Identify the type of collision and state the law of conservation of momentum. Calculate final velocities after collision.

ITEM 1: Collisions and Conservation of Momentum

a) Help Patience and Pius identify the type of collision that the pool balls moved with and state the law of conservation of momentum.

• Patience's collision: The first ball collides with a stationary ball, and both balls move separately after the collision. This is an elastic collision (or partially elastic).

• Pius's collision: The two balls collide and move together with the same final velocity. This is a perfectly inelastic collision.

• Law of Conservation of Momentum: For a closed system, the total momentum before a collision is equal to the total momentum after the collision, provided no external forces act on the system. Mathematically, for two objects:

  $$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$$

  where $m$ is mass, $u$ is initial velocity, and $v$ is final velocity.

b) Support Patience and Pius to know the final velocities that the balls moved with after collision.

Patience's Collision:

Step 1: Identify given values and state assumption.
Ball 1: $m_1 = 3 \text{ kg}$, $u_1 = 5 \text{ m/s}$, $v_1 = 2.8 \text{ m/s}$
Ball 2: $m_2 = \text{unknown}$, $u_2 = 0 \text{ m/s}$
Assuming the second pool ball has the same mass as the first, $m_2 = 3 \text{ kg}$.

Step 2: Apply the law of conservation of momentum.

$$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$$ $$(3 \text{ kg})(5 \text{ m/s}) + (3 \text{ kg})(0 \text{ m/s}) = (3 \text{ kg})(2.8 \text{ m/s}) + (3 \text{ kg})v_2$$ $$15 \text{ kg}\cdot\text{m/s} + 0 = 8.4 \text{ kg}\cdot\text{m/s} + 3v_2$$ $$15 = 8.4 + 3v_2$$

Step 3: Solve for $v_2$.

$$3v_2 = 15 - 8.4$$ $$3v_2 = 6.6$$ $$v_2 = \frac{6.6}{3}$$ $$v_2 = 2.2 \text{ m/s}$$

The final velocity of the first ball is given as $\boxed{2.8 \text{ m/s}}$ and the final velocity of the second ball is $\boxed{2.2 \text{ m/s}}$.

Pius's Collision:

Step 1: Identify given values.
Ball 1: $m_1 = 4 \text{ kg}$, $u_1 = 3.5 \text{ m/s}$
Ball 2: $m_2 = 3 \text{ kg}$, $u_2 = 6 \text{ m/s}$
Since they move together, their final velocity is $V_{ms}$.

Step 2: Apply the law of conservation of momentum for a perfectly inelastic collision.

$$m_1 u_1 + m_2 u_2 = (m_1 + m_2) V_{ms}$$ $$(4 \text{ kg})(3.5 \text{ m/s}) + (3 \text{ kg})(6 \text{ m/s}) = (4 \text{ kg} + 3 \text{ kg}) V_{ms}$$ $$14 \text{ kg}\cdot\text{m/s} + 18 \text{ kg}\cdot\text{m/s} = (7 \text{ kg}) V_{ms}$$ $$32 = 7 V_{ms}$$

Step 3: Solve for $V_{ms}$.

$$V_{ms} = \frac{32}{7} \text{ m/s}$$ $$V_{ms} \approx 4.57 \text{ m/s}$$

The final velocity of the two balls moving together is $\boxed{\frac{32}{7} \text{ m/s}}$ or approximately $\boxed{4.57 \text{ m/s}}$.

ITEM 2: Physics Phenomena

a) Why the colour of the dress changed.
The color of the dress changed due to a change in the color of the incident light. Objects appear a certain color because they reflect specific wavelengths of light and absorb others. If the lighting conditions change (e.g., from white light to colored light), the wavelengths available for reflection change, altering the perceived color of the dress.

b) Why the sound was reaching her after 0.05s.
Sound travels at a finite speed through a medium (like air). The time delay of 0.05s is the time it took for the sound waves to travel the distance from the loud speaker to the adjudicator's ears. This is because sound does not travel instantaneously.

ITEM 3: Material Properties and Construction

a) Confirm that iron is stronger than wood.

Step 1: Compare the forces withstood and extensions.
Iron: Withstands $6000 \text{ N}$ for an extension of $2.0 \text{ cm}$.
Wood: Withstands $4000 \text{ N}$ for an extension of $2.5 \text{ cm}$.
Both pieces have the same diameter ($0.7 \text{ mm}$) and length ($10 \text{ m}$), meaning their cross-sectional area ($A$) and original length ($L$) are identical.

Step 2: Calculate Young's Modulus ($E$) for both materials.
Young's Modulus is a measure of stiffness and strength, defined as $E = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L}$.
For iron:

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