Physics 5054/31 — October/November 2024

Walkthrough

The candidate is asked to measure the length of a straw using a 30 cm ruler. The mark scheme requires the reading to be given to the nearest 0.1 cm. A typical straw length is around 20 cm, so a reading of 20.0 cm is realistic.

Key Takeaways

When measuring with a ruler, always record to the precision of the smallest division. If the smallest division is 1 mm (0.1 cm), record one decimal place.

Common Mistakes

Forgetting the trailing zero (e.g. writing 20 cm instead of 20.0 cm) or reading to the wrong precision (e.g. 20.05 cm).

Things to Be Careful About

Always read the ruler from the zero mark or subtract the starting position. Ensure the eye is directly above the marking to avoid parallax error. The value must be within ±0.2 cm of the supervisor's actual measurement.

Walkthrough

Five pieces of straw are lined up side by side. The total width $D$ is measured. A typical straw diameter is about 0.6 cm, so five pieces would be about 3.0 cm wide.

Key Takeaways

When measuring the combined width of multiple identical small objects, the total width is larger and easier to measure accurately than a single object's width, reducing percentage error.

Common Mistakes

Reading the scale incorrectly or not lining the pieces up tightly together.

Things to Be Careful About

The pieces must be touching and aligned straight. Read to the nearest 0.1 cm. Ensure the ruler is parallel to the line of straws.

Walkthrough

The diameter $d$ of one straw is found by dividing the total width $D$ of 5 straws by 5. Using the representative value $D = 3.0$ cm, $d = 3.0 / 5 = 0.60$ cm.

Key Takeaways

Averaging over multiple identical objects (here, dividing by 5) gives a more accurate measurement of a single object's dimension than measuring one object directly.

Common Mistakes

Dividing by the wrong number or forgetting to use the candidate's own value from (a)(ii).

Things to Be Careful About

Carry forward the candidate's value from (a)(ii). The answer should have an appropriate number of significant figures (2 or 3).

Walkthrough

To measure $D$ accurately, the ends of the straw pieces must be read against the ruler. If reading directly, the eye might not be directly above the edge, causing a parallax error. By placing a set square at each end with one edge against the ruler and the other touching the straw, the position is projected vertically onto the ruler, eliminating parallax error.

Key Takeaways

Set squares are used as right-angled guides to project measurements onto a scale accurately.

Common Mistakes

Saying "to make it straight" without explaining the alignment with the ruler. Not mentioning both set squares.

Things to Be Careful About

The explanation must mention using the set squares at each end and aligning them with the ruler. A diagram showing the set squares is also acceptable.

Walkthrough

The volume of the straw (treated as a solid cylinder) is calculated using the formula $V_1 = \frac{3.14 d^2 l}{4}$. Substituting $d = 0.60$ cm and $l = 20.0$ cm gives $V_1 = 5.652$ cm$^3$. The mark scheme requires the answer to be given to 2 significant figures, so it is rounded to 5.7 cm$^3$.

Key Takeaways

When using a given formula, substitute the values carefully and round the final answer to the required number of significant figures.

Common Mistakes

Forgetting to square the diameter, using the radius instead of the diameter, or not rounding to 2 significant figures.

Things to Be Careful About

Use the candidate's own values from (a)(i) and (a)(iii). The value 3.14 is an approximation of $\pi$, so use 3.14 exactly as given.

Walkthrough

The water transferred from the straw is collected in a measuring cylinder. The volume $V_T$ is read from the cylinder. A typical internal volume of a straw is about 4 cm$^3$, so 6 transfers would give about 24 cm$^3$. The mark scheme requires the reading to be to the nearest 1 cm$^3$ and less than 35 cm$^3$.

Key Takeaways

When reading a measuring cylinder, read the bottom of the meniscus at eye level. The precision depends on the smallest division of the cylinder.

Common Mistakes

Reading the top of the meniscus or not reading at eye level.

