2017 RI H2 Physics P1 solutions

2017 Year 6 H2 Physics Prelim Paper 1 Mark Scheme © Raffles Institution 9749/01 1 A 2 D Option A: The mass of a typical car is about 1500 kg. The volum e of a typical car is about 3.0m×0.7m×1.5m (lower part) + 2.0m×0.7m×1.5m (upper part)  5 m3. So, the density of a car is about 300 kg m3. Option B: Typical A4 paper is labelled as 80 gsm (or grams per square metre). So an A4 paper of dimension 210 mm  297 mm has a mass of 80gm –20.210m0.297m = 5 g. Or, a pack of 500 sheets of A4 paper weighs about 2 kg. Hence, each sheet is about 4 g. Option C: A safe lifting speed is about 1 m s 1. So, the useful power is about mgv  103×10×1 = 104 W. Option D: Volume of a class room is about 10×10×3=300 m 3. Room temperature is about 27 degrees = 300 K. One atmospheric pressure is about 10 5 Pascal. Hence, n = pV/RT = (105×300)/(8.31×288)  105 moles. 3 A Only in region A does the speed change, i.e. there is acceleration. In regions B, C and D, the velocities are constant, i.e. there is no acceleration. 4 D 13.5sin20 10sin5.0 2.0686 m syu   Taking upward displacement as positive, 2 2 1 2 1300 2.0686 (9.81)2 8.03 s , 7.61 s (N.A.) yy ysu t a t tt tt      5 B 11Area under - graph (1 3)(2 ) (2)( ) 322 30 (2.0)(3 ) 5.0 at v x x x pm v x x          6 D By applying conservation of momentum and conservation of kineti c energy to an elastic collision, one would arrive at the conclusion that the relative speed of approach is equal to the relative speed of separation. Refer to notes for derivation. 7 A With A intact, the system would be in equilibrium without F. With B removed, and the system requires F to be in equilibrium. Hence, the moment due to F (and the resulting friction) should be equal to that due to the weight of B. With F in place, translational equilibrium in the horizontal directio n requires a friction f equal in magnitude to F but acting in the opposite direction to F. F and f form a couple. Weight of section removed (B) = 48  36 = 12 N. Method 1: Taking centre of mass of cylinder A as pivot Moment due to the weight of B = 12 × r/3 = 4r Moment due to the couple of F and f = F × 2r Equating the two moments, F = 2 N

© Raffles Institution 9749/01 2 Method 2: Taking point of contact between cylinder and ground as pivot Friction f does not contribute a moment, since it passes through the point of contact. Moment due to the weight of B = 4r Moment due to F = F × 2r It yields the same answer as the previous method. 8 C The scale will measure the following forces: i. weight of beaker of water (Z) ii. weight of object at bottom of beaker (X) iii. force exerted by suspended object on water (Y) Hence, the total force = X+Y+Z 9 A 22 12 122 1 12 2 11 2 2 2 22 1 1 , , 22m 2m mm m4 m ppKK p p Kp Kp   10 C At the same distance d from the sphere, the electrostatic force F experienced by both particles is the same because they carry the same charge. Since time of action t is equal, the change in momentum p = Ft must also be equal for both particles. From Ek = p2/2m, the lighter positron gained much more kinetic energy than the proton. Since the work done by force F on the particles is converted to a gain in kinetic energy of the particles, the work done on the positron is more than that on the proton. Alternatively: Since the positron is much lighter than the proton, the positro n has a larger acceleration a than the proton. In the same duration t, the positron travels a further distance s and hence the work done on the positron is more than that on the proton. 11 C Since the car is travelling at a constant speed, there is no ta ngential acceleration. Therefore, one of the components of the frictional force of the road on the car must point to the right. For the car to execute a uniform circular motion, another compo nent of the frictional force of the road on the car must point towards the centre of the circle. Hence, the net frictional force of the road on the car acts along C. 12 N 3 r F 2r A B f

© Raffles Institution 9749/01 3 12 B         2 Q Q 22 22 Q Q P 22 22 Q PQ 2 22 2P Q 4 22 2 2 mvFor mass Q: Tsin , and Tcos m g r mvTm g + r For mass P: T m g mvmg mg + r m vggmr 2.3 2g g 3 9.81r r 0.311 m                 13 A Above the equator, the gravitational force on the satellite all ows it to move in the same direction of rotation as the Earth. At any other latitude, the satellite would either require a con stant input of energy to move in the same direction as the Earth or the satellite would be mo ving in an orbit which changes latitudes as it circles around the Earth. 14 C Since p and V are constants, from pV = nRT, nT is also constant. Since n is proportional to mass m,  m1T1 = m2T2.  20  (20 + 273.15) = 15  ( + 273.15)   = 117.7 = 120 C 15 B At 30.0 C:  213 30 273.1522 Amc k   (1) At 300 C:  213 300 273.1522 BBmc k   (2) 2 2 (2) 573.15:, w h e r e 2(1) 303.15 0.972 BB B AA B mc m mc m cc   16 D Energy supplied to melt the solid from 2 min to 4 min: Q = Pt = mL  (1000)(2) = (1)(L)  L = 2000 J kg–1. To compare the specific heat capacities of the solid and liquid states:   11 11 From , 441000 1 and 1000 124 500 J kg K and 1000 J kg K solid liquid solid liquid QT mctt cc cc        Hence, option C is wrong.

© Raffles Institution 9749/01 4 17 B The rate of heat transfer between 2 bodies is proportional to the tempe rature difference between them. Since the temperatures of the boiling water and t he environment remain unchanged for both experiments, the rate of heat loss from the boiling water to the environment is the same for both experiments (regardless of the power supplied by the heater or the duration of the experiment) and can therefore be eliminated in calculations. Option A: To reduce random errors in the measurements of mass p er unit time, the experiments should be repeated using the same power and finding the average of the readings. Option C: To determine the average value of the latent heat of vaporisation from two sets of measurements without taking into account the heat loss to th e surrounding will not produce an accurate result. Option D: To check for reproducibility of the measurements, the experiments should be repeated usin g the same power and same duration. 18 C max max 2 0 2 1 0.6001.800 ( ) , where period 0.240 s22 . 5 121.800 (0.050 )2 0.240 2.1 kg PE KE mx T m m        19 B When the first wavefront has moved 6 squares, the dipper has on ly moved 3 squares. Hence, horizontal speed of dipper is ½(20) = 10 cm s1. Wavelength = radius of 1st wave  radius of 2nd wave = ½ (12  8) = 2 cm. f = v /  = 20 / 2 = 10 Hz. 20 C I = I0 cos2 30 – (1), where I0 is the intensity of light incident on Q I' = I0 cos2 120 – (2) (2)/(1): I' = I (cos2 120 / cos2 30) = 0.33 I 21 B Pipe open at both ends: L =  / 2  f = v / 2L Pipe open one end: L = ’ / 4  f ’ = v / 4L = f / 2 22 D Add the electric field strength vector due to each charge at the respective points. 23 C Refer to the definition of an ohmic conductor. 24 B P.d. V across both cells is zero. 1 3 3.0 0.50 mA6.0 10 I 2 3 3.0 1.0 mA3.0 10 I

© Raffles Institution 9749/01 5 25 D Four wires attract P towards the left, while the furthest wir e repels P towards the right. Hence, the net force is towards the left. 26 A As the magnet is pushed towards the coil, an induced current is set up in the coil which produces a magnetic field around it. According to Faraday’s and Lenz’s laws, this increasing magnetic field will produce an induced current in th e ring which will oppose its increase, and hence the ring will repel towa