VCE Physics Units 1 and 2: 10 Practice Questions

VCE Physics Units 1 and 2 are the foundation for everything you’ll learn in Year 12 and beyond. While the topics you cover in Year 11 might not directly appear in your Year 12 VCE exam, they are still very important. You should take this year seriously so you’re ready for what Year 12 brings.

In Year 11, you’ll learn about key topics like:

• electromagnetic radiation
• thermal energy
• motion
• nuclear physics

These introduce important ideas like how waves and particles behave, how energy is conserved, and how forces work. These concepts are the building blocks for the more advanced Year 12 topics, such as electromagnetism, quantum mechanics, and advanced mechanics.

Even if you don’t plan to study science after school, understanding how Physics connects to real-world issues like climate change, medical breakthroughs, and technology can give you a new way to think about the world.

Here are 10 quick questions to test your knowledge of Units 1 and 2!

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VCE Physics Units 1 and 2: Practice questions

Electromagnetic radiation and thermal energy

1. A light ray passes from air (n = 1.00) into water (n = 1.33) at an angle of incidence of 30°. Calculate the angle of refraction. (3 marks)

2. An object emits radiation with a peak wavelength of 2.90 μm. Calculate its temperature using Wien’s Law: \(\lambda_{max} T = 2.898 \times 10^{-3}\). (2 marks)

3. Calculate the energy required to heat 500 g of water from 25°C to 75°C. Assume \( c_{water} = 4.18 J/g/K \). (2 marks)

Nuclear energy

4. A sample of radioactive material has a half-life of 5 years. If the initial mass is 80 g, how much remains after 15 years? (2 marks)

5. Write the nuclear decay equation for the beta-minus decay of carbon-14. (2 marks)

Electricity

6. A 12 V battery is connected to two resistors, \(R_1 = 4 \Omega\) and \(R_2 = 6 \Omega\), arranged in series. Calculate:

      a. The total resistance (1 mark)

      b. The current in the circuit (2 marks)

7. Explain why household circuits are generally connected in parallel rather than in series. (3 marks)

8. A 100 W lightbulb operates on 230 V DC. Calculate the current through the bulb and the resistance of the filament. (3 marks)

Motion

9. A ball is thrown vertically upwards with an initial velocity of 15 m/s. Calculate:

      a. the time it takes to reach its highest point. (1 mark)

      b. the maximum height reached. (2 marks)

10. A 2.0 kg object moving at 6.0 m/s collides elastically with a stationary object of mass 3.0 kg. The first object rebounds at 1.2 m/s.

      a. Calculate the velocity of the 3.0 kg object after the collision. (2 marks)

      b. Is the collision elastic? Justify your answer. (2 marks)

VCE Physics Units 1 and 2: Solutions

Electromagnetic radiation and thermal energy

1. Calculate the angle of refraction (3 marks) 

Using Snell’s Law:

\(n_1 \sin \theta_1 = n_2 \sin \theta_2\)

\(1.00 \times \sin(30^{\circ}) = 1.33 \times \sin \theta_2\)

Solving for the unknown angle, we find \(\theta_2 = 22^{\circ}\)

2. Object temperature using Wien’s law (2 marks) 

Wien’s law states: \(\lambda_{max} T = 2.898 \times 10^{-3}\)

\(T = \frac{2.898 \times 10^{-3}}{\lambda_{max}}\)

Hence the temperature is 1000 K.

3. Energy to heat water (2 marks) 

The equation to calculate the required heat is: \(Q = mc \Delta T\).

The mass is given in units of grams, which matches the units of specific heat capacity; the increase in temperature is the same when expressed in Kelvin as in Celsius. Hence:

\(Q = mc\Delta T = 500 \times 4.18 \times (75 – 25)\)

The heat required is 104,500 J, or 104.5 kJ.

Nuclear energy

4. Remaining radioactive material (2 marks) 

After three half-lives, the remaining amount is one-eighth of the original value.

There are 10 grams left.

5. Beta-minus decay (2 marks) 

In beta-minus decay, the products must include the electron and antineutrino: 

\(^{14}_6 C \rightarrow ^{14}_7 N + e^- + \bar{\nu}\) 

Electricity

6. Resistors in series (3 marks)

      a. Resistances add together in series for a total \(R = 10 \Omega\)

      b. The current through the entire circuit can be determined using Ohm’s Law: \( I = \frac{V}{R} = \frac{12}{10} = 1.2 \)

The current through the circuit is 1.2 A.

7. Household circuits (3 marks)

Parallel circuits allow each device to operate independently, with a consistent voltage across each device even when others are connected or removed. A fault in one branch of the circuit does not affect the others. Overall, parallel circuits are more reliable and deliver more power to connected appliances than series circuits.

8. Lightbulb calculations (3 marks)

      a. The power dissipated by a circuit component is given by \(P=IV\)

\(I = \frac{P}{V} = \frac{100}{230} = 0.44\)

The current is 0.44 A.

      b. Resistance can be calculated using Ohm’s Law.

\(R = \frac{V}{I} = \frac{230}{0.44} = 520\)

The resistance is 520 \(\Omega\).

Motion

9. Equations of motion (3 marks)

      a. Taking upwards as positive, the initial velocity is 15 m/s and the acceleration is -9.8 m/s2.

At the highest point of its motion, the ball reaches a velocity of zero. Hence using \(v = u + at\)

\(0 = 15 – 9.8t\)

This gives the time of flight as 1.53 seconds.

      b. Using the same logic as before, we can solve for the displacement travelled when it reaches a velocity of zero. This will be the highest point. \(v^2 = u^2 + 2as\)

\(0 = 15^2 – (2 \times 9.8 \times s)\)

This gives a maximum upwards displacement of 11.5 metres.

10. Momentum and collisions (4 marks)

      a. We use the conservation of momentum equation: \(m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2\)

We will take object 1’s initial velocity to be positive. After it rebounds, the new velocity will be negative.

Knowing both masses, and the fact that object 2 is initially stationary:

\(2.0 \times 6.0 + 3.0 \times 0 = 2.0 \times -1.2 + 3.0 \times v_2\)

Solving for the final velocity of object 2, we find \(v_2 = 4.8 m/s.\)

      b. The collision is elastic if kinetic energy is conserved, that is, if:

\(\frac{1}{2} m_1 u_{1}^{2} + \frac{1}{2} m_2 u_{2}^{2} = \frac{1}{2} m_1 v_{1}^{2} + \frac{1}{2} m_2 v_{2}^{2} \)

The initial kinetic energy is: \(\frac{1}{2} \times 2.0 \times 6.0^2 + \frac{1}{2} 3.0 \times 0^{2} = 36 J\)

The final kinetic energy is: \(\frac{1}{2} \times 2.0 \times (-1.2)^2 + \frac{1}{2} 3.0 \times 4.8^{2} = 36 J\)

Hence the collision is elastic.

Conclusion

Year 11 VCE Physics Units 1 and 2 is an important step toward your future. By working hard now, you’ll prepare yourself to succeed in Year 12 and beyond. It’s your chance to build the knowledge and skills you’ll need for both academic success and careers that rely on Physics.

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