EJC 2023 J1 H2 Promo P1 MS

©EJC 2023 9749/J1H2Promo/2023 EUNOIA JUNIOR COLLEGE JC1 Promotional Examination 2023 General Certificate of Education Advanced Level Higher 2 PHYSICS MARK SCHEME 9749 Sep/Oct 2023 Paper 1 Multiple Choice Question Key Question Key Question Key 1 C 6 A 11 B 2 C 7 C 12 C 3 A 8 B 13 A 4 D 9 C 14 B 5 C 10 D 15 D 16 B 21 B 26 B 17 C 22 D 27 B 18 B 23 A 28 D 19 B 24 B 29 C 20 A 25 D 30 C 1 Typical length = 1 m = 100 cm Typical breadth = 3 cm Typical height = 0.5 cm Hence, estimate volume = 100 × 3 × 0.5 = 150 cm3 2 Option A: Incorrect based on definition of accuracy and precision. Option B: Checking for zero error prevents systematic error. Option C: 5 5 20 20 5 20 5 5 20 20 5 5 vs 20 20 vs = ⇒∆ = ∆ = ⇒∆ = ∆ ∆∆ ∆∆∆∆ = = = = t Tt T t Tt T tt ttTT Tt t T t t 5 20 5 20 Note: 1. The uncertainty for a stopwatch timing is fixed, whether it is timing 5 or 20 osc. Hence 2. Since fractional uncertainty of is smaller w hen the number of osc is larger. ∆= ∆ = ∆ < T t t t. t t, Option D: 1 123 4 123 4 C e ompare 1 vs 4 measurements, take uncertainty of the instrument as Case 1 (1 measurement): Case 2 4 Hence taking more m (4 measure s me ea u n r ts): 4 ∆ ∆= ∆ ∆ +∆ +∆ +∆⇒ + ∆= = ∆ ++=ave ave d d ,d ,d ,d dddd d d d d d d dddd ments to find average has no impact on uncertainty. 3 Using v2 = u2 + 2as 702 = 402 + 2(a)(300) a = 5.5 m s-2

2 ©EJC 2023 9749/J1H2Promo/2023 4 Taking downward as positive: s = ut + ½ at2 5 = 0 + ½ (9.81) t2 t = 1.00964 Consider the horizontal: sx = vx × t = 0.8 × 1.0 = 0.8 m 5 The sandbag is travelling upwards with a velocity of 4.0 m s-1 when it is first released. Hence the displacement of the sandbag will increase in the upwards direction before it eventually starts falling (since acceleration of the sandbag is 9.81 m s -2 downwards). Since s is defined as the displacement of the sandbag from the point of release, s = 0 at the point of release. 6 Method 1: Each mass is a different system ( ) 11 22 12 12 12 12 (1) (2) (1) (2): ( ) ( ) ( ) mg T ma T mg ma m mg m ma m mga mm −= −= + −= + −= + Method 2: Both masses belong to the same system Mass of system= (m1 + m2) Net force on system = (m1g – m2g) Acceleration of the two objects, ( )1 2 12 12 12 = −= + −= + netF ma mg mg m m a mg mga mm 7 Since total momentum before the collision is 0, total momentum after the inelastic collision should also be 0. Total kinetic energy before the collision is 27 J. Since collision is inelastic, total kinetic energy after collision should be less than 27 J. Students who thought KE = 0 assumed incorrectly that it was a perfectly inelastic collision. 8 At equilibrium, let U be the amount of upthrust required to balance the weight W of the vehicle. ( ) 42 0 10 4 12 5 1000 0 40 m ρ ×× = = = × = × W V . .h h. U mg g

