2022 RI Promo Sect ABC Soln

© 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 gT L gTL          2 B  22 2 1 2 31 2 1 1 u n i t s o f N m( k g m s ) mk g m s mk g 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 su t a t tt tt su t           4 D 2 12 (1.0 9.81) 1.0 2.19 m s Yc cTm g m a a a       () () (1.0 2.0)(9.81) (1.0 2.0) 2.19 36.0 N Xc B c B x X Tm m g m m 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 AX A 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 50rr r Anticlockwise moments      20 30 50rr r 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 PF v D D    sin 357.1 (1800 9.81 sin5.0 ) 1896 N 1896 14 26.5 kW S S FD W P           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 cos30 02 0.2679 0.2679 1.3 mv mg L L vg L mvTm g r mg LTm g L Tm g         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     10 B 2 2 211 GPE22 2 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 mx     Thus, E vs x is an inverted parabola, with 22 0 1 2Em x  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 driv ing 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          II I I 14 C   2 0 22 2 0 220 0 4 cos cos cos cos cos cos5 1cos 5 48            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             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 by weight minus normal contact force. The minimum value of the normal contact force is zero. Hence, the maximum downward force is equal to the weight and the maximum value of q is equal to g. B1 B1 B1 B1 Some candidates did not clearly state that normal contact force can only act upwards on the man standing in the lift. Some even stated th e man is in equilibrium which he is clearly not. (b) Since the man starts from rest, and comes to rest at t = 10 s, thus sum of the positive and negative areas on the graph must be zero.  2 11 52 2 022 5.0 m s q q     M1 A1 Many candidates gave the magnitude of q rather than its value. (c) (i) 1 mark – correct shape of curve from t = 0 to t = 5.0 1 mark – correct shape of curve from t = 8.0 to t = 10 1 mark – correct values (2.5 and 5.0) on vertical axis Most candidates answered this part well. v / m s1 t / s 10 8 4 2 0 6 5.0 2.5

5 © Raffles Institution [Turn over (ii) total distance travelled area under - graph 52 . 5 12.5 m vt   B1 Some candidates did not read the question ca refully and gave the total distance travelled in 10 seconds. (iii) 2 total work done by normal contact force total increase in energy of man increase in KE increase in GPE 1 50 5 50 9.81 12.52 6760 J         Or   2 1 5.0 2.02average acceleration 1.0 m s 5.0 work done by normal contact force distance travelled by lift 50 (1.0 9.81) 12.5 6760 J ave v t ma g         M1 A1 M1 A1 This part was poorly attempted. Many students used force multiply by distance travelled but failed to realise that the force acting on t he man is constantly changing. Thus, merely taking the magnitude of the man’s weight multiply by 12.5 m is in correct. 17 (a) (i) Considering the man, lion and rope as a system, tension is considered as an internal force and is not taken into consideration. The external forces acting on the syst em would be the friction acting on the man as well as the friction acting on the lion. There is no winner only if the friction ac ting on the lion is equal to that acting on the man, resulting in no net force. Or The winner will the one that experience a larger friction from the ground. B1 M1 A1 Alternative The tension acts on the man and the lion separately and does not cancel out. Considering the man, if the frictional force acting on the man due to the ground is less than 500 N, the man will lose as the net force will be towards the lion. B1 M1 A1 Many students failed to define the system that they were referring to in their explanation, so a lot of this was taken to be implied. There was a lot of answers that stated “even though tension acts on the man/lion, the force on the rope is not the same as the tension”. This is not correct as the force on the rope by the man = the force on the man by the rope. A B area A = area B

6 © Raffles Institution Many students then went on to incorrectly conclude that the resultant force = (force on the rope by the man) – (the force on the man by the rope) This statement is wrong because: 1. Force on A cannot cancel out force on B. 2. The resultant force here is not clearly defi ned, is it the resultant force on A or B or A+B? The students that managed to state that there were external forces like friction, did not clearly state how tension plays into the conclusion as to whether there will be a winner. Some students mentioned that weight plays a part without relating this to friction. Weight by itself acts perpendicular to tension, and hence does not affect the horizontal resultant force on the system. (ii) 1. Since the man is still at equilibrium, by taking moments about his toes,  500 0.80 0.50 81.5 kg mg m   M1 A1 2. The man can lower the rope to reduce the clockwise moment about his toes. (Decrease perpendicular distance between tension and his toes.) The man can lean backwards to increase the anti-clockwise moment about his toes. (Increase horizontal distance between CG and his toes.) A1 Common mistakes here include: - equating weight to 500 N (these forces are perpendicular and do not add up that way) - mixing up moments with work done - resolving the tension and weight to be perpendicular to the line joining the c.g. and the tip of the man’s toes. (not necessary) (b) Since the rope is stationary, Fnet = 0 In the x direction, 1000 1000sin25 577.38 NXR    In the y direction, 1000cos25 906.307 NyR      2222 577.38 906.307 1070 N xyRR R    577.38tan 32.5 906.307 x y R R    OR M1 M1 A1 A1 222 1000 1000 2(1000)(1000)cos65 1070 N R R    180 65 57.52 25 32.5        