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AQA Physics A Level Unofficial Mark Scheme, 7408/1, 4 June 2018

Will start it below :smile:

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441 Hz for new frequency of string (I think)
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Avatar for Integer123
Integer123
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AQA Physics A Level Paper 1, Unofficial Mark Scheme

Please feel free to help and correct the mark scheme if it's wrong :smile:

Written questions (60 marks)

01

An engineer produces a graph which predicts the speed of a vehicle, whose brakes are faulty, over a certain time. The vehicle has a mass of 1.8×104  kg1.8\times 10^4\;\text{kg}.

01.1 Find the decelerating force on the vehicle, after 2.0 s. [3 marks]

Use Newton's second law F=maF=ma.

Acceleration is gradient of tangent at 2.0 s, I think I got −0.293  m s−2-0.293\;\text{m s}^{-2}.

Using Newton's second law, the magnitude of the decelerating force is F=5300  NF=5300\;\text{N}.

01.2 Deduce whether an 85 m track will be long enough. [3 marks]

Total distance travelled = area under speed-time graph

Counting squares: approximately 23 squares of dimensions 2.5  m s−22.5\;\text{m s}^{-2} by 1  s1\;\text{s}.

So distance travelled is about 58 m.

Hence an 85 m track is long enough.

01.3 Discuss the energy changes as the vehicle moves along the track. [2 marks]

The kinetic energy of the vehicle is transferred into thermal energy of the surroundings by resistive forces, air resistance and friction

01.4 Determine whether a ramp of length 85 m at angle of 25 degrees to the horizontal is suitable [3 marks]

By Newton's second law the weight is 1.8×104×9.81=176580  N1.8\times10^4\times 9.81 =176580\;\text{N}. Hence component of weight parallel to slope is 176580sin⁡25=74625.9327  N176580\sin 25 = 74625.9327\;\text{N}.

By Newton's second law, component of acceleration down the slope is therefore 74625.93271.8×104=4.15  N kg−1\frac{74625.9327}{1.8\times 10^4} = 4.15\;\text{N kg}^{-1}

Using SUVAT, since constant acceleration, the distance travelled when the vehicle stops is v2−u22a=02−17.522×−4.15=37  m\displaystyle \frac{v^2-u^2}{2a}= \frac{0^2-17.5^2}{2\times -4.15} = 37\;\text{m} (a negative acceleration is used since the force, and hence the acceleration by Newton's second law F=maF = ma, is in the opposite direction to velocity; and u=17.5  m s−1u=17.5\;\text{m s}^{-1} was read off the graph).

Hence the ramp is sufficient.

01.5 Discuss whether a track or ramp is safer [1 mark]

There are probably other valid points to put here

A track is safer because on a ramp, when the vehicle stops, it may begin to roll down the hill

A ramp is safer because the gravel could mean the driver loses control of the car

02

The de Broglie wavelength, λ\lambda, and the velocity, vv, of some wavelengths is measured.

02.1 Show that the data are consistent with the relationship λ∝1v\lambda \propto \frac{1}{v}. [2 marks]

If de Broglie wavelength is inversely proportional to wavelength, then λv=constant\lambda v = \text{constant}.

For two data pairs, the constant was 7.35×10−4  Jkg−17.35\times 10^{-4}\;\mathrm{J kg}^{-1} and for the other pair, the constant was 7.25×10−4  Jkg−17.25\times 10^{-4}\;\mathrm{J kg}^{-1}. Since these constants are approximately equal, the inverse proportionality is supported by the data.

02.2 Calculate a value for Planck's constant using the data [2 marks]

de Broglie wavelength, λ\lambda, and Planck's constant, hh, are related by λ=hmv⇒h=λmv\displaystyle \lambda = \frac{h}{mv} \Rightarrow h = \lambda mv

Using λv=7.25×10−4  J kg−1\lambda v = 7.25\times 10^{-4}\;\text{J kg}^{-1}, h=6.60×10−34  J sh = 6.60\times 10^{-34}\;\text{J s}

Using the average of the de Broglie wavelengths and the velocities gives h=6.67×10−34  J sh = 6.67 \times 10^{-34}\;\text{J s}.

