Past paper(June 2018)
- 1. Horizontal escape lanes made of loose gravel have been constructed at the side of some roads on steep hills so that vehicles can stop safely when their brakes fail.
- Figure 1 shows an engineer’s prediction of how the speed of an unpowered vehicle of mass [math]1.8 \times 10^{4}\ kg[/math] will vary with time as the vehicle comes to rest in an escape lane.
- Figure 1
- 1.1) Determine the force decelerating the vehicle [math]2.0\ s[/math] after entering the escape lane.
Explanation:
Step 1: Calculate mean deceleration from tangent:
The mean deceleration (d) can be calculated using the tangent line. Given that the slope of the tangent is [math]2.7\ m/s^{2}[/math], we can use this value - [math]d=\frac{\Delta V}{\Delta t}[/math]
[math]d=\frac{0-15}{0-5.5}[/math]
[math]d=2.7\ m/s^{2}[/math] - Step 2: Use [math]F=ma[/math] with mean deceleration:
Now, we can use Newton’s second law to find the force (F): - [math]F=ma[/math]
- Substitute the values:
- [math]F=(1.8\times 10^{4})(2.7)[/math]
[math]F=4.86\times 10^{4}\ N[/math] - 1.2) Deduce whether a lane of length [math]85\ m[/math] is long enough to stop the vehicle, assuming that the engineer’s graph is correct.
- Explanation:
According to the given information:
Total number of squares: [math]22[/math]
Distance per square: [math]2.5\ m \times 0.1\ m[/math]
So,
The total distance covered by [math]22[/math] squares is: - [math]\text{Total distance}=22\times 2.5[/math]
[math]\text{Total distance}=55\ m[/math] - The given length of the lane is [math]85\ m[/math], and the distance covered by the vehicle is [math]55\ m[/math]. Since the length of the lane is greater than the distance covered by the vehicle, the lane is long enough to stop the vehicle.
- 1.3) Discuss the energy transfers that take place when a vehicle is decelerated in an escape lane.
- I. Kinetic Energy (KE) of the lorry to KE of Gravel:
- – Initially, the lorry possesses kinetic energy due to its motion.
- – As the lorry enters the escape lane and decelerates, its kinetic energy decreases.
- – The decrease in the lorry’s kinetic energy is transferred to the gravel in the form of kinetic energy.
- This is because the gravel is pushed aside or moved by the lorry.
- II. Potential Energy (PE) of Gravel to other forms:
- – The gravel, as it is pushed aside or moved, may gain potential energy. This potential energy could be in the form of elevation if some of the gravel is lifted or thrown upward during the deceleration process.
- III. Transfer to thermal energy / Internal Energy / Heating:
- – Some of the energy from the kinetic energy of the lorry and potential energy of the gravel may be transferred to thermal energy or internal energy.
- – This can occur due to friction between the lorry and the gravel, as well as between individual gravel particles. The mechanical energy is converted into heat.
- IV. Work done on the gravel / Vehicle increasing internal energy:
- – Work is done as the lorry decelerates and pushes the gravel aside. This work contributes to an increase in the internal energy of the system.
- – The internal energy of the gravel, as well as the lorry, increases due to the work done.
- In summary, the energy transfers involve the conversion of the lorry’s kinetic energy into kinetic energy of the gravel, potential energy of the gravel (if lifted), and the conversion of mechanical energy into thermal energy or internal energy. The work done on the gravel contributes to raising the internal energy and potentially increasing the temperature of the system.
- 1.4) An alternative to an escape lane containing gravel is an escape lane that consists of a ramp. An escape ramp is a straight road with a concrete surface that has a constant upward gradient.
One escape ramp makes an angle of [math]25^{\circ}[/math] to the horizontal and is [math]85\ m[/math] long. Deduce whether this escape ramp is sufficient to stop the vehicle.
Assume that any frictional forces and air resistance that decelerate the vehicle are negligible.
Explanation:
Let’s go through the steps using the trigonometric relationship - [math]\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}[/math]
- Given:
– Length of the ramp (H) = [math]85\ m[/math]
– Angle of the ramp ([math]\theta[/math]) = [math]25^{\circ}[/math]
Using: - [math]\sin\theta=\frac{O}{h}[/math]
[math]h=85\times \sin 25^{\circ}[/math]
[math]h=35.9\ m[/math] - Now, Let’s calculate the height (h) using the energy approach:
- [math]h=\frac{(17.5)^{2}}{2\times 9.8}[/math]
[math]h=15.6\ m[/math] - Yes ramp is sufficient.
- 1.5) Discuss whether an escape lane containing gravel or an escape ramp would provide the safer experience for the driver of the vehicle as it comes to rest.
