题型分析

Question Type 1: Plotting Graphs with Error Bars

如何识别

Part (c)(i) of Q2 — "Plot a graph of Y against X. Include error bars for Y." 题目会提供一个数据表，包含物理量的测量值及其不确定度。

标准解题方法

标准解题方法

1. 选择刻度，使图至少占网格一半（水平和垂直）
2. 刻度只用 1, 2 或 5 对应 2 cm 方格
3. 坐标轴标注物理量（或符号）和单位
4. 用细 × 或 ⊙（直径 < 1 mm）描点，精确到半小格
5. Error bars：对称，中心在数据点，总长度 = $2 \times \Delta Y$
6. 两端画 cap（小横线）

评分标准（MS 模式）

• M1: All points plotted correctly within half a small square
• M1: Error bars plotted correctly in Y-direction, all points, symmetrical

Example 1 — Thermistor Resistance against Temperature

A student investigates how the resistance $R$ of a thermistor varies with temperature $T$. The thermistor is placed in a water bath and its resistance is measured at different temperatures. The student suggests that $R = R_0 e^{kT}$.

$T / ^{\circ}\text{C}$	$R / \Omega$	$\Delta R / \Omega$
20.0	5200	$\pm 200$
30.0	3200	$\pm 150$
40.0	2000	$\pm 100$
50.0	1300	$\pm 70$
60.0	850	$\pm 50$
70.0	550	$\pm 30$

The student calculates $\ln R$ and $1/T$.

(c)(i) Plot a graph of $\ln R$ against $1/T$. Include error bars for $\ln R$.

MS 展开查看

Plotting:

• M1: All 6 points plotted correctly within half a small square
• M1: Error bars in Y-direction ($\ln R$) plotted correctly
  • $\Delta(\ln R) = \Delta R / R$: e.g. 200/5200 = 0.0385, etc.
  • Symmetrical, all points, with caps

Scales:

• x-axis: $1/T$ from 0.0029 to 0.0034 K$^{-1}$ (sensible range)
• y-axis: $\ln R$ from 6.0 to 8.6

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Example 2 — Cooling of a Liquid

A student investigates the cooling of a hot liquid. The liquid is placed in a beaker and allowed to cool. The temperature $\theta$ is measured at different times $t$. It is suggested that $\theta = \theta_0 e^{-kt}$.

$t / \text{s}$	$\theta / ^{\circ}\text{C}$	$\Delta\theta / ^{\circ}\text{C}$
0	80.0	$\pm 0.5$
30	67.5	$\pm 0.5$
60	57.0	$\pm 0.5$
90	48.5	$\pm 0.5$
120	41.0	$\pm 0.5$
150	35.0	$\pm 0.5$
180	30.0	$\pm 0.5$

(c)(i) Plot a graph of $\ln \theta$ against $t$. Include error bars for $\ln \theta$.

MS 展开查看

Plotting:

• M1: All 7 points plotted correctly to within half a small square
• M1: Error bars in Y-direction ($\ln \theta$) plotted
  • $\Delta(\ln \theta) = \Delta\theta / \theta$: 0.5/80.0 = 0.00625, 0.5/67.5 = 0.00741, etc.
  • Caps drawn at both ends of each bar

Scales:

• x-axis: $t$ from 0 to 200 s
• y-axis: $\ln \theta$ from 3.4 to 4.4

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Example 3 — Period of a Simple Pendulum

A student investigates how the period $T$ of a simple pendulum varies with length $L$. A stopwatch is used to measure the time for 10 oscillations. The period $T$ is then calculated. It is suggested that $T = 2\pi \sqrt{L/g}$.

$L / \text{m}$	$T / \text{s}$	$\Delta T / \text{s}$
0.20	0.90	$\pm 0.02$
0.30	1.10	$\pm 0.02$
0.40	1.27	$\pm 0.02$
0.50	1.42	$\pm 0.02$
0.60	1.55	$\pm 0.02$
0.70	1.68	$\pm 0.02$
0.80	1.79	$\pm 0.02$

(c)(i) Plot a graph of $T^2$ against $L$. Include error bars for $T^2$.

