PAPER 1 EXAM QUESTIONS 28 Octobe014 Lesson Description In this lesson we revise: Forces and Newton s Laws Light Electrostatics Electric Circuits Challenge Question Two point charges, Q and Q, a distance 3 m apart, are shown below. The charge on Q is - 14 μc 1 2 1 and the charge on Q is + 20 μc. 2 Calculate the net electric field at point P situated 2 m from Q 2. Question 1 Exam Questions (Adapted from Eastern Cape Grade 11 Paper 1, Nov 2013, Question 3) Thato applies of force of 180 N on a block at 55 to the horizontal as shown in the diagram. The block remains stationary. 1.1 Draw a labelled free-body diagram showing all the forces acting on the block 1.2 Calculate the magnitude of the force of friction 1.3 Find the co-efficient of static friction between the block and the surface 1.4 Using Newton s laws to explain why the block remains stationary even though Thato applies a force on it P a g e 1
Question 2 A light ray moving from water to air, strikes the air-water surface with an angle of incidence equal to 47. Determine if the light ray undergoes refraction or total internal reflection by find the angle of deviation after striking the surface. [Given data: Refractive index for air = 1,00 and refractive index for water = 1,33] Question 3 (Adapted from Eastern Cape Grade 11 Paper 1, Nov 2013, Question 6) Consider the diagram below, which is not drawn to scale Calculate the magnitude of the gravitational force on the moon due to the earth and sun in the position shown in the diagram Question 4 Two identical small metal spheres, P and Q, each with a mass of 1 g, are placed on two insulated stands, 20 mm apart. The charge on P is +3 x 10-11 C and the charge on Q is -2 x 10-11 C 4.1. Draw the electric field pattern between P and Q. 4.2. Calculate the electrostatic force of attraction between the two spheres. 4.3. The two spheres now make contact, exchange charge, and return to their original positions. Is the electrostatic force after contact a force of attraction or repulsion? Question 5 (Adapted from Eastern Cape Grade 11 Paper 1, Nov 2013, Question 6) A number of meters and resistors are connected as shown in the diagram. A 3,0 V battery is connected to the terminals X and Y 5.1 Determine the reading on V 1 5.2 Calculate the reading on A. P a g e 2
5.3 Calculate the reading on V 2 Answers Challenge Question Electric field at P due to Q 1 E 1 =kq/ =(9x10 9 )(14x10-6 )/(1) 2 =1.26x10 5 N.C -1 Electric field at P due to Q 2 E 2 =kq/ = (9x10 9 )(20x10-6 )/(2) 2 =4.5x10 4 N.C -1 Nett electric field at P due to Q 1 and Q 2 E nett =E 1 +E 2 =1.26x10 5 +4.5x10 4 =1.71x10 5 N.C -1 left Exam Questions Question 1 1.1 F A = Force applied by Thato F N f f F A F g Note: Do not include components with F A 1.2 f f = F Hor = F.cosθ = 180 cos55 0 = 103,24N f f = Frictional force F g = mg = weight of the block = force of the earth on the block F N = Normal force = force of the surface on block P a g e 3
1.3 F N = F vert + F g = F.sinθ + mg = 180. sin 55 o + (30)(9,8) = 441,45N f f = μ s F N 103,24 = μ s. 441,45 μ s = 0,23 1.4 The block does not move since the horizontal component of the force Thato applies is equal in magnitude to the frictional force. There is no net force applied on the block. According to Newton s 1 st Law, an object will remain at rest or move at constant velocity unless there is a net or resultant force acting on the object Question 2 n 1. sinθ i = n 2. sinθ r (1,33)(sin 47 ) = (1)(sinθ r ) sin θ r = 0,973 θ r = 76,58 The light ray is refracted and moves away from the normal The angle of deviation = 76,58-47 = 29,58 away from the normal (speed of light increases in air) Question 3 F sun on Moon =. sun. moon F Earth on Moon =. earth. moon 10-11 1 10 0 10 22 2 10 11 2 10 20 N 10-11 10 24 10 22 4 10 2 1 10 20 N F Sun on Moon and F Earth on Moon are perpendicular to each other So F net on moon = = 3,05 x 10 20 N 10 20 1 10 20 P a g e 4
Question 4 4.1. 4.2. F k. 1.. 2 10 10-11 2 10-11 20 10-2 4.3. Repulsion Question 5 5.1 V 1 = 3V 5.2 V across branch with 2Ω and 3Ω resistors = 3V I branch 1 = V R = 3 (2 + 3) = 0,6A 5.3 I branch 2 = V R = 3 (3 + 4) = 0,43A V 2 = I 4Ω R 4Ω = (0,43)(4) = 1,72V P a g e 5