RI 2022 Promo H2 Physics Solutions

© Raffles Institution [Turn over 2022 Year 5 Promotion Examination H2 Physics Solutions 1 C ( ) 2 2 2 4 2 2 0.020 0.030 0.070 LT g gL T g TL gT L π π = = ∆ ∆∆= += + = 2 B ( ) 2 2 2 12 3 12 1 1 units of N m (kg m s ) m kg m s m kg m s units of kg s m (m s ) p Q − − − −− −− − − = = ×= × = = × 3 D ( ) ( ) ( )( ) 2 2 1 2 10.80 12sin25 9.81 2 Using GC, 1.173 or 0.13905 12cos25 1.173 12.8 m yy y xx s ut at tt tt s ut = + − = °+ − = = − = = °= 4 D 2 12 (1.0 9.81) 1.0 2.19 m s Yc cT mg ma a a − −= −× = × = () () (1.0 2.0)(9.81) (1.0 2.0) 2.19 36.0 N X cB cB x X T mm gmm a T T −+ =+ −+ =+× = OR Since B and C have the same acceleration, (2 1) (2 1)(12) 36 NXYTT = += + = ( 9.81) 36 2.19 4.72 kg A XA AA A mg T ma mm m −= × −=× = 5 D Net force ( ) ( )2 30sin45 2 20sin45 14.1 N= °− °= (SW direction) By taking moments about the c.g., clockwise moments ( )( ) ( )( )20 30 50r rr=+= Anticlockwise moments ( )( ) ( )( )20 30 50r rr=+= Hence, net torque = 0 6 C At constant speed on horizontal road, driving force F is equal to drag force D. 5000 14 357.1 N P Fv D D = = × = sin 357.1 (1800 9.81 sin5.0 ) 1896 N 1896 14 26.5 kW S S F DW P θ= + = + × × °= = ×= θ W FS D

2 © Raffles Institution 7 A dUF dx=− The force F acting on the particle is the negative of the potential energy U gradient. 8 B ( ) ( ) 2 2 2 gain in KE loss in GPE 1 0 c os30 02 0.2679 0.2679 1.3 mv mg L L v gL mvT mg r m gLT mg L T mg = −= − − = −= −= = 9 A Considering the star of mass 2 ,M gravitational force provides centripetal force. ( )( ) ( ) ( ) 2 2 3 32 23 5 0.20 GM M Mx x GM x ω ω = = 10 B 2 2 211 GPE2 22 Mm vGm rr GMmmv r = = =− ( ) 1 2 22 22 GPE 5.6 MJ GPE 2 .8 MJ2 1KE (GPE ) 1.4 MJ2 KE GPE 2.8 1.4 1.4 MJ GMm r GMm r = −= − = −= − = −= + = −+= − 11 B 22 22 22 2 00 11 1 1 ()22 2 2E m v m x x mx mxω ωω= = −= − Thus, E vs x is an inverted parabola, with 22 0 1 2E mx ω= at x = 0 and E = 0 at x = x0. 12 A If the mass decreases, the natural frequency of the system k m increases, hence the frequency response curve peaks at a higher driving frequency. The new curve intersects with X.

3 © Raffles Institution [Turn over 13 C ( ) 2 2 21 2 2 2 2 1 2 2 and 4 4 0.5 43 32 0.24 P A r P r P r A r Ar AA π π π = ∝ =  =  = II I I 14 C ( ) ( ) 2 0 2 22 0 220 0 4 cos cos cos cos cos cos5 1cos 5 48 θ θ= θ θ θθ θ= θ = = = = ° I' I I'' I' I I I 15 C ( ) 1 2 3 4 3 4 5 4 21 4 51.3 4 1.04 m n L L L nL λ λ λ λ λ λ = = = −= = = Hence, distances that will hear loud sound are: 0.26 m, 0.78 m, 1.30 m, 1.82 m 1st resonance 2nd resonance 3rd resonance

4 © Raffles Institution 16 (a) In order for the man to accelerate downwards with a magnitude larger than g, there must be an additional downward force acting on the man other than its weight. There are no other forces acting downwards. Or The resultant force acting downwards is given