JJC_H1_PHY_P1_Ans

2016 H1Phy Prelim Exam P1 Soln Page 1 of 4 JURONG JUNIOR COLLEGE PHYSICS DEPARTMENT 2016 Preliminary Examination 8866 H1 Physics Paper 1 Solutions Qn Ans Suggested solution 1 C The area of an A4-sized paper is (0.210 m)(0.297 m) = 0.0624 m2. Since A4-sized paper is typically 70 to 100 gsm (gram-per-square-metre), the mass of a single sheet of A4 paper is between 4 g and 7 g, the order-of-magnitude is 1 g. 2 B 2kr v   2 -4 -2(93.7)(0.83 /1000) 9.93 x 10 N s m0.065  -5 -2 -4 -2 2 (0.1) 2(0.01) (0.002) (93.7) (0.83) (0.065) 5.55 x 10 N s m 0.6 x 10 N s m kr v krv          

-4 -2(9.9 0.6) x 10 N s m  3 B Area X represents the displacement as the ball travels upwards and area Y represents the displacement as the ball travels dow nwards. From one impact to the next, the displacement of the ball must be zero so areas X and Y must be equal but with opposite sign. (A is a wrong statement. C and D are correct statements with only respect to this graph in which air resistance is negligible and do not explain why the two areas are the same. B is correct regardless of air resistance being negligible or and explains why the two areas are the same.) 4 D At the instant when the stone passes the edge of the cliff on its way down, its velocity will be 10 m s -1 downwards. Taking downwards positive, consider only the motion of the stone in the last 1.2 s, as shown on the right. vua t v = 10 + (9.81)(1.2) = 21.8 m s-1 22 2vu a s 475 = 100 + 2(9.81)s s = 19 m (2 s.f.) 5 B Consider the resultant force on the whole system. (54 – 6) = (6 + 2)a a= 6.0 m s-2 resultant force on the 6 kg mass = ma = (6)(6) = 36 N 6 A For all collisions, total momentum (p = mv) is conserved  A, C or D. For an elastic collision, total kinetic energy ( 21 2kEm v ) is conserved  A. 7 C When the ball is in contact with the surface, resultant vertical force = 0 When it leaves the surface, resultant vertical force = W

2016 H1Phy Prelim Exam P1 Soln Page 2 of 4 JURONG JUNIOR COLLEGE PHYSICS DEPARTMENT 2016 Preliminary Examination 8866 H1 Physics Paper 1 Solutions Qn Ans Suggested solution 8 C By definition of a couple, the couple is the product of one of the parallel and equal forces and the perpendicular distance between the two forces. 9 B Taking moments about X, (weight of load)(0.8) + (180)(1.5) = (220)(3) Weight of load = 490 N 10 C By conservation of energy, Increase in KE = WD by constant force = Fs 11 A For the first fall: 1. The kinetic energy must be zero at the beginning and end of the fall  A or C or D 2. The kinetic energy cannot be zero at the middle of the fall  A or C 3. The total energy must be equal to 120 kJ at all positions  A or B or D 4. There is no extension of the bungee cord in the initial part of the fall and therefore there is only very little extension, hence very little EPE, at the middle of the fall  A. The EPE should not quadruple from the middle