MS 展开查看

Plotting:

• M1: All 7 points ($T^2$ against $L$) plotted correctly within half a small square
• M1: Error bars in Y-direction ($T^2$) plotted
  • $\Delta(T^2) = 2T \times \Delta T$: $2 \times 0.90 \times 0.02 = 0.036$, etc.
  • All 7 bars present, symmetrical, with caps

Scales:

• x-axis: $L$ from 0.15 to 0.85 m
• y-axis: $T^2$ from 0.6 to 3.4 s$^2$

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Question Type 2: Best Fit and Worst Acceptable Lines

如何识别

Part (c)(ii) — "Draw the straight line of best fit and a worst acceptable straight line." 题目要求在已有的图上画两条线。

标准解题方法

标准解题方法

1. Best fit: 用直尺画直线，使数据点大致均匀分布在直线两侧（忽略明显异常值）
2. WAL: 过所有 error bars 的最陡（steepest）或最缓（shallowest）直线
3. 两条线有明显区别
4. 用标签或不同线型区分（如 best fit 实线、WAL 虚线）

常见陷阱

• 不要连接第一个点和最后一个点
• WAL 必须穿过所有 error bars
• 如果 error bars 只在一个方向，WAL 可以在 error bars 内旋转
• 两条线不能重合

Example 1 — Thermistor Graph (continued)

(c)(ii) On the graph drawn in (c)(i), draw the straight line of best fit and a worst acceptable straight line.

MS 展开查看

Best fit:

• M1: Straight line drawn with a ruler
• Points roughly equally distributed above and below the line
• The line passes through the general trend of the data

WAL:

• M1: Steepest possible straight line that passes through all error bars
• The WAL is clearly different from the best fit line
• Both lines are labelled (or distinguishable)

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Example 2 — Cooling Graph (continued)

(c)(ii) On the graph of $\ln \theta$ against $t$, draw the straight line of best fit and a worst acceptable straight line.

MS 展开查看

Best fit:

• M1: Straight line through the general trend
• More points lie above and below in roughly equal numbers

WAL:

• M1: Shallowest straight line that still passes through all error bars
• Distinct from best fit, labelled clearly

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Example 3 — Pendulum Graph (continued)

(c)(ii) On the graph of $T^2$ against $L$, draw the straight line of best fit and a worst acceptable straight line.

MS 展开查看

Best fit:

• M1: Straight line through all points (theory predicts $T^2 \propto L$, so line should pass near origin)
• Balanced distribution of points

WAL:

• M1: Steepest OR shallowest line through all error bars
• Must be labelled

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Question Type 3: Gradient Determination

如何识别

Part (c)(iii) — "Determine the gradient of the line of best fit. Include the absolute uncertainty." 有时也会要求给出单位。

标准解题方法

标准解题方法

1. 在 best fit 线上选两个点（不是数据点）
2. 两点间隔 > 所画直线长度的一半
3. 在点旁标注坐标 $(x_1, y_1)$ 和 $(x_2, y_2)$
4. $\text{gradient} = \frac{y_2 - y_1}{x_2 - x_1}$
5. 用相同方法计算 WAL 的 gradient
6. $\text{uncertainty} = |\text{gradient}_{\text{best}} - \text{gradient}_{\text{worst}}|$
7. 结果表示为 $\text{gradient} \pm \text{uncertainty}$，带单位

Example 1 — Thermistor Gradient

The best fit line for $\ln R$ against $1/T$ has the equation $\ln R = \ln R_0 + k(1/T)$.

(c)(iii) Determine the gradient of the best fit line and its absolute uncertainty.

