3. Further Mechanics

Ignoring air resistance, horizontal acceleration is zero and vertical acceleration is −g. With initial speed u at angle θ, x=u cosθ·t and y=u sinθ·t−½gt².

Treat the two components independently, then eliminate t or use symmetry. The launch and landing heights must be stated before using range or time-of-flight formulas.

For level ground, time of flight is 2u sinθ/g and range is u²sin2θ/g. The maximum range occurs at 45° only under this level-ground, no-drag model.

The velocity is not constant as a vector; only its horizontal component is constant, and gravity acts throughout the flight.

Eliminating time from the component equations gives y=x tanθ−gx²/(2u²cos²θ), a quadratic trajectory when gravity is uniform and air resistance is neglected.

Use the equation to find height, range or intersection with a target, but check that the chosen root corresponds to a future time and that the launch/landing geometry matches the question.

At a fixed horizontal distance, the quadratic may give two launch angles: a low path and a high path. Both can reach the point, but they have different flight times and maximum heights.

A parabolic path is an idealisation; drag, varying gravity or wind changes it, and an algebraic x-root is not automatically a physically valid time.

For a projectile launched with speed u at angle θ, x=u cosθ·t and y=u sinθ·t−½gt² still describe the motion. A target at a different height changes the time and range equations, not the component model.

Write the target condition in x and y, eliminate t, and solve only for values consistent with t≥0. Do not use the level-ground range formula unless launch and landing heights are equal.

A ball launched from a platform can hit a lower target on the descending path; the second root of the height equation represents a later intersection, while a negative time is discarded.

The 45° maximum-range result is not universal: it assumes equal heights, uniform gravity and no air resistance.

The moment of a force about a point is force × perpendicular distance from the point to the force line: M=Fd. Choose a sign convention, usually clockwise and anticlockwise, and keep it throughout.

For several forces, take moments about a convenient point so unknown reactions through that point have zero moment. Resolve angled forces before finding their perpendicular contribution.

A 20 N force acting 0.30 m from a pivot produces a 6.0 N m moment. A 10 N force on the opposite side at 0.60 m balances it if the senses are opposite.

Distance along the rod is not automatically the moment arm; use the perpendicular distance to the force line, and do not mix force and moment units.

For particles of masses mᵢ at positions xᵢ, the centre of mass satisfies x̄=Σmᵢxᵢ/Σmᵢ, with the same weighted-average rule applied independently to y-coordinates.

Heavier masses pull the centre towards them. Set an origin and sign convention first; coordinates may be negative, and the centre can lie outside the material for a separated system.

Masses 2 kg at x=0 and 6 kg at x=4 m have x̄=(2·0+6·4)/8=3 m, much nearer the larger mass.

The centre of mass is not always the geometric centre and is not necessarily a point inside an irregular or multi-part object.

A uniform rod, rectangle, circle or sphere has its centre of mass at its geometric centre. Symmetry means equal mass is distributed at equal distances on opposite sides of the symmetry line or plane.

For a composite lamina, split the shape into standard pieces, assign each area as its mass when density and thickness are uniform, and use an area-weighted coordinate average.

A uniform L-shape can be treated as a large rectangle minus a cut-out rectangle; subtract the cut-out’s area moment rather than averaging the two visible arms.

The cut-out is negative area only in the calculation; it is not a negative physical mass, and the uniform-density assumption must be stated.

If components have masses m₁,m₂,… and centres at r₁,r₂,…, the combined centre is r̄=(Σmᵢrᵢ)/(Σmᵢ). Each component’s mass must be proportional to its area or volume only when density is common.

Choose a convenient reference point, include holes or removed pieces with negative area when appropriate, and check that the result lies between the extreme positions of positive masses.

A 3 kg block at x=1 m joined to a 1 kg block at x=5 m gives x̄=2 m, not 3 m: the lighter block should shift the centre only modestly.

Equal-sized parts do not imply equal masses when densities differ; never replace mass weights by geometry without checking the model.

A rigid body is in equilibrium when the vector sum of all forces is zero and the sum of moments about any point is zero. Translational balance alone does not prevent rotation.

Draw all external forces, resolve components, then take moments about a point that removes unknown reactions. Use the force equations afterwards to determine the remaining reactions.

A horizontal beam supported at its ends carries a central load. Equal upward reactions satisfy force balance; taking moments about either support confirms the reactions and prevents a net turn.

A pair of equal opposite forces can form a couple with zero resultant force but non-zero moment, so the body can still rotate.

A body slides when the available friction is insufficient for equilibrium; it topples when the line of action of the resultant reaction moves beyond the supporting base. Both are checked after force and moment balance.

Find the friction required and compare it with μR for sliding. For toppling, take moments about the edge that would become the pivot and set the limiting reaction there to zero.

A crate on a rough floor may remain at rest while a horizontal force increases. If the force line creates a moment that lifts one edge first, toppling occurs before the friction limit is reached.

A large friction coefficient does not prevent toppling, and “about to topple” does not mean the body has already started rotating.

Angular speed ω measures the angle swept per unit time, in radians per second. For a particle at radius r, its tangential speed is v=rω, so equal angular speed does not mean equal linear speed at different radii.

