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 + 540 - 884 - 884

2014 H2P3 Suggested Solutions (ii) The gradient of the graph is the value of the hc. Thus, sJ 6.7x10 34− − =−= h6 19 10x)30.275.3( 10x90.2hc (c) The sketch is shown as the dotted line in (b) (i). A higher value of work function corresponds to a shorter threshold wavelength of o with a bigger horizontal intercept value of 1/o. 6 (a) (i) A field of force refers to a region of space where an object or particle will experience a force due to the specific property possessed by the object or particle, e.g. mass or charge of the object. (ii) The charged particle may be moving at a non-zero angle to a magnetic field, in which case it will experience a magnetic force by virtue of its charge. It may also be in a gravitational field such that, whether it is stationary or moving, it will experience a gravitational force by virtue of its mass. (b) (i) The two regions are within the metal spheres where the field should be zero under electrostatic equilibrium conditions. (ii) 1. Radius of sphere A, rA = 4.0 cm; and radius of sphere B, rB = 2.0 cm. 2. The two spheres have charges of opposite sign. The graph shows that the electric field is always positive between the two spheres, implying that the direction of the electric field is constant (always pointing towards the right). From the definition of the electric field, this means that the left sphere is positively charged while the right sphere is negatively charged. (The above point is sufficient as the answer, but the following additional point is also acceptable, though there is not enough space to write it down.) Furthermore, the graph shows that the electric field is never zero along the line joining the two spheres. If the charges on the two spheres are of the same sign, their electric fields would be opposite in direction along the line between the spheres, and so there will be a point where the resultant electric field is zero. Hence, the spheres must have charges of opposite signs. (Alternate explanation: The two spheres have charges of opposite sign and of different magnitudes. The resultant field in the region between them is given by the vector sum of the two fields contributed by each. The individual fields decrease in magnitude inversely to the square of distance from the centre of the sphere, resulting in the smallest magnitude in between them.) (iii) Note: the electric field is practically constant in the region from x = 16 cm to x = 21 cm. 1 The energy gain is given by the approximate work done by the electric force given by W =FE x = (qE) x x = (3 x 1.6 x10-19)(1.2 x 105)(0.210 – 0.160) = 2.88 X 10-15 J

2014 H2P3 Suggested Solutions 2 At x = 25.0 cm, Ex = 3.0 x 105 V m-1. The acceleration of the nucleus at this point is given by a = F/m = (qE/m)= (3 x 1.6 x 10-19 x 3.0 x 105)/(7 x 1.67 x 10-27) = 1.23 x 1013 m s-2 3 Acceleration  electric field. The acceleration will decrease sharply at first and subsequently more gently from x = 4.0 cm to a minimum value around x = 18.0 cm. Thereafter, it will increase gently at first and more sharply subsequently up to x = 28.0 cm. The acceleration is in the same direction throughout. 7 (a) (i) The resistance R of a wire is defined as the ration of the potential difference V across it over the current I passing through it, i.e. R = V/I. (ii) The resistivity of a wire is the characteristic property of the material of the wire given by L RA= where R, A and L are the resistance, cross-sectional area and length of the wire respectively. (b) (i) The total circuit resistance, Rtot = (0.10 + Rcoil )= 6.0/1.2. Hence, the resistance of the coil, Rcoil = 5.0 – 0.1 = 4.9  (ii) The resistance of the coil is given by 9.4== A NlRcoil  , where N is the number of turns of the coil and l is the length one coil turn. Thus, the number of turns of the coil is given by )220.0 x )(10x7.1( )0003.0 x )(9.4( 8 2 118=== −   l ARN coil (c) T 1.2x10 3−− == 110.0 )2.1)(118()10 x 4(72.0 7B (d) (i) The K.E. of the electron , K is given by m peVK 2 2 1== . Thus, the momentum p of the electron is given by s N x10 8.54 24−−− === )250)(10x6.1)(10x11.9(22 1931meVp (ii) For the circular motion of the electron inside the B field, Fc =FB. Hence, evBr mv = 2 , or eBr p = . Thus, the radius r = p/(eB) = (8.54 x 10-24)/[(1.6 x 10-19)(1.2 x 10-3) = 0.044 m (e) If the electron enters the B field at an oblique angle as shown, its velocity component parallel to the field will cause the electron to move forward along the field lines. The component of the velocity perpendicular to the filed lines will cause it to move in a circular path. The forward and circular motions together will result in the electron moving in a helical path in the field. Since the electron is projected from outside the field, it will describe only half a helix before it exi