Tutoring

I tutor students at every stage of the path toward the Sixth Term Examination Papers (STEP), from younger students building strong foundations in mathematical problem solving to older students working directly on exam material. STEP is the post-offer entrance examination used in almost all conditional offers for Mathematics at Cambridge. It covers pure mathematics, mechanics, and statistics.

My lessons follow the Soviet math-circle tradition: students develop mathematical insight by exploring challenging problems, with guidance rather than ready-made solutions.

To ask about tutoring, email fen@spark-fly.net.

Sample problems

Blocks on an incline

Problem.

A small square block of mass 2 kg and side length 0.15 m is placed on top of a larger square block of mass 5 kg and side length 0.30 m. Both blocks are initially at rest on a rigid incline of length 2.5 m at an angle of 30 degrees. The smaller block is positioned with its uphill edge aligned with the uphill edge of the larger block.

The coefficient of friction between either block and the incline is 0.35, and the coefficient of friction between the two blocks is 0.20. Determine the time required for the smaller block to reach the bottom of the incline, treating it as a point particle once it leaves the larger block.

Objective. This is a significantly more difficult extension of a standard kinematics problem. It requires Newton's second law, friction, relative motion, and careful matching of three stages of motion.

Show solution

Take distance down the incline as positive and use $g = 9.8\,\mathrm{m\,s^{-2}}$ . If the blocks moved together, their acceleration would be

$g(\sin 30^\circ - 0.35\cos 30^\circ).$

The friction needed to hold the upper block in place exceeds $0.20mg\cos 30^\circ$ , so it slides over the lower block. While they remain in contact, their accelerations are

$\begin{aligned} a_s &= g(\sin 30^\circ - 0.20\cos 30^\circ) = 3.203\,\mathrm{m\,s^{-2}}, \\ a_L &= \frac{5g\sin 30^\circ + 0.20(2g\cos 30^\circ) - 0.35(7g\cos 30^\circ)}{5} \\ &= 1.420\,\mathrm{m\,s^{-2}}. \end{aligned}$

The smaller block must move $0.30 - 0.15 = 0.15\,\mathrm m$ relative to the larger block before leaving it. Therefore

$t_1 = \sqrt{\frac{2(0.15)}{a_s-a_L}} = 0.410\,\mathrm s.$

At this instant its speed is $a_st_1 = 1.314\,\mathrm{m\,s^{-1}}$ , and its centre is

$0.075 + \tfrac12 a_st_1^2 = 0.344\,\mathrm m$

down the incline from the starting point. It then falls the 0.30 m height of the larger block. Resolving perpendicular and parallel to the incline gives

$t_2 = \sqrt{\frac{2(0.30)}{g\cos 30^\circ}} = 0.266\,\mathrm s.$

During this time it travels another

$(1.314)t_2 + \tfrac12 g\sin 30^\circ t_2^2 = 0.523\,\mathrm m$

down the incline, reaching the ramp 0.867 m from the top with speed $1.314 + g\sin 30^\circ t_2 = 2.617\,\mathrm{m\,s^{-1}}$ parallel to it. On the ramp its acceleration is

$a_r = g(\sin 30^\circ - 0.35\cos 30^\circ) = 1.929\,\mathrm{m\,s^{-2}}.$

The remaining distance is $2.5 - 0.867 = 1.633\,\mathrm m$ . Solving

$1.633 = 2.617t_3 + \tfrac12(1.929)t_3^2$

gives $t_3 = 0.523\,\mathrm s$ . Hence the total time is

$t_1+t_2+t_3 = \boxed{1.20\,\mathrm s}.$

Skater in a half-pipe

Problem.

A skater starts from rest at the top of a smooth, frictionless half-pipe. The shape of the track is described parametrically by

$x = 0.50(\theta - \sin \theta), \qquad y = 0.50(1 - \cos \theta),$

where $x$ and $y$ are measured in meters, $y$ is measured downward from the starting point, and $\theta$ is measured in radians, with $0 \leq \theta \leq \pi$ .

How long does it take the skater to reach the bottom of the half-pipe?

Objective. This problem uses the tricky geometry and algebraic manipulations expected in STEP: finding an arc-length element from a parametrization, applying conservation of energy, and recognizing a useful cancellation.

Show solution

Write $a=0.50\,\mathrm m$ . Differentiating the parametrization gives

$\frac{dx}{d\theta}=a(1-\cos\theta), \qquad \frac{dy}{d\theta}=a\sin\theta.$

Therefore the arc-length element is

$\begin{aligned} \frac{ds}{d\theta} &=a\sqrt{(1-\cos\theta)^2+\sin^2\theta} \\ &=2a\sin(\theta/2). \end{aligned}$

Conservation of energy gives

$v=\sqrt{2gy} =\sqrt{2ga(1-\cos\theta)} =2\sqrt{ga}\sin(\theta/2).$

The apparently awkward factors cancel:

$dt=\frac{ds}{v}=\sqrt{\frac{a}{g}}\,d\theta.$

It follows that

$t=\int_0^\pi \sqrt{\frac{a}{g}}\,d\theta =\pi\sqrt{\frac{0.50}{9.8}} =\boxed{0.710\,\mathrm s}.$

Mass on a damped spring

Problem.

A 1.0 kg mass hangs from a vertical spring with spring constant 10 N/m. As the mass moves, it experiences air resistance proportional to its velocity, described by

$F_{\mathrm{air}}=-2.0\frac{dy}{dt},$

where $y$ is measured in meters downward from the equilibrium position and $t$ is measured in seconds. The mass is pulled 0.20 m downward from equilibrium and released from rest.

Derive an equation for the displacement $y(t)$ , and determine how long it takes the mass to reach its equilibrium position for the first time.

Objective. This is a physical exploration of a standard textbook technique: translating a damped oscillation into a linear differential equation, solving its characteristic equation, and applying initial conditions.

Show solution

Because displacement is measured from equilibrium, gravity is already accounted for. Newton's second law gives

$y''+2y'+10y=0, \qquad y(0)=0.20, \qquad y'(0)=0.$

The characteristic equation

$r^2+2r+10=0$

has roots $r=-1\pm3i$ . Thus

$y(t)=e^{-t}(A\cos3t+B\sin3t).$

The initial conditions give $A=0.20$ and $B=1/15$ , so

$y(t)=0.20e^{-t}\left(\cos3t+\frac13\sin3t\right).$

At the first return to equilibrium, $y(t)=0$ . The first positive solution satisfies

$3t=\pi-\tan^{-1}(3).$

Therefore

$t=\frac{\pi-\tan^{-1}(3)}{3} =\boxed{0.631\,\mathrm s}.$