Standing wave questions

Feb Assesment y13

1.       (a)     appropriate shape; lines perpendicular to and touching plate and sphere; (2)
arrows towards negative sphere (1)                                                                  3

  (b)   (i)      By moments, e.g F cos 20 = W sin 20 / by triangle of forces /
by resolution of forces / other suitable method; i.e. justification needed (1)
F = 1.0 × 10–5 tan 20; = 1.0 × 10–5 × 0.364; (= 3.64 × 10–6 N) (2)
triangle of forces gives W/F = tan 70, etc (1)                                           3

  (ii)   E = F/Q; = 3.64 × 10–6 / 1.2 × 10–9 = 3.0 × 103;N C–1 / V m–1                3

  (c)   E = (1/4πεo)Q/r2; 3.0 × 103 = 9 × 109 × 1.2 × 10–9/r2; (2)
or use F = (1/4πεo)Q2/r2; r2 = 3.6 × 10–3 giving r = 6 × 10–2 (m) (1)                 3

  (d)   field line sketch minimum of 5 lines symmetrical about line joining
centres with arrows; (1)
Fig 1 sketch matches RHS of Fig 2/plate analogous to mirror/AW
relating to symmetry (1)                                                                                   2

[14]

 

  2.     (a)     (i)      Q = VC; W = ½ VC.V ( = ½ CV2) (2)

(ii)     parabolic shape passing through origin (1)
plotted accurately as W = 1.1 V2 (1)                                                       4

  (b)   (i)      T = RC; = 6.8 × 103 × 2.2 = 1.5 × 104 s = 4.16 h                                     2

(ii)     ΔW = ½ C(V12 –V22) = 1.1(25 – 16) ; = 9.9 (J)                                       2

  (iii)  4 = 5 exp(–t/1.5 × 104) ; giving t = 1.5 × 104 × ln 1.25 = 3.3 × 103 (s)    2

(iv)    P = ΔW/Δt = 9.9/3.3 × 103 = 3.0 mW            ecf b(ii) and (iii)                 1
allow P = Vav2 /R = 4.52/6.8 × 103 = 2.98 mW

[11]

 

  3.     nature and features:
α-particle is 2p + 2n/ mass 4 u (1)
charge of +2e (1)
very short range/heavily ionizing/absorbed by paper (1)
spontaneous; and random nature of radioactive decay (2)
energetically more favourable to eject four particles together than a single
one/other comment about energy minimisation/mainly occurs from higher A
 nuclei/AW (1)
small mass decrease/loss provides kinetic energy of α-particle (1)
particle energy of a few MeV; particular decay is monoenergetic (2)
α-particle scattering:
suitable diagram and/or description to illustrate experiment up to 2 marks (2)
most particles have little if any deflection (1)
large deflection of very few shows nucleus is small; and very massive (2)
(Coulomb’s law enables closest approach to) estimate nuclear size
(in case of α-particle back scattering with conservation of energy argument)     max 7
Quality of Written Communication                                                                            2

[9]

 

4.       (a)     (i)      5.0 (V) (1)                                                                                               1

(ii)     10.0 (V) (1)                                                                                             1

  (b)   (i)      Q = CV;= 1.0 × 10–3 (C) (2)                                                                    2

(ii)     The total capacitance of each circuit is the same (namely 100 μF); (1)
because capacitors in series add as reciprocals/ in parallel add/
supply voltage is the same and Q = VC, etc. (1) max 2 marks                 2

  (c)    (i)      A1 will give the same reading as A2; because the two ammeters are (1)
connected in series /AW (1)                                                                    2
answer only in terms of exponential decrease for a maximum of 1 mark

  (ii)   A4 will show the same reading as A2 at all times; (1)
A3 will show half the reading of A2 initially;
and at all subsequent times (2)                                                                3

[11]

 

  5.     (a)     Positive as E-field is downwards/top plate is positive/like charges repel/AW (1)  1

  (b)   (i)      k.e. = QV; = 300 × 1.6 × 10–19 = (4.8 × 10–17 J) (2)                                2

  (ii)   1/2mv2 = 4.8 × 10–17; = 0.5 × 2.3 × 10–26 × v2 so v2 = 4.17 × 109;
(giving v = 6.46 × 104 m s–1) (2)                                                             2

  (c)   E = V/d; so d = V/E = 600/4 × 104 = 0.015 m (2)                                             2

