9702/34

Physics 9702/34October/November 2013

Cambridge AS Level · Advanced Practical Skills · worked solutions for every part, with the mark scheme

2
questions
40
marks
120
minutes

Topics Manipulation, Measurement and Observation · Presentation of Data and Observations · Analysis, Conclusions and Evaluation

Q1Manipulation, Measurement and ObservationPresentation of Data and ObservationsAnalysis, Conclusions and EvaluationFree sample

In this experiment, you will investigate how the forces supporting a wooden strip change as the position of a suspended mass is moved.

(a)
(i)

Suspend the single spring and the mass hanger from the rod of one of the clamps. Measure and record the height h1h_1 of the mass hanger above the bench, as shown in Fig. 1.1.

h1h_1 = ______

DifficultyEasy
Worked solution

Answer

Measure h1h_1 vertically from the bench to the bottom of the mass hanger using a ruler (eye level with the scale).

Example (to nearest 1 mm1\ \text{mm}):

h1=32.0 cmh_1 = 32.0\ \text{cm}
Final answer

h1 = 32.0 cm (example, to nearest 1 mm)

Detailed explanation

Background Concept

A height measurement is a length measurement, so you use a ruler/metre rule and quote the value to the smallest reliable division (typically 1 mm1\ \text{mm} on a standard rule). The dominant issue is often parallax: if your eye is not level with the mark, the reading appears shifted.

Understanding the Question

You are asked to measure h1h_1, the vertical distance from the bench surface to the bottom of the mass hanger when only the hanger is attached to the spring.

Approach

  1. Set up the spring and hanger exactly as in the diagram.
  2. Place a ruler vertically with its zero at bench level (or measure from bench to hanger directly).
  3. Read the position of the bottom of the hanger, with your eye level with the scale.
  4. Record h1h_1 with a unit and appropriate precision.

Step-by-Step Reasoning

  • Suspend the spring from the clamp rod.
  • Attach the mass hanger and allow it to come to rest.
  • Hold the ruler next to the hanger so it is vertical.
  • Identify the reference point (bottom of the hanger).
  • Read the ruler at that point, ensuring your eye is level with the mark.
  • Record as, e.g. h1=32.0 cmh_1 = 32.0\ \text{cm} (or in mm\text{mm}), matching the smallest division.

Key Takeaways

  • Always state the reference points for a length.
  • Avoid parallax and quote a sensible precision.

Common Mistakes

  • Measuring to the top of the hanger or to the spring end instead of the bottom of the hanger.
  • Not including units.
  • Reading the scale with your eye above/below the mark (parallax error).

Things to Be Careful About

  • Keep the ruler vertical (otherwise you measure a slanted distance).
  • Let the hanger come to rest before reading.
  • Record to the nearest 1 mm1\ \text{mm} (or nearest division of the ruler you actually use).
Techniques used
set up the apparatus as shownmeasure a vertical height with a ruleravoid parallax when taking a scale readingrecord a reading to the instrument precision
(ii)

Add the 50 g slotted mass to the hanger. Measure and record the new height h2h_2 of the mass hanger above the bench.

h2h_2 = ______

1M
DifficultyEasy
Worked solution

Answer

Add the 50 g50\ \text{g} mass, allow the spring to come to rest, then measure h2h_2 in the same way as h1h_1.

Example (to nearest 1 mm1\ \text{mm}):

h2=30.0 cmh_2 = 30.0\ \text{cm}
Final answer

h2 = 30.0 cm (example, to nearest 1 mm)

Detailed explanation

Background Concept

Adding mass increases the load on the spring so the spring extends and the hanger sits lower, so the height above the bench decreases. The measurement technique is identical to part (a)(i).

Understanding the Question

After adding a 50 g50\ \text{g} slotted mass to the hanger, you must measure the new height h2h_2 of the bottom of the hanger above the bench.

Approach

Use the same reference points and the same ruler position as for h1h_1 so that h1h2h_1 - h_2 is meaningful.

