ANSWERS TO QUESTIONS ON THE WHOLE CHAPTER · Salters Horners AS/A level Physics 1 Additional Sheet...

Salters Horners AS/A level Physics 1 Additional Sheet 11

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© 2015 University of York, developed by University of York Science Education Group. This sheet may have been altered from the original. of 2

Chapter 1

HFS

ANSWERS TO QUESTIONS ON THE WHOLE CHAPTER

Q 55 One approach is to use energy conservation:

Ek lost = work done on ball = F x

F = x

mv

2

2

= m) 0.4 (2

)s m (15 kg 0.160

21

= 45 N (Alternatively, you could find the acceleration using Equation 7 and then use Equation 8 to

find the force.) Q 56 Using Equation 7, v 2 = u 2 + 2as, with a = –9.81 m s–2, v = 0 and s = +1.20 m, gives

u = √(2 × 9.81 m s–2 × 1.20 m) ≈ 5 m s–1 Q 57 You first need to estimate the mass m of his fist and arm. If you take this mass to be 5 kg, then,

using Equation 23,

Ek = 21 mv2 = 40 J

v =

m

Ek2=

kg

J 40

5

2= 4 m s–1

Q 58 (a) Initial pcue = mcueucue = 0.45 kg × 0.95 m s–1 = 0.43 kg m s–1 The ball must be initially at rest, so its momentum is zero. Initial total momentum = 0.43 kg m s–1 ≈ 0.4 kg m s–1 (b) Final pcue = 0.45 kg × 0.4 m s–1 = 0.18 kg m s–1 Final pball = 0.43 kg m s–1 – 0.18 kg m s–1 = 0.25 kg m s–1

vball = pball/mball = 0.25 kg m s–1/0.16 kg = 1.56 m s–1 (1.6 m s–1) Q 59 Moment due to Ingrid’s weight = 300 N × 1 m = 300 N m Agnetha and Novalie must provide a moment of 300 N m in the opposite direction – they must

sit on the opposite side of the pivot. If their weights are each W (= 300 N) and their distances from the pivot are xA and xN,

300 N = WxA + WxN = W(xA + xN) so xA + xN = 1 m Any number of combinations will produce equilibrium; for example, they could sit 0.6 m and

0.4 m from the pivot, 0.7 m and 0.3 m, and so on. Q 60 It is useful first to draw a diagram as in Figure 11.1. If a skier is pulled at constant velocity,

then the force F exerted upwards along the slope must be equal and opposite to the component of the skier’s weight that acts downwards along the slope; that is, F = W cos, where the angle between the weight and velocity is = 70°.

Figure 11.1 Diagram for answer to Question 60

1



© 2015 University of York, developed by University of York Science Education Group. This sheet may have been altered from the original. of 2

Chapter 1

HFS

For one skier: weight = mg ≈ 650 N acting downwards (g ≈ 10 N kg–1).

Using Equation 24 to get P = Fv cos,

P = 650 N × 3 m s–1 × cos 70°

= 667 W

For one hundred skiers P = 66 700 W

= 66.7 kW

Q 61 (a) Ek = 21 mv2 = 2

1 × 20 × 103 kg × (50 m s–1)2 = 2.5 × 107 J

(b) Assuming all Ek is transferred as Eel into the wire, then we can use Equation 27:

Eel = 21 kx2 = Ek

so

x =

k

Ek2

=

18

7

1033

10522

mN.

J .

= 0.39 m (≈ 0.4 m)

Q 62 (a) Displacement vertically y = 1.2 m

Using Equation 28,

y = 21 gt2

so

t =

g

y2

=

2

212

s m 10

m .

= 0.49 s

Using Equation 29,

u = t

x

= s 0.49

m 4.0

≈ 8 m s–1

(b) The actual speed on leaving the road is likely to have been greater because the car would have slowed while skidding. Also, the motion of the glass would be affected by air resistance, reducing its ux while it was in motion.

2



© 2015 University of York, developed by University of York Science Education Group. This sheet may have been altered from the original. Page 1 of 1

Chapter 2

MUS


Q 49 (a) 300 nm.

(b) See Figure 9.1.

Figure 9.1 Answer to Question 49

Q 50 Ruari is right that light can travel through space (a near-vacuum).

But he is wrong about sound, since that requires a material in which to set up compressions and rarefactions. Rachel is right: the sounds in films such as Star Wars do not represent reality.

