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Here are 10 fundamental Physics questions you need to practice before your VCE exams!
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To succeed in your Physics VCE exam, you need to do more than memorise formulas. You need to tackle complex problems in both short answer and extended response questions with clear, logical reasoning.
Here are 10 fundamental VCE Physics questions that cover key topics like mechanics, electromagnetism, and nuclear physics to help you practise structured responses that address all parts of the question.
Table of contents:
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Fundamental topics like circular motion, Newton’s laws, and energy conservation remain crucial.
These changes mean you need to apply physics concepts to real-world situations and solve problems across different topics.
Doing practice questions during your revision will help you succeed in VCE Physics. It helps:
Many students find certain VCE Physics topics difficult, including:
The exam covers Units 3 and 4 content but you’ll still need the foundational knowledge from Units 1 and 2. Key science skills from the study design may also be tested.
Here are 10 short answer Physics questions on key topics in the VCE Physics study design.
1. Mechanics: Projectile Motion (4 Marks)
A ball is launched from the top of a 25 m high cliff with an initial velocity of 20 m/s at an angle of 35° above the horizontal.
a) Calculate how far the ball lands from the foot of the cliff.
b) Explain how air resistance would affect the ball’s range if it were taken into account.
2. Special Relativity: Frames of reference (4 marks)
A train passes through a station at 70% of the speed of light. According to a passenger on the train, the length of the train carriage is 20 metres from front to back.
a) A light switches on inside the carriage. Compare the velocity of the emitted light as seen by the passenger and by a worker standing on the platform.
b) What length of the carriage is observed by the worker?
3. Electromagnetism: Magnetic Force (5 Marks)
A proton travels at 4 × 10⁶ m/s through a magnetic field of strength 0.8 T.
a) Calculate the magnetic force acting on the proton if it enters at right angles to the field.
b) Describe the path the proton will follow as a result of this force, and justify your answer.
4. Special Relativity: Mass and Energy (4 Marks)
A sample of a radioactive element is weighed every hour and found to be decreasing in mass over time.
a) Outline what has happened to the “lost” mass.
b) If the mass is measured to decrease by 0.020 µg over 30 days, calculate the average power emitted.
5. Data Analysis: Projectile Motion Graph (5 Marks)
The graph below shows the vertical displacement of a ball over time as it is launched into the air and returns to the ground.
a) Using the graph, determine the time the ball reaches its maximum height and the maximum vertical displacement.
b) Can the range of the projectile be determined from this graph? Why or why not?
6. Mechanics: Conservation of Momentum (6 Marks)
A 1500 kg car travelling at 20 m/s collides head-on with a 1000 kg car travelling at 15 m/s in the opposite direction. The two cars stick together as a result of the collision.
a) Calculate the final velocities of both cars.
b) Explain how the principles of momentum conservation and energy conservation apply in this situation.
7. Electromagnetic Induction: Faraday’s Law (6 Marks)
A coil with 50 turns is placed in a magnetic field that decreases from 0.6 T to 0.1 T over 0.1 s.
a) Calculate the magnitude of the induced electromotive force (emf) in the coil.
b) Explain how Faraday’s law and Lenz’s law apply to this situation.
8. Mechanics: Circular Motion (6 Marks)
A 0.5 kg object moves in a circular path with a radius of 2 m at a speed of 10 m/s.
a) Calculate the centripetal force acting on the object.
b) Discuss how the required force would change if the mass or speed of the object increased.
9. Wave Interference: Double-Slit Experiment (7 Marks)
The diagram shows the interference pattern produced by a double-slit experiment.
a) Using the diagram, calculate the wavelength of light if the distance between the slits is 0.5 mm and the distance between the bright fringes on the screen is 2 cm. The screen is located 1 m from the slits.
b) Explain how changing the slit separation would affect the interference pattern.
