Physics · OCR H556 · Topic 3.4

Materials

Springs, mechanical properties of matter.

  • 2 sub-topics
  • Taught in Year 12
  • 50 practice questions
  • 2 animated lessons
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What the specification says

Every sub-topic OCR can examine in this topic, with its reference number from the official specification. The Y12 and Y13 tags show when each sub-topic is usually taught.

  1. 3.4.1SpringsY12
  2. 3.4.2Mechanical properties of matterY12

Try 3 questions

Easy, medium and hard questions from this topic, each with a worked solutions and the mark for every step. All 50 questions in this topic, free, no account needed.

Materials

Easy

Hooke's law · Year 12 · 1 mark

  1. A spring has a spring constant of 50 N m⁻¹. What is its extension when it supports a load of 4.0 N?
    (1)

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  1. F=kΔLF = k\Delta L, so ΔL=F/k\Delta L = F/k, in metres.D: 8.00 cm8.00\ \mathrm{cm}

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Materials

Medium

Stress, strain and the Young modulus · Year 12 · 4 marks

A wire of length 1.5 m and diameter 0.3 mm extends by 0.8 mm under a tension of 120 N.
  1. (a)
    Calculate the stress in the wire.
    (2)
  2. (b)
    Calculate the strain.
    (1)
  3. (c)
    Calculate the Young modulus of the material.
    (1)

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Part (a)2 marks

  1. AA=7.07×10−8 m2{}= 7.07 \times 10^{-8}\ \mathrm{m^2}
  2. σ\sigma=1.70×109 Pa{}= 1.70 \times 10^{9}\ \mathrm{Pa}

Part (b)1 mark

  1. ε=5.33×10−4\varepsilon = 5.33 \times 10^{-4}

Part (c)1 mark

  1. EE=3.18×1012 Pa{}= 3.18 \times 10^{12}\ \mathrm{Pa}

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Materials

Hard

Springs and elastic energy · Year 12 · 10 marks

Figure 1 shows the force–extension graph for a spring that obeys Hooke's law up to an extension of 8 cm.
extension / cmF / N(8, 3.20)
Figure 1
  1. (a)
    State Hooke's law.
    (1)
  2. (b)
    Use the graph to calculate the spring constant.
    (2)
  3. (c)
    Calculate the elastic strain energy stored when the extension is 8 cm.
    (2)
  4. (d)
    The spring is used to launch a 50 g ball horizontally. Calculate the greatest speed the ball can reach, and state one assumption.
    (4)
  5. (e)
    Two identical springs are joined in series and the same force is applied. State the total extension, compared with one spring.
    (1)

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Part (a)1 mark

  1. Include the limit.The extension is directly proportional to the force applied, up to the limit of proportionality.

Part (b)2 marks

  1. The gradient of the line, with the extension in metres.kk=FΔL{}= \frac{F}{\Delta L}=3.200.08{}= \frac{3.20}{0.08}
  2. k=40.0 N m−1k = 40.0\ \mathrm{N\,m^{-1}}

Part (c)2 marks

  1. The area under the graph.EE=12FΔL{}= \tfrac{1}{2}F\Delta L=12×3.20×0.08{}= \tfrac{1}{2} \times 3.20 \times 0.08
  2. E=0.128 JE = 0.128\ \mathrm{J}

Part (d)4 marks

  1. All the stored energy becomes kinetic energy.12mv2=E\tfrac{1}{2}mv^2 = E
  2. Rearrange, with the mass in kg.vv=2×0.1280.05{}= \sqrt{\frac{2 \times 0.128}{0.05}}
  3. v=2.26 m s−1v = 2.26\ \mathrm{m\,s^{-1}}
  4. Any one sensible assumption.No energy is lost to friction or air resistance (or the spring has negligible mass).

Part (e)1 mark

  1. In series the force is the same in each.Twice as large: each spring carries the full force and extends by the same amount.

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