(a)(i) State Hooke's law. (ii) A spring has a length of 0.20 m when a mass of 0.30 kg hangs on it, and a length of 0.75 nm when a mass of 1.95 kg hangs on it. Calculate the: (i) force constant of the spring; (ii) length of the spring when it is unloaded. [g = 10m/s\(^2\)]
(b)(i) What is diffusion? (ii) State two factors that affect the rate of diffusion of a substance. (iii) State the exact relationship between the rate of diffusion of a gas and its density.
(c) A satellite of mass, m orbits the earth of mass. M with a velocity, v at a distance R from the centre of the earth. Derive the relationship between the period T, of orbit and R.
(a)(i) Hooke's law. Provided the elastic limit is not exceeded, the extension of an elastic material is directly proportional to the force (load) producing it. \(F = k e\).
(a)(ii) Spring calculation. (The second length is 0.75 m.)
Force when 0.30 kg hangs: \(F_1 = 0.30\times10 = 3\ \text{N}\), length \(L_1 = 0.20\ \text{m}\).
Force when 1.95 kg hangs: \(F_2 = 1.95\times10 = 19.5\ \text{N}\), length \(L_2 = 0.75\ \text{m}\).
Force constant:
\[ k = \frac{F_2-F_1}{L_2-L_1} = \frac{19.5-3}{0.75-0.20} = \frac{16.5}{0.55} = 30\ \text{N/m} \]
Unloaded (natural) length \(L_0\): using \(F_1 = k(L_1-L_0)\),
\[ 3 = 30(0.20 - L_0) \;\Rightarrow\; 0.20 - L_0 = 0.1 \;\Rightarrow\; L_0 = 0.10\ \text{m} \]
Force constant \(= 30\ \text{N/m}\); natural length \(= 0.10\ \text{m}\).
(b)(i) Diffusion. Diffusion is the net movement of particles (molecules or ions) of a substance from a region of higher concentration to a region of lower concentration until they are evenly spread.
(b)(ii) Two factors affecting rate of diffusion. Temperature (higher temperature gives faster diffusion); the density or molar mass of the substance (lighter/less dense substances diffuse faster). (Also the concentration gradient.)
(b)(iii) Relationship with density. The rate of diffusion of a gas is inversely proportional to the square root of its density (Graham's law):
\[ \text{rate} \propto \frac{1}{\sqrt{\rho}} \]
(c) Period-radius relationship for a satellite. The gravitational pull provides the centripetal force:
\[ \frac{GMm}{R^{2}} = \frac{mv^{2}}{R} \;\Rightarrow\; v^{2} = \frac{GM}{R} \]
The satellite covers the circumference \(2\pi R\) in one period, so \(v = \dfrac{2\pi R}{T}\). Substituting:
\[ \left(\frac{2\pi R}{T}\right)^{2} = \frac{GM}{R} \;\Rightarrow\; \frac{4\pi^{2}R^{2}}{T^{2}} = \frac{GM}{R} \]
\[ \boxed{\,T^{2} = \frac{4\pi^{2}}{GM}\,R^{3}\,} \]
Hence \(T^{2} \propto R^{3}\) (Kepler's third law).
