(a) Explain what is meant by the accuracy of a measured value.
(b) Two solid cubes, A and B, are measured to determine the density of their materials.
Table 1.1 shows the measurements for cube A.
| quantity | measurement |
|---|---|
| length of side | \((1.53\pm 0.01)\) cm |
| mass | \((31.3\pm 0.5)\) g |
(i) Show that the calculated density of the material of cube A is \(8.7 \times 10^3\) kg m\(^{-3}\).
(ii) Calculate the percentage uncertainty in the density of the material of cube A.
(iii) The density of the material of cube B is determined to be \(9.2 \times 10^3\) kg m\(^{-3}\) ± 6%.
State and explain whether cube A and cube B could be made from the same material.
▶️ Answer/Explanation
(a) Accuracy refers to how close the measured value is to the true value of the quantity being measured.
(b)(i) Density = mass/volume = \(31.3 \times 10^{-3}\) kg / \((1.53 \times 10^{-2})^3\) m\(^3\) = 8730 kg/m\(^3\) ≈ \(8.7 \times 10^3\) kg m\(^{-3}\).
(b)(ii) % uncertainty in mass = (0.5/31.3)×100 = 1.6%, in length = (0.01/1.53)×100 = 0.65%. Total % uncertainty = 1.6% + 3×0.65% = 3.55% ≈ 4%.
(b)(iii) Cube A’s density range (8.7±0.3)×10³ kg/m³ overlaps with Cube B’s range (9.2±0.6)×10³ kg/m³, so they could be the same material.
(a) State the principle of moments.
(b) A solid plastic cylinder floats in water. It is used to support one end of a horizontal uniform beam AB as shown in Fig. 2.1.
Fig. 2.1 shows a diagram of the setup (not to scale)
The beam has length 6.0 m and weight 1700 N. The beam is attached to solid ground with a hinge at end A.
The cylinder is floating vertically in the water. The top of the cylinder is attached at its centre to the beam at a horizontal distance of 5.0 m from end A. The cylinder applies a vertical force of 1300 N to the beam.
A person of weight 660 N stands on the beam at point P.
The beam AB is in equilibrium.
(i) By taking moments about end A, determine the distance x from A to P.
(ii) The bottom of the cylinder is submerged in the water to depth y as shown in Fig. 2.2. The beam is still attached to the cylinder but not shown.
The cylinder has mass 11 kg and diameter 0.78 m. The beam exerts a vertical force of 1300 N on the cylinder. The cylinder is in equilibrium.
Show that the upthrust acting on the cylinder is 1400 N.
(iii) The water has density 990 kg m\(^{-3}\).
Calculate the depth y.
(iv) The person can stand anywhere between A and B.
On Fig. 2.3, sketch the variation of the depth of the bottom of the cylinder with the distance of the person from A, for distances between 0 and 6.0 m. Numerical values are not required.
▶️ Answer/Explanation
(a) For rotational equilibrium, the sum of clockwise moments about any point equals the sum of anticlockwise moments about the same point.
(b)(i) Taking moments about A: 1700N × 3.0m (beam’s weight) + 660N × x = 1300N × 5.0m. Solving gives x = 2.1 m.
(b)(ii) Upthrust balances the cylinder’s weight (11kg × 9.81 ≈ 108N) plus the beam’s force (1300N), totaling ≈1400N.
(b)(iii) Using upthrust = ρgV, with V = πr²y = π(0.39)²y, and upthrust=1400N, we get y ≈ 0.30m.
(b)(iv) line with a non-zero value of depth at distance = 0
A line from distance = 0 to distance = 6.0 m with a gradient that is always positive
(a) A truck R of mass 9400 kg moves with constant acceleration in a straight line down a slope, as illustrated in Fig. 3.1.
At point A the speed of the truck is \(13 \, \text{ms}^{-1}\) and at point B the speed of the truck is \(22 \, \text{ms}^{-1}\). A and B are a distance of \(180 \, \text{m}\) apart.
(i) Calculate the acceleration of the truck between A and B.
(ii) Determine the gain in kinetic energy of the truck between A and B.