Q 51 (a) (i) The waves add with crest + crest and trough + trough to give a large amplitude.

(ii) The waves must be (exactly) in phase.

(b) The wave has shifted by 2

so crests become troughs and vice versa.

(c) There must be a path difference of an odd number of half wavelengths; that is,

2d = 2

)( c12 n =

2

1n c

where c is the wavelength in the cuticle.

(d) As the angle is increased, the path difference between adjacent layers changes, so light of a different wavelength will satisfy the condition for constructive interference.

Q 52 (a) Using Equation 7, f = E/h, we have

highest to lowest level: f = sJ1063.6

J1050.334

18

= 5.28 × 1015 Hz

middle to lowest level: f = sJ.

J.34

18

10636

10952

= 4.45 × 1015 Hz

highest to middle level: f = sJ.

J)..(34

18

10636

10952503

= 8.30 × 1014 Hz

(b) From Equation 3, f = c

From Equation 7, E = hf = hc

= m

sm.sJ.9

1834

10447

1000310636

= 4.45 × 10–19 J

3




Chapter 3

EAT


Q 28 (a) (i) 10.4 ± 0.4 mm

(ii) 3.9% for both.

(iii) Volume V = 582.21 mm3

= 0.5822 cm3

(b) 9.0% for both.

(c) Mass of 1 ball m = 1.116 g

Density = 3cm0.58221

g1.116 = 1.920 g cm−3

= 1920 kg m−3

(d) (i) Total percentage uncertainty = (3 3.9) + 9 = 20.7%

The actual uncertainty is therefore 0.207 1.920 g cm−3 = 0.4 g cm−3

(ii) As the second figure is uncertain, only the first two figures are significant.

(iii) Density = (1.9 ± 0.4) g cm−3 or (1.9 ± 0.4) 103 kg m−3

Q 29 (a) (i) From division 1 to division 8, time interval

t = 7 2 ms = 14 ms = 14 10−3 s

(ii) The distance is found from

s = ut = 14 10−3 s 335 m s−1 = 4.69 m

This is twice the distance from detector to wall as the sound travels ‘there and back’.

Detector–wall distance = 2.345 m

(b) (i) Uncertainty in time interval = 0.2 2 ms = 0.4 ms

(ii) Percentage uncertainty in time interval =

ms14

ms0.4 100

= 2.86% ≈ 3%

(c) The percentage uncertainty in sound speed is 1

1

10 ms

335 ms

100 = 3.0%

(d) (i) Total percentage uncertainty in calculated distance is 3% + 3% = 6%

So the actual uncertainty in the distance is 0.06 2.345 m = 0.14 m

(ii) As the second figure is uncertain, only the first two figures are significant and the distance should be written as 2.3 ± 0.2 m. (The uncertainty is rounded up in order to include values 0.14 m away from 2.345 m.)

4




Chapter 3

EAT

Q 30 (a) Initially, a net downward force (weight − upthrust) accelerates the object. There is a viscous drag force that is velocity dependent. When it is so large that the net force on the drop becomes zero, there is no further acceleration, that is, the velocity remains constant – this is the terminal velocity.

(b) Use Equation 2 from the Student Book:

drag force + upthrust = weight

6πrv + 3

4 3 gr air=

3

4 3 gr water

(c) v =

r

gr

63

4 3

water =

9

2 2 gr water

(d) Hailstones have a larger radius r, so reach a larger terminal velocity v. Compared with a small raindrop, the drag force on a hailstone is much smaller relative to the weight. (Terminal velocity might not be achieved before the hailstone hits the ground.)

Q 31 (a) Look at the reflective surface through the filter. As you rotate the filter, the light changes its brightness – the amount of light absorbed by the filter varies with orientation, so the light must be polarised.

(b) (i) 1.5 = tan θ and so θ = tan−1 1.5 = 56.3°

(ii) The angle of reflection is also 56.3°

(c) Arrange a number of glass surfaces one below another, but with a small air gap in between, all with their top surfaces such that the light is incident at the Brewster angle. From the first surface, 8% of the light will be reflected and (assuming no absorption) 92% will continue to the next glass surface. From the next surface, a further 8% of the 92% passing through will now be reflected, and so on.

5




Chapter 4

SPC


Q 34 (a) (i) A (ii) B (iii) B, C, D (iv) A, C, E

(b) (i) B, C, D (ii) A, E (iii) B, C, D (iv) A

Q 35 (a) All six batteries should be connected in series.

