The Graduate Australian Medical School Admissions Test (GAMSAT) is a standardized exam that assesses your aptitude for a range of skills, including physics. The physics section of the GAMSAT tests the prerequisite knowledge of Year 12 physics and evaluates skills such as mathematical reasoning, graphical analysis, and data interpretation.
This guide aims to provide you with a roadmap on how to prepare for GAMSAT Physics in 2024. We'll delve into key topics, supplemented with worked-out examples. Before starting your GAMSAT Prep journey also check out our article to know What will the March 2024 GAMSAT look like.
One often overlooked aspect of preparation is practicing mental maths. While it may seem trivial, honing your mental arithmetic skills can be a game-changer. Under the stress of the exam, even simple mathematical functions like addition, subtraction, and logarithmic calculations can become stumbling blocks. Therefore, it's advisable to practice mental maths rigorously, even if it means revisiting basic multiplication and addition exercises from Year 3.
Physics has always been the science subject that most heavily tests mathematical skills, and this is unlikely to change in the foreseeable future. However, you can expect the 2024 GAMSAT to feature more calculations involving basic formulations and multi-step problems. This shift will underscore the importance of being proficient in mental maths, as you'll need to quickly and accurately perform calculations to solve these problems.
1. Build a Strong Foundation with Mathematics
Before diving into advanced topics, ensure you have a solid grasp of foundational concepts of Physics as well as Mathematics. Mathematics plays a pivotal role in understanding and solving GAMSAT physics questions. For a deeper dive into the importance of mathematics in GAMSAT, check out this article: GAMSAT Maths.
2. Adopt a Practical Approach to Learning
As you progress, break down the vast physics syllabus into manageable sections. Tools like the GAMSAT Physics checklist can be invaluable in this regard. Such structured approaches not only make the learning process more digestible but also ensure that you cover all necessary topics.
3. Prioritize Conceptual Understanding Over Memorization
Once you've gained a grasp of the fundamental concepts, focus on understanding the underlying concepts rather than merely memorizing formulas. Covering each topic by attempting GAMSAT practice questions can be a great way not only to test your knowledge but also to reinforce your understanding and application of these concepts.
4. Importance of Reflection
Quality trumps quantity in GAMSAT preparation. It's not just about the hours you put in, but how effectively you use them. After each study session, take a moment to introspect. Identify areas where you stumbled and actively work on rectifying those gaps.
Continuously assess your strengths and weaknesses, placing emphasis on fortifying areas where you're less confident. This reflective approach ensures that your preparation is targeted and efficient, maximizing your chances of success. Check out our Section 3 Question Log which will come handy here.
5. Dive into the Interdisciplinary Nature of GAMSAT Physics
Once you've laid a solid groundwork in GAMSAT Chemistry, Biology, and Physics, it's time to embrace the interconnected nature of these subjects. Start attempting more Section 3 questions, which often combine elements from all three disciplines.
The GAMSAT exam doesn't test these subjects in isolation but integrates them, emphasizing the importance of a holistic understanding. This approach underscores the necessity of seeing how physics principles, along with chemistry and biology concepts, intertwine and apply in various contexts.
6. The Importance of Timed Practice
While practicing, ensure you simulate exam conditions by timing yourself. This not only helps you get a feel for the actual exam but also improves your time management skills, which are crucial for the GAMSAT.
7. Utilize Available Resources, Including Fraser's Learning Academy
Throughout your preparation journey, make the most of the resources at your disposal. Fraser’s GAMSAT offers a range of tools, including bridging courses and updated Physics checklist, to aid in your preparation.
Additionally, Fraser's Learning Academy, a free Learning Management System (LMS), provides a wealth of invaluable resources. Within the Academy, you'll find a comprehensive range of free materials such as the Pre-Module Assessment for GAMSAT physics, modules on Kinematics & Motion and Energy, Motion, as well as an extensive Section Question bank. These resources, curated to reflect the latest in the GAMSAT syllabus, ensure you're always up-to-date and well-prepared.
The velocity of an object changes only when there is a net force acting on an object. The relationship between the acceleration of the object and the net force (in N) acting on it is given by Newton’s 2nd Law of motion:
Where, m is the mass of the object in kg. Moreover, when a force is applied on an object and it causes a displacement of the object that is parallel to the force applied, work (in J) will be done on the object. Work is calculated using the equation:
Where W is work; F is force; d is displacement.