3 ©EJC 2023 9749/J1H2Promo/2023 9 Method 1: To find force on chassis at front axle, take moment about the rear axle. Let N be the normal contact force by the front axle on the truck. By principle of moments, Initially, load w is not present: ( ) ( ) 2 2 = = Wx N x WN Finally, let N’ be the new contact force of the front axle on the truck when w loaded: ( ) ( ) ( )32 3 2 W x wx N 'x WwN' += += Hence contact force of front axle on truck increases by3 2 w . Method 2: With w loaded, there is an additional anti - clockwise moment w (3x) about the rear axle. This must be counter-balanced by the extra clockwise moment provided by the extra contact force ∆N of the front axle. ( ) ( )32 3 2 wx Nx wN =∆ ∆= 10 Using Hooke’s Law, tension is directly proportional to extension. When tension is T, extension is (x – L). When tension is T’, extension is (y – L). −= − −= − ' ' T (y L) T (x L) T(y L)T (x L) 11 2 engines of 80% efficiency provides power for the plane. Let P be the power for each engine, then ( )( ) ( ) ( )( ) 6 80% 2 0.8 2 200000 250 31.3 10 W 31.3 MW = = = ×= P Fv P P 12 For first 3.0 s, acceleration is constant at 9.81 m s-2 since drag is insignificant. Given v = at, 2 22 ½½= =KE mv ma t i.e. EK is a quadratic function in t. Once parachute opens, air resistance increases tremendously and net force is upwards. Hence velocity falls and E K falls. Eventually, parachutist will reach terminal velocity and kinetic energy reaches a constant value. w W N

4 ©EJC 2023 9749/J1H2Promo/2023 13 Method 1: Tension provides the centripetal force. When radius of circle is L+ e, extension = e 2 1 2 1 () (1) mvke Le ke L ev m = + += When radius of circle is 2(L+ e), extension = L+ 2e 2 2 2 2 ( 2) 2( ) 2 ( )( 2 ) (2) mvkL e Le kL e L ev m += + ++= 12 1 2 1 Equate (1) and (2): () 2 () ( 2 ) 2 ( )( 2 ) k( ) 2 ( 2) + ++ = ++= + += k e Le k Le L e mm k L e L emm eL e mL e e Method 2: Proportionality Method Tension (Elastic force provides the centripetal force.) 2 since , are constant = = ∝ vF ke m r m re k v 2 22 1 11 1 2 . 2 ( 2) = += m re m re mL em e 14 Resultant of tension and weight provides the centripetal force At top: ( ) 22 1() topmL f T mgπ = + At bottom: ( ) 22 – 2() botmL f T mgπ = (2) – (1) Ttop – Tbot = 2mg 15 2 21 22 1 2 2 1 10 5 2 where is the radius of earth = ⇒∝ ⇒ =   ∴= ⇒ = grGMgg grr r xx rr r 16 Option A: incorrect The period of an equatorial satellite can be a geostationary satellite (period 24 h) if the radius is correct Option B: correct Determine formula for kinetic energy 2 2 2 22 2 = = = K GMm mv RR GMm mv R GMmE R When R increases, EK decreases Option C: incorrect 2 23 22 22 44() ππω= = ⇒=GMm RmR mR T GMRT T independent of m (Can also use Kepl er’s law ( 23Tr ∝ ) here though it is not officially in syllabus and equation must be proved if used in P2 & P3) Option D: incorrect Total Energy 2= += −PK GMmEE R When M increase, total energy decreases (becomes a more negative number) 17 A uniform gravitational field has a constant gravitational field strength in a uniform direction. i.e. same magnitude and same direction (along the direction of the gravitational force ). The direction of the field lines is the direction of motion of a test mass released in it. The gravitational potential at each point may vary.

5 ©EJC 2023 9749/J1H2Promo/2023 18 Since kinetic energy is max at t = 0 s, the mass started at the equilibrium position, and the x-t graph is a sine graph, v-t graph is a cosine graph, and a-t graph is a negative sine graph. From given E-t graph: ( ) 2 max 2 1 2 0 0 - 20.800 s 7.85 Eiminate A & C 1 = 2 12.0 = 0.50 2 2.83 m s cos( ) 2.83cos(7.85 ) & sin( ) sin( ) sin(7.85 )22.2 ω πω ω ω ω = ⇒= = ⇒ = ∴= = =− =− =− o o o o o T T E mv v v vv t t at x vx t t 19 ( ) ( ) ( ) 12.0 5 rad s0.40 22 1 1 1.26 s 0.40 s 544 cos 0.40 cos 5 0.40 0.17 m distance travelled 0.40 0.17 0.57 m ω ππ ω ω −= = = = = = ⇒< < = = = − =+= o v r T TT xx t 20 No damping occurs in a vacuum, in contrast to a small amount of damping in air. Hence at natural frequency of the pendulum, maximum amplitude increases. 21 Phase difference o0.22 sin252 2 0.34 rad1.7ππλ ∆××= ×=s 22 Speed of wave, 1 3 11 (0.6) 300 m s2 10λλ − − = = = = × vf T 23 It is crucial to note that the θ in the diagram is not between the polarisation axis