Using λv=7.35×10−4  J kg−1\lambda v = 7.35\times 10^{-4}\;\text{J kg}^{-1}, h=6.70×10−34  J sh = 6.70\times 10^{-34}\;\text{J s}

02.3 Explain how the screen shows that electrons exhibit a wave behaviour [3 marks]

If the electrons exhibited particle behaviour, they would be deflected, or pass straight through the grating, and so a single spot would be seen

However, an interference pattern is seen

Which means the electrons must have diffracted around the atoms in the grating, and so exhibit wave behaviour

02.4 Explain how the light emitted shows that electrons exhibit a particle behaviour [3 marks]

For atoms in screen to emit light, the atoms inside must be excited

If electrons were acting as waves, then a specific energy is required to excite the atoms (which would be very unlikely to happen, and so no light would be emitted

Hence electrons must behave as particles as they transfer enough energy to move the orbital electrons to a higher energy level, and the remaining energy is kinetic energy of the colliding electrons

03

Stationary waves of frequency 370 Hz at a tension of 25 N after formed on a string of a violin, which consists of strings fixed at an end A.

03.1 Explain how stationary waves are formed on the string [3 marks]

A progressive wave travels down the string
And reflects off the end A
These two waves superpose
They have the same frequency and wavelength

03.2 Show that the mass of a 1.0 m length of string is 4.0×10−4  kg4.0\times 10^{-4} \;\text{kg} [2 marks]

f=12lTμ⇒μ=T(2lf)2\displaystyle f=\frac{1}{2l}\sqrt{\frac{T}{\mu}} \Rightarrow \mu=\frac{T}{(2lf)^2}

μ\mu is the mass per unit length, i.e. mass of a 1.0 m length

μ=25(2×0.33×370)2=4.2×10−4  kg m−1\mu = \frac{25}{(2 \times 0.33 \times 370)^2} = 4.2 \times 10^{-4}\;\text{kg m}^{-1}

So mass of a 1.0 m length of string is 4.2×10−4  kg4.2\times 10^{-4} \;\text{kg}

03.3 Calculate the speed of the wave [1 mark]

v=fλ=370×(2×0.33)=240  m s−1v = f \lambda = 370 \times (2 \times 0.33) = 240\;\text{m s} ^{-1}

03.4 Find the new frequency of the string [4 marks]

When tension is 25 N, extension is ? (read off the graph).

Radius of turning thing is 3.51 mm.

Gain in extension is rθ=3.51×10−3×75180×π=4.59  mmr\theta = 3.51\times 10^{-3} \times \frac{75}{180}\times \pi = 4.59\;\text{mm}

Add the extension when the tension is 25 N to this extension. Then look on the graph to find the new tension, in Newtons.

Hence new frequency, using f=12lTμ\displaystyle f=\frac{1}{2l}\sqrt{\frac{T}{\mu}}, is 440  Hz440\;\text{Hz} (or 441 Hz to 3sf)

04

A circuit consists of a 0.25  kΩ0.25\;\text{k}\Omega fixed resistor, a thermistor with a resistance of 750Ω750 \Omega at room temperature, as well as a variable resistor R Ω\text{R } \Omega.

04.1 The wire used in the 0.25  kΩ0.25\;\text{k}\Omega resistor has a resistivity of x  Ωmx\;\Omega\text{m}. 50 turns of the wire are wrapped around a cylinder of diameter y  mmy\;\text{mm}. Calculate the cross-sectional area of the wire [3 marks]

ρ=RAL⇒A=ρLR\rho = \frac{RA}{L} \Rightarrow A=\frac{\rho L}{R}

L=π×d×50L = \pi \times d \times 50

04.2 The 0.25  kΩ0.25\;\text{k}\Omega resistor is rated with a power of 0.36  W0.36 \;\text{W}. Deduce whether the 0.25  kΩ0.25\;\text{k}\Omega resistor is suitable for the circuit. [2 marks]

P=V2R⇒V=RP=9.49  V\displaystyle P=\frac{V^2}{R} \Rightarrow V = \sqrt{RP} = 9.49\;\text{V} and so the resistor is not suitable since when operating at maximum power, the voltage across it must be larger than the voltage of the voltage source.