Momentum is conserved in isolated systems (i.e. where the only forces involved in the collision are the action reaction forces). The momentum of an object is given by:
Where p is momentum; m is mass; v is velocity. Momentum is a vector quantity, and the direction of momentum is equal to the direction of the velocity of the object. The total momentum can be calculated by simply adding the individual momentums involved, as you would add vectors.
A skateboard of mass 5 kg, was given an initial push along a floor and allowed to roll freely. Its change in velocity with time was measured, as described in Figure 1. At time t = 4s, a lump of dirt was dropped vertically down onto the skateboard.
Question 1: What is the magnitude of the frictional force acting on the skateboard before the lump of dirt was dropped on it?
Question 2: What is the work done by the frictional force in the first 4 s?
Solution to 1: B. Frictional force is the one opposing motion. We use F=ma but we need to find a. We use the gradient of the graph to do this where a = Δv/t = (3−4)4 = − 0.25. We then sub into the F = ma = 5 kg * -0.25 m/s2 = -1.25 N. The answer asks for magnitude, hence the sign doesn’t matter.
Solution to 2: A. Work is the product of force and displacement. Whilst we have force, 1.25 N, we need the displacement. This can be found from the area under the curve, because if we multiply velocity by time, we get displacement. The shape is a trapezium (Area = ½(a+b) x h) OR you can add a triangle and a rectangle. The result is 14 m so the work is 14 m ×1.25 N=17.5 J.
When springs are compressed or extended by a force, work is being done on the spring, and as a result, the spring stores elastic potential energy. Work done on a spring is directly proportional to the force applied by the spring, as well as to its extension or compression.
There are many different forms of energy (measure in J) one would come across when analyzing the motion of an object. These include but are not limited to: heat energy, gravitational potential energy (energy due to a gravitational field), spring potential energy (energy in objects that are stretched or compressed), and kinetic energy (energy of moving objects).
The figure shows how the gravitational potential energy, and spring potential energy of the spring are converted from one form into the other when the mass attached is allowed to oscillate up and down. The formulas for gravitational potential energy and the energy stored in a spring are:
Gravitational potential energy=m×g×Δh
Where m is the mass of the object (kg), g is the magnitude of the gravitational field strength (10 Nkg-1 ); and Δh is the change in vertical height measured from the lowest point.
The energy stored in a spring is equal to ½ × k × Δx2
Question 1. How much spring potential energy is stored from when the 50 g mass is attached as shown in Figure 2, until it settles.
Question 2. In the Figure , two weights are attached to spring 3, corresponding to total length of spring of 100 cm. What would happen to the spring potential energy if the mass of each of the attached weights were doubled?
Solution to 1: D. The spring potential energy is the work done which is directly proportional to theforce as well as the change in length. This is also given by the area under the graph similar to that in Figure 1. This is a triangle so we use 1⁄2 x base x height = 1⁄2 x 0.5N x 0.4m = 0.010 J.
Solution to 2: B. When you double the mass, you double the weight force acting on the spring. This must be opposed by an equal and opposite force exerted by the spring itself. The elastic force of the spring increases as it stretches. The elastic potential energy depends on how much the spring stretches – from question 23, the spring potential energy is equal to 1⁄2 Fx = 1⁄2 kx2. Therefore option C and option D are incorrect, as they imply no change in the elastic potential energy. Therefore, the potential energy of the mass of each spring is proportional to the square of the displacement. If the weight force acting on the spring is proportional to the displacement, then it cannot be proportional to the elastic potential energy stored by the spring. Option A is clearly incorrect. Hence, when the mass is doubled, then the spring potential energy must be quadrupled. Option B is therefore correct.
Buoyancy is the phenomenon that causes separation of fluids based on their density (ρ). It also assists in explaining why some objects float, some sink, and objects such as icebergs have varying degrees of protrusion above the surface of the fluid in which they are immersed. The specific gravity of an object also helps define the buoyancy of an object relative to any fluid. The specific gravity (S.G) of an object is given by: given by:
The specific gravity can then be used to determine the percentage of the object submerged in a fluid when multiplied by one hundred. The density of some common fluids and solids are ice (900 kg/dm^3), brine (1100 kg/dm^3), water (1000 kg/dm^3), cooking oil (750 kg/dm^3), wood (300 kg/dm^3), and metal (2000 kg/dm^3).
