[v] = L T^-1, [u] = L T^-1, and [at] = (L T^-2)(T) = L T^-1. Every term has the same dimension L T^-1, so the equation is dimensionally consistent (homogeneous).

[a] = L T^-2, [t] = T

Three significant figures (4, 2 and the trailing 0). Leading zeros are not significant; a trailing zero after the decimal point is significant.

From F = G M m / r^2, we get G = F r^2 / (M m). Dimensions: [G] = ([F][r^2]) / ([M][m]) = (M L T^-2 * L^2) / (M * M) = M^-1 L^3 T^-2.

G = F r^2 / (M m)

At maximum height the velocity is zero. (a) h = u^2 / (2g) = (20)^2 / (2*10) = 400/20 = 20 m. (b) Time of flight T = 2u/g = (2*20)/10 = 4 s.

h = u^2/2g ; T = 2u/g

Resolve u into ux = u cos(theta) and uy = u sin(theta). Vertical motion has acceleration -g. Time of flight: uy - g(T/2) = 0 at the top, so T = 2u sin(theta)/g. Maximum height: H = uy^2/2g = u^2 sin^2(theta)/2g. Horizontal range: R = ux * T = u cos(theta) * 2u sin(theta)/g = u^2 sin(2 theta)/g. R is maximum when theta = 45 degrees.

T = 2u sin(theta)/g ; H = u^2 sin^2(theta)/2g ; R = u^2 sin(2theta)/g

Impulse is the product of a force and the time for which it acts: J = F * (delta t). By the impulse-momentum theorem it equals the change in momentum: J = delta p = m v - m u. SI unit: N s (= kg m/s).

J = F * delta t = delta p

Force of friction f = mu * m g = 0.2 * 2 * 10 = 4 N. Net force = applied - friction = 10 - 4 = 6 N. Acceleration a = net force / mass = 6 / 2 = 3 m/s^2.

f = mu m g ; a = (F - f)/m

The work-energy theorem states that the work done by the net force on a body equals the change in its kinetic energy: W_net = (1/2) m v^2 - (1/2) m u^2. If net work is positive the body speeds up; if negative, it slows down.

W_net = (1/2)m v^2 - (1/2)m u^2

By conservation of energy, KE at the ground = loss in potential energy = m g h = 5 * 10 * 20 = 1000 J. (Check: v = sqrt(2gh) = 20 m/s, KE = (1/2)(5)(20^2) = 1000 J.)

KE = m g h

Moment of inertia is the rotational analogue of mass: I = sum of (m_i r_i^2). It measures a body's opposition to a change in its rotational motion. SI unit: kg m^2. It depends on (i) the mass and its distribution and (ii) the position/orientation of the axis of rotation.

I = sum(m_i r_i^2)

Parallel axes theorem: I = I_cm + M d^2, where I_cm is the moment of inertia about a parallel axis through the centre of mass and d is the distance between the axes. For a rod, I_cm = M L^2 / 12 and d = L/2, so I_end = M L^2/12 + M (L/2)^2 = M L^2/12 + M L^2/4 = M L^2/3.

I = I_cm + M d^2

Since g = G M / (R + h)^2, increasing the height h increases the distance from the centre of the Earth, so g decreases. For small heights, g' = g (1 - 2h/R) approximately.

g' = g (1 - 2h/R)

Escape velocity v_e = sqrt(2 g R) = sqrt(2 * 9.8 * 6.4 x 10^6) = sqrt(1.2544 x 10^8) = 1.12 x 10^4 m/s = 11.2 km/s.

v_e = sqrt(2 g R)

Young's modulus is the ratio of longitudinal (tensile) stress to longitudinal strain, within the elastic limit: Y = (F/A) / (delta L / L). SI unit: N/m^2 (pascal, Pa).

Y = (F L) / (A * delta L)

For the streamline (steady, non-viscous, incompressible) flow of a fluid, the sum of the pressure energy, kinetic energy and potential energy per unit volume is constant: P + (1/2) rho v^2 + rho g h = constant. Thus where the speed is high, the pressure is low.

P + (1/2) rho v^2 + rho g h = constant

The first law is the law of conservation of energy for a thermodynamic system: delta Q = delta U + delta W. Here delta Q is the heat supplied to the system (positive if absorbed), delta U is the increase in internal energy, and delta W is the work done by the system (positive if the gas expands).

delta Q = delta U + delta W

Isothermal: temperature stays constant, so delta U = 0; obeys P V = constant; occurs slowly with good thermal contact; heat can flow in or out. Adiabatic: no heat is exchanged (Q = 0); obeys P V^gamma = constant; occurs quickly or with perfect insulation; any work done changes the internal energy (delta U = -delta W).

Isothermal: PV = const ; Adiabatic: P V^gamma = const

A Carnot cycle has four reversible steps: (1) isothermal expansion at the source temperature T_h (heat Q_h absorbed); (2) adiabatic expansion (temperature falls from T_h to T_c); (3) isothermal compression at the sink temperature T_c (heat Q_c rejected); (4) adiabatic compression (temperature rises from T_c back to T_h). The efficiency is eta = 1 - Q_c/Q_h = 1 - T_c/T_h, where temperatures are in kelvin. Efficiency depends only on the two temperatures and is always less than 1.

eta = 1 - T_c/T_h

v_rms = sqrt(3 R T / M) = sqrt(3 * 8.31 * 300 / 0.032) = sqrt(7479 / 0.032) = sqrt(233719) = 483 m/s (approximately).

v_rms = sqrt(3 R T / M)

For a bob displaced by a small angle, the restoring force is F = -m g sin(theta). For small theta, sin(theta) ~ theta = x/L, so F = -(m g / L) x. This is of the form F = -k x with k = m g / L, hence the motion is SHM. Then omega^2 = k/m = g/L, so the time period T = 2 pi / omega = 2 pi sqrt(L/g).

T = 2 pi sqrt(L / g)