04.3 Calculate the resistance of R. Give your answer to two significant figures [5 marks]

Voltage across parallel section is 5.0 V

Using potential dividers, and sharing the voltage according to resistance, 9RPRP+250=5⇒RP=0.3125  Ω\displaystyle \frac{9R_P}{R_P + 250} = 5 \Rightarrow R_P = 0.3125\;\Omega.

Hence 1RP=3.2  Ω−1⇒1R+1750=3.2  Ω−1\displaystyle \frac{1}{R_P} = 3.2\; \Omega ^{-1} \Rightarrow \frac{1}{R}+\frac{1}{750} = 3.2 \;\Omega ^{-1}.

Hence R=540  ΩR = 540 \;\Omega, to two significant figures.

04.4 Explain what happens to the output voltage as the thermistor's temperature increases [2 marks]

As the temperature of the thermistor increases, the resistance across it decreases.

Hence the resistance of the parallel section decreases, and its share of the voltage is lower - therefore a lower output voltage.

05

Two acrobats jump off a platform. They move in a horizontal circle with a time period of 5.2 s, 5 m below the platform, with their ropes making an angle of 28.5 degrees with the vertical.

05.1 Show that the linear speed is about 4.5  m s−14.5\;\text{m s}^{-1}. [2 marks]

v=ωrv = \omega r

ω=2π5.2=1.21  rad s−1\omega = \frac{2\pi}{5.2} = 1.21\;\text{rad s}^{-1}

r=5tan⁡28.5+1=3.714  mr=5\tan 28.5 + 1 = 3.714\;\text{m}

v=4.49m s−1v=4.49 \text{m s}^{-1}

05.2 Find the tension in the cable supporting the acrobats [3 marks]

For this question there's a bit of a debate over whether to resolve horizontally or vertically, so I've included both methods

By Newton's second law W=85×9.81=833.85  NW = 85 \times 9.81 = 833.85\;\text{N}

Resolving vertically, Tcos⁡28.5=833.85⇒T=950  NT \cos 28.5 = 833.85 \Rightarrow T = 950\;\text{N}

Other answer: by Newton's second law, centripetal force is mω2r=85×9.81=461  Nm\omega^2r = 85 \times 9.81 = 461\;\text{N}

Resolving horizontally, Tsin⁡28.5=461⇒T=970  NT \sin 28.5 = 461 \Rightarrow T = 970\;\text{N}.

05.3 Explain what happens to the forces on the pole if the mass of one acrobat is significantly larger than the other [3 marks]

There is a larger downwards force on the platform and hence pole

By Newton's third law, the ground will exert a larger support force on the pole

This large downwards force acts on one side of the platform and so will create a much larger clockwise/anticlockwise moment

The pole may hence topple

06

Two trains are moving towards each other. A has a mass of 16000 kg and a speed of 2.8  m s−12.8\;\text{m s}^{-1} and B has a mass of 12000 kg and a speed of 3.1  m s−13.1\;\text{m s}^{-1}.

06.1 State one quantity not conserved in an inelastic collision. [1 mark]

Kinetic energy

06.2 Show that the speed the combined trains move with after collision is
Unparseable latex formula:

0.3\; \text{m s}^{-1}}

[3 marks]

Assume moving east has a positive velocity, so moving west has a negative velocity.

Conservation of momentum 16000×2.8−12000×3.1=28000×v16000\times 2.8 - 12000 \times 3.1 = 28000 \times v

v=0.27  m s−1v = 0.27\;\text{m s}^{-1}

06.3 Calculate the impulse exerted on each truck, give a unit. [2 marks]

Newton's second law F=Δ(mv)Δt⇒FΔt=Δ(mv)F = \frac{\Delta (mv)}{\Delta t} \Rightarrow F\Delta t = \Delta(mv)

Trains move off east (as answer to 06.1 was a positive velocity, and the method took moving east as being a positive velocity) so impulse is 4.0×104  N s4.0\times10^4\;\text{N s}.