Step-by-Step Reasoning

  • Carefully place the 50 g50\ \text{g} mass onto the hanger.
  • Wait until oscillations have stopped (or gently damp them without changing the equilibrium position).
  • Measure from the bench to the bottom of the hanger.
  • Record h2h_2 to the same precision as h1h_1 (e.g. nearest 0.1 cm0.1\ \text{cm}).

Key Takeaways

  • Consistency of method (same reference points, same precision) is vital for differences like h1h2h_1-h_2.

Common Mistakes

  • Measuring h2h_2 from a different point (e.g. top of hanger).
  • Recording h1h_1 and h2h_2 to different decimal places.

Things to Be Careful About

  • Ensure the ruler has not moved relative to the bench reference.
  • Read the scale at eye level to reduce parallax.
Techniques used
add a known mass without changing the measuring referencemeasure a new equilibrium positionrecord a reading to consistent precision
(iii)

Calculate the change in height CC, where C=h1h2C = h_1 - h_2.

CC = ______

DifficultyEasy
Worked solution

Working

C=h1h2C = h_1 - h_2

Using the example values h1=32.0 cmh_1 = 32.0\ \text{cm} and h2=30.0 cmh_2 = 30.0\ \text{cm},

C=32.030.0=2.0 cmC = 32.0 - 30.0 = 2.0\ \text{cm}

Answer

C=2.0 cmC = 2.0\ \text{cm}
Final answer

C = 2.0 cm (example)

Detailed explanation

Background Concept

When you calculate a change in a measured length, you subtract the two readings. The unit stays as a length unit (cm or mm). For exam tables, you should quote the calculated change to a sensible precision consistent with the raw readings.

Understanding the Question

You have measured h1h_1 (before adding the 50 g50\ \text{g} mass) and h2h_2 (after adding it). You must calculate:

C=h1h2C = h_1 - h_2

Approach

Subtract the later height from the initial height. Since adding mass makes the hanger lower, typically h2<h1h_2 < h_1 so CC should be positive.

Step-by-Step Reasoning

  • Write the definition C=h1h2C = h_1 - h_2.
  • Substitute your measured values.
  • Perform the subtraction.
  • Keep the unit (cm, if h1h_1 and h2h_2 were in cm).
  • Quote to appropriate precision (often the same decimal place as h1h_1 and h2h_2).

Key Takeaways

  • A difference keeps the same unit as the original quantity.
  • The sign of the result should make physical sense.

Common Mistakes

  • Reversing the subtraction and getting a negative value.
  • Dropping the unit for CC.

Things to Be Careful About

  • Use consistent units for h1h_1 and h2h_2 before subtracting.
  • Round only at the end of the calculation.
Techniques used
calculate a difference of two measured valueskeep consistent decimal places in a derived quantityinclude the correct unit for a derived quantity
(b)
(i)

Assemble the apparatus as shown in Fig. 1.2, with the mass M near to the middle of the wooden strip.

DifficultyEasy
Worked solution

Answer

Set up two vertical stands with clamps and suspend the wooden strip by two springs as in Fig. 1.2, with the mass MM hung from a string loop positioned near the middle of the strip.

Final answer

Apparatus assembled with M near the middle (as Fig. 1.2)

Detailed explanation

Background Concept

In practical work, correct assembly matters because misalignment (tilted strip, angled springs) changes the forces and therefore changes measured spring extensions. The diagram shows a strip supported at two points by springs; the load position is adjustable.

Understanding the Question

You must assemble the apparatus as shown: two springs support the wooden strip, and a mass MM hangs from a string loop that can slide along the strip. Initially the mass should be near the middle so both springs share the load.

Approach

Replicate the geometry in the diagram: springs vertical, attachment points secure, and ensure you can measure xx from the left-hand hole to the string loop.

Step-by-Step Reasoning

  • Fix two stands on the bench with clamps holding horizontal rods.
  • Suspend the left spring from one rod and connect it to the left side of the strip (via the hanger as shown).
  • Suspend the right spring from the other rod and connect it to the right side of the strip.
  • Place the loop/string around the strip and hang mass MM from it.
  • Slide the loop so MM is close to the centre of the strip.

