2024 EJC J2 H1 PRELIM P1-2 Answer

©EJC 2024 8867/J2H1PRELIM/2024 EUNOIA JUNIOR COLLEGE JC2 Preliminary Examination 2024 8867 PHYSICS MARK SCHEME Paper 1 – Multiple Choice Questions Question Key Question Key Question Key 1 D 11 A 21 D 2 B 12 B 22 A 3 B 13 D 23 D 4 C 14 C 24 A 5 B 15 B 25 D 6 B 16 D 26 C 7 D 17 D 27 B 8 A 18 C 28 C 9 B 19 D 29 C 10 A 20 B 30 C

2 ©EJC 2024 8867/J2H1PRELIM/2024 1 Answer: D 2 units of energy = J =(kg m s )(m)  -2 2 -3 kg m s munits of intensity = m s kg s 2 Answer: B 2 0.0534 2 1 0.1 1268 5.0 25 0.0947 0.0534 0.0947 0.005057 0.005 (5.3 0.5) 10 m LR A RA L R D L R D L                               3 Answer: B Option A gives a radius of 6 cm Option B gives a radius of 13 cm Option C gives a radius of 29 cm Option D gives a radius of 62 cm B is the most sensible answer. Intensity = Power/Area = Rate of Work done/Area

3 ©EJC 2024 8867/J2H1PRELIM/2024 4 Answer: C 2 2 1 gts  on earth, 2'62 1 tgs      on moon  22 6' tt  and tt 6' 5 Answer: B As the projectile rises, the vertical component of the velocity of the projectile decreases to zero at the top of its trajectory, then increases to its initial value in the downward direction as the projectile comes back down to its original launch level. The projectile’s horizontal velocity component, however, remains constant throughout. Speed v is the magnitude of velocity, which is a vector sum of the horizontal and vertical component. The graph for v should be a curve that decreases to a non-zero value (when projectile is at its highest), then increases back to its initial value u. 6 Answer: B Change in momentum,     1 1 2 5 2 2 12 9.0kgms p Area under F t graph          1 ( 2.0) 9.0 2.0( 2.0) 9.02.0 2 6.5 p m v v v ms         

4 ©EJC 2024 8867/J2H1PRELIM/2024 7 Answer: D 8 8 To find the force on X by Y, consider X: ( )8 7 8 To find the force on Y by Z, consider Z: 4 4 ( )8 1 2 By Newton's 3rd Law: 7 8Ratio: 1 2 7 4 XY XY ZY ZY ZY YZ XY YY F ma Fa m F F ma FF F m m F F ma FF m m F F F FF F F              8 Answer: A Total momentum of the system is conserved because no external force acts on the gliders. Option B: Before collision, total momentum of the system is zero. Option C: Kinetic energy is not conserved when both magnets are rest. Option D: Both magnets have the same mass, the magnetic force on each magnet is the same, no friction on track, total kinetic energy of the system is conserved before and after collision.

5 ©EJC 2024 8867/J2H1PRELIM/2024 9 Answer: B Using relative speed of approach = relative speed of separation Eliminate Option A: 5-(-2) = 5-2 (false) Eliminate Option C: 5-(-2) = 4-4 (false) Using logical deduction, eliminate D as after collision, the speed of the large mass cannot remain the same. Option B: 5-(-2) = 10 – 3 (True), and after colliding large mass’s speed is smaller as it has transferred some of its momentum to the smaller mass, resulting in increased in speed of the smaller mass. Mathematical approach using conservation of linear momentum: (Let Large mass = M, Small mass = m) Option D: 5M- (2m) = 5M +12m 0 = 14 m (impossible) Option B: 5M- (2m) = 3M +10m 2M = 12 m M= 12 m (possible) 10 Answer: A P and Q are stretched by a force = W/2. Each of their extension = (W/2)/k = W/(2k) R is stretched by a force = W and its extension = W/(3k) Total extension = W/(2k) + W/(3k) = 5W/6k 11 Answer: A 2 2 Using point when = 14 N and =14 mm 14 140010( )1000 at (20,28) the elastic limit of spring i s reached 1 1 20work done (1400)( ) 0.28 J2 2 1000 F x F kx k kx      

