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# M2

When a particle is projected with speed u, at an angle a to the horizontal, it will move along a s[] curve symmetrical--.
On a level plane, if a particle is projected from the ground of the plane, it strikes the plane when the vertical displacement is zero.
You should never give your answer to a greater accuracy than the used data.
If an object has velocity (Ai+Bj) m/s, where i and j are unit vectors, the horizontal component of velocity is [], while the vertical component of velocity is [] A--B--.
For a particle moving in a straight line, with acceleration that varies with time, you can use calculus.
Velocity is the rate of change of [] with [] displacement--time--. To find the velocity from the displacement, you [] with respect to [] differentiate--time--.
Acceleration is the rate of change of change of [] with [] velocity--time--. To find the acceleration from velocity, you [] with respect to [] differentiate--time--.
To obtain the velocity from the accelration, you [] with respect to [] integrate--time--.
To obtain the displacement from the velocity, you [] with respect to [] integrate--time--.
Where a particle moves in a straight line with acceleration a, displacement x and velocity v (with differentiation): v = dx/dt; a = []/dt = []/[] dv/dt = d²x/dt².
Where a particle moves in a straight line with acceleration a, displacement x and velocity v (with integration): v = ∫a dt, x = ∫v dt. Note: when you integrate, it is important to include a constant of integration; the constant of integration often represents the [] [] or [] [] of a particle initial displacement--initial velocity--; if, for example, you have an equation with t and x as variables and you know that at t = t1, x = x1 (1 ss): [] those values into the equation containing c (the constant) substitute--, then find the value of c. Every time that you integrate, even if it's part of a series of integrations: find the constatant of integration.
Is acceleration a scalar quantity? No; it is a vector quantity. As acceleration is a vector quantity: when specifying an acceleration, you should give both the direction and the magnitude; generally, where O is the origin, you state whether is directed towards or from the origin O.
If a particle started X0 m (0 ss) away from O, and at time t, the particle is Xn m (n ss) away from O: at time t, the particle is [] m away from its starting point Xn - X0 (n,0 ss).
Where v = f(t), the least value of v is when dv/dt = 0; the value of t at that point should thereafter be substituted into f(t). Note: this is obviously adaptable.
When a particle is moving in a plane, you can describe its position r, its velocity v and its acceleration a using v[] vectors. Taking x'=dx/dt, x''=d²x/dt² etc.: if r = xi + yj, then v = dr/dt = r' = []i + []j x'--y'--, and a = dv/dt = d²r/dt² = r'' = []i + []j x''--y''--. Note: instead of succeeding apostrophes, dots above are often used as a short notation for differentiation with respect to time.
When you integrate a vector with respect to time, the constant of integration is a vector.
If P has position vector xi + yj (where i an j are unit vectors), the distance from O to P is √(x²+y²).
An object's direction of motion is its [] vector velocity--.
If an object has velocity vector xi+yj, the angle between the direction of motion of the particle and the horizontal (or i - the unit vector) is tan^-1(y/x).
The centre of mass of a body is the point at which the whole mass of the body can be considered to be concentrated.
[x (overbar) represent the x-value at the centre of mass of the system.] (1,2,n ss) A system of n particles, with masses m1, m2, ..., mn are placed along the x-[] axis at the points (x1,0), (x2,0), ..., (xn,0) respectively. If M = m1+m2+...+mn: m1x1+m2x2+...+mnxn = M * x(overbar); to find x(overbar), on both sides, divide by M; the coordinates of the centre of mass of the system is (x[overbar],0).
(i ss) For a set of particles arranged along a straight line Σ^n(i=1) mixi = x(overbar) * Σ^n(i=1) mi, thus x(overbar) = Σ^n(i=1) mixi / Σ^n(i=1) mi.
[Let y(overbar) represent the y-value at the centre of mass of the system.] (1,2,n ss) If a system of n particles m1, m2, ..., mn are placed along the y-[] axis at the points (0,y1), (0,y2), ..., (0,yn) respectively, then Σ^n(i=1) miyi = y(overbar) * Σ^n(i=1) mi; the position of the centre of the mass of the system is (0, y[overbar]).