- Explanation:
- Escape Lane containing gravel:
- I. Deceleration profile:
- – Initial deceleration may be larger and more variable.
- II. Stability Issues:
- – Traveling through gravel could lead to instability and erratic bouncing.
- III. Predictability:
- – The behavior of the vehicle on gravel may be less predictable.
- IV. Surface Effects:
- – The uneven surface of gravel may affect the control the driver has over the vehicle.
- Escape Ramp with uniform gradient:
- I. Uniform Deceleration:
- – Provides a consistent and uniform deceleration.
- II. Stability during deceleration:
- – Less likely to experience erratic bouncing or instability.
- III. Predictability:
- – The deceleration process is more controlled and predictable.
- IV. Brake – dependent consideration:
- – If the vehicle lacks brakes, it may roll backward after stopping.
- Conclusion:
- Considering the factors listed, the escape ramp with a uniform gradient is generally considered safer due to its controlled deceleration, stability, and predictability.
- However, specific characteristics of the vehicle, such as the availability of brakes, should be taken into account when determining the most suitable option for a particular situation.
- 2) Table 1 shows results of an experiment to investigate how the de Broglie wavelength of an electron varies with its velocity v.
| [math]v / 10^{7}\ \text{m s}^{-1}[/math] | [math]10^{-11}\ \text{m}[/math] |
|---|---|
| 1.5 | 4.9 |
| 2.5 | 2.9 |
| 3.5 | 2.1 |
- 2.1): Show that the data in Table 1 are consistent with the relationship
- [math]\lambda \propto \frac{1}{v}[/math]
- Explanation:
- The de Broglie wavelength (λ) of an electron is given by the relationship:
- [math]\lambda \propto \frac{1}{v}[/math]
- where v is the velocity of the electron. To show that the data in Table 1 is consistent with this relationship, let’s calculate [math]\lambda v[/math] for each entry in the table and check for proportionality.
- [math]\lambda v=1.5\times 10^{7}\times 4.9\times 10^{-11}[/math]
[math]\lambda v=7.35[/math] - For:
- [math]v=2.5\times 10^{7}\ \text{m/s}[/math]
[math]\lambda v=2.5\times 10^{7}\times 2.9\times 10^{7}[/math]
[math]\lambda v=7.25[/math] - Conclusion:
The calculated values for [math]\lambda v[/math] are approximately 7.35, 7.25, and 7.35 for the given velocities, respectively. Since these values are consistent and relatively close Two valuesare closed and one is different, it supports the relationship - [math]\lambda \propto \frac{1}{v}[/math]
- The inverse proportionality between [math]\lambda[/math] and [math]v[/math] is evident from the data in Table 1.
2.2) Calculate a value for the Planck constant suggested by the data in Table 1.
Explanation:
To calculate the Planck constant (h) using the de Broglie wavelength (λ) and the mass-velocity product, you can use the formula: - [math]h=\frac{\lambda m v}{2\pi}[/math]
- Given that [math]\lambda v[/math] has been calculated for the first and third data set :
- For:
- [math]v=1.5\times 10^{7}\ \text{m/s}[/math]
[math]\lambda v=7.25[/math] - For:
- [math]v=3.5\times 10^{7}\ \text{m/s}[/math]
[math]\lambda v=7.35[/math] - Using the first data set:
- [math]h=\frac{\lambda m v}{2\pi}[/math]
[math]h=\frac{7.35\times 9.11\times 10^{-31}\times 1.5\times 10^{7}}{2\pi}[/math]
[math]h=6.7\times 10^{-34}[/math] - Using the third data set:
- math]h=7.35\times 9.11\times 10^{-34}\times 3.5\times 10^{7}[/math]
[math]h=6.7\times 10^{-34}[/math] - Using the second data set:
- [math]v=2.5\times 10^{7}\ \text{m/s}[/math]
[math]\lambda v=7.25[/math]
[math]h=7.25\times 9.11\times 10^{-34}\times 3.5\times 10^{7}[/math]
[math]h=7.59\times 10^{-34}[/math] - Conclusion:
- The calculated values for the Planck constant using the first and third data sets are approximately [math]6.70\times 10^{-34}[/math]and using the second data set is approximately [math]6.59\times 10^{-34}[/math].
- These values are consistent with the expected range for the Planck constant ([math]\hbar\approx 6.626\times 10^{-34}\ \text{J}\cdot\text{s}[/math] indicating that the data in Table 1 supports the determination of the Planck constant.
2.3) Figure 2 shows the side view of an electron diffraction tube used to demonstrate the wave properties of an electron. 
- An electron beam is incident on a thin graphite target that behaves like the slits in a diffraction grating experiment. After passing through the graphite target the electrons strike a fluorescent screen.
