Section A

Question A1

a)   Assuming upwards is the positive direction v2  = u2  + 2as 0  =  15.62 – 2 x 9.81 s1

s1  = 15.62/(2 x 9.81)  = 243.4/19.62  = 12.40 m

d = s + h = 12.40 + 1.23  =  13.6 m

b) s2  = ut + ½ at2          - 1.23 = 15.6t - ½ 9.81t2 4.91t2 – 15.6t – 1.23 = 0

t = {15.6 ± √ [15.62  + (4 x 4.91 x 1.23)]} / 9.81

= {15.6 ± √267.5} / 9.81  = {15.6 ± 16.36} / 9.81

t is positive, so t ={15 .6 + 16 .36}/9 .81 = 31 .96/9 .81 = 3.26 s

Give one accuracy mark for obtaining the correct equation for solving for t and the second for the correct answer.

(The calculations can also be done assuming downwards is the positive direction)

Question A2

a) m1u1  + m2u2  = m1v1  + m2v2

4.65 x 3.26 + 0  = 4.65v1 + 2.41 x 4.12

15.16 = 4.65v1  + 9.929

v1  = (15 . 16 – 9.929) / 4.65  = 5 .231/4.65  = 1.13 m s- 1

b) Ekinitial  = ½ m1u12   =  ½ x 4.65 x 3.262   = 24.71 J Ekfinal   = ½ m1v12  + ½ m2v22

= ½ x 4.65 x 1.1252  + ½ x 2.41 x 4.122  = 2.943 + 20.45 = 23.39 J Eklost  = 24.71 – 23.39 = 1.32 J

Question A3

Heat lost by brass = heat gained by ice, water and can (1M)

mbcbΔθb   = miLi  + (mi  + mw)cwΔθw  + mCucCuΔθCu (1M)

0.500 x 370 x (100- θ) = [0.100 x 334] + [(0.100+0.250) x 4180 θ] + [0.350 x 385θ] (1A)

185(100 – θ) = 33.4 + 1463 θ + 134.8θ

1.85 x 104  - 185θ = 33.4 + 1598 θ

1.85 x 104 – 33.4  = (1598 – 185)θ

θ = (1.847 x 104)/1413  =  13.1 °C (1A)

Question A4

a) E = stress/strain      strain = stress/E (1M) = mg/AE (1M) = 3.50 x 9.81 / {[π x (0 .125 x 10-3)2/4] x 2.11 x 1011 }              = 34.34/2589 = 0.0133

b)   strain = e/L e = L x strain  =  2.40 x 0.01326 = 0.0318 m

Question A5

p1 V1/T1   = p2 V2/T2 (1M) V2   = p1 V1 T2/T1p2 (1M)

= 1.01 x 105 x 0.562 x (25.5 + 273.2) / (273 .2 x 1.78 x 105)

= 1.695 x 107 / (4.863 x 107)  = 0.349 m3

Question A6

a)    1/C = 1/C1  + 1/C2  + 1/C3   =  1/5 + 1/3 + 1/15  = (3 + 5 + 1)/15 = 9/15 C = 15/9  = 1.67 μF

b) V = Voe-t/(RC)

20 = 40.5 e-t/(4.85x 1.667)            20/40.5 = e-t/8.085  = 0.4938 ln(0.4938)  = -t/8.085

[RC = 4.85 x 106 x 1.667 x 10-6]

t = -8.085 x ln(0.4938)  = 5.71 s

Question A7

a) Vs/Vp  = Ns/Np        Ns   = Vs Np/Vp   = 12 x 2000/240  =  100

b) P = V2/R Rbulb   = V2/P =  122/20 = 144/20 = 7.2 Ω 1/Rtotal  = 4/Rbulb   =  4/7.2 Rtotal  = 7.2/4  =  1.8 Ω

c)    Since transformer is 100% efficient, Pp  = Ps   =  4 x 20 = 80 W

Question A8

F = BILsinθ  =  0.357 x 6.58 x 1 x sin70°   =  2.21 N

Question A9

a) eV = ½ mv2 v2   = 2eV/m

= 2 x 1.60 x 10- 19 x 1.80 x 103 / (9.11 x 10-31)

=  5.760  x 10- 16 / (9.11 x 10-31)  =  6.323 x 1014

v = √(6.323 x 1014)  = 2.51 x 107  m s- 1

b) λ = h/mv =  6.63 x 10-34 / (9.11 x 10-31 x 2.515 x 107)

= 6.63 x 10-34 / (2.291 x 10-23)  =  2.89 x 10- 11  m

Question A10

a) hf = hc/λ = Ф + Ekmax (1M) Ф = hc/λ - ½mvmax2 (1M) = {[6.63 x 10-34 x 3.0 x 108] /(166.7 x 10-9)}

- [½ x 9.11 x 10-31 x (9 .602 x 105)2 ]

= [1.989 x 10-25 / (166 .7 x 10-9)] – [8.399 x 10- 19 / 2]