Things to Be Careful About

The volume must be less than 35 cm$^3$ to stay within the capacity of the 50 cm$^3$ cylinder and avoid spillage. Read to the nearest 1 cm$^3$.

Walkthrough

The average volume of water in one straw $V_2$ is found by dividing the total volume $V_T$ by the number of transfers (6). Using $V_T = 24$ cm$^3$, $V_2 = 24 / 6 = 4.0$ cm$^3$.

Key Takeaways

Averaging over multiple measurements reduces the effect of random errors.

Common Mistakes

Dividing by the wrong number (e.g. 5 instead of 6).

Things to Be Careful About

Carry forward the candidate's value from (b)(i). The answer should have an appropriate number of significant figures.

Walkthrough

The candidate closes the switch briefly, reads the voltmeter across X and the ammeter in the circuit, and records both values with the correct units. The switch must be opened immediately after reading to prevent the resistors from overheating and the cell from draining.

Key Takeaways

Candidates must read analogue or digital meters to the precision demanded by the scale and record values with their units. Switches should be opened quickly during electrical measurements to avoid altering the components or the power supply.

Common Mistakes

• Recording readings without units.
• Reading the ammeter to only 1 decimal place when the scale requires 2.
• Leaving the switch closed for too long, causing the resistors to heat up and their resistance to change.

Things to Be Careful About

The mark scheme requires $V_X$ to at least 1 decimal place and $< 1.2 \text{ V}$, and $I_S$ to at least 2 decimal places and $< 0.5 \text{ A}$. Ensure trailing zeros are recorded if the scale demands them (e.g., $0.20 \text{ A}$, not $0.2 \text{ A}$).

Walkthrough

The resistance of X is calculated using the equation $R_X = V_X / I_S$. The candidate substitutes the values recorded in part (a)(i). Using the example values $V_X = 0.8 \text{ V}$ and $I_S = 0.20 \text{ A}$, the calculation is $0.8 / 0.20 = 4.0 \text{ } \Omega$.

Key Takeaways

Ohm's law $R = V/I$ is used to find resistance from measured potential difference and current. The result must include the unit ohms ($\Omega$).

Common Mistakes

• Forgetting to include the unit $\Omega$ in the final answer.
• Carrying forward an incorrect value from part (a)(i) without showing the substitution (though ecf is usually allowed in marking schemes).

Things to Be Careful About

Ensure the values substituted are from the candidate's own readings. If the candidate's $V_X$ or $I_S$ is outside the acceptable range, the calculation will be marked incorrect.

Walkthrough

When a current flows through resistors, electrical energy is transferred to thermal energy stores. If the switch is left closed for too long, the resistors can overheat, potentially changing their resistance or becoming a burn hazard. Additionally, the cell will discharge more quickly than necessary.

Key Takeaways

In practical electrical experiments, switches should be opened as soon as readings are taken to preserve apparatus and ensure accurate, repeatable results.

Common Mistakes

• Saying "to save energy" without specifying what is being saved (the cell or the resistors from overheating).
• Saying "to avoid electric shock" which is not relevant for a 3 V cell.

Things to Be Careful About

Acceptable answers must mention overheating of resistors or the cell running down. "Heat" alone is often rejected in favour of "thermal energy" or "overheating".

Walkthrough

The voltmeter is disconnected from X and connected in parallel across Y. The switch is closed briefly, and $V_Y$ is recorded. Since X and Y are in series, the current $I_S$ is the same. The resistance $R_Y$ is calculated using $R_Y = V_Y / I_S$. The mark scheme requires $R_Y > R_X$, which is consistent with the example values ($5.0 \text{ } \Omega > 4.0 \text{ } \Omega$).

Key Takeaways

In a series circuit, the current is the same through all components, but the potential difference across each depends on its resistance. A larger resistance has a larger potential difference across it.

Common Mistakes

• Forgetting that the current $I_S$ is the same for both resistors and using a different value.
• Not ensuring that the calculated $R_Y$ is greater than $R_X$ as required by the mark scheme.

