YIJC 2024 JC2 PRELIM H2 Phy P2 Soln

1 ©YIJC 9749/02/YIJC/23 2024 JC2 Prelim Exam H2 Physics Paper 2 Solution 1 (a) (i) 1Distance area under the graph (8.8)(0.90) 4.0 m2vt= − = = B1 (ii) ( ) 2 change in velocityaverage acceleration time 4.4 8.8 0.50 26 m s − = −−= =− (Circle but not penalise for missing -ve sign) C1 A1 (b) (i) The accelerations are the same as the gradients of the graph before and after the rebound are the same. (the girl is in a state of free fall before and after the rebound). B1 (ii) The area under the lines represents height and the second area is smaller than the first area, hence, the rebound height is less than the initial height. Thus the GPE of the girl at t = 0 is larger than the GPE at t = 1.85 s. OR The kinetic energy (KE) of the girl at trampoline equals to the GPE at maximum height. The speed after rebound is smaller and hence the KE becomes smaller. Thus the gravitational potential of the girl after rebound is less. M1 A1 (M1) (A1) 2 (a) Using hinge as pivot: Clockwise moment due to weight = anti-clockwise moment due to tension ( )Moment due to cos60 50 Moment due to (sin60)10 y x TT TT = = ( )   cos60 50 (sin60)10 10 5.0 50(cos60) 10(sin60) 16.8 17 cm Wx T T x x =+ =+ == Note: if either moment vertical/horizontal is not considered, max 1 mark. M1 A1

2 ©YIJC 9749/02/YIJC/23 (b) (i) Both vectors drawn and labelled 3 forces have common point B1 B1 (b) (ii) 2 2 2 2 5.0sin60 4.33 N 10 5.0cos60 7.5 N 4.33 ( 7.5) 8.7 N xx yy yy xy RT R T W R W T R R R = = = += = − = − = = + = + − = M1 M1 A0 3 (a) (i) ( ) ( ) 6 3 3 80 6.4 10 1.5 10 10 earth final GMGPE =−  +   Total energy on Earth’s surface = total energy at LEO ( ) ( ) ( ) ( )( )( ) 9 6 6 3 3 9 11 24 66 9 energy supplied 80 80 3.0 106.4 10 6.4 10 1.5 10 10 113.0 10 6.67 10 6.0 10 80 7.9 10 6.4 10 2.1 10 J i i f f earth earth f f f KE GPE KE GPE GM GM KE KE KE − + + = +  − +  = + −   +    =  +   −   = C1 C1 A1 (ii) Centripetal force is provided by gravitational force ( )( ) ( ) 2 11 24 26 6.67 10 6.0 10 80 7.9 10 510 N cG C c c FF GMmF r F F − = =  =  = (1m for both working and answer.) B1 B1 (b) (i) Gravitational potential at a point is the work done per unit mass by external agent in bringing a small test mass from infinity to that point. B1 wall hinge R x 10 N T cord 60° 10 cm 50 cm L-shaped beam

3 ©YIJC 9749/02/YIJC/23 (ii) Since dFm dr =− , the negative of the gradient of the ϕ-r graph gives the force. Positive gradient means gravitational force (vector) is negative Since gravitational force is negative when displacement is positive, it is in the opposite direction as the displacement from Ear