Figure 3 shows the appearance of the fluorescent screen when the electrons are incident on it. 
- Figure 3
- Explain how the pattern produced on the screen supports the idea that the electron beam is behaving as a wave rather than as a stream of particles.
Explanation: - The pattern produced on the fluorescent screen in Figure 3 supports the idea that the electron beam is behaving as a wave rather than as a stream of particles. Here’s how:
I. Particle behavior vs. Wave behavior:
– If the electrons were behaving purely as particles, you would expect to see a patch, circle, or small spot of light on the screen. This is because particles tend to travel in straight lines and create localized impacts on the screen.
II. Random scattering vs. Diffraction / Interference:
– If the electrons were behaving as particles, their impacts on the screen would scatter randomly.
– However, the observed pattern suggests diffraction and interference, which are characteristics of waves. Waves exhibit interference patterns when they overlap, supporting the wave-like nature of electrons.
III. Effect of graphite target:
– The thin graphite target in the setup behaves like the slits in a diffraction grating experiment.
– The graphite causes the electron waves or beam to spread out and travel in particular directions. This spreading out is a characteristic behavior of waves, supporting the wave nature of electrons.
IV. Bright rings and constructive interference:
– The presence of bright rings or areas of maximum intensity on the screen indicates constructive interference. In a diffraction grating, maxima occur when the sine of the angle ([math]\sin\theta[/math]) is equal to [math]\frac{n\lambda}{a}[/math], where n is the order of the maximum, [math]\lambda[/math] is the wavelength of the wave, and d is the separation between the slits.
– This relationship is a key feature of wave interference and further supports the wave nature of the electron beam. - In summary, the diffraction and interference patterns observed on the screen are consistent with the wave properties of electrons, providing strong evidence for the wave-like behavior of electrons in this experiment.
2.4) Explain how the emission of light from the fluorescent screen shows that the electrons incident on it are behaving as particles.
Explanation:
I. Excitation and de – excitation of electrons / atoms:
– The emission of light from the fluorescent screen is a result of the excitation and subsequent de-excitation of electrons in the atoms of the fluorescent material.
– When high-energy electrons from the incident beam collide with the atoms in the material, they provide enough kinetic energy instantly to excite the electrons in the atoms to higher energy levels. - II. Collisions and energy transfer:
– The idea of collisions between incident electrons and electrons in the atoms is crucial.
– As particles, electrons in the incident beam collide with the electrons in the atoms, causing them to move to higher energy levels or orbits. This transfer of energy occurs in discrete amounts, indicating a particle-like behavior.
III. Photon emission during de – excitation:
– When the excited electrons in the atoms de-excite or move back to lower energy levels, they release the excess energy in the form of light or photons.
– This emission of light is characteristic of particles behaving in a quantized manner, providing energy in discrete amounts rather than continuously over time, as would be expected in a wave-like behavior.
IV. Instant energy transfer:
– The notion that electrons must provide enough kinetic energy instantly to cause excitation supports the particle nature.
– This is in contrast to the continuous distribution of energy that would occur in a wave-like behavior.
In summary, the emission of light from the fluorescent screen is consistent with the behavior of electrons as particles. The discrete transfer of energy during collisions and the subsequent emission of photons during de-excitation support the particle-like nature of the electrons.
- 3) Figure 4 shows the structure of a violin and Figure 5 shows a close-up image of the tuning pegs.
- The strings are fixed at end A. The strings pass over a bridge and the other ends of the strings are wound around tuning pegs that have a circular cross-section. The tension in the strings can be increased or decreased by rotating the tuning pegs.
- 3.1) Explain how a stationary wave is produced when a stretched string is plucked.
- Explanation:
I. Generation of waves:
– When the string is plucked, it creates a disturbance that travels along the length of the string. This disturbance generates waves in the string. - II. Reflection at boundaries:
– The waves travel towards the boundaries of the string, which are fixed at end A and around the tuning pegs. At these boundaries, the waves are reflected. - III. Interference of opposite traveling waves:
– As a result of the reflection, two waves are now traveling in opposite directions along the string—one towards end A and the other towards the tuning pegs. - IV. Superposition and interference:
– The two waves interfere or superpose as they travel along the string. This interference results in the formation of a stationary wave. - V. Fixed Boundaries as Nodes:
– The fixed boundaries (end A and around the tuning pegs) act as nodes. Nodes are points on a standing wave where the amplitude is always zero. At these points, the string does not vibrate. - VI. Constructive and destructive interference:
– In some positions along the string, the waves always interfere destructively. This means that the waves cancel each other out, resulting in zero amplitude or no vibration at those positions. These points correspond to nodes. - – In other positions, the waves interfere constructively, producing positions of maximum amplitude or maximum vibration. These points correspond to antinodes, where the string undergoes maximum displacement.