Key Takeaways

  • A correct initial setup reduces systematic errors later.

Common Mistakes

  • Attaching the string loop at an angle so the force on the strip is not vertical.
  • Springs not in line with the attachment points.

Things to Be Careful About

  • Ensure the strip can move freely and is not touching the stands.
  • Ensure the string loop can slide smoothly but does not snag.
Techniques used
assemble apparatus according to a diagramensure correct positioning of a movable loadcheck that support points are secure
(ii)

Adjust the apparatus so that the wooden strip is horizontal and the springs are vertical.

DifficultyEasy
Worked solution

Answer

Adjust the clamp/rod positions until the wooden strip is horizontal (use a set square/spirit level if available) and ensure both springs hang vertically before taking readings.

Final answer

Strip horizontal; springs vertical

Detailed explanation

Background Concept

The relationship you later test assumes the forces are vertical and that the strip is in static equilibrium. If the strip is tilted, the measured spring lengths no longer represent the intended force balance in the same way.

Understanding the Question

Before measuring xx, LAL_A and LBL_B, you must make sure the strip is level and the springs are vertical.

Approach

Make small adjustments to the clamp heights/rod positions. Confirm the strip is horizontal by eye (or with a set square/spirit level) and confirm springs are not angled.

Step-by-Step Reasoning

  • Slide the stands slightly or adjust clamp heights so the strip looks level.
  • If available, place a set square against the strip to check horizontality.
  • Ensure each spring hangs straight down (not rubbing against anything).
  • Wait for oscillations to stop.

Key Takeaways

  • Good alignment reduces systematic error and improves consistency.

Common Mistakes

  • Taking measurements while the strip is still oscillating.
  • Leaving a spring slightly angled, which changes its effective extension.

Things to Be Careful About

  • Do not change the position of the mass along the strip while levelling.
  • Ensure both attachment points are at the same horizontal level once adjusted.
Techniques used
adjust clamp heights to level the stripcheck vertical alignment of springsverify the equilibrium condition before measuring
(iii)

Measure and record the distance xx from the left-hand hole in the wooden strip to the string loop supporting M, as shown in Fig. 1.2.

xx = ______

DifficultyEasy
Worked solution

Answer

Measure xx from the left-hand hole to the string loop along the strip using a ruler aligned with the strip.

Example:

x=15.0 cmx = 15.0\ \text{cm}
Final answer

x = 15.0 cm (example)

Detailed explanation

Background Concept

A distance along a straight object should be measured with the ruler aligned parallel to that object. If the ruler is at an angle, you introduce a cosine error (measuring a longer diagonal rather than the intended horizontal distance).

Understanding the Question

You need the distance xx from the left-hand hole in the strip to the position of the string loop supporting MM.

Approach

Use the left hole as the zero/reference point, align the ruler with the length of the strip, and read the position of the loop.

Step-by-Step Reasoning

  • Place the zero of the ruler at the centre of the left-hand hole (or align a known mark with the hole and subtract).
  • Ensure the ruler edge is parallel to the strip.
  • Read the position of the string loop (choose a consistent point on the loop, e.g. its centre).
  • Record xx to the nearest mm (or nearest 0.1 cm).

Key Takeaways

  • Define and keep the same reference points for every reading of xx.

Common Mistakes

  • Measuring from the end of the strip rather than the left-hand hole.
  • Measuring to the edge of the loop inconsistently as it shifts.

Things to Be Careful About

  • Ensure the strip is horizontal before measuring, as specified in (b)(ii).
  • Keep the loop stationary while reading.
Techniques used
measure a horizontal distance between two reference pointsalign a ruler with the strip to avoid cosine errorrecord readings with appropriate precision
(iv)

Measure and record the length LAL_A of the coiled section of the left-hand spring, and the length LBL_B of the coiled section of the right-hand spring.

LAL_A = ______
LBL_B = ______

1M
DifficultyMedium-Easy
Worked solution

Answer

Measure the length of the coiled section of each spring with a ruler.

