Physics · OCR H556 · Topic 3.1

Motion

Kinematics and linear motion, projectile motion.

  • 2 sub-topics
  • Taught in Year 12
  • 47 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. 3.1.1-3.1.2Kinematics and linear motionY12
  2. 3.1.3Projectile motionY12

Try 3 questions

Easy, medium and hard questions from this topic, each with a worked solutions and the mark for every step. All 47 questions in this topic, free, no account needed.

Mechanics

Easy

Falling from rest · Year 12 · 1 mark

  1. A stone is dropped from rest. Air resistance is negligible. How far does it fall in the first 2.5 s?
    (1)

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  1. From rest, s=12gt2s = \tfrac{1}{2}gt^2.C: 30.7 m30.7\ \mathrm{m}

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Mechanics

Medium

Motion graphs · Year 12 · 8 marks

Figure 1 shows the velocity–time graph for a cyclist who accelerates uniformly from rest, travels at a constant 15 m s⁻¹, then decelerates uniformly to rest.
t (s)v (m s⁻¹)1552024
Figure 1
  1. (a)
    Find the acceleration in the first stage and the deceleration in the last stage.
    (2)
  2. (b)
    Find the total distance travelled.
    (3)
  3. (c)
    Find the average speed for the whole journey.
    (1)
  4. (d)
    Sketch the acceleration–time graph for the journey.
    (2)

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Part (a)2 marks

  1. a1=3.00 m s−2a_1 = 3.00\ \mathrm{m\,s^{-2}}
  2. deceleration\text{deceleration}=3.75 m s−2{}= 3.75\ \mathrm{m\,s^{-2}}

Part (b)3 marks

  1. distance\text{distance}=area under the graph{}= \text{area under the graph}
  2. 12(15)(5)\tfrac{1}{2}(15)(5)+15(15){}+ 15(15)+12(15)(4){}+ \tfrac{1}{2}(15)(4)
  3. s=292 ms = 292\ \mathrm{m}

Part (c)1 mark

  1. vˉ=12.2 m s−1\bar{v} = 12.2\ \mathrm{m\,s^{-1}}

Part (d)2 marks

  1. Constant 3.00 for 0–5 s, zero for 5–20 s,
  2. constant −3.75 for 20–24 s.

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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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