2024 RI Prelim H2 Phy Paper 1 Answers

Raffles Institution Year 5-6 Physics Department 1 2024 H2 Physics Preliminary Examination Solution Paper 1 Qn Ans Solution 1 A units of 2 12 2 kg m s kg m smP − −− = = units of 3kg m −= units of ( ) −− − −  ==  12 22 3 kg m s m skg m nn nP units of −= 1m sv For equation to be homogeneous, units of    n P = units of v 2 2 1m s m snn −− = comparing indices of m: 12 1 2nn=  = 2 D 21 16.24 12.78 3.46 mmd d d= − = − = 21 0.03 0.02 0.05 mmd d d = + = + = 0.05100% 100% 1.4451 1.4%3.46 d d   =  = = 3 C From a-t graph: From = 0t to = 1tt , acceleration is constant which implies that the object’s velocity is increasing at a constant rate. From = 1tt to = 2tt , acceleration is decreasing which implies that the object’s velocity is increasing at a decreasing rate. From = 2tt to = 3tt , acceleration is zero which implies that the object’s velocity is constant. Since = dva dt , the gradient of the v-t graph, which gives acceleration, in Option C follows the description above. 4 D Option A: Possible, if lift is decelerating / decreasing in speed on its way up. Option B: Possible, if lift is moving upwards at a constant speed. Option C: Possible, if lift is accelerating / increasing in speed on its way up. Option D: Hence, all the options above are possible, depending on the lift’s acceleration.

Raffles Institution Year 5-6 Physics Department 2 5 C Applying Newton’s second law on the system of both crates, ( )( ) ( ) ( )( ) ( ) − = − + = +  −+== + , 2 100 2.0 3.0 9.81 2.0 3.0 100 2.0 3.0 9.81 10.19 m s2.0 3.0 net both bothF m a a a Applying Newton’s second law on the 2.0 kg crate, ( ) ( ) = − − = = ,2 2 100 2.0 9.81 2.0 10.19 60 N net kg kgF m a T T OR ( ) ( ) = −= = ,3 3 3.0 9.81 3.0 10.19 60 N net kg kgF m a T T 6 B Motorcycle travels in the same direction during the whole duration. Impulse or the change in momentum is the area under the force-time graph. ( )( ) ( )( ) ( )( )11 22 1 400 4.5 1.0 400 2.0 800 4.5 1.5 4.5 1.5 3.0 m s p Fdt v v v  − = − = − − =− = − =  7 D Since cube is floating, there is vertical equilibrium. ( ) ( ) ( )( ) 12 1 1 2 2 1 2 1 1 2 1 1 2 1 2 2 1 1 21 1 2 32 3 2 2 1 cube c U U W V g V g V V g V V V V V V V V VV V V       += + = + + = + − = − = = 8 A Work done by the force to extend the spring is given by the area under force-extension graph i.e. the area bo unded by the graph and the vertical axis of the graph given. This work done goes to increase the potential energy of the spring. The potential energy represented by area P is released upon the removal of the force. The potential energy represented by area Q is retained in the spring that is permanently stretched i.e. the energy used to separate the particles of the spring further apart.

Raffles Institution Year 5-6 Physics Department 3 9 D At constant speed, engine force = resistive force rate at which energy is delivered = rate at which energy is dissipated 3 3 72 1012 10 60 60 600 N P Fv F F =  =   = ( )( )( ) ( )( ) from fuel tocar 630 30 40 10 12 10 1 60 60 3 6 kg EE .m m. =  =    = 10 C The minute hand takes 1 hour to go round the clock once. 12 rad s60 60 m  −=  The hour hand takes 12 hours to go round the clock once. 12 rad s12 60 60 h  −=  1.5 12 60 60 1860 60 m m m h h h h h v r r v r r    = = =    11 C Option C (correct): dg d g drdr  =−  =−  Hence, the area under the g-r graph gives the change in the gravitational potential  . Options A and B (incorrect): The total gravitational potential between the two planets is always negative. Gravitational potential is zero only at infinity. Option D (incorrect): The gradient of the graph does not give any meaningful quantity. 12 A The gravitational force on each star provides the centripetal force for the star to orbit about the common centre of mass of the system. For two stars, mass M and m, at a distance d apart, 22 2 22 GMm mr MRd mr MR mr MR   == = = R and r are the orbital radii of the stars of masses M and m respectively. The gravitational force on each star is always directed towards the common centre of mass of the system as the stars orbit. Hence the stars should be on opposite sides of their orbital path, lying along the same straight line through the common centre between them. To maintain this, the stars must also have the same angular velocity  . Hence star X having a larger mass should have a smaller orbital radius.

