Applied mathematics · UK A-level

Edexcel Statistics & Mechanics Year 1

Choose statistical models, interpret data and translate motion into equations. Find your chapter, open the matching worked solutions, and try a short practice check.

Chapters & worked solutions

Use your textbook for the questions. Every exercise button below opens the corresponding SolutionBank PDF.

1. Data Collection
2. Measures of Location and Spread
3. Representations of Data
4. Correlation

Define the population parameter and state hypotheses clearly. Give the conclusion in context, keeping correlation separate from causation.

5. Probability
6. Statistical Distributions

State the random variable and check that the model fits: fixed trial count, independence and constant success probability.

Further practice: Binomial-probability practice check.

7. Hypothesis Testing

Define the population parameter and state hypotheses clearly. Give the conclusion in context, keeping correlation separate from causation.

8. Modelling in Mechanics
9. Constant Acceleration
10. Forces and Motion
11. Variable Acceleration

Review & practice-paper solutions

Review Exercises: worked solutions

Practice Exam Paper: worked solutions

Textbooks and linked SolutionBank materials: Pearson Education, accessed through Physics & Maths Tutor’s Edexcel SolutionBank. Original practice questions, hints and explanations on this page: Arij Asad. This is an independent companion to the UK Edexcel 2017 series; match the chapter and exercise labels to your book. International A-level and older modular books use different numbering.

Original practice by Arij Asad

A short practice check

Try these three questions before opening the hints. They sample a few useful skills; use them to choose what to revisit, rather than as a complete assessment of the book.

1. Binomial probability

A biased coin has probability 1/31/3 of landing heads on each toss. Four tosses are independent. Find the probability of at least one head.

Show a hint

The complement of at least one head is no heads.

Show the full solution

The probability of a tail on one toss is 2/32/3. Independence means that the probability of four tails is (2/3)4=16/81(2/3)^4=16/81.

Therefore P(at least one head)=116/81=65/81P(\text{at least one head})=1-16/81=65/81.

Answer: 6581\dfrac{65}{81}.

2. Coding and measures of spread

A data set has mean 1212 and standard deviation 33. Every value xx is replaced by y=2x5y=2x-5. Find the mean and standard deviation of the new data set.

Show a hint

A translation changes the centre but not the spread; multiplying every value scales the spread.

Show the full solution

The mean transforms in the same way as each value: yˉ=2xˉ5=2(12)5=19\bar y=2\bar x-5=2(12)-5=19.

Subtracting 55 does not change deviations from the mean. Multiplication by 22 doubles their sizes, so the standard deviation is 2(3)=62(3)=6.

Answer: Mean 1919; standard deviation 66.

3. Newton's second law and constant acceleration

A particle of mass 3kg3\,\mathrm{kg} starts from rest on a smooth horizontal surface. A constant horizontal force of 12N12\,\mathrm{N} acts on it. Find its displacement after 4s4\,\mathrm{s}.

Show a hint

Find the acceleration from the horizontal resultant force, then use a constant-acceleration equation.

Show the full solution

The surface is smooth, so there is no friction. Vertically, weight and the normal reaction balance. Horizontally, F=maF=ma gives 12=3a12=3a, so a=4ms2a=4\,\mathrm{m\,s^{-2}}.

With u=0u=0 and t=4t=4, s=ut+12at2=0+12(4)(42)=32ms=ut+\tfrac12at^2=0+\tfrac12(4)(4^2)=32\,\mathrm{m}.

Answer: 32m32\,\mathrm{m} in the direction of the force.

Find the mistake

A particle has velocity v=(62t)ms1v=(6-2t)\,\mathrm{m\,s^{-1}} for 0t50\leq t\leq5, with tt in seconds. A student finds a displacement of 5m5\,\mathrm{m} and calls this the distance travelled. Find the actual distance.

Show the correction

The particle changes direction when 62t=06-2t=0, at t=3t=3. Its displacement over the whole interval is indeed [6tt2]05=5m[6t-t^2]_0^5=5\,\mathrm{m}.

From 00 to 33 seconds it travels [6tt2]03=9m[6t-t^2]_0^3=9\,\mathrm{m}. From 33 to 55 seconds its displacement is [6tt2]35=4m[6t-t^2]_3^5=-4\,\mathrm{m}, so it travels a further 4m4\,\mathrm{m}.

Total distance is 9+4=13m9+4=13\,\mathrm{m}.

Take the practice into a lesson

The printable sheet contains the questions first, followed by hints and full worked solutions. The editable LaTeX source is ready for Overleaf.