Mechanics: Kinematics, Newtonian Mechanics, Energy and Momentum Conservation, Circular and Gravitational Motion
Mechanics carries the largest share of tasks in the written Niedersachsen Abitur. Expect a moving body - a cart on a ramp, a horizontally launched ball, a satellite - described first kinematically, then analysed with Newton's second law, conservation of energy or momentum, and finally judged: does the frictionless model hold, and what changes if it does not?
Kinematics
For constant acceleration along one axis use v = v₀ + a·t, s = v₀t + ½at² and v² = v₀² + 2as. Two-dimensional motion splits into independent components: for a horizontal launch the x-motion is uniform while the y-motion is free fall with a = g = 9.81 m/s².
Dynamics
F_res = m·a is a vector equation - draw the free-body diagram, choose axes, resolve. Friction: F_R = μ·F_N. On an incline the downhill component is m·g·sin α, the normal force m·g·cos α.
Conservation
W = F·s·cos θ; E_kin = ½mv², E_pot = mgh, E_spann = ½Ds². Mechanical energy is conserved only without friction; otherwise ΔE_mech = W_friction. Momentum p = mv is conserved when external impulses are negligible; inelastic collisions conserve p only, elastic collisions conserve both p and E_kin.
Circular and gravitational motion
a_z = v²/r = ω²r, so F_z = mv²/r; the resultant points to the centre and is never an extra force. Equating gravitational and centripetal force yields Kepler's third law: T²/r³ = 4π²/(G·M).
Worked example
A ball leaves a 1.80 m table at 8.0 m/s. From h = ½gt², t = √(2·1.80/9.81) = 0.606 s. Range x = 8.0·0.606 = 4.85 m. Vertical speed at impact v_y = g·t = 5.94 m/s, so v = √(8.0² + 5.94²) = 10.0 m/s at 36.6° below the horizontal. Note that the impact speed is not 8.0 + 5.94 - the components add as vectors, not as numbers.
Traps
- Treating centripetal force as an additional force instead of the resultant.
- Mixing ω (rad/s) and f (Hz): ω = 2πf.
- Using energy conservation when friction is present.
- Rounding g to 10 mid-calculation while quoting 9.81.
- Forgetting that momentum is a vector, so a rebound reverses sign.