For use only in [the name of your school] 2014 M2 Note. M2 Notes (Edexcel)

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1 For use onl in [the name of our school] 014 M Note M Notes (Edecel) Copright - For AS, A notes and IGCSE / GCSE worksheets 1

2 For use onl in [the name of our school] 014 M Note Copright - For AS, A notes and IGCSE / GCSE worksheets

3 For use onl in [the name of our school] 014 M Note Copright - For AS, A notes and IGCSE / GCSE worksheets 3

4 For use onl in [the name of our school] 014 M Note Copright - For AS, A notes and IGCSE / GCSE worksheets 4

5 For use onl in [the name of our school] 014 M Note Copright - For AS, A notes and IGCSE / GCSE worksheets 5

6 For use onl in [the name of our school] 014 M Note Copright - For AS, A notes and IGCSE / GCSE worksheets 6

7 For use onl in [the name of our school] 014 M Note Short State how/ what Questions Modelling Assumption Jan 009 Q6 State an additional phsical factor which ma be taken into account in a refinement of the above model [A cricket ball being modelled as a projectile] to make it more realistic. Variable g, Air resistance, Speed of wind, Swing of ball, ball is not a particle. Particle Jan 009 Qc State how ou have used the modelling assumption that [the object] is a particle. So that [the object] s weight acts directl at the point [as labelled in the question]. Rod Jan 004 Q4c State how ou have used the modelling assumption that the ladder is a rod. It does not bend. or. has negligible thickness Copright - For AS, A notes and IGCSE / GCSE worksheets 7

8 For use onl in [the name of our school] 014 M Note Copright - For AS, A notes and IGCSE / GCSE worksheets 8

9 For use onl in [the name of our school] 014 M Note Projectiles Motion in a vertical plane with constant acceleration, eg under gravit. Simple cases of motion of a projectile. In M1 we met the formulae shown below: 1 s ut at 1 s = vt at v = u + at v u as = + 1 s u v t = + = ( + ) The motion we dealt with was in 1D that is with, for eample, a ball moving along a line. We are now considering motion in D, that is, for eample, when a ball is thrown up at an angle. The path shown opposite is for a ball which was fired with velocit u at an angle of with the horizontal. u There are four quantities to consider. Two in each of the horizontal and vertical directions, that is: velocit and displacement at time t after the projectile has been fired. Horizontall The object has no horizontal acceleration, so a = 0 The initial horizontal velocit is simpl the horizontal component of the initial velocit, which is calculated using the M1 method. u is equivalent to. u sin u cos That is, the initial horizontal velocit is u u =. u cos. So cos The equation v = u + at gives that v ucos = at time t. The equation s ut at ut 1 = + gives that cos = at time t. Copright - For AS, A notes and IGCSE / GCSE worksheets 9

10 For use onl in [the name of our school] 014 M Note Verticall u The object is in freefall verticall so a = g u sin is equivalent The initial vertical velocit is simpl the vertical to. component of the initial velocit, which is calculated using the M1 method. That is, the initial vertical velocit is u sin. So u = usin. The equation v = u + at gives that v = u cos gt at time t. u cos The equation s ut at 1 = + gives that ut gt 1 = sin at time t. So the four quantities to consider are: v = ucos v = u sin gt = ut cos = ut sin gt 1 Some ke things to understand about projectiles: Greatest Height When the projectile reaches its greatest height, it is travelling horizontall so to find the time at which the projectile reaches its greatest height solve v = 0. Range The range of the projectile is the horizontal distance it has travelled when it lands. To find the time when the projectile lands solve = 0. Speed The speed of the projectile at an time is the resultant of its horizontal and vertical components of its velocit. So the speed is v= v + v. v v v Direction The direction in which the projectile is travelling is calculated from the horizontal and vertical components of its velocit. So in the triangle, the angle that the projectile makes with the horizontal is β which can be calculated using trigonometr. Time for which projectile is above a certain height To find the time for which the projectile is above a height of hm, first of all find the times 1 when the projectile is at that height b solving = h. This gives ut sin gt = h which is β v v Copright - For AS, A notes and IGCSE / GCSE worksheets 10

11 For use onl in [the name of our school] 014 M Note a quadratic leading to two solutions. If these are t 1 and t where t > t 1 then the time for which the projectile is above a height of hm is t t1. Eample A cricket ball P is projected with speed 6 m s 1 from a point A on a cliff which is hm above horizontal ground. The angle of projection is to the horizontal, where tan =. The ball moves freel under gravit and hits the ground at the point B. O is at the foot of the cliff on the same horizontal level as B. (a) Find the greatest height of P above the level of A. 5 1 A hm The horizontal distance from A to B is 96m. (b) Find the height of A above the ground. O B Since Hence 5 tan = 1 it follows that is in this triangle 5 cos = and sin = Preliminar Work As stated before, there are four quantities to consider. Two in each of the horizontal and vertical directions, that is: velocit and displacement at time t after the cricket ball has been thrown. Horizontall a = 0 and, from the above triangle, u = 4. 1 v = u + at gives that v = 4 s = ut + at gives that = 4t. 1 6 ms is equivalent to. 6 sin = 10ms 1 6 cos = 4ms 1 Verticall a = g and, from the above triangle, u = 10 v = u + at gives that v = t using gives that 10t 4.9t h = +. s ut at 1 = + and the fact that A is hm above O v =. (a) As was seen, greatest height occurs when 0 So solve t = 0 to give t = 1.0 s (to 3sf) At this time its height above the level of O is the level of A is = = 5.10 m (to 3sf) = h so its height above (b) The horizontal distance from A to B is 19m so the ball is at B when = 96. Solving = 96 gives 4t = 96 and so t = 4. Copright - For AS, A notes and IGCSE / GCSE worksheets 11

12 For use onl in [the name of our school] 014 M Note = 10t 4.9t + h so when t = 4, So h = 0 and so h = = h = 0. Hence O is 38.4m verticall below A. B is on the ground, so the height of A above the ground is 38.4m. Copright - For AS, A notes and IGCSE / GCSE worksheets 1