  (d)   (i)      semicircle to right of hole (1) ecf(a); (a) and d(i) to be consistent           1

(ii)     mv2/r; = BQv; (2)
giving r = mv/BQ = 2.3 × 10–26 × 6.5 × 104/(0.17 × 1.6 × 10–19); (1)
r = 55 mm;so distance = 2r = 0.11 m (2)                                                 5

[13]

 

  6.     (i)      leptons;                                                                                                             1

(ii)     neutrino / muon / tau(on);                                                                                 1

[2]

 

 

Y12 Homework on Divided Circuits

Answers are below

Divided Circuits

Answers

1 3 ohm 1.4A
7 ohm 0.6A

2 2 ohm 2A
4 ohm 1A

3 4 ohm 0.5A
8 ohm 0.25A
total 1.75A

4 4 ohm 0.5A
5 ohm 0.4A
20 ohm 0.1A

5 A 4A
B 1A
C 1A
D 4A

6. 3.5V

Capacitors EUT markscheme

A* - 31

A – 29

B – 25

C – 22

D – 18

E – 14

 

1.       (i)      Cp = C + C = 6 μF; 1/Cs = 1/2C + 1/C; = 3/2C giving Cs = 2C/3 = (2 μF)          3

(ii)     2 sets of (3 in series) in parallel/ 3 sets of (2 in parallel) in series                     2

[5]

 

  2.     (a)     (i)      Cp = 2 + 4 = 6 μF                                                                                  A1

  (ii)   1/C = 1/2 + ¼                                                                                        C1
Cs = 4/3 =1.33 μF                                                                                 A1

  (b)   (i)      6.0 V                                                                                                    A1

  (ii)   Q = CpV                                                                                                C1
= 6 × 6 = 36 μC                                                                                    A1

  (c)   E = ½ CsV2                                                                                                    C1
= 24 × 10–6                                                                                                    A1

  (d)   (i)      The capacitors discharge through the voltmeter.                                   B1

  (ii)   V = V0e–t/CR
1/4 =e–t/(6×12)                                                                                        C1
ln 4 = t / 72                                                                                           C1
t = 72 ln 4 ≈ 100 s                                                                                A1

[12]

 

  3.     (i)      C = Q/V or gradient of graph / = 24 μC/3V; = 8.0 (μF)                                     2

(ii)     E = ½ CV2 / = ½ × 8 × 32; = 36 (μJ) ecf a(i)                                                      2
or ½ QV / = ½ × 24 × 3; = 36 (μJ)

  (iii)  T = RC = (0.04); R = 0.04/8.0μ = 5.0 × 103 (Ω) ecf a(i)                                    2

(iv)    idea of exponential/constant ratio in equal times; which is independent of
initial value/AW or argued mathematically in terms of Q/Qo = e–t/RC
give 1 mark for statement that time depends only on time constant/RC            2

[8]

 

  4.     (a)     (i)      Q = VC; W = ½ VC.V ( = ½ CV2) (2)

(ii)     parabolic shape passing through origin (1)
plotted accurately as W = 1.1 V2 (1)                                                       4

  (b)   (i)      T = RC; = 6.8 × 103 × 2.2 = 1.5 × 104 s = 4.16 h                                     2

(ii)     ΔW = ½ C(V12 –V22) = 1.1(25 – 16) ; = 9.9 (J)                                       2

  (iii)  4 = 5 exp(–t/1.5 × 104) ; giving t = 1.5 × 104 × ln 1.25 = 3.3 × 103 (s)    2

(iv)    P = ΔW/Δt = 9.9/3.3 × 103 = 3.0 mW            ecf b(ii) and (iii)                 1
allow P = Vav2 /R = 4.52/6.8 × 103 = 2.98 mW

[11]

 

 

Assessed homework materials

26 A

23 B

20 C

17 D

13 E

1.       (i)      Stress = force / cross-sectional area                                                                B1

(ii)     Strain = extension / original length                                                                 B1

[2]

2.       (a)     Young modulus = stress/strain
(As long as elastic limit is not exceeded)                                                        B1

(b)     Strain has no units because it is the ratio of two lengths.                                B1

[2]

3.       (a)     The extension of a spring is directly proportional to the applied force          M1
as long as the elastic limit is not exceeded)                                                    A1

  (b)   (i)      Correct pair of values read from the graph
force constant = 12/0.080                                                                     C1
force constant = 150 (N m–1)                                                               A1