6 ©EJC 2024 8867/J2H1PRELIM/2024 12 Answer: B The two tension forces T are to be added, hence the tail of one T should be at the head of the other T. The tail of the resultant R should be at the free tail of one T and the head of R should be at the free head of the other T. 13 Answer: D net 2 2 3 2 2 Constant speed, acceleration = 0 F = 0 Hence, - =0 1 2 1 ( )2 1 2 mg kv mg kv mgv k KE mv mgm k m g k      14 Answer: C When the speed is 20 m s1, the driving force F is solved using 3 3 power output 23 10 20 1.15 10 N Fv F F       The frictional force f at this speed is equal to the driving force: 31.15 10 Nf   because the net force on the car is zero. When the speed increases to 40 m s 1 (speed is doubled), because the frictional force is proportional to the square of the speed, the frictional force increases by a factor of 4. By N2L, since the velocity is again constant (net force is 0), the driving force is equal to this new frictional force 3 3 new 4 4 1.15 10 4.6 10 NF f       The new power output is then 3new power output 4.6 10 40 184 kW   

7 ©EJC 2024 8867/J2H1PRELIM/2024 15 Answer: B 0.50 120sin 0.2165 mradius of 2rotatio 2n r   15 2 0.5236 rad sa 6ngular speed 0    2 2 2 0.52 accelera 36 0.21 tion 65 0.05 9 m s a r       16 Answer: D D Tension in string is always perpendicular to velocity, hence no work is done on the tension (external force). By conservation of energy, Gain in KE = Loss in GPE   2 2 1 2 2 mv mgr mv mgr Since resultant of tension and weight provides centripetal force at the bottom,      2 2 3 mvT mg r mvT mg mg r See the diagram. 17 Answer: D 2gravitational acceleration GMg d , where d is the distance to the centre of the Earth, and M is the mass of the Earth. Hence, 2 2 ( ) ( ) constant GMg h R g h R GM      mg T Fc

8 ©EJC 2024 8867/J2H1PRELIM/2024 18 Answer: C             4 4 4 4 16 X Y X X Y Y X Y Y X X Y Y Y X Y X X Y X X Y V V A L A L A L A L A A LR A R L A L A R A L L A 19 Answer: D Potential difference across parallel branches in a circuit are equal. 20 Answer: B 12 2 24 W   P VI 21 Answer: D A and B are wrong because the diode is connected in reverse-bias. When temperature of the thermistor is low, resistance is high and hence p.d. across the thermistor is high. The heater should be connected across the thermistor. 22 Answer: A Positive charge attracted to the lower electric potential by the constant electric force. The path is parabolic. Option C and D are wrong as the path of the positive charge will not be circular. It will travel in a parabolic path into the plane of the paper for both options.

9 ©EJC 2024 8867/J2H1PRELIM/2024 23 Answer: D -22.4 10 0.5 0.10 0.48 A F BIL I     Using FLH Rule, the direction of current is Y to X 24 Answer: A 25 Answer: D When the current is doubled, the magnetic force is doubled. Newton’s 3rd Law will dictate that both forces on one another must be of same magnitude. Hence eliminate option A and C. When the current is doubled, the magnetic field at X due to Y will be doubled. The magnetic field at Y due to X remains the same. Hence answer is D. FB FE v B FB FE -ve +ve

10 ©EJC 2024 8867/J2H1PRELIM/2024 26 Answer: C By FLR, on entry, the magnetic force on the proton is upwards. Hence, to be undeflected, the electric force must be downwards. The direction of E-field must be downwards. Eliminate B and D. Electric force, F=qE Magnetic force, F =Bqv For net force to be zero, qE = Bqv E=Bv = (1.5)(2.0107) = 3.0107 NC-1 27 Answer: B Isotopes have the same number of protons but different number of neutrons. By emitting one α- and two β-particles, the number of protons of the element will be conserved. 28 Answer: C Binding energy per nucleon = (Mass Defect)