Example:

LA=10.6 cmL_A = 10.6\ \text{cm} LB=10.2 cmL_B = 10.2\ \text{cm}
Final answer

LA = 10.6 cm, LB = 10.2 cm (example)

Detailed explanation

Background Concept

The spring’s extension is inferred from how the length of its active (coiled) section changes. To compare two springs, you must measure each in the same way (same definition of the ends of the coiled region).

Understanding the Question

You must measure and record:

  • LAL_A: coiled length of the left spring
  • LBL_B: coiled length of the right spring

Approach

For each spring, identify the top and bottom of the coiled section and measure the distance between these points with the ruler held close and parallel to the spring.

Step-by-Step Reasoning

  • Look at the spring and identify where the tight coils start and where they end (exclude hooks if the question indicates only the coiled section).
  • Hold the ruler next to the spring.
  • Read the positions of the two ends of the coiled section and subtract to get LL.
  • Record LAL_A and LBL_B to the nearest mm (or 0.1 cm) consistently.

Key Takeaways

  • Measuring a defined section (coils only) improves repeatability.

Common Mistakes

  • Including the hook/non-coiled end parts when the instruction says “coiled section”.
  • Ruler not aligned with the spring (gives a slightly longer length if diagonal).

Things to Be Careful About

  • Keep the spring vertical (as required in (b)(ii)) to avoid measuring a slanted length.
  • Read at eye level to reduce parallax.
Techniques used
identify the coiled section of a spring as the measurement regionmeasure the length between two points on a spring with a rulerrecord two related measurements with consistent precision
(c)

Reposition the string loop supporting M.
Repeat (b)(ii), (b)(iii) and (b)(iv) until you have six sets of values of xx, LAL_A and LBL_B.
Include values for (LALB)(L_A - L_B) and for (LALB)C\frac{(L_A - L_B)}{C} in your table.

10M
DifficultyMedium
Worked solution

Answer

Take six sets of readings with different positions of the loop (different xx), each time making the strip horizontal and springs vertical, and record xx, LAL_A, LBL_B.

Use one results table including calculated columns (LALB)(L_A-L_B) and (LALB)C\dfrac{(L_A-L_B)}{C}.

Example table (with C=2.0 cmC = 2.0\ \text{cm}):

x/cmx / \text{cm}LA/cmL_A / \text{cm}LB/cmL_B / \text{cm}(LALB)/cm(L_A-L_B) / \text{cm}(LALB)C\dfrac{(L_A-L_B)}{C}
5.010.410.40.00.00
10.010.510.30.20.10
15.010.610.20.40.20
20.010.710.10.60.30
25.010.810.00.80.40
30.010.99.91.00.50
Final answer

Single clear table with 6 sets of x, LA, LB plus (LA−LB) and (LA−LB)/C (student-dependent).

Detailed explanation

Background Concept

In a practical investigation, you must:

  • choose an independent variable to vary (here xx),
  • measure dependent variables (here LAL_A and LBL_B),
  • take enough readings (here six sets) over a suitable range,
  • present data clearly and calculate any required derived quantities.

A good results table has:

  • clear column headings with quantity and unit,
  • consistent decimal places within a column,
  • derived columns calculated correctly.

Understanding the Question

You must reposition the loop supporting MM and, each time:

  • make the strip horizontal and springs vertical,
  • measure xx, LAL_A, LBL_B,
    until you have six sets.

You must also include two calculated quantities:

(LALB)(L_A - L_B)

and

(LALB)C\frac{(L_A - L_B)}{C}

where CC came from part (a).

Approach

  1. Choose six values of xx spanning a wide range along the strip (not clustered).
  2. For each xx, re-level the strip and ensure springs are vertical.
  3. Measure LAL_A and LBL_B consistently (coiled section only).
  4. Calculate (LALB)(L_A-L_B) and then divide by CC to get the final column.
  5. Record everything in one table with consistent units and precision.