(b) (i) Total internal resistance = 6 × 0.0065 = 0.039

‘Lost volts’ = I × rtotal = 300 A × 0.039 = 11.7 V

Terminal pd V = E − Ir = V = 72.0 V − 11.7 V = 60.3 V

(ii) In each battery, power transferred P = I2rint

Hence P = (300 A)2 × 0.0065 = 585 W

(iii) The batteries will heat up.

(c) At maximum power transfer to the external load there will also be maximum power transfer to the internal resistance of the batteries (when R = r, the internal and external powers are equal), so the batteries would get very hot, which would waste a lot of energy. It is more efficient to design the system so that R >> r.

Q 36 (a) (i) The total charge passing through the LED or lamp since the time of connection.

(ii) In Figure 4.53(a), charge represented by each large square

Q = 0.05 A × 0.1 s = 0.005 C

If the current between 0.5 and 1.0 s remains at 0.2 A, then there are approximately 46 large squares under the curve, so

Q = 46 × 0.005 C = 0.23 C

(iii) In Figure 4.53(b), charge represented by each large square

Q = 0.2 × 10−3 s × 0.1 × 10−3 A = 2 × 10−8 C

There are approximately 20 squares under the curve, so

Q = 20 × 2 × 10−8 C = 4 × 10−7 C

(b) The current rises from zero in both components when they are first connected. The current produces heating. When the metal filament gets hotter, its resistance increases (it has a positive temperature coefficient of resistance). The increased resistance causes the current to fall. The metal reaches a constant temperature (constant resistance and current) when the rate of heating due to the current is equal to the rate at which energy is transferred to the surroundings.

The resistance of the LED does not increase when it is heated (it is not made from metal), so the current does not fall. Instead, the current rises smoothly to reach a constant value.

(c) The filament bulb is most likely to fail when it is first turned on because the power in the filament will reach a peak. Its temperature will reach a maximum, so this is when the metal is most likely to melt and break. A peak in the current does not occur for the LED, and therefore there is no power peak in the LED.

6




Chapter 4

SPC

Q 37 (a) When the cell is new, total resistance of the circuit R + r = 8.0 Ω + 0.5 Ω = 8.5

Current I = R r

= 8.5

V .012

= 1.4 A.

(b) (i) When cell is new, terminal pd V =R

R r

= 12.0 V ×

58

08

.

.

= 11.3 V

(ii) When terminal pd is 80% of ‘new’ value, V = 0.80 Ω × 11.3 V = 9.0 V

(c) When V has fallen to 9.0 V, I = R

V

= 08

09

.

V.

= 1.1 A (1.129 on calculator)

V = ℰ − Ir, so r = I

V

= 12.0 V 9.0 V

1.1 A

= 2.7 Ω

The internal resistance is significantly more than the original internal resistance, at over five times more. This makes the cell much less efficient, with a larger proportion of the cell’s energy being dissipated owing to heating of the internal resistance rather than being supplied to the external circuit.

Q 38 (a) Incident power Pin = IA

= 1400 W m−2 × 5 × 107 m2

= 7 × 1010 W

Pout is 8% of Pin, i.e.

Pout = 0.08 × 7 × 1010 W

= 5.6 × 109 W

= 5.6 × 106 kW

(b) Number of arrays = kW.

kW6

10

1065

10

= 1.8 × 104

Q 39 Energy = 100 eV

= 100 × 1.67 × 10−19 J

= 1.67 × 10−17 J

ℰ

ℰ

ℰ

7




Chapter 4

SPC

Q 40 6.5 × 10−11 A = 6.5 × 10−11 C s−1

No. of electrons per second = C.

sC.19

111

10671

1056

= 3.9 × 108 s−1.

Q 41 A good answer should refer to Equation 36 (I = nAvq) and should include most of the following points:

Copper and tungsten are both metals, so the mobile charged particles are electrons in each case.

A good conductor will have a large number density n of mobile charged particles,

and/or the movement of the particles will not be much impeded by lattice vibrations, so they will reach a relatively high drift speed v for a given applied pd.

So copper may indeed have a greater number density of electrons, but the difference between the metals could also arise from different lattice effects.

The current I must be the same in the filament and the connecting wires.

The filament has a much smaller cross-sectional area than the copper. Assuming their number densities of electrons are similar, if the electrons had the same speed throughout they would pile up at one junction and leave a deficit of charge at the other, which is unlikely to occur.