06.4 Explain what happens if the collision is completely elastic [2 marks]

If completely elastic, kinetic energy is conserved

The trains rebound off each other and move in the opposite direction to each other ([url="https://en.wikipedia.org/wiki/Elastic_collision);"]https://en.wikipedia.org/wiki/Elastic_collision) and train B has a higher speed

Multiple choice questions (25 marks)

07 What is the unit for the quantity given by the area under a force-time graph? [1 mark]

kg m s−1\text{kg m s}^{-1} since the unit is N s\text{N s} (not sure which option)

08 Which gives the greatest uncertainty? [1 mark]

Diameter (I think option C)

26 How long will the battery last for? [1 mark]

48 hours (I think option B)

30 What is the maximum kinetic energy of the oscillating item? [1 mark]

31 Which graph shows the gravitational potential energy of an object performing simple harmonic motion? [1 mark]

The one which resembled the letter U (I think option D)

Ones I can't remember numbers for:

? A current of x  Ax \;\text{A} passes, how many doubly charged ions pass in 1 minute? [1 mark]

Can't remember the answer

? What is the specific charge of a fluoride ion, an atom of the element 919F^{19}_9\text{F} has lost one electron [1 mark]

5.0×106  C5.0\times 10^6\;\text{C} (not sure which option)

? Einstein's theory of special relativity: What is the minimum frequency of a gamma photon, when two photons are released from the annihilation of a muon and an antimuon? [1 mark]

I think 2.5×1016  Hz2.5\times 10^{16}\;\text{Hz} (not sure which option)

? Hooke's law: what is the extension in spring Q? [1 mark]

0.061 m (option D)

? Photons are incidient with an energy of x  Jx\;\text{J}. How many possible wavelengths of emitted EM radiation are there? [1 mark]

3

? Newton's second law: find the thrust of the rocket [1 mark]

I think it was 1.3×105  N1.3\times 10^5 \;\text{N}, I used Newton's second law to get F=m(9.81+a)F=m(9.81+a), for the mass and acceleration I can't remember properly (not sure which option)

? Which set of forces cannot have a zero resultant? [1 mark]

3 N, 6 N, 10 N (I think option D)

? What is the combined resistance in Ω\Omega of the two putty cylinders? [1 mark]

5R  Ω5R\; \Omega (not sure which option)

? What is the maximum kinetic energy of photons emitted from a metal surface with work function 4.1 eV, when photons of 290 mm wavelength are incident?

0.19 eV (option A)

? How to measure the voltage across a component? [1 mark]

Voltmeter across the component, in parallel, with infinite resistance (not sure which option)

? Young modulus: what change in the wire is required for an extension of e4\displaystyle \frac{e}{4} [1 mark]

Radius: 4r4r, length 4l4l (I think option B)

? Young's double slit experiment: what change will increase fringe spacing? [1 mark]

Decreasing slit separation (I think option D)

? What is the correct path of the ray? [1 mark]

The one with total internal reflection (I think option D)

? What is the refractive index of the glass [1 mark]

1.67 (not sure which option)

? Identify the quark composition of particle XX in the equation [1 mark]

us‾\text{u}\overline{\text{s}} (not sure which option)

? What is a property about the missing particle in this beta minus decay equation? [1 mark]

It is an antiparticle (I think option B)

? How long does the projectile's journey take? [1 mark]

2.9 s (I think option C)

? Diffraction grating: what was the wavelength of light used? [1 mark]

590 nm (not sure which option)

? What is the phase difference between displacement and acceleration for an object oscillating with simple harmonic motion? [1 mark]

π\pi radians (not sure which option)

? What is the path difference of the waves? [1 mark]

