2020 Past paper
- Answer all questions in the spaces provided.
- 1).
One strong interaction that occurs when two high-energy protons collide is - [math] P+P \to P+\pi^++\pi^-+X [/math]
- 1.1).
Determine the lepton number, strangeness and charge of particle X. - Explanation:
- [math] P+P \to P+\pi^++\pi^-+X [/math]
- Strangeness Number:
- All particles have a strangeness number of [math] 0 [/math]. Lepton Number: All particles have a lepton number of [math] 0 [/math]
- In the reaction [math] P+P \to P+\pi^++\pi^-+X [/math], the total charge on both sides must be equal to satisfy charge conservation.
- On the left-hand side, we have two protons (p), each carrying a charge of [math] +1 [/math]. So the total charge on the left-hand side is [math] +2 [/math].
- On the right-hand side, we have a proton (p) and a positively charged pion ([math] \pi^+ [/math]), each carrying a charge of [math] +1 [/math], and a negatively charged pion ([math] \pi^- [/math]) carrying a charge of [math] -1 [/math]. Additionally, we have the X particle with a charge number of [math] +1 [/math].
- Strangeness is one of the quantum numbers. Strangeness plays a role in understanding the properties and behavior of particles with strange quarks, such as their decay modes and lifetimes. Strangeness is conserved in strong and electromagnetic interactions, meaning the total strangeness remains constant. (zero).
- Explanation:
– Strangeness Number: All particles have a strangeness number of [math] 0 [/math].
– Lepton Number: All particles have a lepton number of [math] 0 [/math]
– Charge Number: Protons and the positively charged pion have a charge number of [math] +1 [/math], the negatively charged pion has a charge number of [math] -1 [/math], and (X) has a charge number of [math] +1 [/math]. - Conclusion:
- Based on the charge conservation in the reaction, the X particle with a charge number of [math] +1 [/math]
- Lepton number = ……………..[math] 0 [/math]………………..
Strangeness = …………. [math] 0 [/math] …………………
Charge = ………… [math] +1 [/math] ………………… - 1.2) .
- Identify particle X.
- Explanation:
- [math] P+P \to P+\pi^++\pi^-+X [/math]
- Conclusion:
– Based on the charge conservation in the reaction, the X particle with a charge number of [math] +1 [/math] must be a proton. - 1.3).
- A possible decay of a negative pion is
- [math] \pi^- \to e^- + Y [/math]
- What is particle Y?
Tick ([math] \square [/math]) one box.
([math] \square [/math]) Φ_e.
([math] \square [/math]) [math] \nu_{(e̅)} [/math].
([math] \square [/math]) [math] \pi^0 [/math].
([math] \square [/math]) [math] (_0^1)n [/math]. - In the decay of a negative pion ([math] \pi^- [/math]) to an electron ([math] e^- [/math]) and Y, the particle Y is an antineutrino, specifically an electron antineutrino ([math] \nu_{(e̅)} [/math]).
- [math] \pi^- \to e^- + Y [/math]
- L: [math] 0 \to +1 -1 [/math]
Q: [math] -1 \to -1 + 0 [/math] - Answer: [math] \nu_{(e̅)} [/math]
- 1.4).
- Some subatomic particles are classified as hadrons. There are two classes of hadrons.
- Discuss the nature of hadrons
Your answer should include
– the identifying properties of hadrons.
– the structure of a hadron in each class
– a discussion of the stability of free hadrons - Explanation:
- Hadrons are a class of subatomic particles that play a fundamental role in the composition of atomic nuclei and are an important part of the standard model of particle physics.
- Hadrons are characterized by certain identifying properties, are divided into two main classes, and have distinctive structures. Additionally, their stability when free (not bound within atomic nuclei) is an important aspect of their behavior.
- Identifying properties of Hadrons:
- Hadrons are strongly interacting particles, which means they participate in the strong nuclear force, one of the fundamental forces of nature.
- Hadrons have fractional electric charges. This is in contrast to leptons (e.g., electrons and neutrinos), which have integral electric charges.
- Hadrons experience the strong force and electromagnetic interactions.
- Structure of Hadrons in Each Class:
Hadrons are divided into two classes based on their intrinsic properties: - A. Baryons:
- Baryons are one class of hadrons, and they include protons and neutrons, which are the building blocks of atomic nuclei.
- Baryons are made up of three quarks. These quarks are held together by the exchange of particles called gluons, which mediate the strong force. For example, a proton consists of two up quarks and one down quark, while a neutron consists of two down quarks and one up quark.