Question 1: An unknown material is placed in cooking oil and 66% of the object is submerged. The same material is placed in water and only 50% is submerged. What is the density of this unknown material?
Question 2: If an object is submerged completely when placed in water, and only 95% submerged in brine, what is its specific gravity relative to water and brine respectively?
Solution to 1: D. We use our specific gravity ratios here as this equates to the percentage submerged. So, 0.66 * 750kg/dm3, and 0.5 * 1000kg/dm3=500kg/dm3.
Solution to 2: B. Density of the object must be greater than water but less than brine. Specific gravity is >1 when the object is more dense than the fluid and is <1 when the object is less dense than the fluid. A specific gravity of >1 means that the object will be fully submerged, but a specific gravity of <1 will mean that the object is only partially submerged. Hence A or B must be the correct answer, and B predicts the exact SG with respect to brine, because if the object if 95% submerged in brine, then the specific gravity of the object will be 0.95 relative to brine.
Positron Emission Tomography (PET) scanning is a nuclear medicine scan that is used to observe highly metabolic processes in the body such as malignancies. The patient ingests a biologically active molecule (usually an analogue of glucose) which contains a positron-emitting radionuclide (tracer) which also emits gamma rays which are then detected. Three-dimensional images of the tracer concentration within the body are then extrapolated by the PET-scanner with the aid of a CT scan performed at the same time.
Positron emission is a form of beta-particle decay whereby a proton emits a positron (a positively charged β particle) to become a neutron. Radionuclides used in PET scanning usually have short half-lives, such as carbon-11 (20 min), nitrogen-13 (10 min), oxygen-15 (2 min), fluorine-18 (110 min), and gallium-68 (55 min). Ionising radiation dose is measured in Gray (Gy), and 1Gy is defined as the absorption of one joule of radiation per kilogram of matter.
Question 1: If a 2kg sample of gallium-68 is manufactured at 9 am, to be used at 12:40 pm, approximately what amount of the original sample is active and available by this time for use?
Question 2: During the PET scan, gallium emits 1×10-10 J every second. If the scan lasts for 10 minutes for a 120 kg patient, what is the radiation absorbed by the patient, assuming that all the emitted radiation is absorbed?
A. 7.2 × 10-12 Gy.
B. 5.0 × 10-12 Gy
C. 7.2 × 10-12 Gy
D. 5.0 × 10-12 Gy
Solution to 1: B. This time span is 220 min which equates to 4 half-lives. To calculate the proportion of Ga-68 remaining, we can multiply 1⁄2 by itself 4 times. This gives 1/(24) = 1/16 ~ 6.25 %. This means that approximately 6.25 % of the original sample will remain. Given the original sample was 2 kg, this leaves 0.125 kg.
Solution to 2: D. For the duration of the scan (10min) the total energy gallium emits is (1× 10-10 × 60 × 10), divide this by the mass of the patient (120 kg) to solve for the radiation absorbed in Gy. Option C is obtained by multiplying the total energy by the mass of the patient, but this approach is incorrect because we have already calculated the total energy emitted.
Certain cardiac conditions called arrhythmias can cause the heart to beat less efficiently. One of the most life-threatening arrhythmias is known as ventricular fibrillation and has a high mortality. A defibrillator can deliver electric energy to the heart with the goal of depolarising the myocardium and reverting the heart back to a normal sinus rhythm.
A defibrillator consists of a power source, a capacitor to store charge from the power source, and a pair of electrodes called paddles that are placed either side of the chest to deliver the capacitor’s stored charge through the heart. Note that an inductor is simply the equivalent of a variable resistor.
In Figure 1, when the switch is placed at point A, the capacitor is connected to the power supply and will charge, and when the switch is placed at point B the capacitor will discharge. As the capacitor is being charged, the current changes with respect to time, and is described by the following relationship:
Where RC (also known as the time constant) is the product of the circuit resistance and the capacitor’s capacitance. The capacitance of a capacitor is given by:
Where Q is the charge (in coulombs, C), V is the voltage (in volts, V), and C is the capacitance (in Farads, F). Additionally, the energy stored on a capacitor is given by 0.5CV2. In one particular defibrillator, the capacitance is 20.0 μF and is charged to 4 kV.
