SPM Paper 2 Section B High Priority (10 Marks)

Reduction to $Y = mX + c$, Best Fit Line & Constants

Linear Law converts non-linear experimental curves into straight lines. Master the algebraic transformation into $Y = mX + c$, plotting best-fit lines with balanced residual points, computing gradient and $Y$-intercept without reading data points off the line, and deducing experimental constants.

1. Classic Non-Linear to Linear Transformation Table

Non-Linear Form Algebraic Manipulation Linear Form ($Y = mX + c$) $Y$-axis $X$-axis Gradient ($m$) $Y$-Intercept ($c$)
$y = ax^2 + bx$ Divide by $x$ $\frac{y}{x} = ax + b$ $\frac{y}{x}$ $x$ $a$ $b$
$y = \frac{a}{x} + b$ Multiply by $x$ $xy = bx + a$ $xy$ $x$ $b$ $a$
$y = ax^b$ Apply $\log_{10}$ to both sides $\log y = b\log x + \log a$ $\log_{10} y$ $\log_{10} x$ $b$ $\log_{10} a$
$y = ab^x$ Apply $\log_{10}$ to both sides $\log y = (\log b)x + \log a$ $\log_{10} y$ $x$ $\log_{10} b$ $\log_{10} a$
$\frac{p}{y} - \frac{q}{x} = 1$ Rearrange for $\frac{1}{y}$ $\frac{1}{y} = \frac{q}{p}\left(\frac{1}{x}\right) + \frac{1}{p}$ $\frac{1}{y}$ $\frac{1}{x}$ $\frac{q}{p}$ $\frac{1}{p}$

2. SPM Examiner Pitfalls in Linear Law

Trap 1: Using Table Data Points to Calculate Gradient:

SPM marking schemes award ZERO marks for gradient calculation if you use data points directly from the original table! You MUST pick two points directly located on your drawn line of best fit that are far apart (with coordinates clearly indicated).

Trap 2: Forgetting to Convert Logarithmic Constants:

If the $Y$-intercept is $c = 1.30$, and $c = \log_{10} a$, then $a = 10^{1.30} \approx 19.95$. Stating $a = 1.30$ loses the final answer mark!

Trap 3: Awkward Graph Scales ($3\text{ cm to } 10$):

Always use standard multiples: $1, 2, 4, 5, 10, 20$. Never use odd increments like 3 or 7 per $2\text{ cm}$, which violates the SPM graphing scale rule and penalizes plotting marks.

3. Full 10-Mark SPM Paper 2 Section B Model Worked Example

SPM Paper 2 Section B Full 10-Mark Blueprint [10 Marks]
Table below shows the experimental values of two variables, $x$ and $y$, obtained from an physics laboratory experiment. The variables $x$ and $y$ are related by the equation $y = \frac{p}{x} + qx$, where $p$ and $q$ are constants.
x 1.0 2.0 3.0 4.0 5.0
y 14.0 10.0 9.3 9.5 10.2

(a) Plot $xy$ against $x^2$, using a scale of $2\text{ cm to } 5\text{ units}$ on the $x^2$-axis and $2\text{ cm to } 10\text{ units}$ on the $xy$-axis. Hence, draw the line of best fit. [5 marks]

(b) Using the graph in (a), find the value of:

(i) $p$ and $q$. [3 marks]

(ii) The value of $y$ when $x = 3.5$. [2 marks]

Step-by-Step Marking Scheme:
(a) Transform equation: $y = \frac{p}{x} + qx \implies xy = qx^2 + p \implies Y = qX + p$, where $Y = xy$ and $X = x^2$. [1m: K1: Correct linear transformation]
Table of transformed values: $X = x^2: [1.0, 4.0, 9.0, 16.0, 25.0]$. $Y = xy: [14.0, 20.0, 27.9, 38.0, 51.0]$. [1m: N1: All 5 pairs calculated to $\ge 1$ d.p.]
Uniform scale chosen on both axes ($2\text{ cm to } 5$ on $X$, $2\text{ cm to } 10$ on $Y$). [1m: K1: Standard axis scale]
All 5 points plotted accurately within $\pm 1\text{ mm}$ grid tolerance. [1m: K1: Accurate plotting]
Line of best fit drawn passing through points with balanced residuals. [1m: N1: Best fit line]
(b)(i) Gradient $q = \frac{Y_2 - Y_1}{X_2 - X_1} = \frac{51.0 - 12.5}{25.0 - 0} = \frac{38.5}{25.0} = 1.54$. (Accept $1.50 \pm 0.05$). [2m: K1, N1: Value of $q$]
$Y$-intercept $p = 12.5$. (Accept $12.5 \pm 0.5$). [1m: N1: Value of $p$]
(b)(ii) When $x = 3.5 \implies X = x^2 = 3.5^2 = 12.25$. From graph at $X = 12.25$, read $Y = xy \approx 31.4$. [1m: K1: Read $xy$ from graph]
$y = \frac{Y}{x} = \frac{31.4}{3.5} = 8.97$. (Accept $8.95 - 9.05$). [1m: N1: Correct $y$ value]
Open Form 4 Chapter 6 Worksheet