2023 IJMB paper 3 real

1A. Dynamical Friction is the force tending to prevent one body sliding over another. In order to move an object from rest along a surface, a force of inertia needs to be overcome. This results from static friction between the body and the surface on which it rests. If once moving the body is to continue sliding along the surface, a force F needs to be applied to overcome the dynamic friction. This force is smaller than the initial force needed to overcome the inertia caused by static friction as the degree of contact between the sliding body and the surface beneath is less. Neither of these forces are dependent on the area in contact, instead being determined primarily by the types of materials and the roughness of the surfaces in contact. They are also proportional to the force that is pushing the surfaces together in a plane perpendicular to that of the surfaces themselves.

Assume the empirical formula that relates dynamic frictional force F and the inclination θ, can be expressed as:

$$F = Ae^{-\mu\theta} \text{ ......................................(1)}$$

Where A and μ are constant. Some experimental results to verify equation (1) are given in table 1.

Table 1

| F/N | 5.52 | 5.53 | 4.56 | 4.09 | 3.45 | 1.58 | 0.61 | 0.58 | 0.29 | 0.21 | 0.14 | 0.10 |

|—|---|—|---|—|---|—|---|—|---|—|---|—|

| θ/° | 0.45 | 0.67 | 1.01 | 1.24 | 1.87 | 4.23 | 6.12 | 7.28 | 9.41 | 11.22 | 11.6 | 12.14 |

a. (i) Transform equation (1) into a straight line graph to find the values of the constants A and μ.

(ii). Prepare a composite table with all the necessary parameters to determine the constants A and μ.

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(iii). Determine the constant A and μ graphically.

(iv). From your graph, find the value of F when θ = 5.4°.

(v). Also use equation (1) to find the value of F when θ = 5.4°. Compare your answer in (iv) above.

b. Given that: x = 45.4 ± 1.23 and y = 4.2 ± 0.07. Find the error in calculating the value of

$$m = x^2 - y$$

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2A. You are provided with beam balance, meter rule, knife edge and some slotted masses. Balance the uniform meter rule and determine its center of gravity (c.g). An unknown mass $M_o$ is suspended at the end point 2 cm from B as shown in figure 1.

(Figure 1 — diagram showing a meter rule balanced on a pivot, with M₁ on the left side at point A (100 cm end) and M₀ suspended 2 cm from B on the right, with C.G marked in the middle)

i. Suspend another mass $M_1$ = 10g at the other side A of the rule, such that by adjusting $M_1$ the meter rule balances horizontally at a distance x cm from the c.g. Read and record the corresponding value of x. Repeat the whole procedure by increasing $M_1$ = 20, 25, 28, 32 and 38 g respectively. In each case read and record the corresponding values of x. Prepare a composite table.

ii. Plot a graph of $M_1$ against $x^{-1}$. Find the slope S of the graph with its appropriate unit.

iii. If $M_1$ is related to $M_o$ according to the relation, $M_1 = \dfrac{M_o(c.g - 2)}{x}$, from your slope, find the value of $M_o$.

iv. State any two precautions taken to ensure accurate results.

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3A. Use the circuit in figure 2 as a guide to carry out the following instructions:

Connect the circuit as shown in figure 2 with E = 4 or 4.5 volts, R is a variable resistor, X is an unknown resistor, V and A are voltmeter and ammeter respectively. Move the slider to the one end such that the voltmeter reading is in its lowest point. Read and record this lowest voltage and its corresponding current reading. Repeat the procedure by slightly adjusting the variable resistor to obtain five more successive values of Voltmeter and its corresponding ammeter readings respectively. Prepare a composite table. Plot a graph of ln(i) against ln(V). Find the slope S of the graph. What is the physical meaning of the slope S? Hence, Evaluate S⁻¹.

(Figure 2 — circuit diagram showing EMF source E, resistor R, unknown resistor X, voltmeter V and ammeter A)

(ii). State any two precautions taken to ensure accurate results.

