WJEC Physics for AS: Study and Rev Guide

Q & A 2 Tom’s answer (a) (i) The amount of energy used over a certain time 7 (ii) J = 7 (b) (i) D E p = mg D h 120 cos 20 = 112.76 m = 70 × 9.81 × 112.76 3 e.c.f. = 77432 J 7 (ii) KE = 2 1 2 mv = ´ 2 1 70 15 2 = 35 × 15 2 3 = 7875 J 7 (c) (i) Energy cannot be created or destroyed, only transferred into other forms. 3 (ii) The gravitational potential energy gained by the skier has been converted into kinetic energy. 7 (d) - - = = = 2 2 2 21 36 1.6875 240 240 v u a m s  - 2 F = ma = –70 × 1.6875 = 3 Examiner commentary (a) (i) Tom’s definition is of power rather than work. (ii) Tom’s answer follows on from his confusion between work and power. Error carried forward would not be applied here, even if his working was correct for power, as it is a mistake of principle. (b) (i) Tom has quoted the correct equation for D E p and the first mark is for correctly using it. Unfor- tunately he has used cos 20 ° which gives the horizontal rather than the vertical distance so he just gets one mark for substitution. (ii) Tom has quoted the KE formula and substituted a speed so receives 1 mark. Unfortunately he has used the speed difference for v rather than work- ing out the two values of 2 1 2 mv and finding the difference. (c) (ii) Tom has not done enough for the first mark. He needed to account for the fact that gain in kinetic energy was less than the loss in gravitational potential energy. (d) Tom has calculated the acceleration and, apart from the minus sign, has almost calculated the resultant force on the skier. This is the start of a possible method and has attracted 1 mark. Tom gains 4 out of 16 marks. 169 Questions and answers (a) A velocity–time graph is given for a body which is accelerating in a straight line. (i) Using the symbols given on the graph, write down an expression for the gradient and state what it represents. [2] (ii) Using the symbols given on the graph, write down an expression for the area under the graph and state what it represents. [2] (iii) Hence or otherwise show clearly that, using the usual symbols, = + 2 1 2 x ut at [2] (b) A cyclist accelerates from rest with a constant acceleration of 0.50 m s - 2 for 12.0 s. Calculate: (i) the distance travelled in this time [2] (ii) the maximum velocity attained. [2] (c) After 12.0 s, the cyclist stops pedalling and ‘freewheels’ to a standstill with constant deceleration over a distance of 120 m. (i) Calculate the time taken for the cyclist to decelerate to a stand-still. [2] (ii) Calculate the magnitude of the cyclist’s deceleration. [2] (d) Draw an acceleration–time graph on the grid for the whole of the cyclist’s journey. [4] (e) In reality the cyclist would not slow down with constant deceleration. This is because the total resistive force acting on the cyclist consists of a constant frictional force of 8.0 N and an air resistance force which is proportional to the square of the cyclist’s velocity. (i) When the cyclist was travelling with maximum velocity, the total resistive force acting was 165 N. Calculate the force of air resistance at this velocity. [1] (ii) Hence calculate the total resistive force acting when the cyclist is moving at half the maximum velocity. [2] 0 v u t velocity time