MS 展开查看

Best fit gradient:

• M1: Two points on best fit line selected: $(0.00294, 8.55)$ and $(0.00330, 6.95)$
• $\text{gradient}_{\text{best}} = \frac{6.95 - 8.55}{0.00330 - 0.00294} = \frac{-1.60}{0.00036} = -4440 \text{ K}$
• Correct substitution and calculation

WAL gradient:

• Two points on WAL: $(0.00294, 8.70)$ and $(0.00330, 6.75)$
• $\text{gradient}_{\text{WAL}} = \frac{6.75 - 8.70}{0.00330 - 0.00294} = \frac{-1.95}{0.00036} = -5420 \text{ K}$

Uncertainty:

• M1: $\Delta m = |{-4440} - ({-5420})| = 980 \text{ K}$
• Final answer: $m = (-4440 \pm 980) \text{ K}$

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Example 2 — Cooling Gradient

The best fit line for $\ln \theta$ against $t$ has the equation $\ln \theta = \ln \theta_0 - kt$.

(c)(iii) Determine the gradient of the best fit line and its absolute uncertainty.

MS 展开查看

Best fit gradient:

• M1: Two points on best fit line: $(0, 4.38)$ and $(180, 3.40)$
• $\text{gradient}_{\text{best}} = \frac{3.40 - 4.38}{180 - 0} = \frac{-0.98}{180} = -0.00544 \text{ s}^{-1}$

WAL gradient:

• Two points on WAL: $(0, 4.40)$ and $(180, 3.30)$
• $\text{gradient}_{\text{WAL}} = \frac{3.30 - 4.40}{180 - 0} = -0.00611 \text{ s}^{-1}$

Uncertainty:

• M1: $\Delta m = |-0.00544 - (-0.00611)| = 0.00067 \text{ s}^{-1}$
• Final answer: $m = (-0.00544 \pm 0.00067) \text{ s}^{-1}$

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Example 3 — Pendulum Gradient

The graph of $T^2$ against $L$ is expected to be linear with $T^2 = (4\pi^2/g)L$.

(c)(iii) Determine the gradient of the best fit line and its absolute uncertainty. Hence determine a value for $g$.

MS 展开查看

Best fit gradient:

• M1: Two points on best fit line: $(0.20, 0.81)$ and $(0.80, 3.22)$
• $\text{gradient}_{\text{best}} = \frac{3.22 - 0.81}{0.80 - 0.20} = \frac{2.41}{0.60} = 4.02 \text{ s}^2\text{m}^{-1}$

WAL gradient:

• Two points on WAL: $(0.20, 0.78)$ and $(0.80, 3.32)$
• $\text{gradient}_{\text{WAL}} = \frac{3.32 - 0.78}{0.80 - 0.20} = \frac{2.54}{0.60} = 4.23 \text{ s}^2\text{m}^{-1}$

Uncertainty:

• M1: $\Delta m = |4.02 - 4.23| = 0.21 \text{ s}^2\text{m}^{-1}$
• Final answer: $m = (4.02 \pm 0.21) \text{ s}^2\text{m}^{-1}$

Determining $g$:

• $g = 4\pi^2 / \text{gradient} = 4\pi^2 / 4.02 = 9.82 \text{ m s}^{-2}$
• Uncertainty: $\Delta g = g \times (\Delta m / m) = 9.82 \times (0.21 / 4.02) = 0.51 \text{ m s}^{-2}$
• $g = (9.82 \pm 0.51) \text{ m s}^{-2}$

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Question Type 4: y-intercept Determination

如何识别

Part (c)(iv) — "Determine the y-intercept." 有时会要求不确定性。注意不能直接从 y 轴读取。

标准解题方法

标准解题方法

1. 使用 $y = mx + c$
2. 代入 best fit 线上一个点 $(x, y)$ 和 gradient $m$
3. $c = y - mx$
4. 用 WAL 重复计算得到 $c_{\text{worst}}$
5. $\text{uncertainty} = |c_{\text{best}} - c_{\text{worst}}|$
6. 不能直接从图坐标轴读值（false origin 时无效）

评分标准（MS 模式）

• M1: Correct y-intercept calculated using $c = y - mx$
• M1 (if required): Correct uncertainty in y-intercept

Example 1 — Thermistor y-intercept

The graph of $\ln R$ against $1/T$ should give a straight line with equation $\ln R = \ln R_0 + k(1/T)$.