Raffles Institution Year 5-6 Physics Department 4 13 D 2 22 1 3 3From 22 rms rms kT Tm c kT m m cc=  =   22 ,, ,, 3 9 4.522 rms Y rms XX x x Y y x rms X rms X ccm T T m T T cc     = = = =            14 B Both the inlet and outlet temperatures and the room temperature must be kept the same so that the rate of heat loss to the surrounding is kept constant for both experiments and can be eliminated. 15 D 0 0 1 0 cos 24.0 cos 3.0 9.0 8.0 m s v v t v v   − = =  =− ( ) 21 0.020 8.0 0.64 J2 KE =   − = 16 C Oil is more viscous than water hence has a greater damping effect on the oscillating mass compared to water. With greater damping, t he frequency response curve when the mass is in oil will have smaller amplitudes at all frequencies and the frequency at which resonance occur s will be smaller. 17 B 18 B For astronaut to see the light sources, source received pupil min 24 PP A P r=   ( ) ( ) 2 source pupil 13 min max 10 0.0050 2 8838.8 m4 4 2.0 10 r 8800 m PAr P    − = =  = 19 A Diffraction is pronounced when the wavelength of the wave is comparable to the width of the obstacle. Sound waves with a longer wavelength than the diameter of the pillar can bend around the pillar. Light waves with a much shorter wavelength than the diameter of the pillar cannot bend around the pillar. wave at earlier time wave at later time y P Q x

Raffles Institution Year 5-6 Physics Department 5 20 A For light sources to be resolved, angle of separation of the 2 sources  minimum angle of separation by Rayleigh criterion min S Dd  where S is the distance between the 2 sources DS d  The best combination is the one that can resolve the smallest distance S between the two sources i.e. shorter  and D and larger d. 21 D Charge of sphere is Q. 0 0 44 QV Q RV R =  = 0 2 2 2 00 4 44 RVqQq qVRF r r r   = = = 22 B ( ) 2 00 magnitude of at P due to +64 64 4 44 4.0 EQ QQ  == ( ) 2 00 magnitude of at P due to 125 125 5 44 5.0 EQ QQ  − == These two E vectors form a right-angle triangle, with the resultant E pointing upwards with magnitude 22 0 0 0 5 4 34 4 4 Q Q Q        −=        . 23 A ( ) 2 22 4 EdE E EA d E LR L L L A      = = = =I Hence 2d I since L and E across the wires in parallel are constants. 2 2 2 2 1 214 1 16 8 2 1 9 YY X X Y YX Y YY X total dd d d     =  = = =  = I I I I resultant E

Raffles Institution Year 5-6 Physics Department 6 24 D Since ammeter reading is zero, there is also no current in the middle wire joining the circuits on the left and right. There is no potential difference between the two ends of this wire and there is no current exchange between the two circuits. 50  and 100  resistors are in series. R and 200  resistors are in series. Potential difference across the 100  and 200  resistors is the same. 200 100 24 24 10024 24 12200 100 50 24 16200 400 R R VV VV R R R R R =− =−  = − ++ =+ = 25 C The current in X produces a magnetic field along the circumference of coil Y in the clockwise direction. This magnetic field produced is parallel to the current in each part o