:
1. Strict Unit Notation: Always express SI units in index form. Write \(\text{m s}^{-1}\) and \(\text{m s}^{-2}\), never solidus notation like m/s or m/s^2.
2. The Value of \(g\): Use \(g = 9.81\text{ m s}^{-2}\) (or \(\text{N kg}^{-1}\)) as listed on your CCEA Data Sheet, not \(9.8\) or \(10\).
3. Consistency with Signs: If you define upwards as positive, remember that vertical gravitational acceleration must be entered as negative (\(-9.81\text{ m s}^{-2}\)). If an object falls below its starting point, its displacement \(y\) is also negative.
4. Horizontal Independence: Never substitute \(g = 9.81\text{ m s}^{-2}\) into horizontal motion calculations! Horizontal acceleration is strictly zero in projectile problems.
Quick Summary Checklist
Before moving on to the next chapter, ensure you can:
• State the difference between solid friction and fluid drag.
• Draw fully labelled free-body diagrams with forces resolved on a slope (\(W_\parallel = mg \sin\theta\) and \(W_\perp = mg \cos\theta\)).
• Explain the step-by-step transition from initial release to terminal velocity using \(F_{\text{net}} = ma\).
• Split initial launch velocities into \(u_x = u \cos\theta\) and \(u_y = u \sin\theta\).
• Calculate the time of flight, maximum height, range, and instantaneous resultant velocity for any projectile.