Sixth Term Examination Paper (STEP) Mechanics Syllabus

State the friction inequality in the model of limiting equilibrium.

$F \leq \mu R$. Friction takes whatever value (up to $\mu R$) is needed to maintain equilibrium; sliding begins when $F$ would have to exceed $\mu R$.

Define the angle of friction $\lambda$ and relate it to $\mu$.

The angle of friction is the angle between the total contact force and the normal at the point of slipping: $\tan\lambda = \mu$.

For a block on a rough inclined plane at angle $\alpha$, what is the condition for it to remain at rest without applied force?

It stays at rest provided $\tan\alpha \leq \mu$; it is on the point of sliding when $\tan\alpha = \mu$ (i.e. $\alpha = \lambda$).

For two particles connected by a light inextensible string over a smooth pulley, what is true about their accelerations and the tension?

They have equal-magnitude accelerations; the string tension $T$ is the same throughout (light string, smooth pulley) and acts to pull each particle toward the pulley.

Two masses $m_1 > m_2$ hang either side of a smooth pulley on a light inextensible string. Give the acceleration and tension.

Acceleration $a = \dfrac{(m_1 - m_2)g}{m_1 + m_2}$; tension $T = \dfrac{2 m_1 m_2 g}{m_1 + m_2}$.

In connected-particle problems, what assumptions do the words 'light' and 'inextensible' string and 'smooth' pulley each justify?

'Light' means the string/pulley is massless so tension is uniform; 'inextensible' means connected bodies share the same speed and acceleration magnitude; 'smooth' means no friction so tension is unchanged across the pulley.

For a particle of mass $m$ moving in a horizontal circle of radius $r$ at constant speed $v$, state the magnitude and direction of the required force.

A centripetal force directed toward the centre of magnitude $F = \dfrac{mv^{2}}{r} = m r \omega^{2}$, where $\omega = v/r$ is the angular speed.

Write the centripetal acceleration in terms of $v$ and $r$, and in terms of $\omega$ and $r$.

$a = \dfrac{v^{2}}{r} = r\omega^{2}$, directed toward the centre of the circle.

For a particle on the inside of a vertical circle of radius $r$, what is the minimum speed at the top to maintain contact?

At the top, gravity alone can provide the centripetal force when the reaction is zero: $mg = \dfrac{mv^{2}}{r}$, giving minimum speed $v = \sqrt{gr}$.

For motion in a vertical circle, how do you relate speeds at different heights?

Use conservation of energy: $\tfrac{1}{2}mv_1^{2} + mgh_1 = \tfrac{1}{2}mv_2^{2} + mgh_2$, since tension/reaction do no work (perpendicular to motion).

Define the impulse of a constant force and state its relation to momentum.

Impulse $\vec{J} = \vec{F}\,t$ (or $\int \vec{F}\,dt$ for variable force). The impulse–momentum theorem states $\vec{J} = \Delta(m\vec{v}) = m\vec{v} - m\vec{u}$.

State the principle of conservation of linear momentum and its condition.

If no external resultant force acts on a system, total momentum is conserved: $m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$ (momentum is a vector, so signs/directions matter).

State Newton's experimental law of restitution for a direct impact.

$\text{speed of separation} = e \times \text{speed of approach}$, i.e. $v_2 - v_1 = -e(u_2 - u_1)$, where $0 \leq e \leq 1$ is the coefficient of restitution.

What do the limiting values $e = 1$ and $e = 0$ of the coefficient of restitution correspond to?

$e = 1$: perfectly elastic collision (kinetic energy conserved). $e = 0$: perfectly inelastic collision (bodies coalesce / move off together with no separation speed).

How is kinetic energy treated in collisions compared with momentum?

Momentum is always conserved (no external impulse), but kinetic energy is generally lost in a collision unless it is perfectly elastic ($e=1$); the loss equals KE before minus KE after.

For a ball dropped onto the ground and rebounding with coefficient of restitution $e$, relate rebound height to drop height.

Speeds satisfy $v = eu$, so by energy the rebound height is $h' = e^{2} h$ where $h$ is the drop height.

