RVHS_H2_Physics_P3_Soln

River Valley High School Page 1 of 8 J2 H2 Physics 9749 Preliminary Examinations 2020 RVHS JC2 H2 Physics Preliminary Examinations Paper 3 Mark Scheme 1 (a) Distance travelled before hitting the ground = Area under graph from 0 to 0.6 s. = ½ (4 + 10) * 0.60 = 4.2 m M1 A1 (b) Distance that the ball bounce upwards = ½ (1.0 * 10.0) = 5.0 m Height above original = 0.8 m M1 A1 (c) Acceleration = 10/1.0 = 10 m s-2 A1 (d) Time taken to travel upwards = time taken to travel downwards Next time the ball will reach ground = 2.6 s A1 2 (a) (i) conservation of linear momentum gives 10𝑚 + 2(5𝑚) = 𝑚𝑣 + 4.5(5𝑚) C1 𝑣 = −2.5 m s-1 A1 (ii) kinetic energy before collisioin 1 2 𝑚102 + 1 2 (5𝑚)22 = 60𝑚 B1 kinetic energy after collision 1 2 𝑚(−2.5)2 + 1 2 (5𝑚)4.52 = 53.75𝑚 B1 percentage loss = 60−53.75 60 × 100% = 10% B1 (b) conservation of momentum gives 10𝑚 + 2(5𝑚) = 𝑚𝑣1 + (5𝑚)𝑣2 -------(1) B1 for elastic collsion, relative speed of approach = relative speed of separation. So 10 − 2 = 𝑣2 − 𝑣1-------- (2) Or conservation of kinetic energy gives 1 2 𝑚22 + 1 2 (5𝑚)2.02 = 1 2 𝑚𝑣1 2 + 1 2 (5𝑚)𝑣2 2 -------(2) B1 solving simultaneous (1) and (2) M1 𝑣1 = −3.3 m s-1 , 𝑣2 = +4.7 m s-1 A1 3 (a) loss in gravitational potential energy = gain in kinetic energy + gain in elastic potential energy B1 (b) (i) 4.0 × 9.81 × 0.090 = 1 2 × 200 × 0.092 + 𝐸𝑘 B1 𝐸𝑘 = 2.7 J B1 (ii) Since both blocks have same acceleration and hence same velocity, 𝐸𝑘 ∝ 𝑚 So taking ratio, 𝐸𝑘,𝑄 𝐸𝑘 = 4 6 or other appropriate method M1 𝐸𝑘 = 1.8 J A1 (c) When blocks come to rest 𝐸𝑘 = 0. Let distance required by 𝑥, then conservation of energy gives 4 × 9.81 × 𝑥 = 1 2 × 200 × 𝑥2 M1 𝑥 = 39 cm A1

River Valley High School Page 2 of 8 J2 H2 Physics 9749 Preliminary Examinations 2020 4 (a) 1 mark for correct direction of arrows 1 mark for similar length of arrows in vertical direction B1 B1 (b) Since frictional force provides for the centripetal force, 2 2 -1 (80)70 55 6.9 m s mvf R v v = = = C1 A1 (c) (i) For a surface which is banked, the horizontal component of the normal reaction is an additional source of the centripetal force. Since Centripetal Force = r mv 2 , an increase in the centripetal force would allow the rider to turn the corner at a higher speed without slipping, provide the mass of cyclist and bicycle and radius of turn remain constant. B1 B1 (ii) Considering forces acting on the rider/bicycle in the horizontal direction: 2 sin 20 cos 20 mv Nf r  +  = ---------------------- (1) For the equilibrium in the vertical direction: += 20sinfmg20cosN --------------------(2) Solving (1) & (2), ( ) 2 tan 20 sin 20 cos 20 mv mg f f r  +  = −  →max velocity during turning v = 15.7 m s-1 C1 for bot h C1 A1 reaction force we