Question 1. How much charge is stored in the capacitor when it is fully charged?
Question 2: The capacitor is allowed to discharge and release all of its energy. What is the magnitude of the energy released?
Solution to 1: C. Use the given equation C = Q/V and solve for Q taking care to account for units. Voltage must be in Volts and capacitance must be in Farads (F).
Q = C x V
Q = 20.0µF x 4kV
Q = 20x10-6 F x 4 x 103 V
Q = 80 x 10-3 C
Q = 8.0 x 10-2 C
Solution to 2: A. Use the given equation E = 0.5CV2 and solve for E taking care to account for units.
E = 0.5 x 20.0µF x (4kV)2
E = 0.5 x 20x10-6 F x (4 x 103 V)2
E = 10 x 10-6 F x 42 x (103 ) 2 V
E = 1 x 10-5 F x 16 x 106 V
E = 16 x 10 = 160 J
A person X’s hearing characteristics are shown in the graph below. Each curve shows the loudness (on the vertical axis in dB) of frequencies that she hears to have the same loudness. The uppermost curve shows the threshold of pain, and lowermost curve shows the threshold of audibility
Question 1: Simon works in a noisy environment. He is exposed to a sound with a frequency of 200 Hz and a loudness of 125 dB. He is required by law to wear earmuffs that reduce the sound such that he can only just hear the sound. How much do the earmuffs reduce the loudness of the sound?
Question 2. If the earmuffs only brought down the loudness to 50 dB, by what factor would the earmuffs reduce the sound intensity?
Solution to 1:B. At 200 Hz the difference between the threshold of pain and of hearing is 125 dB – 25 dB=100 dB. This can be easily seen on the graph.
Solution to 2: B. For a sound difference of 75 dB (found on the graph, difference between pain threshold and 50dB) you would have to change the frequencies by 1 x 107.5. This equates to 3x107
Refraction occurs at the interface between two mediums with different optical densities. The angle of refraction is the angle between the light ray and an imaginary line perpendicular to the interface (the normal). When light travels from a less optically dense to a more optically dense medium, it bends towards the normal. The opposite is true for light travelling to a less dense medium. Snell’s law, shown below, allows us to predict the angle at which light will refract
Where n1, and n2 are the refractive indices of the mediums (a higher refractive index is a denser medium), and θ1 and θ2 are the angles of incidence and refraction respectively (where the angle is between the normal and the light ray).
Question 1. A ray of light travels from air through a perspex prism as shown in Figure 1 (drawn to scale). Which of the following is the correct ray diagram of the path of the light through the medium? Note that the incident ray is shown as the dashed arrow in Figure 1.
Question 2. A person standing on top of the plastic block described above is looking down onto a rectangular object below. From their position of perspective, the object appears to be
Solution to 1: B. Options A and C are both incorrect, because the light itself is traveling perpendicular to the tangent of the semicircle and therefore the angle of incidence at the interface between the air and the perspex prism is zero. Therefore, the left hand side of the equation describing Snell’s law will equal zero. It follows that the right hand side will equal zero as well. Option D is incorrect, as it is not a possible outcome if Snell’s Law is applied.
Solution to 2: D. The block itself does not have a curved surface and all light rays are therefore refracted at identical angles. The image is therefore not inverted. Hence, options A and C are incorrect. Magnification also requires that rays of light passing through the material are refracted at different, that is, divergent angles. Some degree of scatter is expected if the medium is not 100% transparent, so D is the most logical answer.
Preparing for the GAMSAT physics section is an iterative process that involves a cycle of practice, reflection, and adjustment. Start by building a strong foundation in the basic concepts, formulas, and units. Then, identify your weaknesses and develop a targeted strategy to address them. Finally, apply your strategy through continuous practice and adjust as needed.
Remember, the key to acing the GAMSAT physics section is not just understanding the material but also being able to apply your knowledge quickly and accurately under exam conditions. Therefore, mental maths practice and timed exercises should be integral parts of your preparation.
By following these steps, you'll be well on your way to mastering the GAMSAT physics section in 2024. Good luck!
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