- B. Mesons:
- Mesons are the other class of hadrons, and they include particles like pions and kaons. Mesons are composed of a quark and an antiquark. They also interact via the strong force and are held together by the exchange of gluons. For instance, a pion can consist of an up quark and an anti-down quark or vice versa.
- Stability of free Hadrons:
- Free hadrons, when not bound within atomic nuclei, are generally unstable. This means they have a finite lifetime and eventually decay into other particles.
- The relatively short lifetime of free hadrons is due to the strong force, which mediates their interactions. The strong force is very powerful at short distances but weakens rapidly as particles move away from each other.
- When free hadrons exist, they eventually undergo a process called hadron decay, where they transform into other particles, typically leptons (e.g., electrons, neutrinos) or other hadrons. This is a fundamental aspect of the behavior of hadrons, and it is a consequence of their strong interactions.
- 2).
- A spacecraft entering the atmosphere of Mars must decelerate to land undamaged on the surface.
- 2.1)
Figure 1 shows the spacecraft of total mass 610 kg entering the atmosphere at a speed of [math]5.5 km s^{-1}[/math].
Calculate the kinetic energy of the spacecraft as it enters the atmosphere. Give your answer to an appropriate number of significant figures. - Explanation:
- To calculate the kinetic energy of the spacecraft, we can use the formula:
- [math]Kinetic\ energy=\tfrac{1}{2}(mass)(velocity)^2[/math]
- Given:
– Mass of the spacecraft [math](m)=610\ kg[/math]
– Velocity of the spacecraft[math] (v)=5.5\ km/s[/math] - – Converting the velocity from km/s to m/s: [math]5.5\times1000=5500\ m/s[/math]
- Now we can substitute the values into the formula and calculate the kinetic energy:
- [math]Kinetic\ energy=\tfrac{1}{2}(mass)(velocity)^2[/math]
[math]Kinetic\ energy=\tfrac{1}{2}(610)(5500)^2[/math]
[math]KE=9.3\times10^9\ kg\ m^2/s^2[/math]
[math]KE=9.3\times10^9\ kg\ m^2/s^2[/math] - Rounding to an appropriate number of significant figures:
- [math]KE=9.3\times10^9\ J[/math]
- Therefore, the kinetic energy of the spacecraft as it enters the atmosphere is approximately: [math]9.3\times10^9\ J[/math].
- 2.2)
- A parachute opens during the spacecraft’s descent through the atmosphere. Figure 2 shows the parachute–spacecraft system, with the open parachute displacing the atmospheric gas. This causes the system to decelerate:
- Explain, with reference to Newton’s laws of motion, why displacing the atmospheric gas causes a force on the system and why this force causes the system to decelerate.
- Explanation:
1. Newton’s third law:
– When the parachute displaces atmospheric gas, the gas particles exert an equal and opposite force on the parachute according to Newton’s third law of motion. - 2. Momentum Transfer:
– The collision of the parachute with the gas particles results in a transfer of momentum between them. The gas particles gain momentum due to the force imparted by the parachute. - 3. Change in momentum:
– The change in momentum ([math]\Delta p[/math]) of the gas particles is given by - [math]\Delta p = F\times\Delta t[/math]
- – Where [math]F[/math] is the force exerted by the parachute and [math]\Delta t[/math] is the time interval of the collision.
- 4. Force on the parachute:
– Applying Newton’s second law of motion ([math]F=\Delta p/\Delta t[/math]), we find that the change in momentum ([math]\Delta(mv)[/math]) of the gas particles during the collision results in a force ([math]F[/math]) being exerted on the parachute. - 5. Equal and opposite forces:
– Newton’s third law implies that the force exerted by the gas particles on the parachute is equal in magnitude but opposite in direction to the force exerted by the parachute on the gas particles. - 6. Deceleration:
– As a consequence of the equal and opposite forces, the parachute (part of the system) experiences a force in the opposite direction to its motion, leading to deceleration or a reduction in velocity. - 7. Net force and acceleration:
– The net force acting on the system is the difference between the forward force of the spacecraft and the opposing drag force. This net force causes the system to decelerate as described by Newton’s second law - [math]F=ma[/math]
- In summary:
- – the collision between the parachute and gas particles results in a force that causes deceleration, as explained by Newton’s third law and the equation for momentum transfer.
- 2.3)
- As the parachute–spacecraft system decelerates, it falls through a vertical distance of 49 m and loses [math]2.2\times10^5\ J[/math] of kinetic energy.
- During this time, [math]3.3\times10^5\ J[/math] of energy is transferred from the system to the atmosphere.
- The total mass of the system is 610 kg.
- Calculate the acceleration due to gravity as it falls through this distance.