10. Electromagnetism: Magnetic Flux and Induced Current (8 Marks)
A rectangular loop of wire with dimensions 0.2 m by 0.5 m is placed in a magnetic field of 0.4 T. The magnetic field is perpendicular to the plane of the loop.
a) Calculate the magnetic flux through the loop when the field is constant.
b) The loop is pulled out of the magnetic field in 0.25 seconds. Calculate the induced emf and explain the effect of the orientation of the loop on the magnitude of the induced current.
1. Mechanics: Projectile Motion (4 Marks)
Solution:
a) First, resolve the initial velocity into horizontal and vertical components:
Next, calculate the time of flight using the vertical motion equation:
\(s_{y} = u_{y} t + \frac{1}{2} at^2\)
where \(a = g = -9.8\) m/s² and \(s_y = -25\) m (since the ball falls 25 m below its initial height).
Solving for t using the quadratic equation, we take the positive solution of \(t = 3.7\) seconds.
Now, use \(s_{x} = u_{x} t\) to find the horizontal range:
\(s_x = 16.4 \times 3.7 s = 60.7\)
The range of the projectile is 60.7 metres.
b) Air resistance would continuously reduce both horizontal and vertical velocities. The ball would experience deceleration due to drag, leading to both a reduced horizontal velocity and a reduced time before hitting the ground. These factors both result in a shorter range compared to the case without air resistance.
Why this is a strong response:
Tip:
2. Special Relativity: Frames of Reference(4 Marks)
Solution:
a) Both observers must measure the same speed of light according to Einstein’s Special Theory of Relativity.
b) The proper length of the carriage is 20 metres, as measured by the passenger. The worker on the platform observes the train undergoing relativistic length contraction. The value of \(\gamma\) in this case is:
\(\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\)
\(\gamma = \frac{1}{\sqrt{1-\frac{(0.7c)^2}{c^2}}}\)
\(\gamma = 1.4\)
Therefore the contracted length is
\(L = \frac{L_0}{\gamma} = \frac{20}{1.4} = 14.3 m\)
The worker sees the carriage as being 14.3 metres long.
Why this is a strong response:
Tip:
3. Electromagnetism: Magnetic Force (5 Marks)
Solution:
a) The magnetic force acting on the proton is calculated using:
\(F = qvB\)
\(F = 1.60 \times 10^{-19} \times 4 × 10^{6} \times 0.8\)
\(Force = 5.1 × 10^{-13} N\)
b) The proton will follow a circular path due to the magnetic force always acting perpendicular to the velocity and having a constant magnitude. The right-hand rule can be used to predict the direction of the proton clockwise or anticlockwise depending on the direction of the field lines and initial velocity.
Why this is a strong response:
Tip:
4. Special Relativity: Mass and Energy (4 Marks)
Solution:
a) The radioisotope has undergone a number of nuclear decays, which have rearranged the nucleons into a product with lower rest mass. The difference in mass has been converted to the kinetic energy of the products.
b) The equation \(E = mc^2\) can calculate the energy release associated with the decreased rest mass.
Converting to SI units:
\(0.020 \mu g = 0.02 \times 10^-6 kg\)
The total energy emitted is:
\(E = 0.020 \times 10^-6 \times (3 \times 10^8)^2\)
\(E = 1.8 \times 10^9 J\)
The average power in watts is the energy divided by the number of seconds (in 30 days, there are \(30 \times 24 \times 60 \times 60\) seconds
\(P = \frac{E}{t} = \frac{1.8 \times 10^9}{30 \times 24 \times 60 \times 60}\)
\(P = 690 W\)
Why this is a strong response:
5. Data Analysis: Projectile Motion Graph (5 Marks)
Solution:
a) From the graph, the maximum vertical displacement is 15 m, which occurs at 3 seconds (halfway through the trajectory).
b) No. In order to calculate the range of the projectile, we need to know the horizontal velocity in addition to the time of flight. This graph does not contain information about the horizontal motion, which is independent of the vertical motion.
