Mechanics: kinematics, Newton's laws, momentum, energy, circular and rotational motion
Mechanics
This domain tests how you translate motion graphs into equations, apply Newton's laws to systems with friction and tension, use conservation of momentum and energy, and describe circular and rotational motion with torque, moment of inertia and angular momentum. Expect multi-step problems that combine two areas, such as a collision followed by a slide.
Core rules
- Kinematics: for constant acceleration, v=u+at, s=ut+½at², v²=u²+2as. The slope of an x–t graph is velocity; the slope of a v–t graph is acceleration; the area under v–t is displacement.
- Newton's second law: ΣF=ma along each chosen axis. On an incline, use components mg sinθ and mg cosθ; normal force is N=mg cosθ only when no other perpendicular force acts.
- Momentum: p=mv is a vector; total momentum is conserved when external impulse is negligible. In elastic collisions kinetic energy is also conserved; in perfectly inelastic collisions bodies stick together.
- Energy: W=Fd cosθ, KE=½mv², gravitational PE=mgh, elastic PE=½kx². Non-conservative work equals the change in mechanical energy.
- Circular and rotational motion: centripetal acceleration a=v²/r points to the centre; the net inward force is not a new force. For rotation, τ=Iα and L=Iω.
Worked example
A 2.0 kg block slides 5.0 m down a 30° incline with μk=0.20, starting from rest. Find its speed at the bottom. The height drop is h=5.0 sin30°=2.5 m, so ΔPE=2.0×9.8×2.5=49 J. Friction is f=μk mg cos30°=0.20×2.0×9.8×0.866=3.39 N, so work against friction is 3.39×5.0=17 J. Remaining KE=49−17=32 J, hence v=√(2×32/2.0)=5.7 m/s. The reasoning shows why N=mg cosθ, not mg.
Traps
Using mg instead of mg cosθ for the normal force on an incline; forgetting that friction does negative work; treating centripetal force as an extra force; mixing angular and linear quantities; dropping vector signs in momentum problems; assuming mechanical energy is conserved when friction acts.