Applied mathematics · UK A-level

Edexcel Statistics & Mechanics Year 2

Develop conditional probability, normal distributions, forces and kinematics. 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. Regression, Correlation and Hypothesis Testing

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

2. Conditional Probability
3. The Normal Distribution
4. Moments
5. Forces and Friction
6. Projectiles
7. Applications of Forces
8. Further Kinematics

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. Normal distribution

XX is normally distributed with mean 5050 and standard deviation 44. Find P(X>54)P(X>54), giving your answer to three significant figures.

Show a hint

Standardise using the standard deviation, then choose the upper tail.

Show the full solution

With Z=(X50)/4Z=(X-50)/4, the threshold X=54X=54 corresponds to Z=1Z=1.

Hence P(X>54)=P(Z>1)=1Φ(1)0.158655P(X>54)=P(Z>1)=1-\Phi(1)\approx0.158655, where Φ\Phi is the standard normal cumulative distribution function.

Answer: 1Φ(1)0.1591-\Phi(1)\approx0.159 (three significant figures).

2. Projectiles

A particle is projected from level ground at 20ms120\,\mathrm{m\,s^{-1}}, at 3030^\circ above the horizontal. Neglect air resistance and take g=9.8ms2g=9.8\,\mathrm{m\,s^{-2}}. Find its time of flight and horizontal range when it returns to the same level.

Show a hint

Resolve the initial velocity into components. Use zero vertical displacement to find the non-zero flight time.

Show the full solution

The initial horizontal and vertical velocity components are 20cos30=103ms120\cos30^\circ=10\sqrt3\,\mathrm{m\,s^{-1}} and 20sin30=10ms120\sin30^\circ=10\,\mathrm{m\,s^{-1}}.

Taking upward as positive, vertical displacement is y=10t4.9t2y=10t-4.9t^2. At landing, y=0y=0; the non-zero solution is t=10/4.9=100/49s2.04st=10/4.9=100/49\,\mathrm{s}\approx2.04\,\mathrm{s}.

Horizontal velocity is constant, so the range is (103)(100/49)=10003/49m35.3m(10\sqrt3)(100/49)=1000\sqrt3/49\,\mathrm{m}\approx35.3\,\mathrm{m}.

Answer: Time 10049s\dfrac{100}{49}\,\mathrm{s}; range 1000349m\dfrac{1000\sqrt3}{49}\,\mathrm{m} (approximately 2.04s2.04\,\mathrm{s} and 35.3m35.3\,\mathrm{m}).

3. Moments and equilibrium

A uniform horizontal beam ABAB is 4m4\,\mathrm{m} long and weighs 40N40\,\mathrm{N}. It is supported at AA and BB. A further downward load of 20N20\,\mathrm{N} acts 1m1\,\mathrm{m} from AA. Find the upward reactions at the two supports.

Show a hint

The beam's weight acts at its midpoint. Taking moments about one support eliminates its reaction.

Show the full solution

Let the upward reactions be RAR_A and RBR_B. Vertical equilibrium gives RA+RB=40+20=60NR_A+R_B=40+20=60\,\mathrm{N}.

Taking moments about AA, 4RB=40(2)+20(1)=1004R_B=40(2)+20(1)=100. Therefore RB=25NR_B=25\,\mathrm{N}.

It follows that RA=6025=35NR_A=60-25=35\,\mathrm{N}.

Answer: RA=35NR_A=35\,\mathrm{N}, RB=25NR_B=25\,\mathrm{N}.

Find the mistake

Events AA and BB satisfy P(A)=0.4P(A)=0.4, P(B)=0.5P(B)=0.5 and P(AB)=0.3P(A\cap B)=0.3. A student writes P(AB)=P(A)=0.4P(A\mid B)=P(A)=0.4. Explain why this is wrong and find the correct conditional probability.

Show the correction

P(AB)=P(AB)/P(B)=0.3/0.5=0.6P(A\mid B)=P(A\cap B)/P(B)=0.3/0.5=0.6.

The student's equality would hold for independent events. Here P(A)P(B)=0.4(0.5)=0.2P(A)P(B)=0.4(0.5)=0.2, which differs from P(AB)=0.3P(A\cap B)=0.3; the events are not independent.

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.