Mechanics: Kinematics and Dynamics
Scope of the domain
This section covers motion in one and two dimensions and the forces that cause it. The exam asks you to select and apply the constant-acceleration equations, interpret motion graphs, resolve forces on inclines, and analyse connected bodies with Newton's second law.
Core rules
- For constant acceleration only: v = u + at, s = ut + ½at², v² = u² + 2as, s = ½(u + v)t.
- On a velocity–time graph the gradient is acceleration and the area is displacement; on a displacement–time graph the gradient is velocity.
- Newton's first law: no resultant force means no change in velocity. Second law: F_net = ma, with F_net the vector sum. Third-law pairs are equal, opposite, of the same type, and act on different bodies.
- Weight W = mg acts at the centre of gravity; the normal reaction is perpendicular to the surface; kinetic friction f = μN opposes relative sliding.
- On an incline of angle θ, the weight component along the slope is mg sinθ and perpendicular to it is mg cosθ.
- For connected masses, treat the system together to find acceleration, then isolate one body to find the tension.
Worked example
A car travelling at 20 m/s brakes uniformly and stops in 50 m. Find the deceleration and the time. Using v² = u² + 2as with v = 0, u = 20, s = 50: 0 = 400 + 100a, so a = −4.0 m/s². The negative sign shows the acceleration opposes the motion. Then v = u + at gives 0 = 20 − 4.0t, so t = 5.0 s. Check: s = ½(u + v)t = ½(20)(5) = 50 m.
Traps that cost marks
- Using suvat when acceleration is not constant, for example during a collision or with air resistance.
- Forgetting that deceleration is negative, then losing the sign in the next equation.
- Confusing mass with weight, or quoting weight in kg.
- Describing a third-law pair as two forces on the same body.
- Using μ with the wrong normal force on an incline, where N = mg cosθ.
- Mixing units, such as km/h with m/s², without converting first.