Calculate the gravitational force and initial accelerations of two particles with given masses and separation.

Mass of particle 1: $m_1 = 1 \text{ kg}$.
Mass of particle 2: $m_2 = 2 \text{ kg}$.
Separation distance: $r = 50 \text{ cm} = 0.50 \text{ m}$.
Gravitational constant: $G = 6.674 \times 10^{-11} \text{ N m}^2/\text{kg}^2$.

Using Newton's Law of Universal Gravitation:

Using Newton's Second Law ($F = ma$):
For particle 1 ($m_1$):

For particle 2 ($m_2$):

The gravitational force is $\boxed{5.34 \times 10^{-10} \text{ N}}$.
The initial acceleration of the 1 kg particle is $\boxed{5.34 \times 10^{-10} \text{ m/s}^2}$.
The initial acceleration of the 2 kg particle is $\boxed{2.67 \times 10^{-10} \text{ m/s}^2}$.

Question 12. b. (ii): Escape velocity from the moon

Mass of the moon: $M = 7.4 \times 10^{22} \text{ kg}$.
Radius of the moon: $R = 1740 \text{ km} = 1740 \times 10^3 \text{ m} = 1.74 \times 10^6 \text{ m}$.
Gravitational constant: $G = 6.674 \times 10^{-11} \text{ N m}^2/\text{kg}^2$.

The formula for escape velocity is:

Rounding to two significant figures:

The escape velocity from the moon is:

Question 13. a. (i): Viscosity and coefficient of viscosity

• Viscosity: A measure of a fluid's resistance to flow. It describes the internal friction within a fluid when it is in motion. A highly viscous fluid flows slowly, while a low-viscosity fluid flows easily.

• Coefficient of viscosity ($\eta$): A quantitative measure of a fluid's viscosity. It is defined as the ratio of the shear stress to the rate of shear strain in a fluid.

  • SI unit: Pascal-second ($\text{Pa} \cdot \text{s}$) or Newton-second per square meter ($\text{N s/m}^2$).
  • Dimensions: $[ML^{-1}T^{-1}]$.

Question 13. a. (ii): Stokes's Law and Terminal Velocity derivation

Stokes's Law describes the viscous drag force ($F_d$) on a small spherical object moving through a viscous fluid at low Reynolds numbers:

Where:

• $\eta$ (eta) is the coefficient of viscosity of the fluid.
• $a$ is the radius of the spherical object.
• $v$ is the velocity of the spherical object relative to the fluid.

When a metallic sphere falls under gravity in a liquid, three forces act on it:

1. Gravitational force ($F_g$) acting downwards.
2. Buoyant force ($F_b$) acting upwards (due to Archimedes' principle).
3. Viscous drag force ($F_d$) acting upwards (due to Stokes's Law).

At terminal velocity, the net force on the sphere is zero, meaning the upward forces balance the downward force:

Substitute the expressions for each force:

• Gravitational force: $F_g = mg = \rho V g$, where $V$ is the volume of the sphere. For a sphere, $V = \frac{4}{3}\pi a^3$. $$F_g = \rho \left(\frac{4}{3}\pi a^3\right) g$$
• Buoyant force: $F_b = \sigma V g$, where $\sigma$ is the density of the liquid. $$F_b = \sigma \left(\frac{4}{3}\pi a^3\right) g$$
• Viscous drag force (at terminal velocity $v$): $$F_d = 6\pi \eta a v$$

Substitute these into the force balance equation:

Rearrange to solve for $v$:

Divide both sides by $6\pi \eta a$:

Simplify the expression:

Cancel $\pi$ and one $a$:

Reduce the fraction $\frac{4}{18}$ to $\frac{2}{9}$:

Question 13. b. (i): Conditions for liquid to be ideal

Two conditions for a liquid to be considered ideal are:

1. It is incompressible: Its density remains constant regardless of pressure changes.
2. It is non-viscous: It has no internal friction, meaning its coefficient of viscosity is zero.

Question 13. b. (ii): Kinetic and potential energies per unit volume

The density of the liquid is $\rho = \frac{m}{V}$.

The kinetic energy ($KE$) of a liquid of mass $m$ and velocity $v$ is:

To find the kinetic energy per unit volume, divide by the volume $V$:

Substitute $m = \rho V$:

The potential energy ($PE$) of a liquid of mass $m$ at a height $h$ is:

To find the potential energy per unit volume, divide by the volume $V$:

Substitute $m = \rho V$:

Question 14. a. (i): State the zeroth and first laws of thermodynamics

• Zeroth Law of Thermodynamics: If two thermodynamic systems are each in thermal equilibrium with a third system, then they are in thermal equilibrium with each other. This law establishes the concept of temperature.

• First Law of Thermodynamics: The change in the internal energy ($\Delta U$) of a system is equal to the heat added to the system ($Q$) minus the work done by the system ($W$). It is a statement of the conservation of energy.

  $$\Delta U = Q - W$$