y13 answers london eye shm

y12 mock

y12 work, and springs

y13 questions on pendulum bobsleigh and geostationary orbit

y13 capacitors in Series and Parallel answers

y13 Shm questions

Y12 Questions on Energy and work

I have put in ^ to indicate a superscript (power)

1. (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

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2. (a) (i) speed = d / t C1

= 24 / 55

= 0.436 (m s–1) allow 0.44 A1

do not allow one sf

(ii) kinetic energy = ½ m v2 C1

= 0.5 x 20 x (0.436)^2

= 1.9 (J) note ecf from (a)(i) A1

(iii) potential energy = mg h C1

= 20 x 9.8 x 4

= 784 (J) A1

penalise the use of g = 10

(b) (i) power = energy / time or work done / time C1

= (15 x 784) / 55

note ecf from (a)(iii)

= 214 (W) A1

(ii) needs to supply children with kinetic energy B1

air resistance B1

friction in the bearings of the rollers / belt B1

total mass of children gives an average mass of greater than 20 kg B1

Max B2

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y12 Past questions on Hooke's Law and Young's Modulus

Note 1.2 x 10^4 means 1.2 times ten to the power of 4 (etc)
 
1. (a) (i) Stress = force / area C1


force = stress x area

= 180 x 10 ^ 6 x 1.5 x 10 ^ –4

= 27000 (N) A1

(ii) Y M = stress / strain C1

= 180 x 10^6 / 1.2 x 10^–3 or using the gradient C1

= 1.5 x 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

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2. (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 x e C1

= 2 x 0.5 x 2.5 x 17.5 x 10 ^–3

= 0.044 (J) A1

[allow ecf from (b)(i)]

(c) E = stress / strain C1

Stress = force / area and strain = extension / length C1

extension = (F x L) / (A x E)

= (5 x 0.4) / (2 x10^–7 x 2 x 10^11)

= 5.(0) ^ 10–5 (m) A1

(d) strain energy is larger in the spring B1

extension is (very much larger) (for the same force) for the spring B1

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

k = 2.0 / 250 x 10^–3 = 8.0 A1

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

i.e. 0.008 or 8 x 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 x extension) / area under the graph C1

= ½ x 2.0 x 0.250

= 0.25 (J) A1

(b) (i) F = 8 x 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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4. 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

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5. (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 prop v. M1

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