2024 Cambridge IGCSE Science - Combined (0653) Practice Paper with Answers

For (a), the loss in gravitational potential energy is calculated using \(GPE = m \times g \times h\). Substituting the values: \(GPE = 0.50 \times 9.8 \times 0.80 = 3.92\text{ J}\). For (b), by the principle of conservation of energy, the kinetic energy gained equals the GPE lost. Hence, \(\frac{1}{2} m v^2 = GPE\), which gives \(v = \sqrt{\frac{2 \times GPE}{m}} = \sqrt{\frac{2 \times 3.92}{0.50}} = \sqrt{15.68} \approx 3.96\text{ m/s}\) (rounded to \(4.0\text{ m/s}\)). For (c), in a real situation, there is friction between the car and the track, as well as air resistance. Work must be done against these resistive forces, which transfers some of the mechanical energy into thermal energy (heat) that is lost to the surroundings.

(a) 1 mark for the formula \(GPE = mgh\) or correct substitution; 1 mark for the correct calculation of \(3.92\text{ J}\) (accept \(3.9\text{ J}\)). (b) 1 mark for equating KE to GPE or using the formula \(v = \sqrt{2gh}\); 1 mark for calculating \(3.96\text{ m/s}\) or \(4.0\text{ m/s}\) with correct unit. (c) 1 mark for identifying friction or air resistance acting on the car; 1 mark for stating that work is done against these forces; 1 mark for stating that the lost energy is converted into thermal energy; 1 mark for stating that it is dissipated to the surroundings.

For (a), sperm is produced in the testis, travels through the sperm duct (vas deferens), and leaves the male body via the urethra. During intercourse, it is deposited into the vagina, travels through the cervix, across the uterus, and enters the oviduct where fertilization occurs. For (b), the prostate gland secretes an alkaline fluid that makes up part of semen, which nourishes the sperm and activates its movement. For (c), the sperm (male gamete) is very small and has a tail (flagellum) that allows it to swim (motile). The egg (female gamete) is relatively large as it contains food stores (yolk) and is passive (non-motile).

(a) 1 mark for the male pathway (testis to sperm duct to urethra); 1 mark for entry into the female system (vagina to cervix to uterus); 1 mark for identifying the oviduct (fallopian tube) as the site of fertilization. (b) 1 mark for stating it secretes fluid/semen; 1 mark for stating that this fluid nourishes or activates sperm (assists motility). (c) 1 mark for stating that sperm is much smaller than the egg; 1 mark for stating that sperm is motile while the egg is non-motile; 1 mark for referencing structural reasons (e.g., sperm has a flagellum, egg contains cytoplasm food reserves).

For (a), the standard test for unsaturation is adding bromine water. Ethene, being an alkene (unsaturated), undergoes an addition reaction that turns the orange-brown solution colourless. Ethane is an alkane (saturated) and does not react under normal conditions, so the bromine water remains orange-brown. For (b), ethene undergoes addition polymerisation. During this reaction, the double covalent bond between the carbon atoms (C=C) in each ethene monomer breaks, allowing them to form single covalent bonds (C-C) to link up with adjacent monomers in a long polymer chain. For (c), ethanol has the molecular formula \(\text{C}_2\text{H}_5\text{OH}\). Its structure features two carbon atoms bonded together, with five single C-H bonds, one single C-O bond, and one single O-H bond.

(a) 1 mark for naming bromine water (or aqueous bromine) as the reagent; 1 mark for stating that ethene decolourises it (or turns from orange/brown to colourless); 1 mark for stating that ethane shows no reaction (or remains orange/brown). (b) 1 mark for naming addition polymerisation; 1 mark for stating that the carbon-carbon double bonds (C=C) open up or break; 1 mark for stating that new carbon-carbon single bonds (C-C) are formed. (c) 1 mark for showing a chain of 2 carbon atoms with a single bond; 1 mark for showing the -O-H group with the oxygen-hydrogen bond clearly shown; 1 mark for showing all remaining valencies occupied by single bonds to hydrogen (5 C-H bonds in total).