(ii)     extension, x =  × 80 (= 133.33) (mm)                                              C1
(E = ½ Fx)
energy = ½ × 2/ × 133.33 × 10–3
energy = 1.33 (J)                                                                                  A1

(iii)    The spring has not exceeded its elastic limit                                         B1

(iv)    (elastic potential energy = kinetic energy)
                                                                                      M1
m and k are constant, therefore x µ v.                                                  M1

[9]

4.       (a)     A brittle material does not have a plastic region / it breaks at its elastic limit. B1

(b)     Ultimate tensile strength is breaking stress for a material                               B1
Materials can be chosen / tested to prevent collapse of the bridge                 B1

[3]

5.       (i)      1       Elastic as returns to original length (when load is removed)                  B1

2        Hooke’s law is obeyed as force is proportional to the extension          B1

Example of values given in support from table                                    B1

(ii)     Measure (original) length with a (metre) rule / tape                                        B1

Suitable method for measuring the extension e.g.

levelling micrometer and comparison wire or fixed

scale plus vernier or travelling microscope and marker / pointer                    B1

(iii)    E = stress / strain                                                                                            C1

    = (25 ´ 1.72) / (1.8 ´ 10–7 ´ 1.20 ´ 10–3)                                                   C1

    = 1.99 ´ 1011 (Pa)                                                                                       A1

[8]

6.       (a)     P.E. at top = 80 × 9.8(1) × 150 = 118 000 (J) (1)
K.E. at bottom and at top = 0 (1)
Elastic P.E. at top = 0, at bottom = P.E. at top for ecf = 118 000 J (1)              3

  (b)   24 N m–1 × 100 m = 2400 N                                                                            1

  (c)   elastic P.E. is area under F-x graph (1)
graph is a straight line so energy is area of triangle (1)
elastic P.E. = ½ × kx × x = (½kx2) (1)                                                               2

  (d)   loss of P.E. = 100 × 9.8(1) × 150 = 147 000 J (1)
gain of elastic P.E. = ½ × 26.7 × 1052 = 147 000 J (1)                                      2

  (e)   idea that a given (unit) extension for a shorter rope requires a greater force     1

[9]

 

Markscheme RC & the bullet

Answers to principle of moments 2

3. (a) Principle of moments

In equilibrium

sum of clockwise moment (about any point) is equal to sum of

anticlockwise moment (about that point)

2

(b)(i) Weight

Use of “width x thickness x length”

Use of “density = mass”

volume

Correct value 3

V = 1.2 × 0.6 × 200 (cm3) = 144 (cm3)

Using ρ = m/v

m = 8 (g cm-3) × 144 (cm-3) = 1152 g

Weight = mg = 1152 × 10-3 (kg) × 9.81 ( m s-2) = 11.3 (N) / 12

(N)

(ii) Force F

Correct substitution into correct formula

Correct value with correct unit

F × 60 (cm) = 11.3 (N) × 40 (cm) / 12 (N) × 40 (cm) / 11(N) × 40

(cm)

= 7.5 N / 8 N / 7.3 N

2

(iii) Force R

18.3 N / 18.8 N / 20 N

1

(iv) Sketch graph

Any line upwards

Correct shape for F [concave shaped curve]

2

Y12 Young's Modulus

Stress, strain and the Young modulus

1.         A long strip of rubber whose cross section measures 12 mm by 0.25 mm is pulled with a force of 3.0 N. What is the tensile stress in the rubber?

2.         Another strip of rubber originally 90 mm long is stretched until it is 120 mm long.  What is the tensile strain?

3.         The marble column in a temple has dimensions 140 mm by 180 mm.

a.     What is its cross-sectional area in mm2?

b.    Now change each of the initial dimensions to metres – what is the cross-sectional area in m2?

c.     If the temple column supports a load of 10 kN, what is the compressive stress, in N m–2?

d.    The column is 5.0 m tall, and is compressed by 0.1 mm. What is the compressive strain when this happens?

e.     Use your answers to parts 5 and 6 to calculate Young’s modulus for marble.

4.   A 3.0 m length of copper wire of diameter 0.4 mm is suspended from the ceiling. When a 0.5 kg mass is suspended from the bottom of the wire it extends by 0.9 mm.

a.     Calculate the strain of the wire.

b.    Calculate the stress in the wire.

c.     Calculate the value of the Young modulus for copper.

Cpacitance