2024 DHS Promo Physics H2 Answers

DUNMAN HIGH SCHOOL 2024 PHYSICS H2 (YEAR 5) 1 For Internal Use Only Suggested Answers to Promotional Exam Paper 1 1 2 3 4 5 6 7 8 9 10 C D C B A C A D B D 11 12 13 14 15 C D C B A 1 C The size of a hydrogen atom is around 1 Angstrom or 1010− m. 2 D Student D’s value for 𝑔 = 8.46 ± 0.05 m s−2 is far from the actual value of 9.81 but is the most precise of all the students. 3 C The resultant acceleration is due to the resultant force of air resistance and the weight of the ball 4 B N and W1 are not equal when the woman has non-zero acceleration. (A is incorrect.) (W1 + W2) is not equal to T when the lift has non-zero acceleration. (D is incorrect.) If N = W1, the woman (thus the lift) can be moving at constant velocity. (C is incorrect.) 5 A 1 1 Let be the total mass of the car, the d river and fuel; the fixed retarding force; the change in velocity. Applying Newton's 2nd law of motion to t he two situations: --- (1) m F v vFm t  = ( ) ( ) ( )( ) ( ) 2 2 2 1 2 1 when the car is almost out of fuel --- (2) when the car has a a full load of fuel from (1) and (2), we have 30 50130 0.37 s7000 vFm t vt t m m F = −− = − = = −

DUNMAN HIGH SCHOOL 2024 PHYSICS H2 (YEAR 5) 2 For Internal Use Only 6 C The wires are slightly elastic hence they can be treated as springs. The first wire fixed to the wall will experience the same force F and hence will extend by the same length x. The force F is shared equally by the springs in parallel so each spring experiences half the force and extends half as much, 2 x . Thus, the total extension is 3 22 xxx+= . 7 A Torque = Force multiplied by the perpendicular distance between the line of actions of the two forces. 8 D ( )( ) 9 12 minimum 1.3 10 24 60 W kg s 9.81 m s 2.0 m60 300 kmg h mP g h tt −− = =  = =     9 B The area under the force-extension graph is the total work done to stretch the material. 10 D The net force acting on the stone when it is vertically below the centre of the circle is given by 2T W Mr −= , when T is the tension in the string. 2T W Mr  = + 11 C Gravitational field strength is a vector. It has both direction and magnitude. So C is not true. 12 D The graph shown is an elliptical graph and in simple harmonic motion, the graph of velocity against displacement is elliptical.

DUNMAN HIGH SCHOOL 2024 PHYSICS H2 (YEAR 5) 3 For Internal Use Only 13 C When the plasticine is flattened, the surface area increases leading to an increase in air resistance when oscillating. The new system experiences greater damping than before so the new resonance graph is flatter and has the peak shifted slightly to the left. 14 B ll ll    =  =  = = =  ==  = = = =  = = 22 2 2 X Y X X XY 2 X Y Y Y 2 2 XY X Y X Y2 XY XY YX 12 46.0 4 1 because and ,4 1 2 V V R VPP R R R V R A d dR Rd dR dR 15 A ( ) ( ) ( ) X QP QR X When the potentiometer is balanced (i.e. , reading on the galvanometer is zero), e.m.f. of cell X p.d. across length QP o f the wire QPQR 2.00.50 QPQR QR QRQP 0.50 2.0 4 V V   = = =  =   = = 

DUNMAN HIGH SCHOOL 2024 PHYSICS H2 (YEAR 5) 4 For Internal Use Only Paper 2 1 (a) kg m2 A1 (b) Base units of translational KE = kg m2 s−2 [M] = kg [R2] = m2 [ω2] = s−2 M1 Base units of rotational KE, ERotational = kg m2 s−2 = Base units of translational KE A0 (c) ( )RotationalE MR 2 23 2 2 3 5 1 1 20.79 10 236.935 105 5 2 1.2 2.188 10 J  − − −   = =    = ( ) Rotational Rotational Rotational Rotational E M R T E M R T EE 55 2 0.001 0.01 0.2 2 36.935 20.79 1.2 0.334 0.334 0.334 2.188 10 0.7 10 J (to 1 s.f.)−−     = + +  = + +  =  = =  =  ( )RotationalE 52.2 0.7 10 J − =   (d) Systematic error: A1 Zero error from the micrometer screw gauge / electronic balance Random error: A1 Human error when starting and stopping the stopwatch Marks can be given for other reasonable answers in the context of this experiment 2 (a) Any two: - air resistance varies with the speed of the aeroplane - friction between the aeroplane's wheels and the runway changes (lift reduces normal force on the wheels, decreasing friction with the runway) - mass of fuel in the aeroplane decreases during take-off B2 A1 C1 A1 A1

DUNMAN HIGH SCHOOL 2024 PHYSICS H2 (YEAR 5) 5 For Internal Use Only (b) (i) It states that the rate of change in momentum of a body is directly proportional to the resultant force acting on the body and occurs in the direction of the resultant force. (ii) By Newton’s 2nd law, change in momentum of the aeroplane = area under the graph = final momentum – initial momentum = final momentum at take-off – 0 C1 momentum at take-off ( ) ( ) ( ) 5 5 511 22 6 6 5.0 1.13 10 20.0 1.13 10 1.13 0.97 10 30.0 M 1 5.693 10 5.69 10 N s (to 3 s.f.) A1 =    +   + +    =  (c) Using p = mv, max 6 4 1 5.693 10 7.45 10 76.42 76.4 m s (to 3 1 s.f.) A pv m − = =  =  (d) Note: [1] Curve of increasing gradient from origin and line of constant gradient between 5.0 and 25.0 s. [1] Line of decreasing positive gradient after 25.0 s. velocity time / s 0 0 5.0 25.0 55.0 vmax for 2 (e) 76.4 34.1 3.8 [The quadratic curvatures of the curve for the intervals 0 to 5.0 s and 25.0 to 55.0 s are exaggerated to distinguish them from the straight-line segment between 5.0 and 25.0 s.]

DUNMAN HIGH SCHOOL 2024 PHYSICS H2 (YEAR 5) 6 For Internal Use Only (e) Distance = area under graph 1 76.42 55.0 M12 2101 2200 m A1 =   =  Accept 2050 – 2200 m 3 (a) ( )( ) ( ) 1momentum conservation: 0.20 kg 1.8 m s 0.5 0 kg v− = M1 10.72 m sv −= A1 (b) ( )( ) ( )( ) 2211 22initial KE 0.20 1.8 0.32 J, final KE 0.50 0.7 2 0.13 J= = = = B1 final KE < initial KE, collision is ine lastic B1 (c) ( )( ) 1 Change of momentum of 0.20 kg mass 0.20 kg 0.72 1.8 0.216 kg m s −= − =− B1 ( )( ) 1 Change of momentum of 0.30 kg mass 0.30 kg 0.72 0 0.216 kg m s −= − =+ 1 1 0.216 kg m sForce on 0.20 kg mass 1.08 N0.20 s 0.216 kg m sForce on 0.30 kg mass 1.08 N0.20 s − − −= =− += =+ B1 By Newton’s 3rd law, the force acting by 0.2 kg to 0.3 kg is equal and opposite to the force acting by 0.3 kg on 0.2 kg. B1 4 (a) Upthrust is the (resultant) upward force exerted by a fluid on a body when it is partially or fully submerged in a fluid. (it is equal in magnitude and opposite in direction to the weight of the fluid displaced.) B1 (b) (i) M =  V =  × [4/3] (r3) = 1.21 × [4/3] () (11.0 × 10−2)3 C1 = 0.00675 kg = 6.75 g