Things to Be Careful About

The voltmeter must be connected across Y only, not across both X and Y. Ensure the voltmeter terminals are connected correctly (positive to positive, negative to negative) to avoid a negative reading.

Walkthrough

The candidate must complete Fig. 2.2 to show X and Y in parallel. This means drawing two separate branches between W and Z, one containing X and the other containing Y. The ammeter remains in the main circuit. The voltmeter must be connected in parallel across the entire parallel combination (i.e., between W and Z), not across just one resistor.

Key Takeaways

In a parallel circuit, the potential difference across each branch is the same. The voltmeter must measure this common potential difference by being connected across both branches.

Common Mistakes

• Drawing X and Y in series instead of parallel.
• Connecting the voltmeter across only one resistor instead of the combination.
• Forgetting to label the resistors X and Y.

Things to Be Careful About

Use standard circuit symbols. The voltmeter symbol is a circle with a V inside. The resistor symbol is a rectangle (or zigzag, but rectangle is standard for 5054). Ensure the connections cross cleanly without false junctions.

Walkthrough

The candidate rearranges the circuit to match the diagram in (b)(i). The switch is closed, and the potential difference across the parallel combination ($V_P$) and the total current through the combination ($I_P$) are recorded. The total resistance is calculated using $R_P = V_P / I_P$. The mark scheme requires $V_P < 3 \text{ V}$ and $I_P < 1.0 \text{ A}$.

Key Takeaways

In a parallel circuit, the total current is the sum of the currents in the branches, and the total resistance is less than the resistance of any individual branch.

Common Mistakes

• Reading the ammeter as the current through one branch instead of the total current.
• Forgetting that $V_P$ is the same across both branches.

Things to Be Careful About

The ammeter must be in the main circuit to measure the total current $I_P$. Ensure readings are within the specified limits ($V_P < 3 \text{ V}$, $I_P < 1.0 \text{ A}$).

Walkthrough

The theoretical total resistance for two resistors in parallel is calculated using the given formula $R_T = (R_X R_Y) / (R_X + R_Y)$. Substituting the example values $R_X = 4.0 \text{ } \Omega$ and $R_Y = 5.0 \text{ } \Omega$ gives $R_T = (4.0 \times 5.0) / (4.0 + 5.0) = 20.0 / 9.0 = 2.22 \text{ } \Omega$.

Key Takeaways

The formula for two resistors in parallel can be used to calculate the theoretical total resistance. This value can then be compared with the experimental value $R_P$.

Common Mistakes

• Forgetting to add $R_X$ and $R_Y$ in the denominator.
• Multiplying instead of dividing in the formula.

Things to Be Careful About

Use the candidate's own values of $R_X$ and $R_Y$ from parts (a)(ii) and (a)(iv). Round the final answer to a sensible number of significant figures (usually 2 or 3, matching the data).

Walkthrough

The candidate must determine if $R_P$ and $R_T$ are equal within 10% of each other. This is done by calculating the percentage difference: $|R_P - R_T| / R_T \times 100$. Using the example values, $|2.18 - 2.22| / 2.22 \times 100 = 1.8\%$. Since $1.8\% < 10\%$, the statement is that they are equal. The calculation must use both values and the statement must match the result of the calculation.

Key Takeaways

Experimental values often differ slightly from theoretical values due to measurement errors and component tolerances. A comparison within a specified tolerance (e.g., 10%) is used to judge if the results are consistent with theory.

Common Mistakes

• Calculating the percentage difference incorrectly (e.g., using $R_P$ in the denominator instead of $R_T$, or forgetting the absolute value).
• Making a statement that does not match the calculation (e.g., saying 'not equal' when the difference is $< 10\%$).

Things to Be Careful About

The mark scheme requires a correct calculation to support the statement and a statement that matches the results. Ensure the calculation clearly shows the use of both $R_P$ and $R_T$. The percentage can be calculated using either value as the denominator, but the standard approach is to use the theoretical value $R_T$.