- Explanation:
- Given data:
- [math]Vertical\ distance\ fallen\ (h)=49\ m[/math]
[math]Change\ in\ kinetic\ energy\ (\Delta KE)=2.2\times10^5\ J[/math]
[math]Energy\ transferred\ to\ the\ atmosphere\ (\Delta E)=3.3\times10^5\ J[/math]
[math]Mass\ of\ the\ system\ (m)=610\ kg[/math] - Using the principle of conservation of energy, we can write the equation:
- [math]\Delta E(total)=\Delta KE+\Delta PE=0[/math]
- Substituting the given values and rearranging the equation:
- [math]0=2.2\times10^5\ J+mgh-3.3\times10^5\ J[/math]
- Rearranging and solving for g:
- [math]mgh=3.3\times10^5\ J-2.2\times10^5\ J[/math]
[math]gh=\dfrac{3.3\times10^5\ J-2.2\times10^5\ J}{m}[/math]
[math]g=\dfrac{3.3\times10^5\ J-2.2\times10^5\ J}{m\ h}[/math] - Substituting the given values:
- [math]g=\dfrac{3.3\times10^5\ J-2.2\times10^5\ J}{610\times49}[/math]
- Calculate the acceleration due to gravity:
[math]g=\dfrac{1.1\times10^5\ J}{29990}[/math]
[math]g=3.67\ m/s^2[/math] - The positive value indicates that the acceleration due to gravity acts in the downward direction, aligning with the motion of the system.
- Therefore, the acceleration due to gravity as the parachute-spacecraft system falls through a vertical distance of 49 m is approximately [math]3.67\ m/s^2[/math].
- 2.4)
- Dust from the surface of Mars can enter the atmosphere. This increases the density of the atmosphere significantly.
- Deduce how an increase in dust content will affect the deceleration of the system.
- Explanation:
- – Increased mass of dust particles:
- Dust particles in the atmosphere are heavier compared to gas molecules, resulting in a higher mass per unit volume.
- – Greater momentum transfer:
- Collisions between the parachute and heavier dust particles lead to a more significant transfer of momentum and energy from the system to the dust particles.
- – Opposing force:
- The increased dust content increases the density of the atmosphere, resulting in a higher opposing force acting on the parachute. This stronger opposing force contributes to faster deceleration.
- – Need for greater force:
- Heavier dust particles require a greater force to accelerate due to their higher mass. As the parachute collides with these particles, it experiences a stronger force in the opposite direction, leading to faster deceleration.
- In Summary:
- An increase in dust content in the Martian atmosphere, where dust particles are heavier, causes a stronger deceleration of the parachute-spacecraft system due to the increased mass and momentum transfer, enhanced drag force resulting from higher density, and the need for greater force to accelerate the heavier dust particles.
- 3.1)
Figure 3 shows a golf ball at rest on a horizontal surface 1.3 m from a hole. 
- A golfer hits the ball so that it moves horizontally with an initial velocity of [math]1.8 m s^{-1}[/math]. The ball experiences a constant deceleration of [math]1.2 m s^{-2}[/math] as it travels to the hole.
Calculate the velocity of the ball when it reaches the edge of the hole. - Explanation:
Given values:
[math]Initial\ velocity\ (u)=1.8\ m/s[/math]
[math]Deceleration\ (a)=-1.2\ m/s^2[/math]
[math]Displacement\ (s)=1.3\ m[/math] - Substitute the values into the kinematic equation:
- [math]v^2=u^2+2as[/math]
- Calculate the values:
- [math]u^2=(1.8)^2=3.24\ m^2/s^2[/math]
[math]2as=2(-1.2)(1.3)[/math]
[math]2as=-3.12[/math] - [math]v^2=u^2+2as[/math]
[math]v^2=3.24-3.12[/math]
[math]v^2=0.12[/math]
[math]v=\sqrt{0.12}[/math]
[math]v=0.35\ m/s[/math] - 3.2)
- Later, the golf ball lands in a sandpit. The golfer hits the ball, giving it an initial velocity u at 35° to the horizontal, as shown in Figure 4. The horizontal component of u is [math]8.8 m s^{-1}[/math].
- Show that the vertical component of u is approximately [math]6 m s^{-1}[/math].
- Explanation:
Given values:
[math]Horizontal\ component\ of\ U_x=8.8\ m/s[/math]
[math]Angle\ of\ u\ with\ the\ horizontal\ \theta=35^\circ[/math] - Use the trigonometric relationship for the tangent:
- [math]\tan\theta=U_y/U_x[/math]
[math]\tan 35^\circ=U_y/8.8[/math]
[math]U_y=(\tan 35^\circ)(8.8)[/math]
[math]U_y=(0.7002)(8.8)[/math]
[math]U_y=6.1\ m/s[/math] - The vertical component of the initial velocity is approximately [math]6.1\ m/s[/math].
