DHS 2023 Prelim Physics (Paper 2 Solutions)

DUNMAN HIGH SCHOOL 2023 H2 PHYSICS (YEAR 6) 1 Answers to Prelim Exam H2 Physics Paper 2 Qn Answer Marks 1(a) Speed is a physical quantity and can only be defined in terms of other physical quantities. Distance is a physical quantity, but second is a unit for time which cannot be used to define speed. B1 B1 1(b) Vertically, from v = u + gt and taking upward as positive, −u sin = u sin − gT M1 Time of flight, 2 sin2 uTt g == M1 Horizontally, from s = ut 2 ( cos )( ) 2 sin ( cos ) 2 sin cos R u T uu g u g    = =  = A1 A0 Marker’s comments: It is a “show” type of question and thus important that all steps, including substitution must be explained and shown. 1(c) Maximum R occurs when sin2 1 = . Hence 45o = A1 1(d) 2 0 uR g= ( ) ( )( ) 145.36 1000 m45.36 km 12.6 m s1 h 60 60 s −== ( ) 2 0 2 2 12.6 16.3 9.7399 m s ug R − = = = 0 0 2 Rgu g u R  =+ C1 A1 T

DUNMAN HIGH SCHOOL 2023 H2 PHYSICS (YEAR 6) 2 2 3429.7399 100 100 0.97 1 m s (to 1 s.f.) g g −  =+  = = Therefore, ( )10 1g = m s−2 A1 A1 Marker’s comments: Uncertainty should only be left to 1 sig fig. The final answer and its uncertainty should have the same d.p. It is important to convert the speed to m s−1 to obtain g in units of m s−2. Also, there were mistakes in understanding the meaning of percentage uncertainty. 2(a) The upthrust is the resultant upward force due to the larger pressure exerted by the fluid at the bottom compared to the top surface of the body. B1 2(b) ( ) displaced fluidUpthrust 8001050 9.818000 1030 N (shown) fluid cannon W Vg = = =  = M1 A0 Marker’s comments: It is a “show” type of question and thus important that all steps, including substitution must be explained and shown. 2(c) The rate of change of momentum of a body is directly proportional to the resultant force acting on the body and occurs in the direction of the resultant force. B1 2(d) During its upward motion, the cannon experiences viscous drag force which acts in the opposite direction of its motion and increases with the speed of the cannon. Net force on canon = upthrust – (weight + drag force) decreases (since weight and upthrust are constant) Since the net upward force decreases over time, the momentum increases non-linearly (since the net force is directly proportional to the rate of change of momentum) B1 B1 Marker’s comments: Very often, drag force is not mentioned. There were misconception that the upthrust decreases causing net force to decrease or the velocity decreases while the momentum increases.

DUNMAN HIGH SCHOOL 2023 H2 PHYSICS (YEAR 6) 3 2(e) gradient of tangent line 400 0 40 100 N netF= −= − = (M1 for tangent line and computing gradient correctly) Net force on cannon = upthrust of cannon and bag − weight of cannon Fnet = U – W 100 = U − (800)(9.81) U = 7948 N Hence upthrust on bag = 7948 – 1030 = 6918 N For the bag, U = Vg 6918 = V(1050)(9.81) V = 0.672 m M1 C1 A1 Marker’s comments: Poorly attempted. Gradient not drawn to obtain the initial net force. 3(a) friction on the mass provides the centripetal force for the circular motion of the mass B1 For mud to remain on the plate, 0.72W  md2 0.72mg  0.35m2   4.49 rad s−1 M1 C1 Momentum / kg m s -1 time / s 200 400 600 800 0 5.00 15.0 25.0 10.0 20.0 30.0 35.0

DUNMAN HIGH SCHOOL 2023 H2 PHYSICS (YEAR 6) 4 number of revolutions  4.49 2 × 60 = 42 min−1 A1 Marker’s comments: The question says to explain your working. As such, a statement that friction provides centripetal force to explain before equating the two forces. There were mistakes made in units of  or to convert rad s−1 to number (an integer) of revolutions per minute. 3(b) centripetal force increases as r increases/ centripetal force is larger at the edge M1 so mud slips off at the edge first. A1 Marker’s comments: Mud will slip off the plate when friction is not enough to provide the required centripetal force. Students also did not realise that  is the same for mud near the centre or edge. 4(a) acceleration is directly proportional to its displacement from a fixed point B1 acceleration is always in opposite direction to its displacement/always towards the fixed point) B1 Marker’s comments: Majority of candidates got it wrong. 4(b) displacement line correct B1 acceleration line opposite to displacement line B1 velocity line has B D & F as zero with answers to C and E consistent with their displacement line B1 Marker’s comments: Candidates demonstrated understanding of vector concepts. 4(c) (i) phase difference between displacement and velocity is π/2 OR 3π/2 radians OR 90° B1 4(c) (ii) displacement and acceleration are exactly out of phase OR out of phase by π radians OR 180° B1 4(d) (i)1 amplitude = 8.0 cm A1

DUNMAN HIGH SCHOOL 2023 H2 PHYSICS (YEAR 6) 5 4(d) (i)2 T = 2.0 s so f = 0.50 Hz A1 4(d) (ii)1 F = mAω2 = 0.020 × 0.080 × (2π/2.0)2 = 0.0158 N (or 0.016 N) A1 4(d) (ii)2 negative cosine graph of any amplitude for at least 3.5 s B1 Marker’s comments: Poorly and wrongly drawn graph. Some negative sine graph were also drawn. 5(a) electric field strength is the force acting per unit positive charge B1 5(b) equally spaced, parallel straight lines with arrows down over most of space between plates curved lines shown at edges (showing weaker, non-uniform field) B1 B1 Marker’s comments: Common mistakes of unequal spacing between field lines and wrong direction of electric field were observed. 5(c) (i) (W =) VQ B1 5(c) (ii) (W =) Fd B1 5(d) VQ = Fd so F / Q = V / d i.e., force per unit charge = potential gradient B1 (magnitude of) electric field strength between the plates is equal to the potential gradient B1 Marker’s comments: Candidates could not use the answers in the previous part to deduce the relationship. Some used electric field is force per unit positive charge instead of getting the expression for electric field to deduce the relationship to potential gradient. 6(a) As V increases, I increases initially. Further increase of V will cause a less than proportionate increase in I Since resistance is the ratio of potential difference to current, the resistance of the filament lamp increases. B1 B1 Marker’s comments: Common mistakes stated by candidates that the resistance is the reciprocal of the gradient of the I-V graph.

DUNMAN HIGH SCHOOL 2023 H2 PHYSICS (YEAR 6) 6 6(b) The minimum resistance can be determined by drawing the tangent of the graph at the origin. − − − −== − ==    3 3 3 (4.00 0.00)gradient of tangent 2.00 2.00 0.00 1Hence, resistance 500 2. 0 10 1000 1 M1 C1 A1 6(c) From Fig. 6.1, when V