Step-by-Step Reasoning

  • Pick xx values (e.g. evenly spaced along the usable region of the strip).
  • For each xx:
    • adjust until the strip is horizontal;
    • read xx;
    • read LAL_A and LBL_B.
  • Compute the difference:
(LALB)=LALB(L_A-L_B) = L_A - L_B
  • Compute the ratio:
(LALB)C\frac{(L_A-L_B)}{C}
  • Check units:
    • LAL_A and LBL_B are lengths,
    • CC is a length,
    • so (LALB)C\dfrac{(L_A-L_B)}{C} is dimensionless (no unit).

Key Takeaways

  • Practical marks are often earned by presentation: clear headings, units, and consistent precision.
  • Derived columns should be calculated correctly from measured values.

Common Mistakes

  • Not taking six sets of readings.
  • Missing units in headings (e.g. writing just xx rather than x/cmx/\text{cm}).
  • Inconsistent decimal places within a column.
  • Forgetting to include (LALB)(L_A-L_B) and/or (LALB)C\dfrac{(L_A-L_B)}{C}.

Things to Be Careful About

  • Use the same unit for all lengths (e.g. all in cm) before calculating differences/ratios.
  • Re-level the strip each time; otherwise changes in LAL_A and LBL_B include tilt effects.
  • Do not round intermediate values too aggressively; round appropriately at the end.
Techniques used
vary the independent variable across a suitable rangerecord repeated sets of readings in a single tablecalculate derived columns from measured quantitiesuse consistent significant figures and units in a table
(d)
(i)

Plot a graph of (LALB)C\frac{(L_A - L_B)}{C} on the yy-axis against xx on the xx-axis.

3M
DifficultyMedium-Easy
Worked solution

Answer

Plot y=(LALB)Cy = \dfrac{(L_A-L_B)}{C} (no unit) on the yy-axis against xx (with unit, e.g. cm\text{cm}) on the xx-axis using a suitable scale (at least half the grid in each direction) and plot all six points accurately.

Final answer

Graph of (LA−LB)/C vs x plotted (student-dependent).

Detailed explanation

Background Concept

A good graph is one where the axes are clearly labelled and the scale lets you read values accurately. For Cambridge practical papers, credit is typically given for:

  • correct choice of variables on axes,
  • sensible, simple scales,
  • accurate plotting with small crosses,
  • using most of the grid.

Understanding the Question

You must plot a graph with:

  • yy-axis: (LALB)C\dfrac{(L_A-L_B)}{C}
  • xx-axis: xx
    using your six data points.

Approach

Use the table from part (c). Decide your axis ranges from the min/max values. Choose scales such as 1 large square = 1 cm on xx or 0.05 on yy (whatever fits your data well).

Step-by-Step Reasoning

  • Draw axes and label:
    • horizontal: x/cmx/\text{cm} (or whatever unit you used),
    • vertical: (LALB)C\dfrac{(L_A-L_B)}{C}.
  • Choose a scale so your plotted points occupy a large area.
  • Plot each of the six points with a small cross (×\times).
  • Check that each point matches the correct (x,y)(x, y) pair.

Key Takeaways

  • Axes labels must include the quantity and unit (unit omitted if dimensionless).
  • Scales should be easy to use (1, 2, 5, 10 style steps).

Common Mistakes

  • Swapping axes (plotting xx on the yy-axis).
  • Missing units on the xx-axis.
  • Using an awkward scale (e.g. 3 units per large square) or using only a small corner of the graph.

Things to Be Careful About

  • (LALB)C\dfrac{(L_A-L_B)}{C} is dimensionless, so do not add a unit on that axis.
  • Plot points precisely; do not draw large dots.
  • Ensure you plotted the ratio column, not (LALB)(L_A-L_B) itself.
Techniques used
select suitable axes and scaleslabel axes with quantities and unitsplot experimental points accuratelyuse data from a results table to construct a graph
(ii)

Draw the straight line of best fit.

1M
DifficultyMedium-Easy
Worked solution

Answer

Draw a single straight line of best fit through the data so that the points are balanced (approximately equal scatter above and below the line).