Walkthrough

In this step, the candidate balances the metre rule horizontally on the pivot placed at the $25.0\text{ cm}$ mark with a $20\text{ g}$ mass placed on the short side (or as needed to balance the fixed mass at the $5.0\text{ cm}$ mark and the rule's own weight acting at the centre of mass).

The position of the centre of the $20\text{ g}$ mass on the metre rule is noted, and the distance $d$ from the pivot at $25.0\text{ cm}$ is determined by subtracting the mark positions:

The reading must be recorded to the precision of the metre rule scale, which is to the nearest $0.1\text{ cm}$ (or $1\text{ mm}$), and must be less than $50.0\text{ cm}$. A typical value is around $40.0\text{ cm}$.

Key Takeaways

• All length readings taken from a standard metre rule must include one decimal place ($0.1\text{ cm}$ precision), including trailing zeros.
• The distance $d$ is measured relative to the pivot, not from the zero end of the rule.

Common Mistakes

• Forgetting to record the measurement to $0.1\text{ cm}$ (e.g. writing "40" instead of "40.0").
• Recording the mark on the rule instead of the distance $d$ from the pivot.

Things to Be Careful About

• Ensure the rule is balanced horizontally before taking the reading.
• Avoid parallax error by looking directly perpendicular to the ruler markings when reading the position of the centre of the slotted mass.

Walkthrough

To find the exact position of balance on a pivot, the candidate gently slides or shifts the movable mass back and forth in small increments until the metre rule rests horizontally without tilting to either side.

Key Takeaways

• Finding an equilibrium point practically requires fine adjustment around the balance position.
• Clearly describe the action taken ("moving backwards and forwards").

Common Mistakes

• Vague statements such as "just put it on the rule" or "look at it".

Things to Be Careful About

• Ensure the phrase conveys an iterative adjustment (moving back and forth) to pinpoint the balance position.

Walkthrough

1. For each added mass $m$ ($20\text{ g}, 30\text{ g}, 40\text{ g}, 50\text{ g}, 60\text{ g}$), the position of balance is found, and the distance $d$ from the pivot is measured and recorded.
2. As mass $m$ increases, the required distance $d$ to balance the clockwise and anticlockwise moments must decrease ($m \times d = \text{constant}$ approximately).
3. Every entry in the $d\text{ / cm}$ column must be recorded to the nearest $0.1\text{ cm}$ (e.g. $40.0$, $28.5$, etc.).
4. For each row, calculate $1/d$: $$\frac{1}{d} = \frac{1}{40.0} = 0.025\text{ cm}^{-1}$$
5. Record all $1/d$ values consistently to 2 significant figures (or 3 significant figures matching the precision of $d$).

Key Takeaways

• Increasing load requires a shorter distance from the pivot to maintain equilibrium of moments.
• Precision rule: all raw length readings must have consistent decimal places ($0.1\text{ cm}$). Derived quantities like $1/d$ must maintain appropriate significant figures (typically 2 or 3 s.f.).

Common Mistakes

• Inconsistent number of significant figures in the $1/d$ column (e.g. mixing $0.025$ and $0.0350877$).
• Omitting the ".0" for whole-number readings.

Things to Be Careful About

• Ensure that calculated values are rounded correctly to 2 significant figures without truncation errors.

Walkthrough

The pivot is positioned at the $25.0\text{ cm}$ mark on the $100\text{ cm}$ rule. The maximum possible distance from the pivot to the long end of the rule is:

To balance the fixed anticlockwise moment with a smaller mass of only $10\text{ g}$, the required moment arm $d$ would exceed $75\text{ cm}$ (e.g. roughly $2 \times 40.0\text{ cm} = 80\text{ cm}$). Therefore, the mass would fall off the end of the rule before balance could be achieved.

Key Takeaways

• Practical apparatus has physical boundary constraints (here, the length of the metre rule beyond the pivot).
• Smaller loads require larger lever arms to provide the necessary counteracting moment.

