E Fields 3

Electron in E Field

Frog

Pressure Answers

Experiment with Scales and Charged Rods

E Filed Calculations

E Field and a Bunsen Burner

Gravitational Field Lines

Gravitational Field Strength

G and Centripetal Force

Inverse Square Law

Earth - moon

Young's Modulus - Past questions

     (a)     (i)      Stress = force / area                                                                              C1

force = stress ´ area

 = 180 ´ 106 ´ 1.5 ´ 10–4

 = 27000 (N)                                                                                 A1

(ii)     Y M = stress / strain                                                                              C1

= 180 ´ 106 / 1.2 ´ 10–3 or using the gradient                               C1

= 1.5 ´ 1011 N m–2                                                                       A1

  (b)   brittle

elastic/ graph shown up to elastic limit

obeys Hooke’s law / force α extension / stress α strain

no plastic region                                                                                             B3

MAX 3

[8]

     cast iron:     brittle
                   brittle explained as having no plastic region
                   elastic
                   elastic explained as returning to original length when
                   the load is removed / linear graph / Hooke’s law obeyed
                   or equivalent words                                                                       MAX 3

          copper:       ductile
                   ductile explained as can be formed into a wire
                   initially elastic
                   plastic where it stretches more and more with little
                   increase in stress
                   plastic explained as does not return to its original length
                   when the load is removed
                   reference to necking at the end                                                      MAX 3

          polythene:  easy to deform / deformed with a small force
                   plastic
                   ductile
                   polymeric                                                                                      MAX 2

                                                                                                                                  MAX 8

          QWC:        spelling, punctuation and grammar                                                    B1
                   organisation and logic                                                                         B1

[10]

Gravity Log graph Kepler 3

A bit small for the Sun - I missed a zero off should be 10 to the power of 30

Expression for g

y12 homework on Elastic Energy

Gravity questions

 (a)     arrow towards centre of planet                                                                                   1

(b)     (i)      g = GM/R2                                                                                               1

(ii)     gS/gO = R2/25R2; gS = 40/25 (= 1.6 N kg–1)                                             2

(iii)    gC/gO = R2/16R2 giving gS = 40/16 (= 2.5 N kg–1)                                    1

(iv)    average g = (2.5 + 1.6)/2 = 2.(05) (1)
Δp.e. (= mgavR) = 3.0 × 103 × 2.05 × 2.0 × 107; = 1.2 × 1011 (J) (2)       3

  (c)   g = v2/r; = 4π2(5R)/T2 (2)
1.6 = 4 × 9.87 × 1.0 × 108/T2 giving T2 = 24.7 × 108 and T = 5.0 × 104 (s) (2) 4

[12]

Springs Questions

Q1

(a)     One reading from the graph e.g. 1.0 N causes 7 mm                                               C1

Hence 5.0 (N) causes 35 ± 0.5 (mm)                                                             A1

(allow one mark for 35 ± 1 (mm)

(b)     (i)      Force on each spring is 2.5 (N)                                                             C1

extension = 17.5 (mm) allow 18 (mm) or reading from graph              A1

[allow ecf from (a)]

(ii)     strain energy = area under graph / ½ F ´ e                                            C1

                    = 2 ´ 0.5 ´ 2.5 ´ 17.5 ´ 10–3

                    = 0.044 (J)                                                                        A1

[allow ecf from (b)(i)]

Q2

(a)     (i)      F = kx / k is the gradient of the graph                                                             C1

k = 2.0 / 250 ´ 10–3 = 8.0                                                                     A1

Correct unit for value given in (a)(i)

i.e. 0.008 or 8 ´ 10–3 requires N mm–1.

Allow N m–1 / kg s–2 if no working in (a)(i).

Do not allow unit mark if incorrect physics in part (a)(i)                      B1

(ii)     W = ½ (F ´ extension) / area under the graph                                       C1

          = ½ ´ 2.0 ´ 0.250

          = 0.25 (J)                                                                                    A1

  (b)   (i)      F = 8 ´ 0.15 = 1.2 (N)                                                                           A1

(ii)     Hooke’s law continues to be obeyed / graph continues as a straight

line / k is constant / elastic limit has not been reached                          B1

  (c)    (i)      1.     correct time marked on the graph with a V (t = 0.75 s or 1.75 s)   B1

2.       tangent in the correct place for downward velocity or implied
by values                                                                                     B1

          value between 0.95 to 1.1(m s–1)                                                A1

(ii)     1.     X marked in a correct place (maximum or minimum on graph)   M1

2.       relates the extension / compression to F = kx to explain why the

          force is a maximum or maximum extension gives max force or
maximum extension gives max acceleration                               A1

Q3

   (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

Stretching

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