Final answer

Straight line of best fit drawn.

Detailed explanation

Background Concept

A line of best fit represents the overall trend in data with random scatter. It should not be forced through every point; instead it should reflect the most likely linear relationship.

Understanding the Question

After plotting the six points, you must draw the straight line that best represents them.

Approach

Use a ruler to draw a straight line that passes through the middle of the scatter, aiming for roughly equal numbers of points above and below. Do not simply join neighbouring points.

Step-by-Step Reasoning

  • Place the ruler so the line runs through the general centre of the plotted points.
  • Adjust so the deviations look balanced.
  • Draw one clean straight line across most of the graph.
  • If one point is clearly anomalous, the line should still represent the other points.

Key Takeaways

  • Best-fit means balanced scatter, not connecting-the-dots.

Common Mistakes

  • Joining points with a zig-zag line.
  • Forcing the line through the origin when the data do not support it.

Things to Be Careful About

  • Draw the line long enough to allow accurate gradient/intercept reading.
  • Use a sharp pencil for accuracy.
Techniques used
draw a straight line of best fit through plotted pointsbalance points above and below the best-fit lineignore anomalous points when appropriate
(iii)

Determine the gradient and yy-intercept of this line.

gradient = ______
yy-intercept = ______

2M
DifficultyMedium
Worked solution

Working

Choose two well-separated points on the best-fit line (not necessarily data points), e.g.

(x1,y1)=(10.0 cm, 0.10),(x2,y2)=(30.0 cm, 0.50)(x_1, y_1) = (10.0\ \text{cm},\ 0.10),\quad (x_2, y_2) = (30.0\ \text{cm},\ 0.50)

Gradient:

gradient=ΔyΔx=0.500.1030.010.0=0.4020.0=0.020 cm1\text{gradient} = \frac{\Delta y}{\Delta x} = \frac{0.50-0.10}{30.0-10.0} = \frac{0.40}{20.0} = 0.020\ \text{cm}^{-1}

yy-intercept (from y=mx+cy = mx + c using (10.0,0.10)(10.0,0.10)):

0.10=(0.020)(10.0)+cc=0.100.10 = (0.020)(10.0) + c \Rightarrow c = -0.10

Answer

gradient=0.020 cm1\text{gradient} = 0.020\ \text{cm}^{-1} y-intercept=0.10y\text{-intercept} = -0.10
Final answer

gradient = 0.020 cm^-1, y-intercept = -0.10 (example)

Detailed explanation

Background Concept

For a straight-line graph of yy against xx, the equation is:

y=mx+cy = mx + c

where:

  • mm is the gradient m=ΔyΔxm = \dfrac{\Delta y}{\Delta x},
  • cc is the yy-intercept (the value of yy when x=0x=0).

You should find mm using two points far apart on the best-fit line to reduce the effect of reading uncertainties.

Understanding the Question

You must determine:

  • the gradient of your best-fit line,
  • the yy-intercept of your best-fit line,
    from the graph of (LALB)C\dfrac{(L_A-L_B)}{C} against xx.

Approach

  1. Pick two points on the best-fit line that are widely separated.
  2. Read their coordinates carefully.
  3. Compute m=Δy/Δxm = \Delta y/\Delta x.
  4. Find the intercept either by reading where the line crosses the yy-axis, or by using c=ymxc = y - mx.

Step-by-Step Reasoning

  • Draw a large right-angled triangle on the best-fit line.
  • Read x1,y1x_1, y_1 and x2,y2x_2, y_2 at the ends of the triangle.
  • Calculate:
Δy=y2y1,Δx=x2x1\Delta y = y_2 - y_1,\quad \Delta x = x_2 - x_1 gradient=ΔyΔx\text{gradient} = \frac{\Delta y}{\Delta x}
  • Determine the intercept:
    • either read yy when x=0x=0,
    • or use c=y1mx1c = y_1 - m x_1.
  • State the unit of gradient:
    • yy is dimensionless,
    • xx is a length,
    • so gradient has unit (length)1\text{(length)}^{-1} e.g. cm1\text{cm}^{-1}.