(b) A short time after passing point B truck R moves in a straight line on horizontal ground. The driver of the truck applies the brakes.
Fig. 3.2 shows the variation with time of the momentum of the truck.
(i) Define force.
(ii) Show that the average resultant force \( F \) acting on truck R between time \( t = 0 \) and \( t = 15 \, \text{s} \) is \(-1.2 \times 10^4 \, \text{N}\).
(iii) An identical truck S has the same initial momentum as truck R. Truck S experiences a constant force equal to the force F in (b)(ii). State and explain whether truck S will take more, less or the same amount of time to come to rest as truck R.
▶️ Answer/Explanation
(a)(i) Using \(v^2 = u^2 + 2as\): \(a = (22^2 – 13^2)/(2×180) = 0.88 \, \text{ms}^{-2}\).
(a)(ii) KE gain = \(\frac{1}{2}m(v^2-u^2) = \frac{1}{2}×9400×(22^2-13^2) = 1.5×10^6 \, \text{J}\).
(b)(i) Force is rate of change of momentum.
(b)(ii) \(F = \Delta p/\Delta t = (2.5×10^4 – 21×10^4)/15 = -1.2×10^4 \, \text{N}\).
(b)(iii) Truck S stops faster. Both have same momentum change, but S experiences larger constant braking force than R’s average force.
A device containing a microwave emitter and receiver is placed in front of a large metal sheet in a vacuum as shown in Fig. 4.1.
The line XY is perpendicular to the metal sheet. The device emits microwaves of frequency 6.3GHz.
(a) When the device is at position P, a stationary wave is formed between the device and the sheet.
Explain how the stationary wave, including the nodes and the antinodes, is formed.
(b) (i) Calculate the wavelength of the microwaves.
(ii) At point P the receiver detects a maximum amplitude of the stationary wave.
The device is moved slowly from point P along the line XY and the receiver detects a series of minimum and maximum amplitudes. The first time a minimum amplitude is detected by the receiver is when the device is at point Q.
Determine the distance between P and Q.
(iii) The intensity of the microwaves emitted by the device is increased. The frequency of the microwaves is unchanged. The device is moved slowly along the line XY from point Q until the next maximum amplitude is detected at point R.
State and explain whether the distance QR is greater than, less than or the same as distance PQ.
▶️ Answer/Explanation
(a) The microwave reflects at the metal sheet. The incident and reflected waves superpose/interfere. At antinodes, constructive interference creates maximum amplitude. At nodes, destructive interference creates minimum/zero amplitude.
(b)(i) Using \(\lambda = c/f\): \(\lambda = 3 \times 10^8 / 6.3 \times 10^9 = 0.048\) m. The wavelength is 0.048 m.
(b)(ii) The distance PQ is \(\lambda/4 = 0.048/4 = 0.012\) m. This is the distance from a maximum (antinode) to the first minimum (node).
(b)(iii) Distance QR is the same as PQ (both are \(\lambda/4\)). The distance between maxima/minima depends only on wavelength, not intensity, and the wavelength hasn’t changed.
A stationary loudspeaker emits sound of constant frequency. A microphone is placed near to the loudspeaker and connected to a cathode-ray oscilloscope (CRO). The trace on the screen of the CRO is shown in Fig. 5.1.
The time-base of the CRO is set to \(5.0 \times 10^{-4} \, \text{s cm}^{-1}\).
(a) The speed of the sound emitted by the loudspeaker is \(330 \, \text{m s}^{-1}\). Determine the wavelength of the sound.
(b) The loudspeaker now moves in a straight line while emitting the same sound of constant frequency. The period of the trace on the CRO increases continuously. Describe the motion of the loudspeaker.
▶️ Answer/Explanation
(a) Let’s find the wavelength step by step:
- 1. Find the period (T):
From the CRO trace, we see one complete wave cycle spans 5.8 cm.
With the time-base set to \(5.0 \times 10^{-4}\) s/cm, we calculate:
\( T = 5.8 \, \text{cm} \times 5.0 \times 10^{-4} \, \text{s/cm} = 2.9 \times 10^{-3} \, \text{s} \) - 2. Calculate the wavelength:
We know the wave speed \( v = 330 \, \text{m/s} \).