- 3.3)
- The ball is travelling horizontally as it reaches X, as shown in Figure 5.
- Assume that weight is the only force acting on the ball when it is in the air. Calculate the time for the ball to travel to X.
- Explanation:
Given values:
[math]Horizontal\ displacement\ (s)=1.3\ m[/math]
[math]Horizontal\ component\ of\ initial\ velocity\ (U_x)=6.2\ m/s[/math]
[math]Acceleration\ due\ to\ gravity\ (a)=-9.8\ m/s^2[/math]
[math]Vertical\ component\ of\ initial\ velocity\ U_y=0\ m/s[/math] - [math]U_y=U_x+at[/math]
[math]0=6.2+(-9.8)(t)[/math]
[math]t=6.2/9.8[/math]
[math]t=0.6327\ s[/math] - 3.4)
- Calculate the vertical distance of X above the initial position of the ball.
- Explanation:
[math]U_y=0\ m/s[/math] (Vertical component of initial velocity)
[math]t=0.6327\ s[/math]
[math]a=-9.8\ m/s^2[/math] - [math]s=\dfrac{(U_y+U_x)}{2}t[/math]
[math]s=\dfrac{(0+(-6.1926))}{2(0.6327)}[/math]
[math]s=-1.9579\ m[/math]
[math]s=1.9579\ m[/math] - The golfer returns the ball to its original position in the sandpit. He wants the ball to land at X but this time with a smaller horizontal velocity than in Figure 5.
- 3.5)
- Sketch on Figure 6 a possible trajectory for the ball.
- 3.6)
- Explain your reason for selecting this trajectory.
- Explanation:
- Higher initial angle to the horizontal:
– When you launch an object at a higher angle to the horizontal, you are essentially giving it a steeper initial trajectory. This means the angle of launch is closer to vertical (90 degrees), which can lead to specific advantages in certain situations. - Smaller horizontal velocity:
– With a higher launch angle, more of the initial velocity is directed vertically (VY) and less is directed horizontally (VX). This results in a smaller horizontal velocity, meaning the object moves more vertically compared to a lower launch angle where more of the initial velocity is in the horizontal direction. - oParabolic trajectory:
– This is a key characteristic of projectile motion. When you launch an object at any angle, its trajectory will form a parabolic path. However, the steepness of the parabola varies with the launch angle. - – Curve starts below the ‘golf ball’ and ends at point label X: This implies that the trajectory starts from a position lower than the golf ball’s initial height and reaches point X. A higher launch angle could achieve this as it allows for a steeper initial ascent.
- Maximum turning point above the midpoint between the ball and label X:
– A higher launch angle results in a higher peak in the trajectory. This point is where the object momentarily stops moving upwards before it starts descending. A steeper trajectory ensures that this point is above the midpoint between the starting point (golf ball) and the target point (X). - Descent and landing at or near position X:
– If the goal is to hit or reach point X, a higher launch angle helps in ensuring the object reaches or lands near the desired target. - Smaller horizontal velocity at landing:
– Again, due to the distribution of velocity components, a higher launch angle will result in a smaller horizontal velocity at landing, which can be advantageous in controlling the horizontal position. - At a higher initial angle to the horizontal, the vertical component of the velocity (VY) will increase while the horizontal component (VX) will decrease: This is a fundamental principle of projectile motion. By increasing the launch angle, you increase the vertical component of velocity, which can be beneficial in achieving a desired vertical position or reaching a higher peak.
- At a higher initial angle to the horizontal, the vertical component of the velocity (VY) will increase while the horizontal component (VX) will decrease: This is a fundamental principle of projectile motion. By increasing the launch angle, you increase the vertical component of velocity, which can be beneficial in achieving a desired vertical position or reaching a higher peak.
4) - A sample of pure boron contains only isotope X and isotope Y. A nucleus of X has more mass than a nucleus of Y.
4.1)
The sample is ionised, producing ions each with a charge of [math]+1.6\times 10^{-19}\ \text{C}[/math]. The specific charge of an ion of X is [math]8.7\times 10^{6}\ \text{C kg}^{-1}[/math].
Calculate the mass of an ion of X.