Key Takeaways

  • Use the best-fit line, not point-to-point gradients.
  • Use a large triangle to reduce fractional uncertainty.
  • Check units: gradient carries the inverse unit of the xx-axis.

Common Mistakes

  • Using two adjacent plotted points (gives a very uncertain gradient).
  • Calculating gradient as Δx/Δy\Delta x/\Delta y instead of Δy/Δx\Delta y/\Delta x.
  • Forgetting the unit on the gradient.

Things to Be Careful About

  • Do not assume the line passes through the origin.
  • Choose points that lie clearly on the best-fit line and are easy to read from grid intersections.
  • Keep enough significant figures consistent with graph-reading precision.
Techniques used
determine the gradient using two well-separated points on the best-fit lineuse a large triangle to reduce percentage uncertainty in gradientread or calculate the y-intercept from the best-fit line
(e)

The quantities LAL_A, LBL_B, CC and xx are related by the equation

(LALB)C=ax+b\frac{(L_A - L_B)}{C} = ax + b

where aa and bb are constants.

Use your answers from (d)(iii) to determine the values of aa and bb.
Give appropriate units.

aa = ______
bb = ______

2M
DifficultyMedium-Easy
Worked solution

Working

Given

(LALB)C=ax+b\frac{(L_A - L_B)}{C} = ax + b

Comparing with y=mx+cy = mx + c for the graph of y=(LALB)Cy = \dfrac{(L_A-L_B)}{C} against xx:

a=gradient,b=y-intercepta = \text{gradient},\quad b = y\text{-intercept}

Using the values from (d)(iii):

a=0.020 cm1a = 0.020\ \text{cm}^{-1} b=0.10b = -0.10

Answer

a=0.020 cm1a = 0.020\ \text{cm}^{-1} b=0.10 (no unit)b = -0.10\ (\text{no unit})
Final answer

a = gradient (cm^-1), b = y-intercept (no unit); e.g. a = 0.020 cm^-1, b = −0.10

Detailed explanation

Background Concept

If experimental variables satisfy

y=ax+by = ax + b

then a graph of yy against xx is a straight line with:

  • gradient =a= a,
  • intercept =b= b.

Units come from the variables:

  • here y=(LALB)Cy = \dfrac{(L_A-L_B)}{C} is a ratio of lengths, so it is dimensionless,
  • xx is a length,
    so aa has unit (length)1\text{(length)}^{-1} and bb has no unit.

Understanding the Question

You are told the relationship

(LALB)C=ax+b\frac{(L_A - L_B)}{C} = ax + b

and you have already found the gradient and yy-intercept from your graph in (d)(iii). You must now state aa and bb (with units).

Approach

Match your graph variables to the equation:

  • yy-axis is (LALB)C\dfrac{(L_A-L_B)}{C},
  • xx-axis is xx,
    so directly:
  • aa equals the gradient,
  • bb equals the yy-intercept.

Step-by-Step Reasoning

  • From (d)(iii), take your gradient value and write it as aa.
  • Take your yy-intercept and write it as bb.
  • Assign units:
    • if xx was measured in cm, then aa is in cm1\text{cm}^{-1};
    • bb has no unit.

Key Takeaways

  • Constants in a linear equation come directly from the gradient and intercept of the corresponding graph.
  • Unit-checking is a powerful way to confirm your answers.

Common Mistakes

  • Giving bb the same unit as aa.
  • Forgetting that (LALB)C\dfrac{(L_A-L_B)}{C} is dimensionless and writing a unit on the yy-axis/for bb.

Things to Be Careful About

  • Ensure your unit for aa matches the unit you used for xx on the graph (cm vs m).
  • Keep aa and bb to a sensible number of significant figures consistent with your graph-reading precision.
Techniques used
match a straight-line equation to y = mx + cidentify constants from gradient and interceptdeduce units of constants from plotted variables

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  • Q2Manipulation, Measurement and Observation · Presentation of Data and Observations · Analysis, Conclusions and Evaluation20M
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