2013_JJC_H2_Physics_P1 solutions

Page 1 of 7 JURONG JUNIOR COLLEGE PHYSICS DEPARMENT JC2 Prelim Exam 2013 H2 Physics Paper 1 solutions Qn Ans Suggested solution (Font Arial, 11 pt) 1 C         2 12 sin kg m s kg A sA m FB I L FFB I L B IL         2 C The changes in horizontal and vertical components of the ball’s velocity means that the force acting on the ball by the horizontal surface has components in the x and +y directions. 3 D Taking downwards and to the right as positive, ux = 5.0 cos 30 and uy = 5.0 sin 30 Applying v2 = u2 +2as to the vertical component of motion just before the ball hit the floor, v2 = (5.0 sin 30)2 + 2(9.81)(2.5) = 55.3 Speed just before ball hit floor = = 8.61 m s-1 4 A 1F7 5 2 5Fm a a 2 m s m (10 15)      For the 10kg mass, Resultant force=10(2)=20 N 5 D if 12 12 2 1 11 1 pp m(u) m( v ) M(v ).................(1) u v v ......................(2) v u v Subst.(2) (1) m(u) m( v ) M(u v ). (Mv Byconservationof momentum, Relativespeedof approach = Relativespeedof separation into         m) u(m M) 6 C By definition.

Page 2 of 7 JURONG JUNIOR COLLEGE PHYSICS DEPARMENT JC2 Prelim Exam 2013 H2 Physics Paper 1 solutions Qn Ans Suggested solution (Font Arial, 11 pt) 7 D The 3 forces acting on the rod must must all act through the same point. Since there is an upward component provided by F, there should be a upward component provided by the unknown force. Thus the force on the rod by the wall should be pointing in the option A direction. By Newton’s 3rd Law, the force acting on the wall by the rod should be equal in magnitude and opposite in direction, therefore the ans is option D. 8 C Since the car is moving at constant speed, forward force = frictional force WD by friction = Fd = 3800 x 300 = 1140 kJ 9 C KE = ½ mv2 = ½ m(u2 + 2as) = mas since u = 0 This means that KE is directly proportional to displacement (linearly related) Consider when x=0 (At ground level), KE will be maximum Consider when x=H (At top of tower), KE will be zero Hence graph of KE vs x is best represented by C 10 D Consider the forces acting on the body of 1 kg at top T top + mg = mv2/r Consider the forces acting on the body of 1 kg at the bottom T bottom – mg = mv2 /r Note: v is the same when body is at top and bottom. Hence T top + mg = Tbottom – mg Difference in tension = 2mg = 2(1)(10) = 20 N 11 B T sin = mv2/r -----------(1) T cos = mg ------------(2) (1)/(2) tan  = v2/rg (1)2+(2)2 T2 = m2g2+ m2v4 / r2

Page 3 of 7 JURONG JUNIOR COLLEGE PHYSICS DEPARMENT JC2 Prelim Exam 2013 H2 Physics Paper 1 solutions Qn Ans Suggested solution (Font Arial, 11 pt) 12 C A is wrong as the geostationary satellite must be at a fixed distance above the Earth equator. B is wrong as linear speed is proportional to the distance away from the centre of the earth (v=r) hence the speed of the satellite can never b