(1,2,n ss) In general, if a system consists of n particles: mass m1 with position vector r1, mass m2 with position vector r2, ... mass mn with position vector rn, then Σ^n(i=1) miri = r(overbar) * Σ^n(i=1) mi, where the position vector of the centre of mass of the system is r(overbar) = ([],[]) (x[overbar],y[overbar]).
If a question does not specify axes or coordinates, you can choose your own [] and [] axes and origin; if, for example, you are given a right-angle triangle ABC and you choose B (the corner point) as the origin, and BA and BC as a[] axes, and you find the centre of mass to be at the point (x,y), you should then say that the point is x from [] and y from [] AB--BC--(you must NOT leave it in [] form)
An object with thickness that is very small in comparison with its length and width is modelled as a lamina; its thickness it taken to be one-dimensional, hence the object is regarded as being two dimensional - with area rather than volume (e.g. paper).
If a lamina's mass is evenly spread throughout its area, it is uniform.
If a lamina has an axis of symmetry: on its axis of symmetry lies its centre of mass. If a lamina has more than one axis of symmetry, then it follows that the centre of mass must be at the [] of [] of the axes of symmetry point of intersection--. Note: you can find the positions of the centres of mass of standard [] plane laminas, including a rectangle, a rectangle, a triangle and a semicircle uniform--.
In a uniform circular disc (a circle), as every diameter of the disc is a line of symmetry, the centre of mass is at the centre.
In a uniform triangular lamina, there are medians, which each join a [] to the [] of the [...] vertex--midpoint--opposite side-.
The centre of mass of any uniform triangular lamina is at the intersection of the medians. This point is called the [] of the triangle centroid--.
The medians of an equilateral triangle are also its [] of [] axes of symmetry; if the triangle is isoceles, then [] of the medians is an axis of symmetry one--.
(1,2,3 ss) If the coordinates of the three vertices of a uniform triangular lamina are (x1,y1), (x2,y2) and (x3,y3), then the coordinates of the centre of mass are given by taking the [] of the vertices average (i.e. mean). If the centre of mass is G, G is the point (x1+x2+x2/3,y1+y2+y3/3).
(1,2 ss) If the ends of a rod (one-dimensional) are (x1,y1) and (x2,y2), then its centre of mass is the point (x1+x2/2,y1+y2/2).
A uniform sector of a circle of radius r and centre angle 2a, where a is measured in radian, has its centre of mass on the [] of [] axis of symmetry (the line which splits the sector equally into two) at a distance [] from the centre 2rsina/3a--.
If a uniform plane figure consists of multiple standard unifrom shapes, let the mass per area be m kg per cm². Then you can [] the shape into the component pieces split--. For each component shape, find the centre of mass and the shape's total mass. Thereafter, replace each centre of mass with a [] that is at the same [] particle--point-- and has the same [] as the component shape that it corresponds to mass--. You can then find the [...] of those particles centre of mass--. If m1, m2, . .., mn are the masses of the component pieces and (a1,b1), (a2,b2), ..., (an,bn) are the centres of mass respectively: r(overbar)*Σmi = Σmi(ai,bi).
In some situations, standard uniform shapes are taken from standard uniform shapes to produce the lamina. Let the mass per area be m kg per cm². For the primary shape and the shapes subtracted, multiply the total ... mass by the centre of mass. That value, for each of the subracted shapes, should be [] from that value for the primary shape subracted--; that can then be set equal to the [] of the lamina multiplied by [] mass--(x[overbar],y[overbar])--. In such situations, to help solve, you can use an equation. If the primary shape has mass M and centre of mass (A,B), and the subtracted shapes have masses m1, m2, ..., mn and centres of mass (a1,b1), (a2,b2), ..., (an,bn) respectively: the centre of mass of the lamina can be found through the equation: M*(A,B) - Σ^n(i=1) mi*(ai,bi) = (M - Σ^n(i=1) mi) * (x[overbar],y[overbar]); note: (M - Σ^n(i=1) mi) is the lamina's mass (and area).