For (a), the left ventricle is responsible for pumping oxygenated blood through the systemic circulation to the entire body, which offers high resistance and requires high pressure. The right ventricle only pumps deoxygenated blood to the lungs (pulmonary circulation), which is close by and requires much lower pressure to avoid damaging delicate lung tissues. For (b), valves ensure that blood flows in only one direction. When the ventricles contract, the atrioventricular (AV) valves are pushed shut by the high pressure, preventing blood from flowing back into the atria. Similarly, semilunar valves close when the ventricles relax, preventing blood from returning to the ventricles. For (c), the pulmonary vein brings oxygenated blood from the lungs back to the left atrium of the heart, and the pulmonary artery carries deoxygenated blood from the right ventricle to the lungs.

(a) 1 mark for identifying that the left ventricle pumps blood to the whole body; 1 mark for identifying that the right ventricle pumps blood only to the lungs; 1 mark for explaining that a thicker muscle wall is required to generate the higher pressure needed for systemic circulation. (b) 1 mark for stating that the main function is to ensure one-way flow of blood; 1 mark for explaining that valves close in response to pressure differences; 1 mark for referencing specific valves (e.g., AV valves or semilunar valves) and how they close to protect the chambers. (c) 1 mark for naming the pulmonary vein; 1 mark for naming the pulmonary artery.

For (a), because the resistors are connected in series, the total resistance is the sum of the individual resistances: \(R_{\text{total}} = 4.0\ \Omega + 8.0\ \Omega = 12.0\ \Omega\). Using Ohm's law, the current in the circuit is \(I = \frac{V}{R_{\text{total}}} = \frac{12\text{ V}}{12.0\ \Omega} = 1.0\text{ A}\). For (b), the potential difference across the fixed resistor is given by \(V = I \times R = 1.0\text{ A} \times 4.0\ \Omega = 4.0\text{ V}\). For (c)(i), increasing the resistance of the variable resistor increases the total resistance of the series circuit. With a constant EMF of \(12\text{ V}\), this causes the overall current to decrease. For (c)(ii), because the current through the circuit decreases, the potential difference across the fixed \(4.0\ \Omega\) resistor must also decrease, since \(V = I \times R\) and \(R\) is constant.

(a) 1 mark for finding the total resistance of \(12.0\ \Omega\); 1 mark for using the formula \(I = \frac{V}{R}\); 1 mark for the correct answer of \(1.0\text{ A}\) with unit. (b) 1 mark for using the formula \(V = I \times R\); 1 mark for the correct answer of \(4.0\text{ V}\) with unit. (c) 1 mark for stating that the current decreases; 1 mark for explaining that the total resistance of the circuit increases; 1 mark for stating that the potential difference across the fixed resistor decreases because \(V = I \times R\) and current has decreased.

For (a), chemical reactions involve breaking existing bonds in the reactants and forming new bonds in the products. Bond breaking is an endothermic process (requires energy input), while bond forming is an exothermic process (releases energy). In this reaction, the total energy released when forming the C=O and O-H bonds is greater than the total energy absorbed to break the C-H and O=O bonds, leading to a net release of thermal energy to the surroundings. For (b), the reaction pathway diagram plots Energy (or Enthalpy) on the vertical y-axis and the Progress of Reaction on the horizontal x-axis. Since the reaction is exothermic, the reactants have a higher energy level than the products, and are drawn as a higher horizontal line. A curve starts from the reactants, rises to a peak (representing the transition state), and then drops down to the products line. The activation energy (\(E_a\)) is represented by an arrow pointing upwards from the reactants line to the peak. The overall enthalpy change (\(\Delta H\)) is represented by a downward arrow pointing from the reactants line to the products line.

(a) 1 mark for stating that bond breaking is endothermic (requires energy); 1 mark for stating that bond forming is exothermic (releases energy); 1 mark for concluding that the energy released during bond forming is greater than the energy absorbed during bond breaking. (b) 1 mark for identifying the axes (y-axis: energy/enthalpy, x-axis: progress of reaction); 1 mark for placing the reactants line at a higher energy level than the products line; 1 mark for drawing a curve with a peak between reactants and products; 1 mark for showing the activation energy (Ea) as an arrow from the reactants to the peak; 1 mark for showing the enthalpy change (delta H) as a negative/downward arrow from reactants to products.