(c)(iv) Determine the y-intercept of the best fit line and its absolute uncertainty.

MS 展开查看

Best fit y-intercept:

• M1: Using $c = y - mx$
• Point on best fit: $(0.00310, 7.72)$, gradient $m = -4440 \text{ K}$
• $c = 7.72 - (-4440 \times 0.00310) = 7.72 - (-13.76) = 21.48$
• y-intercept $= 21.5$ (3 SF)

WAL y-intercept:

• Point on WAL: $(0.00310, 7.72)$, gradient $m = -5420 \text{ K}$
• $c_{\text{WAL}} = 7.72 - (-5420 \times 0.00310) = 7.72 - (-16.80) = 24.52$

Uncertainty:

• M1: $\Delta c = |21.48 - 24.52| = 3.04$
• Final answer: $c = 21.5 \pm 3.0$

Physical meaning:

• $\ln R_0 = 21.5$, therefore $R_0 = e^{21.5} = 2.18 \times 10^9 \ \Omega$

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Example 2 — Cooling y-intercept

The graph of $\ln \theta$ against $t$ should give a straight line with equation $\ln \theta = \ln \theta_0 - kt$.

(c)(iv) Determine the y-intercept of the best fit line and its absolute uncertainty. State the value of $\theta_0$.

MS 展开查看

Best fit y-intercept:

• M1: Using $c = y - mx$
• Point on best fit: $(90, 3.88)$, gradient $m = -0.00544 \text{ s}^{-1}$
• $c = 3.88 - (-0.00544 \times 90) = 3.88 - (-0.490) = 4.37$
• y-intercept $= 4.37$

WAL y-intercept:

• Point on WAL: $(90, 3.88)$, gradient $m = -0.00611 \text{ s}^{-1}$
• $c_{\text{WAL}} = 3.88 - (-0.00611 \times 90) = 3.88 - (-0.550) = 4.43$

Uncertainty:

• M1: $\Delta c = |4.37 - 4.43| = 0.06$
• Final answer: $c = 4.37 \pm 0.06$

Physical meaning:

• $\ln \theta_0 = 4.37$, therefore $\theta_0 = e^{4.37} = 79.0 \ ^{\circ}\text{C}$
• This matches the measured initial temperature $80.0 \pm 0.5 \ ^{\circ}\text{C}$

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Example 3 — Pendulum y-intercept

The graph of $T^2$ against $L$ should give a straight line with equation $T^2 = (4\pi^2/g)L + c$.

(c)(iv) Determine the y-intercept of the best fit line. Comment on whether your result is consistent with the theoretical prediction.

MS 展开查看

Best fit y-intercept:

• M1: Using $c = y - mx$
• Point on best fit: $(0.50, 2.02)$, gradient $m = 4.02 \text{ s}^2\text{m}^{-1}$
• $c = 2.02 - (4.02 \times 0.50) = 2.02 - 2.01 = 0.01 \text{ s}^2$

WAL y-intercept:

• Point on WAL: $(0.50, 2.02)$, gradient $m = 4.23 \text{ s}^2\text{m}^{-1}$
• $c_{\text{WAL}} = 2.02 - (4.23 \times 0.50) = 2.02 - 2.115 = -0.095 \text{ s}^2$

Uncertainty:

• $\Delta c = |0.01 - (-0.095)| = 0.105 \text{ s}^2$

Comment:

• The theoretical prediction is $c = 0$ (line through origin)
• The experimental value $c = 0.01 \pm 0.11 \text{ s}^2$ includes zero within uncertainty
• Therefore the result is consistent with the theoretical prediction

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题型对比总结

题型	对应 Q2 Part	核心技能	分值
Plotting with error bars	(c)(i)	选刻度、描点、画 error bars	2
Best fit + WAL	(c)(ii)	画两条线，区分标记	2
Gradient determination	(c)(iii)	计算 gradient + 不确定度	2
y-intercept determination	(c)(iv)	代入公式计算	1–2