Title 2021 GCE A level H2 Physics 9749 Ans.pdf ✅

2 1 Ans C (challenging – prefix of c and d) Estimated mass of a smartphone = 200 to 300 g Estimated weight of a smartphone = (200/1000)(9.81) = 2 N = 200 cN = 20 dN Thus best answer is 150 cN Comments: Many students are unfamiliar with prefix of c and d. 2 Ans A Using F = BIL B = F/IL Units of B = (Units of F)/(Units of I)(Units of L) = kgms-2 /Am = kg s-2 A -1 3 Ans B (challenging - resultant force of W and air resistance for projectile motion) Resultant force = Vector sum of all forces = Vector sum of weight and air resistance Air resistance is in opposite direction to the velocity at that position. Comments: Students need to include air resistance for this question (will chose C if ignored air resistance) and distinguish between the resultant force vector and the velocity vector (will choose D if thinking about velocity vector) 4 Ans B For elastic collision, relative speed of approach = relative speed of seperation  : ux – uY = vY – vX v – 0 = 0.67v – v v = - 0.33 v (towards the left) Speed = 0.33 v 5 Ans A weight of brick = force Earth exerts on brick.  3 rd law pair is the force brick exerts on Earth. Support force from floor on brick = force floor exerts on brick.  3 rd law pair is the force brick exerts on floor. weight air resistance resultant force
3 6 Ans B (challenging – careless in reading question and manipulation of symbols) At equilibrium, T + U = W T + Weigh of fluid displaced = Weight of ball 0.75W + 1⁄2 Vg = W (reading on newton meter is the tension T) V = 2(0.25W)/g = 0.5 (100)/[(1.0)(9.81)] = 50 cm3 (units of  and W are consistent) Comments: Students need to work through the algebra carefully and note that the ball is only half submerged thus only displace half the volume of water. 7 Ans D EPE = Area enclosed by F and x-axis = F dx = Shaded area = 1⁄2 x1W1 – 1⁄2 xoWo = 1⁄2 (W1x1 – Woxo) 8 Ans A P = FDv , t = 1000/20 = 50 s WD = Pt = FDv t = 400 (20) (50) = 400 000 J Energy input = (400 000/16) x 100 = 250 0000 J = 2.5 MJ Amt of fuel = (1000/48) x 2.5 = 52 g 9 Ans A r = mv/Be, v = r When B increases  r decreases   must increase (v = constant) When v increases  r increases   can be constant (v increases) 10 Ans B WD in raising 2kg 10 m = 6 x 2 = 12 J WD in raising 2kg 2.5 m = (12/10) x 2.5 = 3 J 11 Ans D FG = FC GMm/r2 = mv2 /r v = (GM/r)1/2 Since g = GM/R2 at surface of Earth 9.81 (6400 x 103 ) = GM For geostationary satellite, r = (36 000 + 6400) x 103