STEP mechanics: guide and practice problems
The two mechanics questions on each STEP paper use A Level mechanics, but they expect you to set up your own equations from a description, with no diagram given.
Key ideas
- Projectiles: eliminate time to get the trajectory, then treat it as a quadratic in tan θ.
- Collisions: momentum and Newton's law of restitution give two linear equations; solve them in general first.
- Circular motion on a surface: the particle leaves when the normal reaction becomes zero.
- Energy is usually quicker than forces when you only need speeds.
Common traps
- Sign errors in velocities after a collision (choose a positive direction and keep it).
- Forgetting the normal reaction can only push, not pull.
- Leaving g in when the question gives a numerical value, or the reverse.
Practice problems
Original problems in the style of STEP. Try each one for at least 20 minutes before you open a hint; write a full solution, then compare.
Problem 1: Which points can a projectile reach?
A particle is projected from the origin \(O\) with speed \(u\) at an angle \(\theta\) above the horizontal, under gravity \(g\) only. Axes: \(x\) horizontal, \(y\) vertically up.
- Show that the particle passes through \((X,Y)\) if \(\displaystyle Y=XT-\frac{gX^2}{2u^2}(1+T^2)\), where \(T=\tan\theta\).
- Show that \((X,Y)\) can be reached for some \(\theta\) if and only if \(\displaystyle Y\le\frac{u^2}{2g}-\frac{gX^2}{2u^2}\).
- Take \(g=10\) and \(u=20\) (SI units). Show that \((30,8)\) can be reached, and find the two possible values of \(\tan\theta\).
Hint 1
Eliminate \(t\) from \(x=ut\cos\theta\), \(y=ut\sin\theta-\tfrac12gt^2\), and use \(\sec^2\theta=1+\tan^2\theta\).
Hint 2
Treat (i) as a quadratic in \(T\). It has a real root iff its discriminant is \(\ge0\).
Full solution
(i) \(t=\dfrac{X}{u\cos\theta}\), so \(Y=X\tan\theta-\dfrac{gX^2}{2u^2\cos^2\theta}=XT-\dfrac{gX^2}{2u^2}(1+T^2)\).
(ii) As a quadratic in \(T\): \(\dfrac{gX^2}{2u^2}T^2-XT+\left(Y+\dfrac{gX^2}{2u^2}\right)=0\). Real \(T\) exists iff \(X^2\ge\dfrac{2gX^2}{u^2}\left(Y+\dfrac{gX^2}{2u^2}\right)\), which (for \(X\ne0\)) is \(Y\le\dfrac{u^2}{2g}-\dfrac{gX^2}{2u^2}\).
(iii) The bound is \(20-\dfrac{900}{80}=8.75\ge8\), so \((30,8)\) is reachable. The quadratic is \(11.25T^2-30T+19.25=0\), or \(45T^2-120T+77=0\), so \(T=\dfrac{120\pm\sqrt{540}}{90}=\dfrac{20\pm\sqrt{15}}{15}\).
Results: \(\tan\theta=\frac{20\pm\sqrt{15}}{15}\).
2 more mechanics problems
A chain of collisions; Sliding off a smooth dome. Each with two hints and a full solution.
Next steps
Related topics: Probability and statistics · Trigonometry. Stuck? Read the problem-solving strategies. Ready for the real thing? Official STEP past papers.
STEP is run by OCR. A Level Math Revision is independent: it is not affiliated with or endorsed by OCR, the University of Cambridge or any university. Every problem here is our own, written in the style of STEP; for real papers use OCR's official past papers.