Further Mathematics · AQA 7367 · Option M · Topic MB

Momentum and collisions

Momentum, impulse and collisions, including the coefficient of restitution.

  • 4 sub-topics
  • Taught in Year 12
  • 62 practice questions
  • 4 animated lessons
Practice questions (62) Lessons Worksheets All AQA Further Mathematics 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. MB1Conservation of momentum, including velocities as vectorsY13
  2. MB2The coefficient of restitution and Newton's experimental lawY13
  3. MB3Impulse and momentum, Ft = mv − muY13
  4. MB4Impulse of a variable force, I = ∫F dtY12

Try 3 questions

Easy, medium and hard questions from this topic, each with a worked solutions and the mark for every step. All 62 questions in this topic, free, no account needed.

Momentum and impulse

Easy

Impulse · Year 12 · 1 mark

  1. A ball of mass 0.5 kg hits a wall at 10 m s⁻¹ and rebounds along the same line at 4 m s⁻¹. What is the magnitude of the impulse on the ball?
    (1)

Show worked solutionHide worked solution
  1. The velocity changes direction, so the change in momentum is m(u+v)m(u + v).BB: 7.00 N s7.00\ \mathrm{N\,s}

Total1

  • M
  • A
  • B
How did it go?

Report a problem with this question

Momentum and impulse

Medium

Momentum and impulse · Year 12 · 6 marks

A ball of mass 0.2 kg moves horizontally with speed 10 m s⁻¹ towards a vertical wall. It hits the wall directly and rebounds with speed 4 m s⁻¹.
  1. (a)
    Find the magnitude of the impulse exerted on the ball by the wall.
    (3)
  2. (b)
    The ball is in contact with the wall for 0.02 s. Find the average force exerted by the wall.
    (1)
  3. (c)
    Find the kinetic energy lost in the impact.
    (2)
Show worked solutionHide worked solution

Part (a)3 marks

  1. Take the rebound direction as positive: the velocity changes sign.I=m(v−u)=0.2(4−(−10))I = m(v - u) = 0.2(4 - (-10))
  2. =2.8 N s= 2.8\text{ N s}

Part (b)1 mark

  1. Impulse = force × time.2.80.02=140 N\frac{2.8}{0.02} = 140\text{ N}

Part (c)2 marks

  1. 12(0.2)(102−42)\tfrac{1}{2}(0.2)(10^2 - 4^2)
  2. =8.4 J= 8.4\text{ J}

Total6

  • M
  • A
  • B
How did it go?

Report a problem with this question

Momentum and collisions

Hard

Impulse of a variable force · Year 12 · 5 marks

A particle of mass 4 kg moves in a straight line. At time tt seconds a force of magnitude 12t212t^2 N acts on it in its direction of motion. When t=0t = 0 its speed is 4 m s⁻¹.
  1. (a)
    Find the impulse of the force over the interval 0≤t≤20 \leq t \leq 2.
    (3)
  2. (b)
    Find the speed of the particle when t=2t = 2.
    (2)
Show worked solutionHide worked solution

Part (a)3 marks

  1. Impulse = ∫F dt.I=∫0212t2 dtI = \int_0^{2} 12t^2\,\mathrm{d}t
  2. =[4t3]02= \left[4t^3\right]_0^{2}
  3. =32 N s= 32\text{ N s}

Part (b)2 marks

  1. Impulse = change in momentum.4v−4(4)=324v - 4(4) = 32
  2. v=12 m s−1v = 12\text{ m s}^{-1}

Total5

  • M
  • A
  • B
How did it go?

Report a problem with this question