A number of rods joined together or a number of pieces of wire joined together make up a framework.
If you have a framwork, the centre of mass of each wire or rod is at the midpoint. Let each unit of length m weigh m kg. Thus for each wire or rod of length M, thus mass M, let their be a particle at their [...] of mass [] centre of mass--M--. The centre of mass of those particles is the centre of mass of the framework.
If a uniform circular arc of radius r and centre angle 2a, where a is measured in radians, has its centre of mass on the axis of symmetry at a distance [] from the centre rsina/a--.
A lamina can be suspended by means of a [] attached to some point of the lamina string--, or can be allowed to pivot freely about a [] [] which passes through some point of the lamina horizontal axis--. When a lamina is suspended freely from a fixed point or pivots freely about an horizontal axis, it will rest in [] in a vertical plane equilibrium--, with its [...] vertically below the point of suspension or the pivot centre of mass--; the resultant of the moments about O (the point of suspension) in both cases is 0. If a lamina is suspended from fixed point, the force acting on it are its weight and tension in the string; both forces pass through ... the point of suspension.
If a lamina is free to rotate about a fixed horizontal pivot, the forces acting on it are its weight and the reaction of the pivot on the [] lamina; both pass through the point of suspension.
If a lamina rests in equilibrium on a rough inclined plane then the line of action of the weight of the lamina must pass through the side of the lamina which is ... in contact with the plane. The weight of the lamina produces a [] moment clockwise--, which causes the lamina to ... remain in contact with the plane. If the angle of the plane's inclination is increased so that the line of action of the weight doesn't pass through the side in contact with the plane, then the weight produces an [] moment about []++ anticlockwise--A--, meaning the lamina will topple over. Note: you can usually assume that the coefficient of friction between the lamina and the plane is large enough to prevent the lamina from ... slipping down the plane.
If a shape has A as a vertice: when the shape is suspended freely from A, the vertical is a continuation of the line from A to the centre of mass.
A lamina is on a plane inclined by a; the bottom side AB is in contact with a rough inclined plane. AM = x(overbar), AN = y(overbar), G =(AM,AN)=(x [overbar],y[overbar]). The angle of the plane is gradually increased; assuming that the lamina does not slide down the plane, what is the angle that the plane makes witht he horizontal when the Lamina is about to topple over? At that point, the moment of W (the weight) about A is zero, thus G would be vertically above the point A (W, A and G would be on the same line). Angle GAM = 90° - a (in this situation, the bottom part of the lamina is a rectangle); triangle AMG is a [] triangle right-angle--, thus angle AGM = a. In triangle AMG, tan(a) = AM/MG = x(overbar)/y(overbar). (The angle a is the answer).
When a lamina on an inclined plane is about to topple, its centre of mass will be vertically above the [] point of the lamina which is in contact with the plane lowest--.
You can calculate the work done by a force when its point of application moves using the fomula: work = force x distance moved by the point of application in the direction of the force, or in short notation: work = Fs, where F is the [] and s is the [...] force--distance moved by the point of application in the direction of the force--. When the force is measured in newtons and the disctance in metres, the work done is measured in joules (J).
Work is done against [] whenever a particle's vertical height is increased gravity-- (this may be because the particle moved vertically or at an angle to the horizontal). Work done against gravity = mgh, where m is the object's [], g is the object's acceleration due to [] and h is the [] [] raised mass--gravity--vertical distance--.
You can calculate the work done by a force acting at an angle to the direction of motion using the formula: work done = component of force in the direction of motion x distance moved in the direction of motion.
The kinetic energy of a moving particle is given by the formula: kinetic energy (K.E) = mv²/2, where m is the particle's [] and v is its [] mass--velocity--.
The potential energy of a particle is given by the formula: potential energy (P.E) = mgh, where h is height of the particle above an arbitrary [] level fixed-- (usually ground).
A particle possesses kinetic energy when it is moving.
When the mass of a particle is measured in kilograms and its velocity is measured in metres per second, the kinetic entergy is measured in joules.