Define the work done by a constant force and give the formula when the force is at an angle to the displacement.

Work $W = \vec{F}\cdot\vec{d} = F d \cos\theta$, where $\theta$ is the angle between force and displacement; measured in joules (J).

State the work–energy principle.

The total work done by all forces on a body equals its change in kinetic energy: $W_{\text{net}} = \Delta KE = \tfrac{1}{2}mv^{2} - \tfrac{1}{2}mu^{2}$.

Write the formulae for kinetic energy and gravitational potential energy.

Kinetic energy $KE = \tfrac{1}{2}mv^{2}$; gravitational potential energy (near Earth) $PE = mgh$ relative to a chosen reference level.

State the principle of conservation of mechanical energy and when it applies.

When only conservative forces (e.g. gravity) act, $KE + PE = \text{constant}$: $\tfrac{1}{2}mu^{2} + mgh_1 = \tfrac{1}{2}mv^{2} + mgh_2$. It fails if friction or other dissipative forces do work.

Define power and give two formulae for it.

Power is the rate of doing work: $P = \dfrac{dW}{dt}$; for a force moving its point of application at velocity $v$, $P = Fv$. Units: watts (W), where $1\,\text{W} = 1\,\text{J s}^{-1}$.

Define the moment of a force about a point.

Moment $= \text{force} \times \text{perpendicular distance from the point to the line of action}$, $M = F d$. It measures turning effect, in newton-metres (N m), with a sense (clockwise/anticlockwise).

State the conditions for a rigid body to be in equilibrium.

The resultant force is zero ($\sum \vec{F} = \vec{0}$) and the resultant moment about any point is zero ($\sum M = 0$); equivalently, total clockwise moments equal total anticlockwise moments.

How is the centre of mass of a system of particles defined?

$\bar{x} = \dfrac{\sum m_i x_i}{\sum m_i}$ (and similarly for $\bar{y}$); it is the mass-weighted average position of the particles.

Give the position of the centre of mass for a uniform triangular lamina and a uniform semicircular lamina of radius $r$.

Uniform triangle: at the centroid, $\tfrac{1}{3}$ of the way from each side's midpoint, i.e. the intersection of the medians. Uniform semicircular lamina: on the axis of symmetry at distance $\dfrac{4r}{3\pi}$ from the straight edge.

For a body on the point of toppling about an edge, where does the line of action of its weight lie?

At the point of toppling the weight acts vertically through the pivoting edge; beyond this the weight's line falls outside the base and the body topples.

For a block on a rough inclined plane, give the conditions that determine whether it slides or topples first as the incline angle increases.

It slides when $\tan\alpha > \mu$; it topples when the vertical through the centre of mass passes outside the base, i.e. roughly when $\tan\alpha > \dfrac{\text{base width}}{2 \times \text{height of C of M}}$. Whichever condition is met at the smaller angle happens first.

State Hooke's law for an elastic string or spring with modulus of elasticity $\lambda$.

Tension $T = \dfrac{\lambda x}{l}$, where $x$ is the extension, $l$ the natural length, and $\lambda$ the modulus of elasticity. (A spring can also be compressed, giving a thrust; a string cannot.)

Give the elastic potential energy (strain energy) stored in a stretched string or spring.

$E = \dfrac{\lambda x^{2}}{2l}$, where $x$ is the extension, $l$ the natural length and $\lambda$ the modulus of elasticity.

State the defining equation of simple harmonic motion and the general expression for its period.

$\ddot{x} = -\omega^{2} x$: acceleration is proportional to displacement from equilibrium and directed toward it. Period $T = \dfrac{2\pi}{\omega}$, frequency $f = \dfrac{\omega}{2\pi}$.

For SHM with amplitude $a$ and angular frequency $\omega$, write the velocity as a function of displacement and state where speed is maximum.

$v^{2} = \omega^{2}(a^{2} - x^{2})$, so maximum speed $v_{\max} = a\omega$ occurs at the centre ($x=0$); speed is zero at the extremes $x = \pm a$.