For (a), xylem is the vascular tissue responsible for the upward transport of water and dissolved inorganic minerals from the roots. For (b), root hair cells have long hair-like projections that greatly increase their surface area-to-volume ratio. This adaptation increases the rate of water absorption by osmosis down a water potential gradient, and allows active transport of mineral ions against their concentration gradient using energy from respiration. For (c), transpiration begins when water evaporates from the wet cell walls of mesophyll cells into the air spaces within the leaf. This water vapour then diffuses out of the leaf down a concentration gradient through the stomata. This constant loss of water creates a tension or pull (transpiration pull) in the xylem vessels, drawing water molecules upwards due to their cohesive forces (cohesion-tension theory).

(a) 1 mark for naming xylem. (b) 1 mark for identifying the hair-like projection/extension; 1 mark for explaining that this increases the surface area; 1 mark for stating that this increases the rate of osmosis/active transport. (c) 1 mark for stating that water evaporates from the surfaces of mesophyll cells; 1 mark for stating that water vapour diffuses out of the leaf through stomata; 1 mark for explaining that this loss of water creates a tension/suction force in the xylem; 1 mark for explaining that cohesion between water molecules allows them to be pulled up as a continuous column.

For (a), sound is a longitudinal wave. This means that as the sound wave passes through a medium like air, the particles of the medium vibrate or oscillate back and forth in a direction parallel to the direction of energy transfer (the direction the wave is travelling), creating areas of compression and rarefaction. For (b), using the wave equation \(v = f \lambda\), we can rearrange to find wavelength: \(\lambda = \frac{v}{f} = \frac{330\text{ m/s}}{440\text{ Hz}} = 0.75\text{ m}\). For (c), sound travels significantly faster in solids than in gases. This is because the particles in a solid are tightly packed in a regular arrangement with strong forces between them, allowing the mechanical vibrations/compressions to be transmitted from particle to particle much faster than in a gas where particles are far apart.

(a) 1 mark for identifying the sound wave as longitudinal; 1 mark for stating that particles vibrate/oscillate; 1 mark for stating that the direction of vibration is parallel to the direction of wave propagation. (b) 1 mark for using the formula \(\lambda = \frac{v}{f}\); 1 mark for substituting the values correctly; 1 mark for the correct answer of \(0.75\text{ m}\) with appropriate unit. (c) 1 mark for stating that the speed of sound increases (or is faster in solids); 1 mark for explaining that the particles in solids are closer together/more tightly packed, allowing quicker transmission of physical vibrations.

**(a)**
Using the formula for acceleration:
\(a = \frac{v - u}{t}\)
\(a = \frac{30 - 0}{2.5} = 12\text{ m/s}^2\)

**(b)**
Using Newton's second law:
\(F = m \times a\)
\(F = 0.25\text{ kg} \times 12\text{ m/s}^2 = 3.0\text{ N}\)

**(c)**
First, calculate the weight of the rocket acting downwards:
\(W = m \times g = 0.25\text{ kg} \times 10\text{ m/s}^2 = 2.5\text{ N}\)

Since the resultant force is upwards:
\(F_{\text{resultant}} = F_{\text{thrust}} - W\)
\(3.0\text{ N} = F_{\text{thrust}} - 2.5\text{ N}\)
\(F_{\text{thrust}} = 3.0 + 2.5 = 5.5\text{ N}\)

**(d)**
- At the moment the engine shuts off, the rocket has a large amount of kinetic energy.
- As it rises and slows down, kinetic energy decreases and is transferred to gravitational potential energy.
- Some energy is also transferred to the thermal energy store of the surroundings and the rocket due to work done against air resistance.

**(a)**
- Formula or substitution: \(a = \frac{30}{2.5}\) [1]
- Correct calculation with units: \(12\text{ m/s}^2\) [1]

**(b)**
- Formula or substitution: \(F = 0.25 \times 12\) [1]
- Correct calculation showing \(3.0\text{ N}\) [1]

**(c)**
- Calculation of weight: \(0.25 \times 10 = 2.5\text{ N}\) [1]
- Calculation of thrust force: \(3.0 + 2.5 = 5.5\text{ N}\) [1]

**(d)**
- Kinetic energy decreases / GPE increases [1]
- Kinetic energy is transferred to gravitational potential energy [1]
- Mention of transfer to thermal energy / work done against air resistance [1]