The work done by a force which accelerates a particle (in terms of direction) [] is connected to the kinetic energy of that particle horizontally--. Work done = change in kinetic entergy. Using the equation of motion (F = ma) and the constant acceleration formula v² = u² + 2as, we can get F = m(v²-u²)/2s by [] them and eliminating [] combining--a--, thus work done = [] = [] Fs = mv²/2 - mu²/2, hence work done = [] K.E - [] K.E final--initial-- = change in K.E.
The short notation for work done = change in kinetic energy is Fs = mv²/2 - mu²/2.
Before calculating a particle's potential energy, choose a zero level.
If a particle moves upwards, its potential energy will increase; if it moves downwards its potential energy will decrease.
When no external forces (other than gravity) do work on a particle during its motion, the sum of the particle's kinetic and potential energies remains constant. This is called the principle of conservation of mechanical energy. As a result, in such a situation, decrease in P.E = increase in K.E.
The total energy possessed by a particle can only change during the particle's motion if some [] [] (except gravity) is doing work on the particle external force--. Any non-gravitational resistance to motion acting on a plane will [] the particle's total energy reduce-- as the particle will have to do [] to overcome the resistance work--. The change in the total energy of a particle is equal to the [] [] on the particle work done-- (including work done against a force): this is called the work-energy principle.
If a force of magnitude A acts against the direction of motion: the work done against the force (that has moved 's') is As. In the work-energy principle, work done against a force counts as work done on the particle.
If the total loss of energy is L and the work done against resistances is W, then the work done by the object is W - L. Generally, a particle's total loss of energy = [...] + [...] loss of K.E + loss of P.E.
The rate of doing work is called power.
Power is measured in watts (W) where 1 watt is 1 joule per second. The power of an engine is often measured in killowatts (kW) - 1000W.
The power developed by the engine of a moving vehicle is calculated with the formula: power = F x v, where F is the [] [] produced by the engine and v is the vehicle's [] driving force--speed--.
If you want to find the maximum speed of a moving vehicle, find the speed at which there is no acceleration, thus where the resultant horizontal force is 0. Where R is the resistance force, P is the known power, F the driving force and v the unknown speed, to find the maximum speed put P = Fv, where F and [] are equal R--.
The impulse-momentum principle states that the impulse of a force is equal to the produced change in momentum: I = mv - mu. Impulse is measured in Newton-seconds (NS).
The principle of conservation of linear momentum states that the total momentum before impact equals the total momentum ... after impact. (1,2 ss) m1u1 + m2u2 = m1v1 + m2v2.
Momentum = mass x velocity. Momentum is measured in Ns.
Impulse and momentum are both [] quantities vector--. You can write the impulse-momentum principle and the principle fo conservation of linear momentum as [] equations vector--, and use them to solve problems involving collisions where the velocities and any impulse are given in [] form vector--.
A direct impact is a collision between particles, of the same size, which are moving along the same [] [] straight line. When two particles collide, their [] after the collision depends upon the material from which they are made speeds--.
Newton's Law of Restitution (/Newton's Experimental Law) defines how the speeds of the particles after the collision depends on the nature of the particles as well as the pre-collision speeds; this law only holds when the collision takes place in [] space or on a [] surface free--smooth--.
Newton's Law of Restitution (sometime called Newton's Experimental Law): e = speed of separation of particles/speed of approach of particles. The constant 'e' is called the coefficient of restitution and e lies between [] and [] inclusive 0 and 1--; 0[]e[]1 =<--=>.
The value of the coefficient of restitution (e) depends on the [] from which the particles are made materials--. Particles for which e = 1 are called [...] particles perfectly elastic; particles for which e = 0 are called [...] particles inelastic--. Inelastic particles [] on impact coalesce--.
With Newton's Law of Restitution, the speed of approach is the [] in the speeds of the particles before impact, and the speed of seperation is the [] in the speeds after impact difference--difference--. For collision and seperation to take place, both differences must be positive.