Use radians, not degrees, in v=rω and in arc-length relations. Keep the direction of v tangent to the circle even though ω describes the rotation about the centre.

At ω=4 rad s⁻¹, a point 0.25 m from the axis moves at v=1.0 m s⁻¹; a point twice as far moves twice as fast.

Angular speed is not the same as revolutions per second: f revolutions per second gives ω=2πf.

For uniform circular motion the velocity direction continually changes, so acceleration is directed towards the centre. Its magnitude is a=v²/r=rω².

The inward acceleration is supplied by the resultant inward force, not by a new separate force. Draw the radial direction first, then apply Newton’s second law along it.

A 0.50 kg mass moving at 3.0 m s⁻¹ on a 2.0 m radius circle needs a=4.5 m s⁻² inward and resultant force 2.25 N.

Constant speed does not mean zero acceleration; only the magnitude of velocity is constant, while its direction changes.

For a particle moving at constant speed in a horizontal circle, the resultant horizontal or radial force must equal mv²/r. Vertical forces must separately balance if the height is constant.

Resolve the actual forces—tension, normal reaction, friction or a component of weight—before setting their radial resultant equal to mv²/r. The centre direction changes around the circle.

For a conical pendulum, the vertical component of tension balances mg while the horizontal component supplies m v²/r; using all of T as centripetal force is wrong.

“Centripetal force” is a role played by the resultant inward force, not an extra force to add to the free-body diagram.

In a vertical circle, speed changes with height. Use conservation of energy between points, then apply radial Newton’s law to find tension or normal reaction at that point.

At the top and bottom, define inward separately: tension or reaction may add to weight at the bottom but oppose it at the top. Complete contact requires the limiting reaction to remain non-negative.

For a particle on a string, the minimum speed at the top occurs when tension is zero, giving mv²/r=mg there; energy then determines the required bottom speed.

The condition T=0 is a limiting contact condition, not “no gravity”, and using one fixed speed around the circle violates energy conservation.

For a spring or elastic string obeying Hooke’s law, the tension or compression is proportional to extension or compression: F=λx/l, where λ is the modulus of elasticity and l is the natural length.

The model applies only within the elastic range and uses extension from the natural length, not the total length. Work done is the area under the force–extension graph.

If λ/l=50 N m⁻¹ and extension is 0.04 m, the force is 2.0 N; the elastic energy is ½Fx=0.040 J.

Returning to the original length does not prove Hooke’s law held throughout, and compression/extension signs must match the chosen convention.

The work done in stretching an elastic string or spring is stored as elastic potential energy: E=∫F dx. For a Hookean region, E=½Fx=½(λ/l)x².

Use the actual force–extension graph when the law changes or the string becomes slack. Extension is measured from natural length, and energy is a scalar even when the force direction changes.

A spring reaches 3.0 N at extension 0.06 m while remaining linear. Its stored energy is ½×3.0×0.06=0.090 J.

The energy is not Fx for a gradually applied load; that would be the rectangle, whereas the linear graph gives a triangle.

For an elastic string, the tension is zero when slack and proportional to extension only while taut and within the elastic range. In a force problem, combine this law with resolved force or moment equations.

Find the current length from the geometry before calculating extension. Check whether the calculated tension is positive; a negative tension means the string cannot supply that force and is slack.

Two equal elastic strings supporting a mass may have equal tensions only by symmetry; each tension’s vertical component, not the full tension, balances the weight.

An elastic string can pull but not push. Do not continue Hooke’s law into compression or beyond the stated elastic limit.

When force depends on position or time, Newton’s law still gives ma=F, but constant-acceleration formulae may fail. Work–energy is often the cleanest route when F is a function of x.

Use v dv/dx=a to convert a position-dependent acceleration into an integrable equation, or integrate F dx for work. State the interval and initial condition before applying limits.

If F=kx on a frictionless line, work from 0 to x is ½kx², so ½mv²=½mu²+½kx²; speed grows with x rather than time uniformly.

A variable force does not justify using v²=u²+2as with an average force unless that average has been derived from work.

For a direct impact, the coefficient of restitution e is relative speed of separation divided by relative speed of approach, measured along the line of impact. For ordinary passive impacts 0≤e≤1.

Choose one positive direction and write velocities immediately before and after impact. Combine the restitution equation with conservation of momentum when external impulse is negligible.

If two particles approach at 5 m s⁻¹ relative speed and separate at 2 m s⁻¹, e=2/5=0.4. The individual velocities still depend on their masses and momentum.

Restitution does not conserve kinetic energy except in the elastic case e=1; it also does not mean each particle reverses direction.

For a system of colliding particles, total momentum before impact equals total momentum after impact when the external impulse during the collision is negligible: Σmu=Σmv.

Keep signed velocities, identify the system, and use restitution only as a second equation. For an explosion or separation, the same momentum principle applies even though kinetic energy may increase.

A 2 kg trolley at 3 m s⁻¹ collides with a stationary 1 kg trolley and they move together: their common speed is (2×3)/3=2 m s⁻¹.

Momentum is a vector and can cancel; kinetic energy is not generally conserved in an inelastic impact.