= 1.193 x 10- 18 – (4.200 x 10- 19)  =  7.73 x 10- 19 J

b) hfmin   = Ф

= 1.17 x 1015  Hz

fmin   = Ф/h =  7.732 x 10- 19 / (6.63 x 10-34)

Section B

Question B1

a)   i .      If the acceleration of a body is directly proportional to its distance [ 2 ]

from a fixed point [1 mark] and is always directed towards that point, [1 mark] the motion is simple harmonic.

ii .     This is the maximum displacement from the equilibrium position. [ 1 ]

b)   i . F =  -kx = mg k = -mg/x = -(0.034 x 9.81) / (0.085 – 0. 121) [1M]

=  0.3335/0.036  =  9.26 N m- 1 [1A]

ii . T =  2π√(m/k)  =  2π√(0.034/9.264)  =  2π√(3.670 x 10-3) [1M] = 2π x 0.06058  =  0.381 s [1A]

iii . f =  1/T =  1/0.3806  =  2.63 Hz [1M] [1A]

iv. v = Aωcosωt (1M) vmax   = Aω =  A x 2πf (1M) [2M] = 0.015 x 2π x 2.627    =  0.248 m s- 1 [1A]

v.     Magnitude of a = ω2x =  (2πf)2x =  (2π x 2.627)2 x 0.012 [1M] =  16.512 x 0.012  =  272.4 x 0.012  =  3.27 m s-2 [1A]

c)   i . T2   = T1  + 0.250          and l2   = l1  + 0.301 [2M] T =  2 π√(l/g) T2  = 4 π2l/g (1M)

T12    = 4 π2l1/g = 4 π2l1/9 .81          so T12  =  4.024l1

T22    = 4 π2l2/g = 4 π2l2/9 .81          so T22  =  4.024l2 (1M)

Substitute for T2  and l2               (T1  + 0 .250)2    =  4.024(l1  + 0 .301)

T12  + 0.5 T1  + 0.2502    =  4.024 l1  + (4.024 x 0.301) [2A]

= T12  + 1.211 (1A)

T12  + 0.5 T1  + 0.0625  = T12  + 1.211

0.5 T1   =  1.211 – 0.0625  =  1.149

T1  = 1.149/0.50  =  2.30 s (1A)

ii . l1   = T12/4.024  =  2.2982/4.024  =  1.31 m

Question B2

a)   i .     The total current entering a junction in a circuit equals the total

current leaving the junction.

ii .     Conservation of charge.

iii .    For any complete loop in a circuit the algebraic sum of the e.m.f.s equals the algebraic sum of the products of current and resistance.

iv.    Conservation of energy.

b)   i .      I       2nd  law applied to right hand loop

12 – 4 = 6I1  + (1.50 x 2)

8  =  6I1 + 3

6I1   =  8-3  = 5

I1   =  5/6  = 0.833 A

II      1st law applied to point A

I1   = I2  + 1.50

I2   = I1  – 1.50  = 0.8333 – 1.50  =  -0.667 A

III     Law 2 applied to left hand loop

ε – 4  =  6I1  + 5I2   =  [6 x 0.8333] + [5 x (-0.6667)]

= 5 – 3.334  =  1.667

ε =  1.667 + 4  =  5.67 V

p.d.BA  = V + IR = 4.0 + 6I1  = 4.0 + (6 x 0.8333)

=  4.0 + 5.000 = 9.00 V

(or use any other paths between B and A)

1/Rp   =  1/R1  + 1/R2   =  1/5 + 1/20  =  (4 + 1)/20  =  5/20 Rp   =  20/5  =  4 Ω

1/Rout   =  1/8 + 1/40  =  (5 + 1)/40  =  6/40

Rout   =  40/6  =  6.667 Ω

Rtotal   = Rp  + Rout   =  4 + 6.667  =  10.67 Ω

Vout   =  (Rout x Vin)/Rtotal   =  (6.667 x 12.0)/10.67  =  80 .00/10.67

=  7.50 V

ii . V20   = Vin – Vout   =  12.0 – 7.498  = 4.502 V P = V2/R =  4.5022/20  =  1.01 W

iii . Q = It = Vt/R =  4.502 x 60 /5  =  54.0 C

Question B3

a)   i .

ii .

b)   i .

ii .

(1A for at least 3 correct entries + 1A for all remaining correct entries)

2 marks for labelling axes correctly with units and 2 marks for [2M] plotting data points correctly. [2A]

iii .   Gradient = -0.1444 (1M for obtaining gradient using a large [1M] triangle on graph) + (1A) Hence λ =  0. 144 min- 1 (1A) [2A] (Allow ± 0.010 min- 1)

iv. t½   =  ln2/λ =  ln2/0.1444  =  4.800 min (Will depend on value of λ .)

v. A = Aoe-λt (1M) =  12.31 e-0.1444 x 20   =  12.31 e-2.888 (1M) =  0.686 MBq

(Will depend on value of λ .)

vi . λ must be in s- 1 (1M) =  0. 1444/60 = 0.002407 s- 1

N  = Ao/λ (1M) =  12.31 x 106/0.002407  =  5.11 x 109 (Will depend on value of λ .)