2021 GCE A Level 9814 H3 P1 SS [HCI]

1 2021 H3 Physics A Level Paper Suggested Solutions 1a) In vacuum, the object will fall at a constant acceleration of g. Taking direction downward as positive, and applying the kinematics equation s = ut + 1 2at2. 100 = (0)t + 1 2 (9.81)t2 t = 4.5152 = 4.5 s (2 s.f.) [1] 1bi) time-of-flight t’ = 2 x 4.5152 = 9.0304 s Taking direction downward as positive, and apply the kinematics equation s = ut + 1 2at2. s = (0)(9.0304) + 1 2(9.81)(9.0304)2 = 399.96 m = 400 m (2 s.f.) [1] 1bii) Instead of dropping the object from the top of the tower, the same double time- of-flight could be achieved by throwing the object up from the ground of the tower with the same speed as the speed of the object just before it reaches the ground from its fall from rest as indicated in (b)(i). [1] 1cii) Let h be the initial height of the safety net and vi be the final velocity of the ‘weightless’ (free-fall) phase and initial velocity at the safety net phase. Final speed at the safety net phase is zero. Taking direction downward as positive, and apply the kinematics equation v2 = u2 + 2as For the ‘weightless’ phase, vi2 = 02 + 2 (g)(100 – h) …… (1) For the safety net phase, 0 = vi2 + 2(-4g)(h) ⇒ vi2 = 8gh……(2) Subst (2) in eqn (1) 8gh = 2(g)(100 – h) ∴ h = 20 m (to 2 s.f.) [1] [2] [1] top of tower bottom of tower safety net h 100-h

2 1ci) Taking direction downward as positive, and apply the kinematics equation s = ut + 1 2at2. For the ‘weightless’ phase, (100 – 20 ) = (0) + 1 2(9.81)t2 t = 4.0386 s Hence the ‘weightless’ time of flight is reduced by 4.5152 - 4.0386 = 0.4766 = 0.5 s (1 s.f.) [1] 2a) By PCOLM, 4 4 44 44 mm mm mv mv mv v v v =+ =− By RSOA = RSOS, 44m m m mv v v v v v= − → = + 4 4 4 4 4 4 3 5 3 / 5 8 / 5 m m m m m v v v v v v v v vv + = − → = → = = % of kinetic energy transferred to m, 2 2 18 6425 1 4(25)42 0.64 64% vm mv   = =→ Method II: Using zero-momentum frame (center-of-mass frame) 4 0 4 55 cm mv m vv m +== The velocities of particle in the zero-momentum frame before the collision, 4 , 4 , , , , , 4 55 440 55 m cm m Earth Earth cm m cm m Earth Earth cm vvv v v v vvv v v = + = − = = + = − =− After the collision, the sign of velocities will change, hence, 4 4 and 55 mm vvvv =− = Rewriting the velocities after the collision in the Earth frame, 4 , 4 , , , , , 43 5 5 5 4 4 8 5 5 5 m Earth m cm cm Earth m Earth m cm cm Earth v v vv v v v v vv v v = + =− + = = + = + = % of kinetic energy transferred to m, [1] [1] [1]

3 2 2 18 6425 1 4(25)42 0.64 64% vm mv   = =→ Note: There is a high chance that students might perform calculation error in solving simultaneous equations in the Earth frame (first method) . The zero-momentum frame does not involve such equations. 2bi) 44 55 cm mvvv m== Using the vector diagram could help you find the direction of velocity of 4m in the center-of-mass frame readily. Note: The diagram may help visualize the direction of velocities in earth and zero-momentum (CM) frames. [3]

4 2bii) [1] [1] [1] 3a) Total energy of mass m in gravitational field by mass M 2 2 1 ()2 1 ()2 T t E KE PE GMmmv r GMmmv r =+ = + − =− Where v is the velocity of mass m and tv is the tangential component of the velocity, noting that tvv= for a circular orbit. Since gravitational force is a central force, tL mrv= , where we can write 2 2 2 1 22 t Lmv mr= Thus, 2 22 T GMm LE r mr=− + [1] [1] [1]