- In summary, the production of a stationary wave on a plucked violin string involves the reflection of waves at fixed boundaries, the interference of waves traveling in opposite directions, and the establishment of nodes and antinodes along the string. This standing wave pattern is what gives rise to the distinct sound produced by a plucked violin string.
- 3.2 The vibrating length of one of the strings of a violin is [math]0.33\ \text{m}[/math]. When the tension in the string is [math]25\ \text{N}[/math], the string vibrates with a first-harmonic frequency of [math]370\ \text{Hz}[/math].
Show that the mass of a [math]1.0\ \text{m}[/math] length of the string is about [math]4 \times 10^{-4}\ \text{kg}[/math]. - Explanation:
– The formula for the frequency is: - [math]f=\frac{1}{2L}\sqrt{\frac{T}{\mu}}[/math]
- Where:
– L is the vibrating length,
– T is the tension,
– μ is the linear mass density. - Given f=370 Hz, L=0.33 m, and T=25 N, we can rearrange the formula to solve for μ:
- [math]\mu = \frac{T}{4f^{2}L^{2}}[/math]
- Now, Plug in the given values:
- [math]\mu = \frac{25}{4(370)^{2}(0.33)^{2}}[/math]
- Calculating this gives
- [math]\mu = 4.2 \times 10^{-4} \ \text{kg/m}[/math]
- Therefore, the mass of a 1.0 m length of the string μ×1.0 m is approximately
[math]4.2 \times 10^{-4} \ \text{kg}[/math]. The given answer of [math]4.19 \times 10^{-4} \ \text{kg}[/math] - 3.3 Determine the speed at which waves travel along the string in question 03.2 when it vibrates with a first-harmonic frequency of 370 Hz.
- Explanation:
The speed (v) of waves traveling along a string is given by the formula: - [math]v = f \lambda[/math]
- Where:
– F is the frequency,
– λ is the wavelength. - For a string vibrating in the first harmonic (fundamental mode), the wavelength is twice the length of the vibrating portion [math]\lambda = 2L[/math]
- Given that the frequency f=370 Hz and the vibrating length L=0.33 m we can find the speed of the waves:
- [math]\lambda = 2(0.33)[/math]
[math]\lambda = 0.66 \ m[/math] - Now, use the formula
- [math]v = f\lambda[/math]
[math]v = (370)(0.66)[/math]
[math]v = 244 \ m/s[/math] - Therefore, the speed at which waves travel along the string when it vibrates with a first-harmonic frequency of 370 Hz is approximately 244 m/s.
- 3.4 Figure 6 shows how the tension in the string in question 03.2 varies with the extension of the string.
- The string with its initial tension of 25 N is vibrating at a frequency of 370 Hz. The diameter of the circular peg is 7.02 mm.
Determine the higher frequency that is produced when the string is stretched by rotating the tuning peg through an angle of 75°.
Assume that there is no change in the diameter of the string. - Explanation:
To find the higher frequency ([math]f_{2}[/math]) when the string is stretched by rotating the tuning peg through an angle of 75°, we can use the relationship between frequency and tension for a vibrating string. The formula is: - [math]f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}[/math]
- Where:
– F is the frequency,
– L is the vibrating length of the string,
– T is the tension in the string,
– μ is the linear mass density of the string. - Since the linear mass density (μ) and the vibrating length (L) of the string remain constant, we can express the frequency as directly proportional to the square root of tension (T).
- [math]f \propto \sqrt{T}[/math]
- Now, let’s calculate the tension ([math]T_{2}[/math]) when the string is stretched by rotating the tuning peg through an angle of 75°.
- Calculate the extra extension:
- [math]\text{Extra extension} = \frac{22}{360} \times 75 \ \text{mm}[/math]
- [math]\text{Extra extension} \approx 4.6 \ \text{mm}[/math]
- Total extension:
- [math]\text{Total extension} = 11 + 4.6[/math]
[math]\text{Total extension} = 15.6 \ \text{mm}[/math] - Calculate the new tension ([math]T_{2}[/math]):
From graph: - [math]T_{2} = 35 \ N[/math]
- Calculate the higher frequency ([math]f_{2}[/math]):
- [math]\frac{f_{2}}{f_{1}} = \sqrt{\frac{T_{2}}{T_{1}}}[/math]
- [math]f_{2} = f_{1}\sqrt{\frac{T_{2}}{T_{1}}}[/math]
- [math]f_{2} = 370 \sqrt{\frac{35}{25}}[/math]
- [math]f_{2} = 438 \ Hz[/math]
- Therefore, the higher frequency produced when the string is stretched by rotating the tuning peg through an angle of 75° is approximately 438 Hz.