Explanation:
Given data:
Specific charge [math]\frac{q}{m} = 8.7\times 10^{6}\ \text{C kg}^{-1}[/math]
Charge of the ion = [math]+1.6\times 10^{-19}\ \text{C}[/math] - We’ll use the same equation as before:
- [math] \frac{q}{m} = \text{Charge/mass} [/math]
- [math] 8.7 \times 10^6 = \frac{1.6 \times 10^{-19}}{mass} [/math]
- [math] mass = (8.7 \times 10^6)(1.6 \times 10^{-19}) [/math]
- [math] mass = 0.183 \times 10^{-19-6} \ kg [/math]
- [math] mass = 0.183 \times 10^{-25} \ kg [/math]
- [math] mass = 1.83 \times 10^{-26} \ kg [/math]
- 4.2)
Determine the number of nucleons in a nucleus of X. - [math]\text{mass of a nucleon} = 1.7\times 10^{-27}\ \text{kg}[/math]
[math]\text{mass of an ion of X} = 1.839\times 10^{-26}\ \text{kg}[/math] - To determine the number of nucleons in a nucleus of X:
Calculate the number of nucleons by dividing the mass of the ion of X by the mass of a nucleon:
[math]\text{number of nucleons} = \frac{\text{Mass of ion of X}}{\text{mass of a nucleon}}[/math] - [math]\text{number of nucleons} = \frac{1.839\times 10^{-26}}{1.7\times 10^{-27}}[/math]
- [math]\text{number of nucleons} \approx 10.823[/math]
- Round the value to the nearest whole number to obtain the final result:
- [math] Number \ of \ nucleons \approx 11 [/math]
- 4.3)
Compare the nuclear compositions of X and Y.
Explanation: - In general, isotopes of the same element have the same number of protons (atomic number) since they belong to the same element. However, they can have different numbers of neutrons (different mass numbers) which accounts for their different masses.
- Therefore, in the case of X and Y, X will have a greater number of neutrons in its nucleus compared to Y, resulting in a heavier mass for isotope X.
- Isotopes of Hydrogen
4.4)
Ions of Y have the same charge as ions of X.
State and explain how the specific charge of an ion of X compares with that of an ion of Y.
Explanation:
The formula for specific charge ([math]\frac{q}{m}[/math]) is given by: - [math]\text{Specific charge}=\frac{\text{Charge}}{\text{Mass}}[/math]
- The specific charge represents the ratio of the charge of a particle to its mass. It is typically expressed in units of coulombs per kilogram (C/kg).
Therefore: We can conclude that there is an inverse relationship between the mass and specific charge of a particle. When the mass is large, the specific charge tends to be small, and when the mass is small, the specific charge tends to be large. - 4.5)
Table 1 contains data about two completely ionised samples of pure boron. Each sample contains only isotopes X and Y.
Table 1Sample number Number of ions in sample Mass of sample / kg Charge on each ion / C 1 3.50 × 1016 6.31 × 10−10 +1.60 × 10−19 2 3.50 × 107 6.20 × 10−19 +1.60 × 10−19 - Deduce which sample, 1 or 2, contains a greater percentage of isotope Y.
- Sample 1:
Number of ions: [math]3.5\times 10^{16}[/math]
Mass of sample: [math]6.31\times 10^{-10}\ \text{kg}[/math]
Charge on each ion: [math]+1.6\times 10^{-19}\ \text{C}[/math] - Sample 2:
Number of ions: [math]3.5\times 10^{7}[/math]
Mass of sample: [math]6.20\times 10^{-19}\ \text{kg}[/math]
Charge on each ion: [math]+1.6\times 10^{-19}\ \text{C}[/math] - For sample 1:
- Mean mass of one ion in sample 1 = [math]\frac{\text{Mass of sample 1}}{\text{number of ions in sample 1}}[/math]
- [math]\text{Mean mass (sample 1)} = \frac{6.31\times 10^{-10}}{3.5\times 10^{16}}[/math]
- [math]\text{Mean mass (sample 1)} \approx 1.803\times 10^{-26}\ \text{kg per ion}[/math]
- For sample 2:
- Mean mass of one ion in sample 2 = [math]\frac{\text{Mass of sample 2}}{\text{number of ions in sample 2}}[/math]
- [math]\text{Mean mass (sample 2)} = \frac{6.20\times 10^{-19}}{3.5\times 10^{7}}[/math]
- [math]\text{Mean mass (sample 2)} \approx 1.771\times 10^{-26}\ \text{kg per ion}[/math]
- Mean mass represents the average mass per ion in each sample.
- Based on the mean mass, we can conclude that Sample 2 has a lower mean mass per ion compared to Sample 1, indicating a greater percentage of the lighter isotope Y in Sample 2.