WJEC Physics for AS Level Student Book 2nd Edition

15 1.1 Basic physics We have changed a subtraction into the addition of a negative number. The same procedure works with vectors. Look at Fig. 1.1.8(a). To find v 2 – v 1 , we add – v 1 to v 2 . The vector – v 1 is the same magnitude as v 1 but has the opposite direction. The dotted lines in (b) show us how we can think of it, using the nose- to-tail way of adding vectors: go backwards along v 1 and forwards along v 2 . So v 2 – v 1 is the vector from the end of v 1 to the end of v 2 . Diagram (c) shows the same calculation using the parallelogram method. It doesn’t matter which method we use: v 2 – v 1 : the red vector obviously has the same length and direction in (b) and (c). Example A car changes velocity from 25 m s – 1 due E , to 20 m s – 1 due N in 8.0 seconds. Calculate the mean acceleration. Answer Step 1: Draw the diagram. Take care with the direction of v 2 – v 1 : Go backwards along the v 1 vector (– v 1 ) then forwards along the v 2 (+ v 2 ). Step 2: Use Pythagoras' theorem to calculate Δ v . ( Δ v ) 2 = 25 2 + 20 2 = 1025 . \ Δ v = 32.0 m s – 1 . Step 3: Calculate θ . tan θ = 25 20 = 1.25 . \ θ = 51.3° Step 4: Calculate a . a = Δ v Δ t = 32.0 8.0 = 4.0 m s – 2 at 51.3° W of N. [NB direction!] (b) Components of vectors Look back at Fig. 1.1.1(a). How much of the force F is pulling the sledge forwards and how much of it is lifting the sledge? In other words, if a force, F , acts at an angle θ to the horizontal, what are its horizontal and vertical components, F h and F v ? Fig. 1.1.9 clarifies the question: F h and F v are the horizontal and vertical forces which add together to give F as the resultant. Using elementary trigonometry: F h = F cos θ F v = F sin θ and F = F h 2 + F v 2 This process is called resolving . Why is this a useful technique? For all sorts of reasons. Here are just two: 1. If the motion is horizontal (like the sledge) the horizontal component of the force multiplied by the distance moved gives the work done, i.e. the energy transferred. 2. When adding several (i.e. more than two) vectors, it is often easier to find the horizontal and vertical components of each and add them. Sometimes it is useful to find the components in directions other than horizontal and vertical, e.g. for the forces on a car on a slope the sensible directions to calculate components of the forces or velocity are parallel to and at right angles to the slope. We will meet this sort of situation frequently. The important thing to remember is that the component of a vector, A , in a direction at angle θ to the direction of the vector is always A c os θ . v 1 = 25 m s – 1 v 2 = 20 m s – 1 Δ v = v 2 – v 1 θ Fig. 1.1.8 Subtracting vectors Fig. 1.1.9 Force components F v F F h θ (a) (b) (c) v 2 – v 1 v 2 – v 1 v 1 v 2 v 2 – v 1 v 2 v 1 – v 1 v 2 Key term Resolving : Finding the component of a force: we resolve a force into its horizontal and vertical components. Knowledge check In Fig. 1.1.9, F = 150 N and θ = 30° . Calculate F v and F h . 1.1.6 Top tip At AS, you will only have to subtract vectors at right angles.