1 of 42 Boardworks Ltd 2009 2 of

1 of 42 © Boardworks Ltd 2009

2 of 42 © Boardworks Ltd 2009

What is a scalar? Scalar quantities are measured with numbers and units. length temperature time (e. g. 16 cm) (e. g. 102 °C) (e. g. 7 s) 3 of 42 © Boardworks Ltd 2009

What is a vector? Vector quantities are measured with numbers and units, but also have a specific direction. acceleration displacement force (e. g. 30 m/s 2 upwards) (e. g. 200 miles northwest) (e. g. 2 N downwards) 4 of 42 © Boardworks Ltd 2009

Comparing scalar and vector quantities 5 of 42 © Boardworks Ltd 2009

Speed or velocity? Distance is a scalar and displacement is a vector. Similarly, speed is a scalar and velocity is a vector. Speed is the rate of change of distance in the direction of travel. Speedometers in cars measure speed. Velocity is a rate of change of displacement and has both magnitude and direction. Averages of both can be useful: average speed 6 of 42 = distance time average displacement = velocity time © Boardworks Ltd 2009

Vector or scalar? 7 of 42 © Boardworks Ltd 2009

Adding vectors Displacement is a quantity that is independent of the route taken between start and end points. If a car moves from A to B and then to C, its total displacement will be the same as if it had just moved in a straight line from A to C. C B Two or more displacement vectors can be added ‘nose to tail’ to calculate a A resultant vector Any two vectors of the same type can be added in this way to find a resultant. 8 of 42 © Boardworks Ltd 2009

Simplifying vectors Because of the way vectors are added, it is always possible to simplify a vector by splitting it into components. Its displacement is the same, but it is now much easier to describe. How would you describe the car’s displacement in component terms? C B y A x x component D The car has a final displacement of x miles east and y miles north. This can be represented by (x, y). 9 of 42 © Boardworks Ltd 2009 y component Imagine that instead of travelling via B, the car travels via D:

Understanding vector calculations 10 of 42 © Boardworks Ltd 2009

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Motion under constant acceleration Calculating vector quantities such as velocity or displacement can be complicated, but when acceleration is constant, four equations always apply. v = u + at s = ½(u + v)t v 2 = u 2 + 2 as s = ut + ½at 2 These are sometimes known as the constant acceleration equations, or the ‘uvast’ or ‘suvat’ equations. What do the symbols u, v, a, s and t represent? 12 of 42 © Boardworks Ltd 2009

Representing motion The symbol a represents acceleration. The symbol t represents time. Displacement, s is always measured relative to a starting position, so it is always true that when t = 0, s = 0. Velocity at time t is represented by v, and u represents the value of v when t = 0. This is the initial velocity. l a = acceleration l t = time l s = displacement l v = velocity l u = initial velocity 13 of 42 © Boardworks Ltd 2009

Two velocity equations Two of the constant acceleration equations are velocity equations. v = u + at v 2 = u 2 + 2 as The first of these can be used to find the velocity at a particular time t. The second can be used to find the velocity at a particular displacement s. When deciding which equation to use, it is good practice to write down what you know about the values of u, v, a, s and t before you start any calculations. 14 of 42 u=? v=? a=? s=? t=? © Boardworks Ltd 2009

Calculating final velocity A cyclist accelerates towards the end of the race in order to win. If he is moving at 6 m/s then accelerates by 1. 5 m/s 2 for the final five seconds of the race, calculate his speed as he crosses the line. First write down what you know about u, v, a, s and t: u=6 v=? a = 1. 5 m/s 2 s= t = 5 s The question gives a value for t. What is the relevant equation? v = u + at v = 6 + (1. 5 × 5) v = 13. 5 m/s 15 of 42 © Boardworks Ltd 2009

Rearranging a velocity equation 16 of 42 © Boardworks Ltd 2009

Rearranging an equation – calculations A coin is dropped from a window. If it hits the ground at 10 m/s, work out the height of the window. First write down what you know about u, v, a, s and t: u = 0 m/s v = 10 m/s a = 10 m/s 2 s=? t= 17 of 42 The question involves the variables u, v, a and s, so the relevant equation is v 2 = u 2 + 2 as. Start by rearranging the equation to find a formula for s: v 2 – u 2 s= 2 a s = 100 20 102 – 02 s= 2 x 10 s = 5 m © Boardworks Ltd 2009