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COMPLETE SOLUTIONS

QUESTION 1A

Part a(i) — Linearization of the equation

Given: $F = Ae^{-\mu\theta}$ …(1)

Taking natural logarithm of both sides:

$$\ln F = \ln A + \ln(e^{-\mu\theta})$$

$$\boxed{\ln F = \ln A - \mu\theta}$$

Comparing with Y = C + mX (straight line):

• Y = ln F

• X = θ

• Intercept C = ln A

• Slope m = −μ

So plotting ln F (vertical axis) against θ (horizontal axis) gives a straight line with:

• Y-intercept = ln A → A = e^(intercept)

• Slope = −μ → μ = −slope

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Part a(ii) — Composite Table

Computing ln F for each value:

| F (N) | θ (°) | ln F |

|—|---|—|

| 5.52 | 0.45 | 1.708 |

| 5.53 | 0.67 | 1.710 |

| 4.56 | 1.01 | 1.517 |

| 4.09 | 1.24 | 1.408 |

| 3.45 | 1.87 | 1.239 |

| 1.58 | 4.23 | 0.457 |

| 0.61 | 6.12 | −0.494 |

| 0.58 | 7.28 | −0.545 |

| 0.29 | 9.41 | −1.238 |

| 0.21 | 11.22 | −1.561 |

| 0.14 | 11.60 | −1.966 |

| 0.10 | 12.14 | −2.303 |

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Part a(iii) — Determining A and μ graphically

Using two well-separated points from the best-fit line through the data:

Taking points: (0.45, 1.708) and (12.14, −2.303)

Slope (= −μ):

$$\text{slope} = \frac{-2.303 - 1.708}{12.14 - 0.45} = \frac{-4.011}{11.69} = -0.3431$$

$$\boxed{\mu = 0.343 \text{ per degree}}$$

Y-intercept (= ln A):

Using point (0.45, 1.708):

$$\ln A = \ln F + \mu\theta = 1.708 + 0.343 × 0.45 = 1.708 + 0.154 = 1.862$$

$$A = e^{1.862} = \boxed{6.44 \text{ N}}$$

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Part a(iv) — Value of F when θ = 5.4° (from graph)

From the best-fit line at θ = 5.4°:

$$\ln F = 1.862 - 0.343 × 5.4 = 1.862 - 1.852 = 0.010$$

$$\boxed{F = e^{0.010} \approx 1.01 \text{ N}}$$

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Part a(v) — Verification using equation (1)

$$F = Ae^{-\mu\theta} = 6.44 × e^{-0.343 × 5.4}$$

$$= 6.44 × e^{-1.852}$$

$$= 6.44 × 0.1572$$

$$= \boxed{1.01 \text{ N}}$$

Comparison: Both graphical (iv) and analytical (v) methods give F ≈ 1.01 N — results are consistent, confirming the validity of the empirical formula. ✓

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Part b — Error Propagation

Given: x = 45.4 ± 1.23, y = 4.2 ± 0.07

$$m = x^2 - y$$

Nominal value:

$$m = (45.4)^2 - 4.2 = 2061.16 - 4.2 = 2056.96$$

Error in x²:

$$\delta(x^2) = 2x \cdot \delta x = 2 × 45.4 × 1.23 = 111.68$$

Error in m:

$$\delta m = \delta(x^2) + \delta y = 111.68 + 0.07 = 111.75$$

$$\boxed{m = 2056.96 \pm 111.75}$$

The percentage error:

$$= \frac{111.75}{2056.96} × 100 = 5.43\%$$

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QUESTION 2A

Part (i) — Procedure and Composite Table

Theory: Taking moments about the knife edge (pivot at c.g):

$$M_1 \cdot x = M_o(c.g - 2)$$

$$M_1 = \frac{M_o(c.g-2)}{x}$$

This can be rewritten as:

$$M_1 = M_o(c.g - 2) \cdot x^{-1}$$

Which is of the form Y = SX, where S = M_o(c.g − 2) is the slope.

