y13 answers london eye shm

y12 mock

y12 work, and springs

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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

[9]

Answers to questions on Young's modulus

Questions on Young's modulus

Calculations on stress, strain and the Young modulus




Practice questions

These are provided so that you become more confident with the quantities involved, and with the large and small numbers.



Try these

A strip of rubber originally 75 mm long is stretched until it is 100 mm long.

1. What is the tensile strain?

2. Why has the answer no units?

3. The greatest tensile stress which steel of a particular sort can withstand without breaking is about 109 N m-2. A wire of cross-sectional area 0.01 mm2 is made of this steel. What is the greatest force that it can withstand?

4. Find the minimum diameter of an alloy cable, tensile strength 75 MPa, needed to support a load of 15 kN.

5. Calculate the tensile stress in a suspension bridge supporting cable, of diameter of 50 mm, which pulls up on the roadway with a force of 4 kN.

6. Calculate the tensile stress in a nylon fishing line of diameter 0.36 mm which a fish is pulling with a force of 20 N

7. A large crane has a steel lifting cable of diameter 36 mm. The steel used has a Young modulus of 200 GPa. When the crane is used to lift 20 kN, the unstretched cable length is 25.0 m. Calculate the extension of the cable.


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.

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

II.

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

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



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



VI.

VII. Use your answers to parts 5 and 6 to calculate the Young 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.

I. Calculate the strain of the wire.







II. Calculate the stress in the wire.







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

Y12 Homework on Sankey Diagrams

Answers to calculations only.

Constructing Sankey Diagrams

1. An energy efficient light bulb is rated at 20W. It produces 5W of light. Calculate its efficiency and draw a Sankey Diagram to scale.

Efficiency = Useful power out/ power in x 100%

Eff= (5/20) x 100 = 25%

2. Paula transfers 40 000J of chemical energy during a race. She transfers 32 000J of heat energy to the surroundings during the race. Calculate her efficiency and draw a Sankey diagram to scale.

Efficiency = useful energy out/ energy in x 100%

Eff = 40 000 -32 000) /40000 = 8000/40000 = 0.2 =20%

3. Bradley does 1600J of work turning the pedals on his bike. 1577.6J is transferred to the rear sprockets. How much heat is lost and what is the efficiency of Bradley’s bike. Draw a Sankey diagram of Bradley’s chain

Efficiency = useful energy out/ energy in x 100%

Eff = 1577.6 / 1600 = 0.986 = 98.6%

4. The coal in Thomas’ boiler contains 40 kJ of energy. He loses 25.76kJ as heat as he puffs along a branch line. How efficient is Thomas and draw a Sankey diagram.

Efficiency = useful energy out/ energy in x 100%

Eff = (40 - 25.76)/40 = 0.356 = 35.6%

5. The fuel in Diesel’s tank contains 60kJ of chemical energy. He does 20.1kJ of work on the mainline. Calculate Diesel’s efficiency and draw a Sankey diagram.

Efficiency = useful energy out/ energy in x 100%

Eff = 20.1/60 = 0335 = 33.5%

6. The electric engine in Jeremy’s car can develop 3kW. If the car develops 2.658kW what is its efficiency.

 Efficiency = Useful power out/ power in x 100%

Eff = 2.658/3 = 0.886 = 88.6%

7. Jeremy dreams of a Ferrari Enzo which can develop a maximum of 700 bhp. (1 bhp = 750 Watts). Sadly for Jeremy cars are not very efficient. Typically, only about 30% of the energy that is available from the combustion of the petrol actually ends up overcoming friction to move the car forwards. Of the 70% of energy is that is not usefully converted, 55% may heat the cooling water that surrounds the engine block whilst  15% may be in the hot exhaust gases. To make car engines more efficient  the fuel has to burn at a higher temperature and the exhaust must be kept cooler. Draw a Sankey diagram of Jeremy’s dinosaur

y13 Homework Gravity

y13 test on Gravity

Y13 Gravity