WSSS 2007 Prelim P2 ANS

2 Section A Answer all the questions in this section. 1 Fig. 1.1 shows the speed-time graph for a sma ll balloon initially at rest, then falling vertically through the air and then attaining constant speed. (a)(i) Use Fig. 1.1 to determine the time interval over which the balloon is accelerating. 0.4 s [1] [note: 0.5 is also acceptable for this diagram] (ii) Is the acceleration constant ? Explain your answer. No. Since the gradient of the graph is decreasing, the acceleration decreases too. Hence, the acceleration is not constant. [2] (b) Explain, using the forces acting, the motion of the balloon for the first second a f t e r r e l e a s e . As the ballon falls from rest, the air resistance increases gradually and the resultant force decreases. This would result in a decreasing acceleration of the balloon. As the balloon continue falling, the air resistance increases until it balances the weight of the balloon. There is no resultant force and thus no acceleration, which will make the balloon to fall a constant speed. [2] For Examiner’s Use Fig. 1.1

3 (c) The mass of the balloon is 60 g. The gravitational field strength is 10 N/kg. (i) Calculate the weight of the balloon, W = mg = (60/1000) (10) = 0.6 N weight =……………………. [1] (ii) State the value of the air resistance on the balloon 1.0 s after release. Explain y o u r a n s w e r . [ 2 ] 0.6 N. Since the balloon is moving with constanct speed after 1.0 s of release, the forces are balanced (i.e. there is no resultant force). Hence, the air resistance has the same magnitude as the weight of balloon. For Examiner’s Use

4 2 A cyclist, together with his bicycle, has a tota l mass of 90 kg and is travelling with a constant speed of 15 m/s on a flat road at A, as shown in Fig. 2.1. He then goes down a small slope to B so descending 4.0 m. Calculate (a) the kinetic energy at A, Ek = ½ mv2 = ½ (90)(15)2 = 10125 J kinetic energy =……………………. [1] (b) the loss of potential energy between A and B, Ep = mgh = (90)(10)(4) = 3600 J loss of potential energy =……………………. [1] (c) the speed at B, assuming that all the lost potential energy is transformed into kinetic energy of the cyclist and bicycle. Using conservation of energy, Energy at the bottom of slope = Energy at the top of slope ½ (m)(v) 2 = ½ (m)(v)2 + mgh ½ (90)(v) 2 = 10125 + 3600 [2] v 2 = (2)(13725) / 90 v 2 = 305 v = 17.5 m/s [1] speed =……………………. [3] Fig. 2.1 For Examiner’s Use

5 3 Sebast