13.3λ13.3\lambda (not sure which option)
(edited 5 years ago)
Reply 3
Avatar for Dat1Guy
Dat1Guy
16
The first question, find the force; acceleration was 2.5 and mass 1.8 x 10^4 so force was 4.5 ×10^4 N
2nd part, 22.5 small boxes, each had 2.5 metres per box, so about 56 metres to stop, hence less than the 85m
Reply 4
Avatar for Dat1Guy
Dat1Guy
16
Original post by mattycoates
441 Hz for new frequency of string (I think)


440 had to be 2sf
the circuits question;

a) dont remember what it was?
b) the resistor was suitable when it took 4V (9v cell - 5v output) by P = V^2 / R
c) 540 was the value of R, only took me about a million tries but im certain now
d) output voltage would decrease, it was a standard potential divider question
Reply 6
Avatar for usf99
usf99
3
Original post by Integer123
Q1

(a) Find the decelerating force on the vehicle, after 2.0 s. [3 marks]

Use Newton's second law F=maF=ma.

Acceleration is gradient of tangent at 2.0 s.

F=5300  NF=5300\;\text{N}.


couldn't u do f=mv-mu/t
Original post by BDunlop
440 had to be 2sf


oh dear, does that mean I'll drop 1 mark every question for doing 3sf 😬
Reply 8
Avatar for Quizlet
Quizlet
19
Original post by mattycoates
oh dear, does that mean I'll drop 1 mark every question for doing 3sf 😬


No. You're only penalised once if the question says for it to be to a specific sf or appropriate sf. You'll only lose a maximum of 1 mark in the whole paper.
Q1.3) Describe the energy changes that occur as the car is slowed by the escape road (gravel).

Q1.4) Another type of escape road is an escape ramp, which is a road with a positive incline.
For an escape ramp of length 85m and 35 degrees, deduce whether or not the ramp will stop the car.

I tried relating Ep = mgh and Ke = 1/2 mv^2 for this and the result was that the ramp WOULD stop the car, but I am not guaranteeing this is correct.

Q1.5) Discuss which out of the escape ramp or the escape road would provide a safer stopping experience to the driver.

Here I spoke about the potential of the car to roll backwards down the escape ramp when it reaches the top, which is a danger.
What did people get for the distance travelled by the vehicle in the loose gravel?
How did you work out R?
Original post by usf99
couldn't u do f=mv-mu/t


it wasnt linear deceleration, the question asked for force at that moment not the average force
Original post by BDunlop
What did people get for the distance travelled by the vehicle in the loose gravel?


I got 55.6m by working out the area under the graph. It probably should’ve been 2 s.f though
Original post by usf99
couldn't u do f=mv-mu/t


I believe not as this is impulse force, I tried this myself and got around 70kN which seems unreasonable but did not try finding a tangent of the graph x-(
Original post by MathsyTsrUser
I got 55.6m by working out the area under the graph. It probably should’ve been 2 s.f though


I got that too to 2sf :smile:
Original post by BDunlop
What did people get for the distance travelled by the vehicle in the loose gravel?


did everyone calculate it? i just drew a box with area 85 and showed that its definitely bigger than the area under the curve
Reply 17
Avatar for Axel004
Axel004
8
Pretty sure only question which specified Sig figs was the resistance 5 marker so for the frequency anything was fine
Original post by MathsyTsrUser
How did you work out R?


parallel resistance of R and the thermistor:
1/pr = 1/R + 1/750

then 5v = 9v * pr/(250+pr), solve for R
it was absolutely horrendous, the other way to go about it was
4v = 9v * 250/(250+pr) but still pretty bad
Q4) Question about two acrobats on ropes attached to a spindle.
(Show that the linear velocity of each acrobat was about 4.5ms^-1)

You had to here use trig (or pythagoras I cant remember) to find the distance from the acrobat and the edge of the spindle.
The diameter of the spindle they are attached to was 2.0m, so you had to half this and add it to the distance between the spindle and acrobat to get around 3.714.
Multiplying the angular velocity by the radius gives you a linear vel. of about 4.5ms^-1

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