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NSW HSC Physics

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Notes for NSW HSC Physics, in a folder for each of the four Year 12 modules (Advanced Mechanics, Electromagnetism, The Nature of Light, From the Universe to the Atom) in the syllabus's order, with one note for each sub-topic or group of sub-topics and the practical investigations in the topics they belong to. They follow the NESA Physics Stage 6 Syllabus (2017). Delete any folder your course leaves out once the notes are yours.

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Có gì bên trong

  • Advanced mechanics
    • Projectile motion5 KB
    • Projectile range, height and launch height5 KB
    • Uniform circular motion5 KB
    • Forces in circular motion5 KB
    • Torque and rotation5 KB
    • Universal gravitation and gravitational fields6 KB
    • Orbits and Kepler's laws7 KB
    • Gravitational potential energy, orbital energy and escape velocity6 KB
  • Electromagnetism
    • Charged particles in electric fields6 KB
    • Charged particles in magnetic fields6 KB
    • The motor effect4 KB
    • Parallel current-carrying wires and the ampere5 KB
    • Magnetic flux and Faraday's law5 KB
    • Lenz's law and induction in practice5 KB
    • Transformers and power transmission6 KB
    • The DC motor6 KB
    • Generators, AC induction motors and magnetic braking7 KB
  • The nature of light
    • Maxwell and electromagnetic waves6 KB
    • Measuring the speed of light5 KB
    • Spectra and spectroscopy7 KB
    • Diffraction and interference of light7 KB
    • Newton, Huygens and polarisation7 KB
    • Black body radiation and Planck5 KB
    • The photoelectric effect7 KB
    • Einstein's postulates, time dilation and length contraction8 KB
    • Relativistic momentum and mass-energy equivalence6 KB
  • From the universe to the atom
    • The Big Bang and the expanding universe6 KB
    • Stellar spectra and the Hertzsprung-Russell diagram6 KB
    • Nucleosynthesis in stars7 KB
    • Evidence for the electron6 KB
    • Evidence for the nuclear atom and the neutron5 KB
    • The Bohr model and the hydrogen spectrum6 KB
    • Matter waves and the Schrödinger model6 KB
    • Radioactive decay and half-life8 KB
    • Fission, fusion, mass defect and binding energy8 KB
    • The Standard Model and particle accelerators8 KB

Ghi chú đầu tiên

Advanced mechanics / Projectile motion

Two independent motions

A projectile is an object moving under the influence of gravity alone after it has been launched. Its path is curved, but the curve is easy to analyse because the motion splits into a horizontal part and a vertical part that do not affect each other. The force of gravity acts straight down, so it changes only the vertical velocity. Nothing pushes sideways, so the horizontal velocity stays as it was.

The analysis rests on two assumptions.

  • The vertical acceleration is constant and equal to the acceleration due to gravity, g=9.8 m s−2g = 9.8\,\mathrm{m\,s^{-2}} downward. This is reasonable when the height gained is small compared with the radius of the Earth.
  • There is no air resistance. In reality drag shortens the range and lowers the maximum height, especially for light objects and high speeds, so the predictions are upper limits for a real projectile.

Choosing the launch point as the origin, with up as positive, a projectile launched with speed uu at an angle θ\theta above the horizontal has initial velocity components

ux=ucos⁡θ,uy=usin⁡θ u_x = u\cos\theta, \qquad u_y = u\sin\theta

uxu_x and uyu_y are the horizontal and vertical components in metres per second, and θ\theta is measured from the horizontal.

Equations for each direction

The horizontal motion is at constant velocity, and the vertical motion is at constant acceleration ay=−ga_y = -g.

HorizontalVertical
vx=uxv_x = u_x (constant)vy=uy−gtv_y = u_y - gt
x=uxtx = u_x ty=uyt−12gt2y = u_y t - \tfrac{1}{2}gt^2
vy2=uy2−2gyv_y^2 = u_y^2 - 2gy
y=12(uy+vy)ty = \tfrac{1}{2}(u_y + v_y)t

These are the equations of uniformly accelerated motion, v=u+atv = u + at, s=ut+12at2s = ut + \tfrac12 at^2, v2=u2+2asv^2 = u^2 + 2as and s=12(u+v)ts = \tfrac12(u+v)t, with the vertical quantities put in. Time is the one quantity that links the two directions: the same tt is used in the horizontal and the vertical equations, so a typical problem finds the time from the vertical motion and then uses it to find how far the projectile travels sideways.

Velocity at any instant

At any moment the velocity is the vector sum of vxv_x and vyv_y, so its magnitude and direction are

v=vx2+vy2,tan⁡ϕ=∣vy∣vx v = \sqrt{v_x^2 + v_y^2}, \qquad \tan\phi = \frac{\lvert v_y\rvert}{v_x}

where ϕ\phi is the angle of the velocity below or above the horizontal. The speed is smallest at the top of the flight, where vy=0v_y = 0 and only uxu_x remains. On the way up the vertical velocity is positive and shrinking; on the way down it is negative and growing in magnitude, while the horizontal velocity never changes.

Features of the path

Because the vertical acceleration is constant and the horizontal velocity is constant, eliminating tt gives yy as a quadratic in xx, so the path is a parabola. For a projectile launched from the origin,

y=xtan⁡θ−gx22u2cos⁡2θ y = x\tan\theta - \frac{g x^2}{2u^2\cos^2\theta}

Other features follow from the symmetry of the parabola when the projectile lands at the height it was launched from.

  • The time to reach maximum height is half the total time of flight.
  • The speed on landing equals the launch speed, and the landing angle below the horizontal equals the launch angle above it.
  • At maximum height the velocity is horizontal and the acceleration is still gg downward. The acceleration is never zero, not even at the top.

The symmetry fails when the landing height differs from the launch height, which is dealt with for launches from a height in the work on range and maximum height.

Worked example

A ball is kicked from level ground at 25 m s−125\,\mathrm{m\,s^{-1}} at 40∘40^\circ above the horizontal. Find the time to reach the highest point, the maximum height, the time of flight and the horizontal distance travelled before it lands.

The components of the launch velocity are

ux=25cos⁡40∘=19.2 m s−1,uy=25sin⁡40∘=16.1 m s−1 u_x = 25\cos 40^\circ = 19.2\,\mathrm{m\,s^{-1}}, \qquad u_y = 25\sin 40^\circ = 16.1\,\mathrm{m\,s^{-1}}

At the highest point vy=0v_y = 0, so 0=uy−gt0 = u_y - gt and

tup=16.079.8=1.64 s t_{\text{up}} = \frac{16.07}{9.8} = 1.64\,\mathrm{s}

The height gained comes from vy2=uy2−2gyv_y^2 = u_y^2 - 2gy with vy=0v_y = 0:

ymax⁡=uy22g=16.07219.6=13.2 m y_{\max} = \frac{u_y^2}{2g} = \frac{16.07^2}{19.6} = 13.2\,\mathrm{m}

The ball lands at the launch height, so the time of flight is twice the time up, 3.28 s3.28\,\mathrm{s}. The horizontal distance in that time is

x=uxt=19.15×3.28=62.8 m x = u_x t = 19.15 \times 3.28 = 62.8\,\mathrm{m}

Comparing projectiles

Two objects released from the same height, one dropped and one thrown horizontally, reach the ground together, because the sideways velocity has no effect on the vertical motion. A horizontal launch is simply the case θ=0\theta = 0, where uy=0u_y = 0 and the vertical motion is a fall from rest.

The same approach works for any situation where the net force on the object is constant, which is why the motion of a charged particle in a uniform electric field is treated in the same way, with the electric force taking the place of weight.

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