SHM Graph Gradient

Resonance of severl pendula tied to a string

Y13 Circular Motion

(a)     (i)      speed v = 2π r / t
v = 2 × π × 122/2 /(30 × 60) (1)
v = 0.21 m s–1 (1) allow 0.2 m s–1                                                           2

  (ii)   F = 12.5 kN × 16 = 200 kN (1)                                                                1

(iii)    W = F × s or
= 200 k × 2 × π × 122 / 2 (1) ecf (ii) allow ecf for distance from (i)
= 7.7 × 107 J (1) allow 8 × 107                                                                2

  (iv)  P = W / t, energy / time or F × v or
= 7.67 × 107 / (30 × 60) (1) or ecf (iii) / (30 × 60)
= 42.6 kW (1) allow 43 kW, only allow 40 kW if working shown             2

(v)     •        Friction force at bearing opposes motion so not useful (1)
•        Friction force of tyres on rim drives wheel, so is useful (1)
•        Electrical energy supplies power to drive wheels /
          useful implied (1)
•        Input energy (electrical or energy supplied to motor)
          is converted into heat (1)

          Last point to do with the idea that once moving with constant speed e.g.
•        All work is done against friction
•        No input energy is converted into Ek
•        All input energy ends up as heat
•        Any other relevant point relating to energy (1)                              5

  (b)         (i)            k = F / x
= 1.8 × 106 / 0.90 (1)
= 2.0 × 106 Nm–1 (1)         

the pendulum bob is travelling in a circle (1)
so it is accelerating towards the centre (1)
(it has a constant speed in the time interval just before vertical to just
after vertical)

          bob is not in equilibrium (1)
so the tension must be (slightly) larger than the weight of the bob (1)                        3

MAXIMUM 3 

Rollercoaster

Mark Scheme for Car on Hump Back Bridge

Markscheme for questions on circular motion

1.       (i)      (v = 2πr/t) t = 2π60/0.26 = 1450 s

Correct answer is 1449.96 hence allow 1.4 × 103 
Do not allow a bare 1.5 × 103

B1

  (ii)   correct substitution into F = mv2/r: eg F = (9.7 × 103 × 0.262)/60

C1

          F = 10.9 N

Allow 11 N

A1

[3]

  2.     (i)      THREE correct arrows at A, B and C all pointing towards
the centre (judged by eye)

Ignore starting point of arrow

B1

  (ii)   1.      Greatest reaction force is at C

This is a mandatory M mark. The second mark cannot be gained unless this is scored.

M1

          because it supports weight of sock AND provides the required
upward resultant (centripetal) force (WTTE)

Any indication that candidates think that the centripetal force is a third force loses this second and possibly the next mark. They must make correct reference to the resultant force that provides the required centripetal force/acceleration.

A1

2.       Least at A because sock’s weight provides part of the required
downward resultant (centripetal) force (WTTE)

Allow answers using the equation F = mv2/r
such as Nc – mg (at C) = centripetal force OR mv2/r
OR mg +NA (at A) = centripetal force OR mv2/r

B1

[4]

Moments and the forearm

y12 Question on spider's web

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]