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The weight of an object is considered to act at it's centre of mass (or centre of gravity). So 'the total effect of gravity on each bit of a body is the same as one force at the centre of mass'.The centre of mass will lie on a line of symmetry of an object - if there are two (or more) lines of symmetry it is where they intersect.For triangles the centre of mass is on the 'centroid'. In right-angled triangles this is 1/3 of the way from the right-angle in both perpendicular directions.For other simple shapes there are formulae in the formula book to work out the position of the centre of mass.To work out the centre of mass of a group of particles, combine moments in an x- and y- direction.This is also the method used to work out the centre of mass of composite shapes:work out the individual centres of mass, and then combine them by taking moments.

Impulse is the change in momentum: \(mv - mu\)Impulse has direction and, due to conservation of momentum, the impulses in a collision will always balance. ie. one object will get a +ve impulse and the other -ve but the actual number will be the same.The impulse 'given to A' or 'put on A' is just the change in momentum of A.Newton's Experimental Law (also called Newton's Law of Restitution) states that when two particles collide the speed they bounce away with depends on the 'coefficient of restitution' (e).e = \(\frac{separation speed}{approach speed}\)e is always between 0 and 1, and varies depending on the material:-if e=0 the collision is 'inelastic'; the particles stick together instead of bouncing back-if e=1 the collision is 'perfectly elastic'; they'll bounce back with no loss of speedThis can be used for direct impacts between two particles, or a wall and a particle.In problems involving two unknowns combine conservation of momentum and Newton's experimental law to create two simultaneous equations.Kinetic energy is only conserved in perfectly elastic collisions:Loss of KE due to impact = Total KE before - Total KE afterAn impulse on an object will cause a change in KE.

A body is in equilibrium if:- The vector sum of the forces is zero-The sum of the moments of the forces (about ANY point) is zeroThis means that when you resolve (in ANY direction) the sum of the forces going one way equals the sum of the forces going the opposite way. AND the sum of the clockwise moments equals the sum of the anti-clockwise moments (about any point)The moment of a force (about a point) = Magnitude of the force x Perpendicular distance from the point to the line of action of the forceHowever you can work with the perpendicular component of a force at an angle, if the perpendicular distance is not obvious.If the line of action of a force goes through the point then it's moment is 0 because in the equation the distance is 0.You can use these methods to work out WHERE a force should be applied or WHAT SIZE force should be applied.If a rigid rod is connected to a wall you can draw the reaction force as a horizontal and a vertical component.Use \(F_{max}\)=\(\mu\)R if objects are on the point of slipping on a rough surface. R is always perpendicular to the surface, and F always opposes the direction of motion (or the direction the thing will try to move in if it's stationary).Combine with centre of mass techniques to work out if an object will slip or topple first.

The projectile model follows these rules:-The projectile is modelled as a particle (point-mass).-Air resistance is ignored-The horizontal speed is constant-There is constant acceleration (due to gravity) in the vertical direction.The weakness of this model is that it ignores air resistance so it's a bit too simple for real life applications.Solve problems by considering the horizontal and vertical components separately and using M1 equations of motion (suvat equations).Problems may ask you to:-Work out the magnitude or direction of velocity at a given time.-Find greatest height (this is when the vertical velocity equals 0)-Find horizontal range (calculate time using vertical equations then sub into horizontal equations to find distance travelled)-Derive and use the Cartesian equation of trajectory (by using algebra if initial speed/angle is unknown)

As well as linear speed,\(v\), (the distance travelled divided by the time taken) a particle moving in a circle also has an angular speed, \(\omega\), (the angle it has moved through, in radians, divided by the time).\(v\)=\(\frac{2\pi r}{T}\) and \(\omega\)=\(\frac{2\pi}{T}\) which leads to the equation \(v\)=\(r\omega\)Because the direction of motion of the particle is always changing, it's velocity is always changing. This means the particle is accelerating (even though the magnitude of it's velocity is constant!). The acceleration of a particle will always be in the same direction as the resultant force on it, and as it is moving in a circle the force is toward the centre of the circle.There are two equations for radial acceleration:\(a\)=\(r\omega^2\) and \(a\)=\(\frac{v^2}{r}\)Using Newton's \(2^{nd}\) Law, F=ma, you get two equations for the force on an object in uniform circular motion:\(F\)=\(mr\omega^2\) and \(F\)=\(\frac{mv^2}{r}\)The force could come from a string attached to the particle or from the reaction force with the inside face of a cylinder but the method for solving the problem is just the same.Another uniform circular motion problem is a conical pendulum; just resolve the tension in the string horizontally and vertically and use the equations above when required.For banked track and bowl problems DRAW a diagram, work out what information you have been given (in terms of the variables above) as well as what you need to find out and resolve if necessary. Remember that in a hemispherical shell the distance of any point on the inner surface from the centre is the radius, you may have to use trigonometry to calculate the radius of the circles that a particle is travelling in.

Centre of Mass

Coefficient of Restitution; Impulse

Equilibrium of a Rigid Body

Motion of a Projectile

Uniform Motion in a Circle

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