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Grade 12 Physics entrance exam, 2006 E.C. (2014)

Real questions from the Grade 12 university entrance exam in Physics, as students sat it in 2006 E.C. Every question comes with the correct answer and a worked explanation.

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  • Natural science stream

    Who sits it

  • 2006 E.C. (2014)

    Exam year

  • 48 questions with answers

    In this set

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Answer each question yourself before opening the answer. The explanations point to the exact place in the textbook, so you know which page to reread when you miss one.

Question 11

Two projectiles are fired from ground level at equal speed but different angles. One is fired at an angle of 30 degrees and the other at 60 degrees. The projectile to hit the ground first will be the one fired at (neglect air resistance)

  1. A.60 degrees
  2. B.30 degrees
  3. C.both hit at the same time
  4. D.cannot be determined from the given condition.
Show the answer and explanation

Answer: B 30 degrees

Why

The time of flight of a projectile launched from ground level is given by T = (2v₀ sin θ)/g. Since sin(30°) = 0.5 is less than sin(60°) ≈ 0.866, the projectile fired at 30° spends less time in the air and hits the ground first.

Question 12

A planet of mass m is circling a star in an elliptical orbital as shown below. Its velocity at point A is 60,000m/s and rb=8rar_b = 8r_a what is its velocity at point B? [Figure: A hand-drawn wide ellipse lying horizontally. A small open circle marks the left-hand end of the major axis, labelled A just outside it; another small open circle marks the right-hand end, labelled B just outside it. A star symbol sits on the major axis, well left of centre, representing the star at one focus. A straight horizontal line runs along the major axis from A through the star to B. The segment from A to the star is labelled rar_a (written above the line, left of the star) and the much longer segment from the star to B is labelled rbr_b (written above the line, right of the star). No numeric distances are marked on the drawing.]

  1. A.7,500m/s
  2. B.8,571m/s
  3. C.60,000m/s
  4. D.788,000m/s
Show the answer and explanation

Answer: A 7,500m/s

Why

By conservation of angular momentum at the apsides of an elliptical orbit, m*v_a*r_a = m*v_b*r_b. Therefore, v_b = v_a * (r_a / r_b) = 60,000 m/s / 8 = 7,500 m/s.

Question 13

A small radio receiver uses a battery that delivers a constant current of 25mA for 3 hours. Calculate the total charge delivered by the battery.

  1. A.1080000C
  2. B.270000
  3. C.540C
  4. D.270C
Show the answer and explanation

Answer: D 270C

Why

Total charge is given by Q = I * t. Converting the units gives Q = (0.025 A) * (3 * 3600 s) = 270 C.

Question 14

A cup of coffee at a temperature of 80°C is placed on table in a 20°C room. What happens to the thermal energy and molecular motion of the coffee when it is left on the table?

  1. A.The thermal energy decreases and the molecules move more slowly.
  2. B.The thermal energy increases and the molecules move more slowly.
  3. C.The thermal energy decreases and the molecules move faster.
  4. D.The thermal energy increases and the molecules move faster.
Show the answer and explanation

Answer: A The thermal energy decreases and the molecules move more slowly.

Why

As the hot coffee cools down toward room temperature, it loses heat to the surroundings, so its thermal energy decreases. Because temperature is directly proportional to the average kinetic energy of the molecules, the coffee molecules move more slowly.

Question 17

A mass suspended from the end of a spring vibrates up and down 20 times in 36s. What are the frequency and period of the vibration?

  1. A.0.55Hz, 1.8S
  2. B.0.67Hz, 1.5S
  3. C.0.0012Hz, 864S
  4. D.864Hz, 0.0012S
Show the answer and explanation

Answer: A 0.55Hz, 1.8S

Why

Frequency is the number of cycles per second: f = 20 / 36 s ≈ 0.55 Hz. The period is the reciprocal of the frequency: T = 36 s / 20 = 1.8 s.

Question 18

Determine the total impedance for the circuit shown below [Figure: A hand-drawn single-loop series AC circuit. Along the top branch, from left to right: a rectangular resistor box labelled 4Ω4\Omega above it, then a coil (zig-zag/solenoid symbol) labelled 27mH above it, then a pair of parallel vertical plates (capacitor symbol) labelled 380μ above it. The top branch runs from the top-left corner to the top-right corner of the loop; vertical wires run down both sides. Along the bottom branch there is a source symbol drawn as two small open circles with a sine-wave symbol between them (an AC generator), labelled 110v in large text just above it. No frequency is marked anywhere on the figure, and the capacitor's unit letter after μ is not printed.]

  1. A.0Ω0\Omega
  2. B.4Ω4\Omega
  3. C.27Ω27\Omega
  4. D.31Ω31\Omega
Show the answer and explanation

Answer: B 4Ω4\Omega

Why

The resonant frequency of the circuit is f₀ = 1 / (2π√(LC)) = 1 / (2π√(0.027 H * 380 μF)) ≈ 49.7 Hz, which corresponds to the standard 50 Hz AC supply frequency. Near resonance, inductive and capacitive reactances cancel (X_L ≈ X_C), leaving the total impedance equal to the resistance, Z ≈ R = 4 Ω.

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Physics papers from other years

2018 E.C. (2026)2017 E.C. (2025)2016 E.C. (2024)2015 E.C. (2023)2014 E.C. (2022)2013 E.C. (2021)2012 E.C. (2020)2011 E.C. (2019)2010 E.C. (2018)2009 E.C. (2017)2008 E.C. (2016)2007 E.C. (2015)2005 E.C. (2013)

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Are these the real 2006 E.C. Physics entrance exam questions?

Yes. They come from the national university entrance exam in Physics that students sat in 2006 E.C. (2014). Our set holds 48 of its questions, each with its answer and an explanation that points back to the textbook.

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Can I download the Physics 2006 E.C. paper as a PDF?

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