Using the relationship \( v = \lambda f \) and \( f = 1/T \):
\( \lambda = v \times T = 330 \times 2.9 \times 10^{-3} \)
\( \lambda = 0.957 \, \text{m} \approx 0.96 \, \text{m} \)
So the wavelength is approximately 0.96 meters.
(b) The increasing period on the CRO means the frequency is decreasing. This is the Doppler effect in action – the loudspeaker must be moving away from the microphone. Since the change is continuous, we can conclude the loudspeaker is accelerating away from the microphone.
A cylindrical copper wire P of length 0.24 m is shown in Fig. 6.1.
The current in the wire is 0.85A.
The resistance of the wire is 3.3 mΩ.
The total number of charge carriers \(N\) in the wire is \(2.6 \times 10^{22}\).
The resistivity of copper is \(1.8 \times 10^{-8} \Omega \, \text{m}\).
(a) Calculate the potential difference between the two ends of the wire.
(b) (i) Show that the cross-sectional area of the wire is \(1.3 \times 10^{-6} \, \text{m}^2\).
(ii) Show that the number density of charge carriers in the wire is \(8.3 \times 10^{28} \, \text{m}^{-3}\).
(iii) Calculate the average drift speed of the charge carriers (electrons) in the wire.
(c) A different copper wire Q has the same volume as wire P, but non-uniform radius, as shown in Fig. 6.2.
The radius \(r_1\) at end X of wire Q is the same as the radius of wire P. Radius \(r_2\) is less than \(r_1\).
(i) State and explain how the resistance of wire Q compares with the resistance of wire P.
(ii) On Fig. 6.3, sketch a graph of the variation of the average drift speed of the charge carriers with distance from end X of wire Q.
▶️ Answer/Explanation
(a) Using \(V = IR\): \(V = 0.85 \times 3.3 \times 10^{-3} = 2.8 \times 10^{-3} \, \text{V}\).
(b)(i) Using \(A = \rho L / R\): \(A = (1.8 \times 10^{-8} \times 0.24) / 3.3 \times 10^{-3} = 1.3 \times 10^{-6} \, \text{m}^2\).
(b)(ii) Number density \(n = N/(A \times L) = 2.6 \times 10^{22} / (1.3 \times 10^{-6} \times 0.24) = 8.3 \times 10^{28} \, \text{m}^{-3}\).
(b)(iii) Drift speed \(v = I/(nAq) = 0.85 / (8.3 \times 10^{28} \times 1.3 \times 10^{-6} \times 1.6 \times 10^{-19}) = 4.9 \times 10^{-5} \, \text{m/s}\).
(c)(i) Wire Q has greater resistance because its average cross-sectional area is smaller while length is greater (same volume but non-uniform radius).
(c)(ii) Graph should show increasing drift speed with distance from X (as area decreases, drift speed must increase to maintain current).
An isolated stationary nucleus Q decays into nucleus R and an α-particle. The α-particle has speed \(1.5 \times 10^7 \, \text{ms}^{-1}\).
(a) Complete the equation for this decay.
\( _{88}^{…..}\textrm{Q} \to _{…..}^{222}\textrm{R} + _{2}^{4}\textrm{α}\)
(b) By considering momentum, calculate the speed of nucleus R after the decay.
(c) State three quantities that are conserved during the decay.
▶️ Answer/Explanation
(a) \(\frac{226}{88}\)Q → \(\frac{222}{86}\)R + \(\frac{4}{2}\)α
The nucleon number (226) and proton number (86) are determined by conservation laws in alpha decay.
(b) \(2.7 \times 10^5 \, \text{m s}^{-1}\)
Using momentum conservation: \(4u \times 1.5 \times 10^7 = 222u \times v\). Solve for v to get the speed of nucleus R.
(c) Any three from: momentum, charge, nucleon number, neutron number, proton number
These quantities are always conserved in nuclear reactions. The specific numbers must balance on both sides of the equation.