2014 H2 P3 solutions vjc

2014 H2P3 Suggested Solutions 1 (a) (i) The horizontal component of the velocity will be constant since there is no horizontal acceleration. (ii) The vertical component of the velocity will increase from uniformly from zero at a constant rate of g or 9.81m s-2 since there is vertical gravitational acceleration. (b) (i) The horizontal component of the velocity will decrease non-uniformly with time at a bigger initial rate of decrease. (ii) The vertical velocity will increase at an initial rate of g (or 9.81 m s-2). However, the final vertical velocity component will be smaller because the gravitational force downwards will now be opposed by air resistance upwards. (c) 2 (a) (i) The radian is the angle subtended by an arc of length equal to the radius of a circle i.e.  = s/r. (ii) For an object in simple harmonic oscillation, angular frequency  is the characteristic constant of the oscillator given by  = 2 f, where f is the natural frequency of the oscillator. (b) (i) The total energy of the sphere is equivalent to its initial gravitational potential energy at its highest point before release. Hence, Etot = mgh = (0.120)(9.81)(0.0040) = 4.7 x 10-3 J (ii) Etot = ½ m2xo2 = ½ (0.120)(2)(0.0802)  = 3.50 = 2 f , i.e. 22 3 2 0 )10x0.8)(120.0( )10x7.4(2 2 12 2 1 − − ==  mx Ef tot = 0.56 Hz 3 (a) The First Law of Thermodynamics states that the internal energy U of a system depends on its state and the increase in internal energy U is equal to sum of the heat supplied Q and work done W on the system; i.e. U = Q + W. (b) (i) Using pV = nRT for state X, the amount of gas, mol 0.0498=== − 290x31.8 10x0.5x10x4.2 45 RT pVn (ii) 1 Work done on the system in the expansion from X to Y is given by W = - (2.4 x 105)([14.4 – 5.0]x10-4) = -226J . Alternatively, work done by the system is 226 J. With air resistance

2014 H2P3 Suggested Solutions (ii) 2 The change in internal energy for one complete cycle XYZX is zero as the system returns to its original state. (iii) 4 (a) The principle states that when waves of the same kind meet at a point in space, the resultant wave displacement is given by the vector sum of the individual wave displacements at that point. (b) (i) The phase difference between the two waves arriving at point P is o57.3 ==== rad 0.1244 72  xxT t (ii) The bright fringe closest to point P has resultant amplitude, Ab = 3.4 + 0.6 = 4.0 units. The dark fringe closest to point P has resultant amplitude Ad = 3.4 - 0.6 = 2.8 units. Intensity of light at a point is directly proportional to the square of the amplitude. Hence, 0.49== 2 2 0.4 8.2 fringebright theofintensity fringedark theofintensity the 5 (a) Emax=hf -  =hf – hf0 =hc(1/ - 1/o) (b) (i) The intercept at the 1/ axis will give the value of 1/o where Emax is zero. From the graph 1/o = 2.3 x106 m-1. Thus, o = 4.35 x 10-7 m - 226 +344