Markscheme for questions on Springs

 

1.       (a)     The graph shows length and not extension of the spring / spring has
original length (of 2.0 cm) (AW)

Allow: ‘length cannot be zero’

B1

  (b)   Straight line (graph) / linear graph / force µ extension / constant
gradient (graph)

Not ‘force µ length’

B1

  (c)    force constant = 2.0/0.04

Note: The mark is for any correct substitution

C1

          force constant = 50 (N m–1)

Allow: 1 mark for 0.5 (N m–1) – 10n error
Allow 1 mark for 5/12 × 10–2 = 41.7 or 4/10 × 10–2 = 40 or
                          3/8 × 10–2 = 37.5 or 2/6 × 10–2 = 33.3 or
                          1/4 × 10–2 = 25

A1

  (d)   work done = 1/2Fx or 1/2kx2 or ‘area under graph’

C1

          work done = 1/2 × 3.0 × 0.06 or 1/2 × 50 × 0.062

Possible ecf

          work done = 0.09 (J)

Note: 1 sf answer is allowed

A1

  (e)    Find the gradient / slope (of the tangent / graph)

B1

          Maximum speed at 1.0s / 3.0s / 5.0s / steepest ‘part’
of graph / displacement = 0

Allow: 2 marks for ‘steepest / maximum gradient’

B1

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  2.     (i)      It has maximum / large / increased stress at this point

Allow: it has ‘same force but thinner/smaller area’
Not: Thin / small area

B1

(ii)     The tape has (permanent) extension / deformation when the
force / stress is removed (AW)

Note: Need reference to force or stress removed
Allow: ‘…does not return to original size / shape / length when force / stress is removed’

B1

[2]

 

  3.     Copper

B1

[1]

 

  4.     extension (or compression) ∞ force (as long as
elastic limit is not exceeded)

Allow: ‘load’ instead of force
Not: x ∞ F, unless the labels are defined

[1]

 

  5.     (i)      force = 75 × 0.085

C1

          F = 6.38 (N) ≈ 6.4 (N)

A1

(ii)     acceleration = 6.38/2.5 x 10^-3
acceleration = 2550 (m s–2)

Note: a = (kx-mg)/m  gives 2540 (m s–2)

Possible ecf

B1

  (iii)  Correct selection of equation: mgh / 1/2x2 not equal to 1/2Fx

C1

          0.0025 × 9.81 × h = 1/2 × 75 × 0.0852

C1

          height = 11 (m)

Note: Bald answer of 11 (m) scores 3/3 marks

A1

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  6.     (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 = 20/12 × 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)
1/2 kx2  = 1/2mv2                                                                                    M1
m and k are constant, therefore x µ v.                                                      M1

[9]

 

  7.     (i)      Graph through origin with (short) linear section then reducing gradient.                1

(ii)     Straight section - elastic; (1)
Curved section - plastic. (1)                                                                                2

[3]

 

  8.     (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

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

[9]

 

  9.     (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)]

 

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  10.   (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

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