Physics · AQA 7408 · Topic 3.4

Mechanics and materials

Vectors, moments, motion graphs, the suvat equations, projectiles, Newton's laws, momentum, energy and the properties of materials.

  • 10 sub-topics
  • Taught in Year 12
  • 217 practice questions
  • 10 animated lessons
Practice questions (217) Lessons Worksheets All AQA Physics topics

What the specification says

Every sub-topic AQA 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. 3.4.1 Force, energy and momentum
  2. 3.4.1.1Scalars and vectorsY12
  3. 3.4.1.2MomentsY12
  4. 3.4.1.3Motion along a straight lineY12
  5. 3.4.1.4Projectile motionY12
  6. 3.4.1.5Newton's laws of motionY12
  7. 3.4.1.6MomentumY12
  8. 3.4.1.7Work, energy and powerY12
  9. 3.4.1.8Conservation of energyY12
  10. 3.4.2 Materials
  11. 3.4.2.1Bulk properties of solidsY12
  12. 3.4.2.2The Young modulusY12

Try 3 questions

Easy, medium and hard questions from this topic, each with a worked solutions and the mark for every step. All 217 questions in this topic, free, no account needed.

Mechanics

Easy

Scalars and vectors · Year 12 · 7 marks

A walker goes 9 m due east and then 12 m due north, taking 50 s in total.
  1. (a)
    Explain the difference between a scalar and a vector quantity, giving an example of each from this situation.
    (2)
  2. (b)
    Find the walker's displacement from the start.
    (3)
  3. (c)
    Find the average speed and the magnitude of the average velocity.
    (2)

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Part (a)2 marks

  1. A scalar has magnitude only (e.g. distance); a vector has magnitude and direction (e.g. displacement).
  2. Distance travelled and time are scalars; displacement and velocity are vectors.

Part (b)3 marks

  1. 92+122\sqrt{9^2 + 12^2}
  2. s=15.0 ms = 15.0\ \mathrm{m}
  3. tan θ = east/north.on a bearing of 037°

Part (c)2 marks

  1. speed\text{speed}=0.420 m s−1{}= 0.420\ \mathrm{m\,s^{-1}}
  2. ∣vˉ∣|\bar{v}|=0.300 m s−1{}= 0.300\ \mathrm{m\,s^{-1}}

Total7

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Mechanics

Medium

Horizontal projection · Year 12 · 3 marks

A ball is thrown horizontally at 5.0 m s−15.0\ \mathrm{m\,s^{-1}} from a height of 1.8 m above level ground. Ignore air resistance.
  1. Calculate the horizontal distance the ball travels before it hits the ground.
    (3)
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  1. Treat vertical motion on its own: it starts with zero vertical velocity and accelerates at gg.1.81.8=12×9.81×t2{}= \tfrac{1}{2} \times 9.81 \times t^2
  2. Solve for the time of flight.t=0.606 st = 0.606\ \mathrm{s}
  3. Horizontal velocity is constant, so distance = speed × time.Distance =5.0×0.606= 5.0 \times 0.606=3.0 m{}= 3.0\ \mathrm{m}

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Mechanics

Hard

Projectile motion · Year 12 · 11 marks

Figure 1 shows a ball kicked from level ground at O with speed 28 m s⁻¹ at 30° above the horizontal. It lands at A. Air resistance is negligible.
28 m s⁻¹30°OA
Figure 1
  1. (a)
    Calculate the horizontal and vertical components of the initial velocity.
    (2)
  2. (b)
    Calculate the time of flight.
    (3)
  3. (c)
    Calculate the horizontal distance OA.
    (2)
  4. (d)
    Calculate the maximum height reached.
    (2)
  5. (e)
    In practice air resistance acts. State and explain its effect on the distance OA.
    (2)

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Part (a)2 marks

  1. Horizontal: u cos θ.ux=24.2 m s−1u_x = 24.2\ \mathrm{m\,s^{-1}}
  2. Vertical: u sin θ.uy=14.0 m s−1u_y = 14.0\ \mathrm{m\,s^{-1}}

Part (b)3 marks

  1. The ball lands when its vertical displacement is back to zero.sy=uyt−12gt2=0s_y = u_y t - \tfrac{1}{2}gt^2 = 0
  2. Solve for the non-zero time.tt=2uyg{}= \frac{2u_y}{g}=2×14.09.81{}= \frac{2 \times 14.0}{9.81}
  3. t=2.85 st = 2.85\ \mathrm{s}

Part (c)2 marks

  1. Horizontal velocity is constant.OA=uxt=24.2×2.85OA = u_x t = 24.2 \times 2.85
  2. OA=69.2 mOA = 69.2\ \mathrm{m}

Part (d)2 marks

  1. At the top the vertical velocity is zero.00=uy2−2gH{}= u_y^2 - 2gH⇒H{}\Rightarrow H=14.022×9.81{}= \frac{14.0^2}{2 \times 9.81}
  2. H=9.99 mH = 9.99\ \mathrm{m}

Part (e)2 marks

  1. State the effect.OA is smaller.
  2. Give the reason in terms of the force.Air resistance opposes the motion, so the horizontal velocity falls during the flight (and the flight is shorter).

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