Schleswig-Holstein Abitur Physics study materials
Schleswig-Holstein Abitur Physics study resources

Schleswig-Holstein Abitur Physics: Mechanics: kinematics, Newtonian dynamics, circular motion, gravitation, conservation laws

Schleswig-Holstein Abitur Physics · Work through your exam, one topic at a time.

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Mechanics: kinematics, Newtonian dynamics, circular motion, gravitation, conservation laws

What this domain tests

Mechanics in the Schleswig-Holstein Abitur covers kinematics in one and two dimensions, Newtonian dynamics, circular motion, gravitation and the conservation laws for energy and momentum. Tasks usually embed the physics in a context - a braking car, a loop track, a satellite - and demand a calculation, a justification of the model and a short evaluation of the result.

Kinematics

  • Average and instantaneous velocity; in the t-x, t-v and t-a diagrams the slope of one curve is the value of the next, and the area under the curve is the change of the previous quantity.
  • Uniform acceleration: v = v₀ + at, s = v₀t + ½at², v² = v₀² + 2as.
  • Two-dimensional motion is a superposition: a horizontal throw is uniform motion in x combined with free fall in y.

Dynamics

  • Newton's three laws; the working equation is always F_res = ma, written after a free-body diagram.
  • Friction F_R = μ·F_N, with static friction generally larger than sliding friction.
  • Circular motion needs a real centripetal force F_Z = mv²/r, a_Z = v²/r; in a loop this is supplied by the normal force plus gravity.

Gravitation

  • F = G·m₁m₂/r², g = GM/r², G = 6.67·10⁻¹¹ N m²/kg².
  • Kepler III: T²/r³ = 4π²/(GM); for circular orbits E_pot = −GMm/r and E_kin = GMm/(2r).

Conservation laws

  • Energy is conserved; with friction ΔE = F_R·s appears as internal energy.
  • Momentum p = mv is conserved in every isolated system; elastic collisions also conserve kinetic energy, inelastic ones do not.

Worked example

A marble runs through a vertical loop of radius r = 0.50 m. At the top the normal force may just vanish, so gravity alone provides the centripetal force: mg = mv²/r gives v = √(gr) = √(9.81·0.50) ≈ 2.2 m/s. Energy conservation from the release height h to the top: mgh = mg·2r + ½mv², hence h = 2r + v²/(2g) = 1.00 m + 0.25 m = 1.25 m. State both the force condition and the energy balance - the marks are for the reasoning, not only the number.

Traps that cost marks

  • Treating the centripetal force as an extra force instead of the resultant of real forces.
  • Changing sign conventions in the middle of a calculation.
  • Claiming momentum conservation in a system with relevant external forces (a wall, a fixed ramp).
  • Using v = √(2gh) when friction is present.