Expected composite table format:

| M₁ (g) | x (cm) | x⁻¹ (cm⁻¹) |

|—|---|—|

| 10 | x₁ | 1/x₁ |

| 20 | x₂ | 1/x₂ |

| 25 | x₃ | 1/x₃ |

| 28 | x₄ | 1/x₄ |

| 32 | x₅ | 1/x₅ |

| 38 | x₆ | 1/x₆ |

(Actual x values to be measured experimentally and recorded)

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Part (ii) — Graph of M₁ against x⁻¹

Plot M₁ (g) on the vertical axis against x⁻¹ (cm⁻¹) on the horizontal axis.

The graph passes through the origin (y-intercept = 0).

Slope:

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The slope represents: $S = M_o(c.g - 2)$

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Part (iii) — Finding M₀

From the relation:

$$M_1 = \frac{M_o(c.g - 2)}{x} = M_o(c.g-2) \cdot x^{-1}$$

Therefore slope S = M_o(c.g − 2)

$$M_o = \frac{S}{(c.g - 2)}$$

where c.g is the measured center of gravity position (in cm from end A), read from the balanced meter rule before suspending M₀.

Example: If c.g = 50 cm (uniform rule) and S = 480 g·cm:

$$M_o = \frac{480}{50 - 2} = \frac{480}{48} = \boxed{10 \text{ g}}$$

(Actual value depends on experimentally obtained slope)

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Part (iv) — Precautions

1. Ensure the meter rule is perfectly horizontal before taking any reading — use a spirit level if available

2. Avoid parallax error when reading the position x on the meter rule — read perpendicularly

3. Ensure the knife edge (pivot) is sharp and positioned exactly at the c.g

4. Allow the rule to fully stabilize before recording each balance position

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QUESTION 3A

Theory

For a circuit with EMF E, variable resistor R, and unknown resistor X:

If the relationship between current i and voltage V follows a power law:

$$i = kV^n$$

Taking natural log:

$$\ln i = \ln k + n \ln V$$

This is linear: ln(i) vs ln(V) gives slope = n and intercept = ln k

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Procedure

1. Connect circuit as in Figure 2 with E = 4 or 4.5 V

2. Set slider to give minimum voltage reading

3. Record V (voltmeter) and i (ammeter) simultaneously

4. Adjust slider to get 5 more readings

5. Record all 6 pairs of (V, i) readings

Expected composite table:

| V (volts) | i (amperes) | ln V | ln i |

|—|---|—|---|

| V₁ | i₁ | ln V₁ | ln i₁ |

| V₂ | i₂ | ln V₂ | ln i₂ |

| V₃ | i₃ | ln V₃ | ln i₃ |

| V₄ | i₄ | ln V₄ | ln i₄ |

| V₅ | i₅ | ln V₅ | ln i₅ |

| V₆ | i₆ | ln V₆ | ln i₆ |

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Graph Analysis

Plot: ln(i) on vertical axis vs ln(V) on horizontal axis

Slope:

$$S = \frac{\Delta \ln i}{\Delta \ln V} = n$$

Physical meaning of slope S:

The slope S represents the power index n in the relationship i = kVⁿ. It describes how the current varies with voltage across the unknown resistor X.

• If S = 1: resistor X is ohmic (linear)

• If S ≠ 1: X is non-ohmic (non-linear device)

S⁻¹:

$$S^{-1} = \frac{1}{n} = \frac{\Delta \ln V}{\Delta \ln i}$$

This represents the reciprocal of the power index, indicating how voltage responds per unit fractional change in current.

For an ohmic resistor: S = 1, so S⁻¹ = 1

$$\boxed{S^{-1} = \frac{1}{S}}$$

(Actual numerical value depends on experimentally plotted graph)

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Part (ii) — Precautions

1. Zero error: Check and correct for zero error on both voltmeter and ammeter before starting

2. Parallax error: Read meters perpendicularly to avoid parallax errors in readings

3. Allow readings to stabilize before recording — avoid taking readings while slider is still being adjusted

4. Do not exceed the rated current/voltage of the components to avoid damage