Common Mistakes

• Vague explanations like "the mass is too small" or "it won't balance" without explaining why (i.e. the balance point is beyond the end of the rule).

Things to Be Careful About

• Explicitly state that the position is beyond the physical length/end of the ruler (or $d > 75\text{ cm}$).

Walkthrough

To plot the graph correctly:

1. Axes and Units: Clearly write $1/d\text{ / cm}^{-1}$ on the vertical (y) axis and $m\text{ / g}$ on the horizontal (x) axis.
2. Origin & Scale: The question explicitly requires starting axes from $(0, 0)$. Choose convenient scales:
   • x-axis: $0$ to $70\text{ g}$, where 1 major grid division ($2\text{ cm}$) represents $10\text{ g}$.
   • y-axis: $0$ to $0.08\text{ cm}^{-1}$, where 1 major grid division represents $0.01\text{ cm}^{-1}$.
     Both scales are linear, easy to read, and occupy well over half of the provided graph grid.
3. Plotting: Mark each point with a small neat cross ($+$ or $\times$) or a dot in a circle, positioned within $\pm 0.5$ small square of its correct value.
4. Line of Best Fit: Use a clear ruler to draw a straight line that has a balanced distribution of points on either side and follows the overall trend.

Key Takeaways

• Always ensure scales use more than half the grid in both directions and avoid awkward ratios (such as 3s, 7s, etc.).
• When instructed to start from $(0, 0)$, the origin must be $(0, 0)$.
• Best-fit lines should be thin, straight (not multiple lines or sketchy), and reflect the trend of all points.

Common Mistakes

• Using non-linear scales or awkward scales (e.g. 1 major division = 3 units).
• Forgetting to include units on axis labels.
• Drawing a thick, feathered, or point-to-point line.

Things to Be Careful About

• Ensure the line is drawn using a sharp pencil and a 30 cm ruler.

Walkthrough

1. Mark the Triangle: Draw a large right-angled triangle on the graph grid with its hypotenuse lying on the best-fit line. The horizontal base $\Delta m$ must span at least half the length of the drawn line ($\Delta m \ge 20\text{ g}$).
2. Read Coordinates: Read the coordinates $(x_1, y_1)$ and $(x_2, y_2)$ directly from the line of best fit (do not use raw data points unless they lie precisely on the line).
3. Compute Gradient: $$G = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\Delta (1/d)}{\Delta m}$$ Substitute the values and compute $G$ (typically around $0.0010 - 0.0012\text{ cm}^{-1}\text{g}^{-1}$).

Key Takeaways

• Gradient is calculated as $\frac{\Delta y}{\Delta x}$ from points on the line, never as a single coordinate $y/x$.
• Use a large triangle covering at least half the drawn line to minimize read-off errors.

Common Mistakes

• Using a triangle that is too small ($< 50\%$ of the line length).
• Inverting the ratio (calculating $\frac{\Delta x}{\Delta y}$ instead of $\frac{\Delta y}{\Delta x}$).
• Choosing original data points that do not lie on the line.

Things to Be Careful About

• Clearly indicate the chosen points on the graph by dashed lines or coordinates.

Walkthrough

1. Use the equation provided in the question: $$M = 160 - \frac{0.040}{G}$$
2. Substitute your calculated gradient $G$ from part (b)(ii): $$\frac{0.040}{0.00108} \approx 37.0$$
3. Subtract from 160: $$M = 160 - 37.0 = 123\text{ g}$$
4. The expected value of $M$ for a typical wooden metre rule falls within the range $110\text{ g} \leqslant M \leqslant 130\text{ g}$.

Key Takeaways

• Accurately substitute values into empirical or derived physics relations.
• Quote the final answer to an appropriate number of significant figures (2 or 3 s.f.).

Common Mistakes

• Arithmetic or rounding errors when computing $\frac{0.040}{G}$.
• Omitting the unit (grams).

Things to Be Careful About

• Ensure error-carried-forward from the gradient value in (b)(ii) is maintained if $G$ was slightly different.