MI 2024 PU3 H2 PHYSICS PRELIM P1 ANS

1 H2 Physics PU3 Preliminary Examination Paper 1 Answers S/N Answer Explanation 1 D [𝐸] = |𝑄|𝑛 |𝑘| kg m2 s-2 = 𝐴𝑛 s𝑛 𝐴2 s4 kg-1 m-2 n = 2 2 C a = b2c c = 2 a b 2c a b c a b   =+ 3 C At point A: object just reached the ground for the first time At point B: object just about the leave the ground after first bounce At point C: object reached the maximum height after first bounce At point D: object just reached the ground for the second time 4 C 21 2=+s ut at u = o since the stone falls from rest. 212 2= at 24 = ta 2=t a ------------ (1) where t is the time taken for stone to fall on Earth Acceleration of free fall of moon = 6 a , where a is the acceleration of free fall on Earth. Sub 6 a into equation (1) ( ) 22 66 6 = = =moontt aa 5 B Area under the force-time graph = change in momentum = impulse Total area = (5 x 0.5) + (15 x 0.5) = 10 p = m v = 10 v =v - 0 ➔ v = 10 / 4 = 2.5 m s–1 2.5 m s–1 + 2.5 m s–1 = 5.0 m s–1

2 6 B A and D obviously wrong. Try taking moments about C, clockwise moment = (3m)(1.5x) = 4.5mx anti-clockwise moment = (3m)(1.5x) + (m)(0.5x) = 4.5mx + 0.5mx = 5 mx Therefore C is also wrong and centre of gravity is slightly to the left side of C. 7 C Since the balloon is in equilibrium, upthrust on balloon = weight of balloon and helium + force by spring 𝜌𝑎𝑖𝑟𝑉𝑏𝑎𝑙𝑙𝑜𝑜𝑛𝑔 = 𝜌ℎ𝑒𝑙𝑖𝑢𝑚𝑉𝑏𝑎𝑙𝑙𝑜𝑜𝑛𝑔 + 𝑚𝑏𝑎𝑙𝑙𝑜𝑜𝑛𝑔 + 𝑘𝑥 (1.29)(5.0)𝑔 = (0.180)(5.0)𝑔 + ( 3.5 1000)𝑔 + (100)𝑥 x = 0.544 m 8 C Force constant, 𝑘 = 5.0 0.1 = 50 N m-1 Initial extension = 𝐹 𝑘 = 3 50 = 0.06 m and final extension = 𝐹 𝑘 = 2.5 50 = 0.05 m Change in E.P.E.= 1 2 𝑘𝑥𝑓2 − 1 2 𝑘𝑥𝑖2 = 1 2 (50)(0.05)2 − 1 2 (50)(0.06)2 = −0.028 J 9 D Frictional force on the object provides the centripetal force mr2. Both objects have the same angular velocity and same mass, but the centripetal force required for Q is larger due to larger radius. When the centripetal force required exceeds the frictional force available, Q starts to slide. 10 D Using COE: 2 2 1 2 2 ------------- (1) if if EE GPE KE mgr mv mgr mv = = = = 2 bottom 2 bottom sub (1) into eqn 2 3 mvT mg r mvT mg r mgr mgr mg −= =+ =+ =

3 11 C 2= GMg r 2 Mg r Let the quantities with subscript ‘J’ represent that of Jupiter while those with subscript ‘E’ represent that of Earth. 2 JJ E 2 E E J =   gM r g M r ( ) 23 J 7 E 6370 10318 7.15 10  =   g g Since gE = 9.81 N kg-1, ( ) ( ) 23 J 7 6370 10318 9.817.15 10  =   g = 24.8 N kg-1 12 B 1 3.05.0 20 3.0 7.0 4.0 J kg BC BC C AB AB C Um U U     −  =     + −= = =  + − + = = − 13 A When two objects A and B are placed in thermal contact, heat flows from the hotter object to the colder object B, until they reached thermal equilibrium. At thermal equilibrium, both objects A and B are at the same temperature. 14 B Average K.E. ( )( ) 23 2133 1.38 10 27 273.15 6.