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Physics Structured Questions (Internal) Exam

Topics to work on

  • Graph skills and calculations
    • Finding the spring constant (k) from different axes
    • Interpreting y-intercepts (total length graphs do not start at 0; always subtract original length to find extension)
    • Graph scale precision
  • Speed-time graphs and motion
  • Energy transfers
  • Properties of matter (Hooke’s law)
    • The limit of proportionality
    • Limit of proportionality vs. elastic limit
  • Volume

Structured questions

  1. A student adds 20 drops of water to the water that is in the measuring cylinder in Fig 1.1 (for reference the volume of water inside is 21cm³). The new volume of water in the measuring cylinder is 25cm³. Calculate the average volume of one drop of water.

    • Correct working:
      • Added vol. = 25cm³ - 21cm³ = 4cm³
      • Avg. = 4 / 20 = 0.2cm³ for one drop
    • Feedback: You wrote 20cm³ when you were trying to find the volume added when the drops were added.

  2. Fig 2.3 shows the speed-time graph (view image in 2026/sq) for another car. Calculate the distance travelled by this car between time = 2.0s and time = 6.0s.

    • Correct working:
      • In the graph, draw two lines that go down to 2.0s and 6.0s. That is your area to calculate.
      • In order to calculate the area of the part between 2.0s and 6.0s, you draw a triangle and a rectangle. You can label either of those as A₁ and A₂. A₁ will be the triangle, and A₂ will be the rectangle for this working out.
        • With this, calculate A₁’s area. The formula is (b * h)/2. To apply the formula for speed, s = d/t, you will replace “b” and “h” with “speed” and “time”, forming (s * t)/2.
        • (12 * 4)/2 = 24m
        • Now, calculate A₂’s area. The formula for calculating this area is simply bh, but replace those with the variables inside the formula for calculating distance to make it st.
        • 6 x 4 = 24m
      • With our values for A₁ and A₂, we can simply add those together to find the total distance travelled within 2.0s and 6.0s:
        • 24m + 24m = 48m
    • Feedback: In your previous working, you only found the area for the triangle, leaving out the rectangle, which caused it to be wrong.

  3. i) Describe two useful energy transfers when the sailor uses the winch to raise the sail.

    • Correct answers:
      • Chemical energy to kinetic energy. The chemical energy stored in the sailor’s muscles are transferred into kinetic energy of the winch handle and the internal gears.
      • Kinetic energy to gravitational potential energy. The kinetic energy of the winch as it rotates is transferred into kinetic energy of the rope attached to the sail and the sail which lifts the sail upward.
    • Feedback:
      • You simply listed “kinetic energy” and “gravitational potential energy”. This is wrong because the questions asks you to DESCRIBE. The question also asks you for ENERGY TRANSFERS, NOT only the types of energy involved.

  4. ii) Describe one non-useful energy transfer when the sailor uses the winch to raise the sail.

    • Correct answer:
      • Kinetic energy to thermal and sound energy. As the sailor turns the winch and the rope tightly wraps around the winch drum, moving parts rub against each other. This causes friction and it also causes heat. Some energy is also transferred as sound waves from the mechanical components and from the rope.
    • Feedback:
      • Same mistakes as before, did NOT describe the energy transfers and only listed one energy type.

  5. State what is meant by the limit of proportionality.

    • Correct answer:
      • The limit of proportionality is the point up to which the extension of an elastic object is directly proportional to the force applied to it.
    • Feedback:
      • You wrote “The limit of proportionality is when the elasticity of an elastic object reaches the limit of the elastic limit”. This no longer obeys Hooke’s law.

  6. Using Fig. 6.2 (refer to the image in the 2026/sq folder), determine the spring constant of this spring.

    • Correct working:
      • On the graph, weight (F) is on the x-axis and length (L) is on the y-axis. We will calculate the gradient of the line on the graph. The formula will be Δy/Δx.
      • The change in length is basically exstension. Therefore, the gradient formula would be extension / weight, x/F.
      • We need to find the spring constant, so the formula for Hooke’s law becomes k = F/x.
        • The gradient is the upside-down version of the spring constant. Therefore, k effectively becomes k = 1/gradient.
      • First, pick two points on the straight-line section of the graph.
        • Point 1 will be (0N, 0.12m).
        • Point 2 will be (10.8N, 0.58m).
      • Second, calculate Δy and Δx.
        • Δy = 0.58 - 0.12 = 0.46m
        • Δx = 10.8 - 0 = 10.8m
      • Third, find the gradient.
        • Δy/Δx = 0.46/10.8 = 0.043N/m
      • Fourth, invert the gradient to find the spring constant k.
        • k = 1/gradient = 1 / 0.043N/m = approx. 23.25N/m.