Show — 4 marks
A student investigates how the extension of a spring changes with the mass hung from it. The student collects data and plots a graph of extension (in cm) against mass (in kg). The relationship between extension and mass is linear.
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(a) The student obtains the following data:
Mass (kg): 0, 0.5, 1.0, 1.5, 2.0
Extension (cm): 0, 2.5, 5.0, 7.5, 10.0
Show that the relationship between extension and mass is linear by calculating the gradient of the line.
[2 marks]
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(b) Using your answer from part (a), write down the equation of the line in the form extension = m × mass + c, where m is the gradient and c is the y-intercept.
[1 mark]
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(c) Show that when the mass is 2.5 kg, the extension would be 12.5 cm using the equation from part (b).
[1 mark]
Show mark scheme
- (a) Correctly calculates gradient as change in extension ÷ change in mass = 5.0 cm ÷ 1.0 kg (or equivalent using any two correct data points) [1 mark]
- (a) States gradient = 5 cm/kg (or 5.0) [1 mark]
- (b) Writes equation as extension = 5 × mass + 0 or extension = 5 × mass (correctly identifies m = 5 and c = 0) [1 mark]
- (c) Substitutes mass = 2.5 into equation and calculates extension = (5 × 2.5) + 0 = 12.5 cm [1 mark]
Calculate — 4 marks
A student investigates the motion of a ball thrown vertically upwards. The height h (in metres) of the ball above the ground at time t (in seconds) is given by the equation: h = 25t - 5t². The student plots a graph of h against t and needs to analyse its properties to determine key features of the motion.
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(a) Calculate the height of the ball at t = 2.0 seconds.
[1 mark]
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(b) By drawing or calculating values from the function h = 25t - 5t², determine the maximum height reached by the ball and the time at which this occurs. Show your working.
[2 marks]
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(c) Calculate the time(s) when the ball returns to ground level (h = 0). Explain which solution is physically meaningful for this context.
[1 mark]
Show mark scheme
- (a) Correct substitution of t = 2.0 into h = 25t - 5t² to give h = 50 - 20 = 30 m (or 30 metres)
- (b) Identifies that maximum occurs at t = 2.5 s (by calculus: dh/dt = 25 - 10t = 0, or by recognising vertex of parabola at t = -b/2a = 25/10 = 2.5)
- (b) Calculates maximum height h = 25(2.5) - 5(2.5)² = 62.5 - 31.25 = 31.25 m (or 31.3 m to 3 s.f.)
- (c) Solves 25t - 5t² = 0 to give t(25 - 5t) = 0, so t = 0 s or t = 5.0 s. Correctly identifies t = 5.0 s as physically meaningful (t = 0 is initial position; t = 5.0 s is when ball lands)
Calculate — 3 marks
A student investigates how the resistance of a thermistor changes with temperature. The relationship between resistance R (in ohms) and absolute temperature T (in kelvin) is given by the function: R = 5000e^(-0.005T) + 100, where e is Euler's number (approximately 2.718).
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(a) Calculate the resistance of the thermistor at a temperature of 300 K.
[1 mark]
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(b) The student plots a graph of R against T. Determine the resistance value that the function approaches as T increases significantly (the asymptotic value).
[1 mark]
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(c) Calculate the temperature at which the resistance equals 2600 Ω. Show your working.
[1 mark]
Show mark scheme
- (a) Correct substitution of T = 300 into R = 5000e^(-0.005×300) + 100 gives R = 5000e^(-1.5) + 100 = 5000(0.223) + 100 ≈ 1215 Ω (accept 1210-1220 Ω)
- (b) As T → ∞, e^(-0.005T) → 0, therefore R approaches 100 Ω (the constant term)
- (c) Rearranging: 2600 = 5000e^(-0.005T) + 100; 2500 = 5000e^(-0.005T); 0.5 = e^(-0.005T); ln(0.5) = -0.005T; T = ln(0.5)/(-0.005) ≈ 139 K (accept 138-140 K)
State — 4 marks
A student investigates how the temperature of a cooling cup of tea changes over time. The tea is left in a room at constant temperature. The graph below shows the relationship between temperature (in °C) and time (in minutes).
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(a) State the initial temperature of the tea at time = 0 minutes.
[1 mark]
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(b) State the temperature of the tea after 10 minutes.
[1 mark]
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(c) State the name of the type of graph shown (for example, linear, quadratic, exponential, etc.).
[1 mark]
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(d) State what happens to the rate of cooling as time increases.
[1 mark]
Show mark scheme
- (a) Correct reading from graph at time = 0 (accept 95–100 °C depending on graph provided)
- (b) Correct reading from graph at time = 10 minutes (accept values within ±2 °C of marked point)
- (c) Exponential decay / exponential cooling (not linear or quadratic)
- (d) The rate of cooling decreases / becomes slower (or equivalent statement that gradient decreases)
Calculate — 2 marks
A taxi company uses the formula C = 2 + 1.5m to calculate the cost, C pounds, of a journey of m miles.
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(a) Calculate the cost of a journey of 4 miles.
[1 mark]
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(b) Calculate how many miles can be travelled for £11.
[1 mark]
Show mark scheme
GCSE Perfect