Physics · OCR H556 · Topic 2.3

Nature of quantities

Scalars and vectors.

  • 1 sub-topic
  • Taught in Year 12
  • 19 practice questions
  • 1 animated lesson
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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. 2.3.1Scalars and vectorsY12

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Easy and hard questions from this topic, each with a worked solutions and the mark for every step. All 19 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}}

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Mechanics

Hard

Forces on a slope · Year 12 · 10 marks

Figure 1 shows a block P of mass 5.0 kg on a slope inclined at 30° to the horizontal.
30°P
Figure 1
  1. (a)
    Calculate the components of the weight of P parallel and perpendicular to the slope.
    (3)
  2. (b)
    State the normal contact force on P, and explain your answer.
    (2)
  3. (c)
    The slope is smooth and P is released from rest. Calculate its speed after it has moved 2.0 m down the slope.
    (3)
  4. (d)
    In practice a frictional force of 12.3 N acts on P. Calculate the acceleration of P.
    (2)

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Part (a)3 marks

  1. Weight = mg.W=49.1 NW = 49.1\ \mathrm{N}
  2. Parallel to the slope.Wsin⁡α=24.5 NW\sin\alpha = 24.5\ \mathrm{N}
  3. Perpendicular to the slope.Wcos⁡α=42.5 NW\cos\alpha = 42.5\ \mathrm{N}

Part (b)2 marks

  1. It balances the perpendicular component.N=42.5 NN = 42.5\ \mathrm{N}
  2. Newton's first law in that direction.There is no acceleration perpendicular to the slope, so the forces in that direction balance.

Part (c)3 marks

  1. Only the parallel component accelerates the block.aa=gsin⁡α{}= g\sin\alpha=4.90 m s−2{}= 4.90\ \mathrm{m\,s^{-2}}
  2. Constant acceleration from rest.v2v^2=2as{}= 2as=2×4.90×2.0{}= 2 \times 4.90 \times 2.0
  3. v=4.43 m s−1v = 4.43\ \mathrm{m\,s^{-1}}

Part (d)2 marks

  1. Resultant force down the slope.ma=Wsin⁡α−F=24.5−12.3ma = W\sin\alpha - F = 24.5 - 12.3
  2. a=2.45 m s−2a = 2.45\ \mathrm{m\,s^{-2}}

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