Why this is a strong response:
Tip:
6. Mechanics: Conservation of Momentum (6 Marks)
Solution:
a) Using the principle of conservation of momentum:
\(m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2}\)
Substitute the known values to find the total initial momentum, noting that velocity is a vector:
\(m_{1} = 1500 kg, u_{1} = 20 ms^{-1}\)
\(m_{2} = 1000 kg, u_{2} = -15 ms^{-1}\)
\(p_{total} = 1500 \times 20 + 1000 \times -15 = 15 000 kgms^{-1}\) in the direction of the more massive car
The cars stick together after the collision, acting like one combined mass of 2500 kg. Hence the final velocity must be
\(v = \frac{p}{m} = \frac{15000}{2500} = 6 ms^{-1}\)
b) Momentum is always conserved. This means that the total momentum before the collision is equal to the total momentum after the collision, which allows us to calculate the final speed of the wreck. Energy is also conserved but in an inelastic collision such as this, a portion of kinetic energy is converted to other types such as sound and heat.
Why this is a strong response:
Tip:
7. Electromagnetic Induction: Faraday’s Law (6 Marks)
Solution:
a) The induced electromotive force (emf) is calculated using Faraday’s Law. The flux changes by -0.5 Wb over 0.1 seconds, and there are 50 turns in the coil.
\(\varepsilon = -N \frac{\Delta \Phi_{B}}{\Delta t}\)
\(\varepsilon = -50 \times \frac{-0.5}{0.1} = 250 V\)
The question requires a magnitude only, so the answer is 250 V.
b) Faraday’s Law shows that the rate of change of the magnetic flux determines the magnitude of emf in the loop. According to Lenz’s Law, the direction of the induced current will oppose the change in magnetic flux.
Why this is a strong response:
Tip:
8. Mechanics: Circular Motion (6 Marks)
Solution:
a) The centripetal force is calculated using:
\(F=\frac{mv^{2}}{r}\)
\(F=\frac{0.5 \times 10^{2}}{2} = 25 N\)
b) If the mass or speed of the object increases, the centripetal force also increases. The centripetal force increases linearly with mass, as the inertia of the object increases. Force is proportional to the square of the velocity. Doubling the speed would quadruple the required force to maintain circular motion.
Why this is a strong response:
Tip:
9. Wave Interference: Double-Slit Experiment (7 Marks)
Solution:
a) The wavelength of light can be calculated starting from this formula on the formula sheet:
\(\Delta x = \frac{\lambda L}{d}\)
And rearranging to make wavelength the subject:
\(\lambda = \frac{d \Delta x}{L}\)
\(\lambda = \frac{0.5 \times 10^{-3} \times 0.02}{1}\)
\(\lambda = 1.0 \times 10^{-5} m\)
b) For a given wavelength, increasing the slit separation d reduces the distance between the bright fringes \(\Delta x[latex], leading to a narrower interference pattern. The interference maxima will be closer to each other.
Why this is a strong response:
Tip:
10. Electromagnetism: Magnetic Flux and Induced Current (8 Marks)
Solution:
a) When field lines are perpendicular to the plane of the loop, the magnetic flux through the loop is calculated using:
[latex]\Phi_{B} = B \perp A\)
\(\Phi_{B} = 0.4 \times (0.2 \times 0.5) = 0.04 Wb\)
b) Removing the loop from the field drops the flux to zero. The induced emf is calculated using Faraday’s Law with N = 1 for a single loop:
\(\varepsilon = -N \frac{\Delta \Phi_{B}}{\Delta t}\)
\(\varepsilon = -1 \frac{-0.04}{0.25} = 0.16 V\)
According to Lenz’s Law, the negative sign indicates that the induced current will oppose the change in the magnetic flux. The orientation of the loop is critical: when the loop is perpendicular to the magnetic field, the flux is maximised, leading to a greater induced emf and therefore current. If the loop is angled or moved out of the field, the maximum flux changes, and the induced emf adjusts accordingly.
Why this is a strong response:
Tip:
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