Physics · Pearson Edexcel 9PH0 · Topic 6

Further mechanics

Momentum in two dimensions, collisions and circular motion.

  • 11 sub-topics
  • Taught in Year 13
  • 112 practice questions
  • 11 animated lessons
Practice questions (112) Lessons Worksheets All Edexcel Physics topics

What the specification says

Every sub-topic Pearson Edexcel 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. 97Impulse, FΔt = ΔpY13
  2. 98CP9: Force exerted on an object and its change of momentumY13
  3. 99Conservation of linear momentum in two dimensionsY13
  4. 100CP10: Use ICT to analyse collisions between small spheresY13
  5. 101Elastic and inelastic collisionsY13
  6. 102Kinetic energy, Ek = p²/2mY13
  7. 103Radians and degreesY13
  8. 104Angular velocity, v = ωrY13
  9. 105Centripetal accelerationY13
  10. 106Centripetal forceY13
  11. 107Equations for centripetal forceY13

Try 3 questions

Easy, medium and hard questions from this topic, each with a worked solutions and the mark for every step. All 112 questions in this topic, free, no account needed.

Further mechanics

Not needed for Year 12Easy

Radians · Year 13 · 2 marks

  1. (a)
    Convert 120° to radians.
    (1)
  2. (b)
    Find the arc length subtended by this angle on a circle of radius 2.0 m.
    (1)

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Part (a)1 mark

  1. θ=2.09 rad\theta = 2.09\ \mathrm{rad}

Part (b)1 mark

  1. s=4.19 ms = 4.19\ \mathrm{m}

Total2

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Further mechanics

Not needed for Year 12Medium

Force and change of momentum · Year 13 · 4 marks

A ball of mass 0.45 kg travelling at 30 m s⁻¹ is hit straight back at 18 m s⁻¹. The contact lasts 0.02 s.
  1. (a)
    Calculate the change in momentum of the ball.
    (2)
  2. (b)
    Calculate the average force on the ball.
    (1)
  3. (c)
    Explain why a longer contact time would reduce the force on the player's racket.
    (1)

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Part (a)2 marks

  1. Take the final direction as positive.0.45(18)−0.45(−30)0.45(18) - 0.45(-30)
  2. Δp\Delta p=21.6 kg m s−1{}= 21.6\ \mathrm{kg\,m\,s^{-1}}

Part (b)1 mark

  1. F=1080 NF = 1080\ \mathrm{N}

Part (c)1 mark

  1. F = Δp/Δt, so for the same change in momentum a longer time means a smaller force.

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Further mechanics

Not needed for Year 12Hard

Momentum in two dimensions · Year 13 · 6 marks

A puck of mass 0.2 kg moving at 3.0 m s⁻¹ hits a stationary puck of mass 0.2 kg. Afterwards the first puck moves at 1.5 m s⁻¹ at 30° to its original direction.
  1. (a)
    Find the velocity of the second puck.
    (4)
  2. (b)
    Determine whether the collision is elastic.
    (2)
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Part (a)4 marks

  1. px:0.2(3.0)p_x: 0.2(3.0)=0.2(1.5)cos⁡30∘{}= 0.2(1.5)\cos 30^\circ+0.2vx{}+ 0.2v_x
  2. Momentum is conserved in each perpendicular direction.py:0p_y: 0=0.2(1.5)sin⁡30∘{}= 0.2(1.5)\sin 30^\circ−0.2vy{}- 0.2v_y
  3. v=1.86 m s−1v = 1.86\ \mathrm{m\,s^{-1}}
  4. at 23.8° to the original direction, on the other side.

Part (b)2 marks

  1. Ek:0.900→0.571 JE_k: 0.900 \to 0.571\text{ J}
  2. Inelastic: kinetic energy is not conserved.

Total6

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