With Newton's Law of Restitution, when one seeks e, pick one direction to be taken as p[] positive. If (before collisions) particle A has speed X m/s in the positive direction, while particle B moves at Y m/s in the opposite direction: the speed of seperation is X+Y (as B is taken to have speed [in reference to the positive direction] -Y).
If a is algebraic and b is a constant greater than n, and a = b, with inequality a >n.
When drawing outcomes of impacts, where the post-collision speeds are known, the arrows representing the unknown speeds are draw to face in the [] direction positive--.
The speed of a particle after impact with a smooth plane surface perpendicular to the direction of motion depends on the speed of the particle [...] and the [] of [] ([]) between the particle and the plane before the impact--coefficient of restitution (e)--. If the particle's speed before impact is u and the speed after is v: e = v/u (as the speed of the plane is zero), thus e = speed of seperation/speed of approach can be rewritten for these cases as e = speed of []/speed of [] rebound--approach--.
If a particle falls A cm from a speed of x m/s onto a smooth horizontal plane, then rebounds to a height of B cm: you can find the speeds before and after impact using the suvat equations. In the suvat equations: the speed before impact (the speed of rebound) can be found by finding v, while the speed after impact (the speed of approach) can be found by finding u.
If P and Q have collided, and after impact P has speed v (algebraic/unknown) and Q has speed w (algebraic/unknown), with Q being further in the positive direction, the speed of separation is given as w-v.
You can solve problems relating to successive impacts involving three particles, or two particles and a smooth plane surface by considering each collision separately.
The total kinetic energy where there are two objects, one with kinetic energy of A and the other with kinetic energy of B, is A+B.
Where velocity is given as a vector: K.E = mass x speed²/2.
If particle A of mass m1 has velocity u1 before impact and velocity v1 afterwards, while particle B of mass m2 has velocity u2 before impact and velocity v2 afterwards: the total kinetic energy lost in impact is (1,2 ss) (m1u1²/2 + m2u2²/2)-(m1v1²/2 + m1v1²/2).
The [] of a force measures the turning effect of the force on the body on which it is acting moment--. The moment of a force is measured in Newton-metres (Nm). When describing the turning effect of the force you need to consider its [] and the sense of [] magnitude--rotation-- (clockwise or anticlockwise).
If a body is resting in equilibrium, then there is zero [] force in any [] resultant--direction--, and the sum of the moments about any point is zero (NB the point doesn't need to be on the body).
Where d m is the perpendicular distance to the line of action of the force, and the force is F N, then the moment of the force about P is Fd Nm [#A47#]. However, if P is D from a point on the line of force (F), then the moment of the force about P is D times the [] of F that is [] to D component--perpendicular--.
If a rod is hinged at A, then the reaction at A can be expressed in terms of its [] and [] [] horizontal and vertical components.
When a rigid body is resting in [] under the action of three non-[] forces, you can solve problems using the triangle of forces equilibrium--parallel--; you could, for example, use the fact that d[] can be expressed in different ways distances--. Note: a triangle can be used in a variety of quetion types, including those to do with friction.
If a rod is resiting in equilibrium: the vector sum of the forces acting on it must be zero.
If a rigid body is resting in equilibrium under the action of three non-parallel forces, the line of action of each force shares a common [] with each of the other forces point, thus you can draw a [] of vectors/forces to represent the three forces triangle--.
The weight of a []-[] beam doesn't have to act at its midpoint non-uniform--.
If a body is in limiting equilibrium, the frictional force takes its [] value maximum--; where fm (m ss) is the maximum value of the frictional force: Fm = μR. To find μ (the coefficient of friction), you need Fm and R; note: to find F and R, you may need to use moments (remembering that the point you take the moments about doesn't need to be on the body) and resolving.
A plane that passes through a vertical line is a vertical plane.
A particle moving in a vertical plane is sometimes called a projectile. In M2, one makes the following modelling assumptions: air resistnce can be ignored, objects can be modelled as particles, the force due to gravity is uniform and acts vertically []wards down--, objects only move in a [] plane vertical-- (i.e. there is no side-to-side movement).
Created by: Toluo
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