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Two displacement equations Two of the constant acceleration equations are displacement equations. s = ½(u + v)t s = ut + ½at 2 The first of these gives the displacement as a function of the initial velocity and velocity at time t. The second gives it as a function of initial velocity, acceleration and time. As with the velocity equations, it is good practice to write down what you know about the values of u, v, a, s and t before you attempt any calculations. 19 of 42 u=? v=? a=? s=? t=? © Boardworks Ltd 2009

Calculating displacement A car travelling at 20 m/s takes five seconds to stop. What is the stopping distance of the car? First write down what you know about u, v, a, s and t: u = 20 m/s v = 0 m/s a= s=? t = 5 s The question gives u, v and t and asks for a value for s, so the relevant equation is s = ½(u + v)t: s = ½ × (20 + 0) × 5 20 of 42 s = ½ × 20 × 5 s = 50 m © Boardworks Ltd 2009

Rearranging a displacement equation 21 of 42 © Boardworks Ltd 2009

Rearranging an equation – calculations Calculate the acceleration of an ambulance if it starts at rest and takes six seconds to travel 50 m. First write down what you know about u, v, a, s and t: u = 0 m/s v= a=? s = 50 m t = 6 s The question involves the variables u, a, s and t, so the relevant equation is s = ut + ½at 2. Start by rearranging the equation to find a formula for a: a= a= 22 of 42 s – ut ½t 2 50 – (0 × 6) ½× 6× 6 50 a= 18 a = 2. 8 m/s 2 © Boardworks Ltd 2009

Using the equations of motion 23 of 42 © Boardworks Ltd 2009

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Describing parabolic motion What are some of the terms that are used to describe projectile motion? l A trajectory is the path of any moving object. l A projectile is an object that is given an initial force, then allowed to move freely through space. l A parabola is the name given to the shape of the curve a projectile follows when gravity is the only force acting on it. 25 of 42 © Boardworks Ltd 2009

Forces on a projectile What are the two forces that act on a freely moving projectile? Air resistance and gravity. Depending on the shape and density of an object, it is often possible to ignore the effects of air resistance. This reduces the forces on the object to one constant force in a constant direction, giving: l constant horizontal velocity l constant vertical acceleration. Horizontal motion is therefore very simple, and vertical motion can be solved with the constant acceleration equations. 26 of 42 © Boardworks Ltd 2009

When is a trajectory a parabola? 27 of 42 © Boardworks Ltd 2009

Understanding projectile motion 28 of 42 © Boardworks Ltd 2009

Forces on a projectile 29 of 42 © Boardworks Ltd 2009

Motion under gravity 30 of 42 © Boardworks Ltd 2009

Solving projectile motion How should you go about solving a problem involving projectile motion? Split the problem into two parts: Horizontal motion at constant velocity: l Use the equation, speed = distance / time (equation 1). Vertical motion under constant acceleration: l As always, start by writing down what you already know about u, v, a, s and t. l Choose a constant acceleration equation (equation 2). The order you use these two equations in will depend on the nature of the problem. Start by writing down everything you know, and the rest should follow! 31 of 42 © Boardworks Ltd 2009

Projectile calculations 32 of 42 © Boardworks Ltd 2009

Momentum is a vector quantity given by the following formula: momentum = mass × velocity To have momentum, an object needs to have mass and to be in motion. Which of these two vehicles do you think has a higher momentum? 34 of 42 © Boardworks Ltd 2009

Conservation of momentum Momentum is always conserved in any event or interaction: initial momentum = final momentum When a snooker ball collides with another, what happens to the momentum of each ball? Momentum is not created or lost, but transferred from one object to the other. What happens when a car brakes and comes to a halt, or when a rock falls to the ground and stops? Momentum is transferred to the Earth. 35 of 42 © Boardworks Ltd 2009

Explosions and recoil What happens to the momentum in the following situations? Remember: momentum is always conserved. 36 of 42 © Boardworks Ltd 2009

Calculating momentum Two vehicles, each weighing 1. 5 tonnes, are driving towards each other at 30 mph. If they collide, what will happen? 10 m/s Both vehicles come to a halt. What has happened to their momentum? Momentum is a vector quantity. The van has a momentum of 1500 × 10 = 15000 kgm/s to the right. The car’s momentum is 15000 kgm/s to the left, or – 15000 kgm/s to the right: initial momentum to the right = 15000 + – 15000 = 0 final momentum to the right = 0 37 of 42 © Boardworks Ltd 2009