Further Mathematics · OCR H245 · Option M · Topic 6.06

Further dynamics and kinematics

Linear motion under a variable force.

  • 1 sub-topic
  • Taught in Year 13
  • 29 practice questions
  • 1 animated lesson

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. 6.06aLinear motion under a variable forceY13

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Medium and hard questions from this topic, each with a worked solutions and the mark for every step. All 29 questions in this topic, free, no account needed.

Further dynamics

Not needed for Year 12Medium

Acceleration as a function of displacement · Year 13 · 1 mark

  1. A force on a particle depends on its displacement xx, and the velocity is wanted in terms of xx. Which form of the acceleration should be used?
    (1)

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  1. dvdt\dfrac{\mathrm{d}v}{\mathrm{d}t}=dvdx×dxdt{}= \dfrac{\mathrm{d}v}{\mathrm{d}x} \times \dfrac{\mathrm{d}x}{\mathrm{d}t}=vdvdx{}= v\dfrac{\mathrm{d}v}{\mathrm{d}x}. Then separate the variables.DD: vdvdxv\dfrac{\mathrm{d}v}{\mathrm{d}x}

Total1

  • MMethod mark: for a correct method, even with a slip.
  • AAccuracy mark: for a correct answer from a correct method.
  • BIndependent mark: for a correct fact or result on its own, with no method needed.
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Further dynamics

Not needed for Year 12Hard

Variable force · Year 13 · 6 marks

A particle PP of mass 0.5 kg moves along the positive xx-axis. When its displacement from OO is xx m, the only force on PP is 6(x+1)2\dfrac{6}{(x + 1)^2} N, directed away from OO. PP passes through OO with speed 3 m s⁻¹.
  1. (a)
    Show that v2=33−24x+1v^2 = 33 - \dfrac{24}{x + 1}.
    (4)
  2. (b)
    Find the speed of PP when x=1x = 1.
    (1)
  3. (c)
    Explain why the speed of PP never exceeds 5.745.74 m s⁻¹.
    (1)
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Part (a)4 marks

  1. a=vdv/dxa = v \mathrm{d}v/\mathrm{d}x.0.5vdvdx0.5v\frac{\mathrm{d}v}{\mathrm{d}x}=6(x+1)2{}= \frac{6}{(x + 1)^2}
  2. Separate and integrate.12v2\tfrac{1}{2}v^2=−12x+1+c{}= -\frac{12}{x + 1} + c
  3. x=0,v=3: c=332x = 0, v = 3: \ c = \frac{33}{2}

Part (b)1 mark

  1. v=4.58 m s−1v = 4.58\text{ m s}^{-1}

Part (c)1 mark

  1. 24x+1>0 so v2<33\frac{24}{x + 1} > 0 \text{ so } v^2 < 33

Total6

  • MMethod mark: for a correct method, even with a slip.
  • AAccuracy mark: for a correct answer from a correct method.
  • BIndependent mark: for a correct fact or result on its own, with no method needed.
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