2024 Cambridge IGCSE Science - Combined (0653) 模擬試題連答案詳解

(a) Acceleration is defined as the change in velocity per unit time. (b) Acceleration \(a = \frac{v - u}{t} = \frac{6.0\text{ m/s} - 0}{4.0\text{ s}} = 1.5\text{ m/s}^2\). (c) Force \(F = m \times a = 0.50\text{ kg} \times 1.5\text{ m/s}^2 = 0.75\text{ N}\). (d) Distance \(d = \text{average speed} \times t = \frac{0 + 6.0}{2} \times 4.0 = 3.0 \times 4.0 = 12\text{ m}\). (e) If resistive forces are present, the net force must still be 0.75 N. Therefore, the applied force must be greater than 0.75 N to overcome resistance.

(a) 1 mark: change in velocity per unit time / rate of change of velocity. (b) 2 marks: 1 mark for correct calculation (1.5), 1 mark for correct unit (m/s^2). (c) 2 marks: 1 mark for correct formula (F=ma) or substitution, 1 mark for correct answer (0.75 N). (d) 2 marks: 1 mark for correct formula or substitution, 1 mark for correct answer (12 m). (e) 1.89 marks: 1.89 marks for stating a larger force is needed to overcome resistive forces (allow 1 mark for saying larger force, 0.89 mark for mentioning overcoming resistance/friction/air resistance).

(a) A voltmeter must be connected in parallel (across) the component being measured. (b) Potential difference \(V = I \times R = 1.2\text{ A} \times 8.0\ \Omega = 9.6\text{ V}\). (c) In a series circuit, total voltage is shared: \(V_{\text{lamp}} = V_{\text{total}} - V_{\text{resistor}} = 12.0\text{ V} - 9.6\text{ V} = 2.4\text{ V}\). (d) Resistance of the lamp \(R = \frac{V}{I} = \frac{2.4\text{ V}}{1.2\text{ A}} = 2.0\ \Omega\).

(a) 2 marks: 1 mark for 'in parallel', 1 mark for 'across the resistor'. (b) 2 marks: 1 mark for correct calculation/formula (V = I * R), 1 mark for 9.6 V. (c) 2 marks: 1 mark for using series circuit rule (12 - V_resistor), 1 mark for 2.4 V. (d) 2.89 marks: 1 mark for correct formula (R = V / I), 1 mark for correct substitution (2.4 / 1.2), 0.89 mark for final answer 2.0 ohms (allow 2 ohms).

(a) According to kinetic particle theory, gas particles are in constant, random motion. They collide with each other and the walls of the tube, spreading out from an area of high concentration to low concentration. (b) The white smoke forms closer to the hydrochloric acid end. This is because ammonia gas molecules have a smaller relative molecular mass (17) compared to hydrogen chloride molecules (36.5). Lighter molecules move and diffuse faster. (c) At a higher temperature, particles have more kinetic energy. This causes them to move faster, increasing the rate of diffusion and decreasing the time taken for the gases to meet and form the smoke.

(a) 2 marks: 1 mark for describing random motion/collisions of particles, 1 mark for movement from high to low concentration. (b) 3 marks: 1 mark for stating 'closer to the HCl end', 1 mark for comparing molecular masses (ammonia is lighter/lower Mr), 1 mark for relating lower mass to faster diffusion speed. (c) 3.89 marks: 1 mark for stating rate of diffusion increases, 1 mark for stating time taken decreases, 1 mark for mentioning particles gain kinetic energy / move faster, 0.89 mark for linking increased temperature directly to average kinetic energy.

(a) An enzyme is a biological catalyst that increases the rate of chemical reactions and is made of protein. (b) At high temperatures above 60 degrees C, the rate of reaction drops to zero. This is because the high temperature breaks bonds holding the enzyme's three-dimensional structure together, denaturing it. The active site changes shape, meaning starch molecules (substrate) can no longer fit into the active site. (c) A control experiment can be set up by repeating the experiment using boiled (and thus completely denatured) amylase, or by replacing the amylase solution with an equal volume of distilled water. This shows that the reaction does not occur without active enzyme.