5 3bi) Consider the turning points such that 21 02 rmv = 2 22 T L GMmE mr r=− 2 2 2 2 2 02 T TT LE r GMmrm GMm Lrr E mE =− + − = Since ar and br are solutions, we can write 2 ( )( ) 0 ( ) 0 ab a b a b r r r r r r r r r − − = − + + = Comparing coefficients, ( ) 2ab T GMmr r a E− + = =− 2 T GMmE a=− [1] [1] [1] [1] 3bii) From part (b)(i), 2 T GMmE a=− Additionally, we know that 21 ()2 TE KE PE GMmmv r =+ = + − Where v is the speed of the exoplanet. Thus, [1]

6 2 2 1 22 2 GMm GMmmv ra GM GMv ra =− =− 11 32 12 12 1 21() 216.67 10 5.4 10 ( ) 2.8 10 3.8 10 5 v GM ra 1.27×10 m s − − =− =     −  = [1] [1] 4a) Length of coil, L dN= Resistance of coil, 22 4 2 R dd ==   ll 2 0 0 0 0 2 7 1 4 44 10 NNB V V dk Vd L Vd L Vd d Vd d k            −   = = = = =   = Il l l l l l Unit of k is ( )( ) ( ) -2 -2 -2 2 (Tesla)( m)m (Tesla)m kgms m kgms AAmpere m Ampere Am  = = = [1] [1] [1] [1] [1] 4bi) ( ) ( )( ) ( )( ) 73 8 4 10 3.0 0.25 10 0.01544 1.7 10 2.9 o VdBT l   −− −  = = =  [2] 4bii) The thickness of the insulation is assumed to be negligible in the calculations. If the thickness of the insulation is considered, the diameter, d, of the wire itself will be less than 0.25 mm. Since the magnetic flux density is proportional to d, a small diameter will give a smaller magnetic flux density. [1]

7 5a) From first law of thermodynamics, onU Q W = + Since volume of gas is constant in the process of heat transfer, W = 0, therefore UQ= Given that ( )VV QC Q C T UT= → =  = For a monatomic ideal gas, 33 22 VVU Nk T C T C Nk =  =  → = [1] [1] 5b) From first law of thermodynamics, onU Q W = + Given that ( )PP QC Q C TT= → =  For a monatomic ideal gas, 3 2U Nk T =  The temperature increases, so the work done on the gas is negative. Rewriting the first law of thermodynamics, 3 2 33 522 2 P P Nk T C T p V Nk T p V Nk T Nk T C Nk TT  =  −   +   +  = = =  [1] [1] [1] 5c) For the same rise in temperature T , under the constant pressure process, additional work must be done by the gas to the surroundings. Hence CP ˃ CV. [1] 6ai) Assume that the upthrust is negligible. [1] 6aii) The oil droplet gained electrons when it went through the atomiser as it is attracted to the positively charged top plate and repelled by the negatively charged bottom plate. [1] weight electric force

8 6aiii) Assume that the droplet has a spherical shape. 3 341 3 2 6 mD m V DV     = → = = =  [1] 6aiv) ( )( ) 33 6 1711 960 0.5 10 6.283 10 kg66mD   −−= =  =  ( ) ( ) 17 17 17 17 17 1 0.13 3 0.61100 0.5 6.283 10 0.61 3.81 10 kg 4 10 kg 6 10 4 10 kg mD mD m m mm   −− − −−   = + = + =  =  =   =   =   [1] [1] [1] 6bi) Assume that the upthrust is negligible. [1] 6bii) At terminal velocity, DFW= terminal terminal3 3 mgDv mg v D = → = ( ) ( ) 2 terminal 2 2 3 terminal 36 17 3 3 0.5 10 3 960 186.238 10 vDB g D g D g DB mgvm D B         − − = = = =  ==  [1] 6biii) When the electric field is switched on and the oil droplet is stationary, Electric force = weight of oil droplet [1] Weight, W drag force, FD

9 ( ) ( ) ( ) ( ) 3 3/2 3/2 terminal 3/23/2 53 1/2 1/2 10 1/2 1/2 1 6 6 18 1.8 106.0 10 66 960 9.81 1.89 10 kg m s V mgd mgd D gdq mg Vd q ne ne gd BVv ne g Bdk g k       −− −− = → = = = =  ==  = [1] [1] 6ci) The Fig. 6.2 shows that there are three clusters of data points with different gradients (k/ne). Hence, three straight lines should be drawn. Since k and e are cons