Show — 3 marks
A student is investigating the relationship between the resistance of a wire and its length. They use the equation R = ρL/A, where R is resistance in ohms, ρ is resistivity, L is length in metres, and A is the cross-sectional area. For a particular wire, ρ = 1.7 × 10⁻⁸ Ω m and A = 2 × 10⁻⁶ m². The student needs to find the length of wire required to achieve a resistance of 8.5 Ω.
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(a) Show that the equation R = ρL/A can be rearranged to L = RA/ρ
[1 mark]
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(b) Using the rearranged equation and the values given in the context, show that the length of wire required is 1000 m.
[2 marks]
Show mark scheme
- (a) Multiply both sides by A and divide both sides by ρ to give L = RA/ρ (or equivalent algebraic steps shown)
- (b) Correct substitution of values: L = (8.5 × 2 × 10⁻⁶) / (1.7 × 10⁻⁸)
- (b) Correct final answer of 1000 m (or 1.0 × 10³ m) with working shown
Define — 3 marks
A student is investigating the motion of a toy car rolling down a ramp. She measures the distance travelled and the time taken at different points. To analyse her results, she needs to solve equations to find the car's acceleration and set up inequalities to determine safe speed limits for the experiment.
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(a) Define what is meant by a linear equation.
[1 mark]
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(b) The student uses the equation v = u + at to find the final velocity of the car. If u = 0 m/s, a = 2 m/s² and t = 3 s, solve this equation to find the value of v.
[1 mark]
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(c) The student wants the car's velocity to remain below 15 m/s for safety reasons. Write down an inequality that represents this constraint.
[1 mark]
Show mark scheme
- (a) An equation where the highest power of the variable is 1 (or equivalent statement such as: a relationship between variables that forms a straight line when plotted)
- (b) v = 6 m/s (accept 6 with or without units for full mark)
- (c) v < 15 or v ≤ 15 (accept either strict or non-strict inequality)
Describe — 3 marks
A student is investigating how the stopping distance of a car depends on its speed. The relationship is given by the equation: d = 0.05v², where d is the stopping distance in metres and v is the speed in metres per second. The student needs to find the speed at which the stopping distance equals 20 metres.
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(a) Describe the first step you would take to solve the equation 0.05v² = 20 to find the speed.
[1 mark]
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(b) Describe how you would rearrange the equation to isolate v² on one side.
[1 mark]
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(c) Describe what mathematical operation you would perform next to find the value of v, and explain why this operation is necessary.
[1 mark]
Show mark scheme
- (a) Divide both sides by 0.05 (or multiply both sides by 20)
- (b) v² = 400 (or equivalent rearrangement showing v² isolated)
- (c) Take the square root of both sides / v = 20 m/s, because v² needs to be reversed to find v (or: to undo the square operation)
Show — 4 marks
A student is investigating the relationship between the extension of a spring and the force applied to it. They apply various forces and measure the extension. The spring obeys Hooke's law up to a limit. The relationship is given by F = kx, where F is the force in newtons, k is the spring constant, and x is the extension in metres.
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(a) A spring has a spring constant of 250 N/m. Show that when a force of 5 N is applied, the extension is 0.02 m.
[1 mark]
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(b) The student wants the spring to extend by at least 0.015 m but no more than 0.04 m. Show that the force F must satisfy the inequality 3.75 ≤ F ≤ 10, using k = 250 N/m.
[2 marks]
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(c) Rearrange the equation F = kx to make x the subject, and use it to verify that a force of 7.5 N produces an extension of 0.03 m when k = 250 N/m.
[1 mark]
Show mark scheme
- (a) Award 1 mark for correctly substituting F = 5 and k = 250 into F = kx to obtain x = 5/250 = 0.02 m (or equivalent working shown).
- (b) Award 1 mark for correctly setting up the lower inequality: F = 250 × 0.015 = 3.75 N (or F ≥ 3.75). Award 1 mark for correctly setting up the upper inequality: F = 250 × 0.04 = 10 N (or F ≤ 10). Accept combined statement 3.75 ≤ F ≤ 10.
- (c) Award 1 mark for rearranging to x = F/k and substituting correctly to obtain x = 7.5/250 = 0.03 m (or equivalent working shown).
Calculate — 2 marks
A rectangle has a length of 3x cm and a width of x cm. The perimeter of the rectangle is 32 cm.
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(a) Write an equation in terms of x for the perimeter of the rectangle.
[1 mark]
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(b) Calculate the value of x.
[1 mark]
Show mark scheme
- (a) 8x = 32 or equivalent equation 6x + 2x = 32 or 2(3x + x) = 32
- (b) x = 4
GCSE Perfect