(a) 2 marks: 1 mark for biological catalyst / speeds up reactions, 1 mark for being made of protein. (b) 3.89 marks: 1 mark for stating rate decreases/stops, 1 mark for mentioning 'denatures', 1 mark for explaining active site changes shape, 0.89 mark for explaining substrate/starch can no longer fit. (c) 3 marks: 1 mark for suggestion of boiled/denatured enzyme or water, 1 mark for keeping all other variables constant, 1 mark for stating this shows the reaction doesn't happen without active enzyme.

(a) Copper(II) oxide is added in excess to ensure that all of the dilute sulfuric acid is completely neutralized, so no unreacted acid remains in the solution. (b) The excess, unreacted solid copper(II) oxide is removed by filtration. The mixture is poured through filter paper in a filter funnel; the copper(II) oxide remains as residue while copper(II) sulfate solution passes through as filtrate. (c) The balanced equation is \(\text{CuO(s)} + \text{H}_2\text{SO}_4\text{(aq)} \rightarrow \text{CuSO}_4\text{(aq)} + \text{H}_2\text{O(l)}\). (d) To obtain dry crystals: 1. Heat the copper(II) sulfate solution to evaporate water until a saturated solution is formed. 2. Allow the solution to cool so crystals form. 3. Filter the crystals to remove excess liquid, and dry them between sheets of filter paper.

(a) 2 marks: 1 mark for ensuring all acid reacts, 1 mark for ensuring the final salt solution is not contaminated with acid. (b) 2 marks: 1 mark for mentioning filtration / filtering, 1 mark for identifying the copper(II) oxide as residue or copper(II) sulfate as filtrate. (c) 2.89 marks: 1 mark for correct reactants and products (CuO + H2SO4 -> CuSO4 + H2O), 1 mark for correct state symbols (s, aq, aq, l), 0.89 mark for balanced equation. (d) 2 marks: 1 mark for heating to crystallization point/evaporating water, 1 mark for cooling and drying crystals with filter paper.

(a) Arteries have a thick wall of muscle and elastic fibers. The thick wall prevents the artery from bursting under high pressure, and the elastic fibers allow the wall to stretch and recoil to maintain blood pressure. (b) The left ventricle pumps blood to the systemic circulation (the entire body), which requires high pressure to overcome high resistance over a long distance. The right ventricle only pumps blood to the lungs (pulmonary circulation), which are nearby and require lower pressure to avoid damage. (c) Valves prevent the backflow of blood. They open to let blood pass through in one direction and close when blood tries to flow backwards, maintaining one-way flow.

(a) 3 marks: 1 mark for thick outer wall, 1 mark for elastic fibers (to stretch and recoil), 1 mark for muscle layer (to withstand/maintain pressure). (b) 3 marks: 1 mark for left ventricle pumping blood to the whole body, 1 mark for right ventricle pumping blood to the lungs only, 1 mark for linking distance/resistance to the need for higher pressure/force. (c) 2.89 marks: 1 mark for stating prevent backflow of blood, 1 mark for ensuring one-way/unidirectional flow, 0.89 mark for explaining that they close when pressure behind them decreases.

(a) To test for starch, add a few drops of iodine solution to the food sample. If starch is present, the color changes from orange-brown to blue-black. (b) Calcium is needed for the development and maintenance of strong bones and teeth (also muscle contraction/blood clotting). A deficiency can lead to rickets (in children) or osteoporosis/brittle bones (in adults). (c) Mechanical digestion is the physical breakdown of food into smaller pieces without chemical change (e.g., chewing by teeth, churning in stomach). Chemical digestion is the chemical breakdown of large, insoluble food molecules into small, soluble molecules using enzymes (e.g., amylase breaking starch into maltose).

(a) 2 marks: 1 mark for using iodine solution, 1 mark for color change to blue-black. (b) 2.89 marks: 1 mark for role (bones/teeth/muscle contraction), 1 mark for deficiency symptom (rickets/osteoporosis/weak teeth), 0.89 mark for linking calcium directly to bone mineralization. (c) 4 marks: 1 mark for defining mechanical digestion (physical breakdown/no chemical change), 1 mark for defining chemical digestion (using enzymes/chemical change), 1 mark for mechanical example (chewing/churning), 1 mark for chemical example (enzyme action, e.g., amylase/protease/lipase).

(a) Differences include: 1. Sound waves are longitudinal while light waves are transverse. 2. Sound waves require a medium to travel, whereas light waves can travel through a vacuum. (b) Using the wave equation: \(v = f \lambda\), where \(v = 330\text{ m/s}\) and \(f = 440\text{ Hz}\). Rearranging gives \(\lambda = \frac{v}{f} = \frac{330}{440} = 0.75\text{ m}\). (c) Sound is a mechanical wave that propagates through the vibration of particles. In gases (like air), particles are very far apart, so collisions are infrequent and the wave travels slowly. In liquids (like water), particles are much closer together, allowing the vibrational energy to be transferred more quickly from particle to particle.

(a) 2 marks: 1 mark for each valid difference (longitudinal vs transverse, requires medium vs travels in vacuum, speed difference). (b) 3 marks: 1 mark for correct formula rearranged (wavelength = speed / frequency), 1 mark for substitution (330 / 440), 1 mark for correct final value (0.75) and unit (m). (c) 3.89 marks: 1 mark for stating sound propagates via particle vibrations, 1 mark for describing particles in water as closely packed/closer than in air, 1 mark for describing particles in air as far apart, 0.89 mark for explaining closer packing allows faster transfer of energy/vibrations.

**(a) Calculate the charge:**
1. Convert time from minutes to seconds:
\t\(t = 5.0\text{ min} \times 60\text{ s/min} = 300\text{ s}\)
2. Use the formula for charge:
\t\(Q = I \times t\)
\t\(Q = 0.40\text{ A} \times 300\text{ s} = 120\text{ C}\)

**(b) Calculate the resistance of the lamp:**
1. Use Ohm's law:
\t\(R_{\text{lamp}} = \frac{V_{\text{lamp}}}{I}\)
\t\(R_{\text{lamp}} = \frac{4.8\text{ V}}{0.40\text{ A}} = 12\ \Omega\)

**(c) (i) Calculate power dissipated by the lamp:**
1. Use the power formula:
\t\(P = I \times V\)
\t\(P = 0.40\text{ A} \times 4.8\text{ V} = 1.92\text{ W}\)

**(c) (ii) Determine the resistance \(R\) of the fixed resistor:**
*Method 1 (using potential differences):*
1. The total voltage in a series circuit is the sum of the potential differences:
\t\(V_{\text{total}} = V_{\text{lamp}} + V_R\)
\t\(12.0\text{ V} = 4.8\text{ V} + V_R \implies V_R = 7.2\text{ V}\)
2. Calculate the resistance \(R\):
\t\(R = \frac{V_R}{I} = \frac{7.2\text{ V}}{0.40\text{ A}} = 18\ \Omega\)

*Method 2 (using total resistance):*
1. Calculate the total resistance of the circuit:
\t\(R_{\text{total}} = \frac{V_{\text{total}}}{I} = \frac{12.0\text{ V}}{0.40\text{ A}} = 30\ \Omega\)
2. Calculate \(R\):
\t\(R = R_{\text{total}} - R_{\text{lamp}} = 30\ \Omega - 12\ \Omega = 18\ \Omega\)

**(a) [Total: 3 marks]**
* \(t = 300\text{ s}\) (conversion of time) [1]
* \(Q = 0.40 \times 300 = 120\) (correct numerical calculation) [1]
* \(\text{C}\) or \(\text{Coulombs}\) (correct unit) [1]

**(b) [Total: 2 marks]**
* \(R = \frac{V}{I}\) or \(\frac{4.8}{0.40}\) (correct substitution of formula) [1]
* \(12\ \Omega\) (correct evaluation with unit; allow ohm) [1]

**(c) (i) [Total: 1.89 marks]**
* \(P = 0.40 \times 4.8 = 1.92\text{ W}\) (correct calculation of power with unit; accept 1.9 W) [1.89]

**(c) (ii) [Total: 2 marks]**
* \(V_R = 12.0 - 4.8 = 7.2\text{ V}\) OR \(R_{\text{total}} = \frac{12.0}{0.40} = 30\ \Omega\) (correct intermediate step showing potential difference across resistor or total circuit resistance) [1]
* \(18\ \Omega\) (correct resistance with unit) [1]