Pile-up and deficit would be avoided if the drift speed were greater in the filament than in the connecting wires.

8




Chapter 5

DIG


Q 28 (a) R =l

A

(Equation 4 in Student Book)

A = r2 = 23

2

1010

m.

so

R =23

8

1010

35121051

2

m.

m.m.

= 23.6

(b) (i) New length = 12.35 m 1.05 = 12.97 m

(ii) Area is divided by 1.05 (to keep the same volume), so

new resistance = (1.05)2 old resistance

= 23.6 (1.05)2 = 26.019 = 26

(c) There will be variations in resistance due to changes in the length of the wire, caused by wind or thermal expansion, for example. The resistance of the wire is also likely to vary with temperature.

Q 29 (a) (i) A, +4.5 V; B, 0 V; C, −4.5 V

(ii) Potential divider.

(b) (i) Vd = 25 V =1000

totalV

so Rd = 1000

totalR

Rd = 39 k (or 3.9 104 )

Rp = Rtotal − Rd = 38.961 M ≈ 39 M

(ii) The total resistance between the cable and the rail is increased, so Rd is a smaller fraction of Rtotal and hence the detector voltage would be reduced.

9




Chapter 5

DIG

Q 30 (a) l = 2πr

(b) Each half has resistance R = A

l =

A

lr

The two halves are connected in parallel, so their total resistance is A

r

2

(c) The two parts have lengths 2

r and

2

3 r

totalR

1 =

A

r

2

1

+

A

r

2

31

= r

A2+

r

A

3

2

= r

A

3

8

Rtotal =A

r

8

3

(d) See Figure 8.1. The resistance is zero when the angle is zero, and maximum when the two probes are diametrically opposite. The graph is symmetrical about 180°.

Figure 8.1 Answer to Question 30(d)

(e) A practical limitation is that each resistance corresponds to two different angles either side of 180°, and the system would not be able to distinguish between those two angles.

(f) With θ in radians, the two parts have resistance A

r and

A

r )( 2

totalR

1 =

r

A +

)( 2r

A

=)2(

)2(

r

AA

=)(

2

2

r

A

Rtotal =A

r

2

2 )(

= θ(2π – θ) × A

r

2

10




Chapter 5

DIG

Q 31 (a) B, C and E

(b) A and D

(c) Waves are noticeably diffracted by objects that are comparable in size to their wavelength, which blurs the image and reduces the resolution. When waves encounter an object much smaller than their wavelength, they pass by it undisturbed so the object does not create a shadow. The smaller the wavelength, the smaller the features that can be imaged. If electrons are accelerated through a few kilovolts, their de Broglie waves have a shorter wavelength than visible light, so smaller features can be studied.

Q 32 (a) See Equations 18 and 19 in the Student Book.

Ek = 10 keV = 10 103 eV 1.60 10−19 J eV−1

= 1.60 10−15 J

(b) Ek = 21 mv2

so

v2 =m

Ek2

and

v =

m

Ek2

=

kg109.11

J.31

15106012

= 5.93 107 m s−1

(c) From Equation 17 in the Student Book,

=mv

h

=1731

3410636

sm105.93kg109.11

sJ.

= 1.23 10−11 m

Q 33 (a) The bone would be dated using different techniques – perhaps fluorine analysis and radiocarbon dating. Other material in the soil, including plant matter and pollen, would be checked to see if it was consistent.

(b) The journal would send the article to a referee, who would be another archaeologist/scientist who was an expert in the same field, asking for specific comments about whether the article was suitable for the journal and whether it appeared to be correct. A journal will say whether it is a peer-reviewed journal in the information it gives about its aims, readership and so on.

11




Chapter 6

SUR


Q 65 (a) A Limit of proportionality/Hooke’s law limit.

B Elastic limit.

C Yield point.

(b) The stiffer a material, the greater its Young modulus, that is, the steeper the gradient of the stress–strain graph, so material X is stiffer. (Note the different scales on the two graphs.)

Q 66 (a) Resolving along the slope:

force due to friction F = 12 500 N

mass of cabin + passengers m = 3500 kg + (10 × 70 kg) = 4200 kg

T ≥ F + mg sin 34°

= 12 500 N + 4200 kg × 9.81 N kg−1 × sin 34° = 35 540 N

so T ≥ 35 540 N

(b) Maximum tension Tmax = 3 × 35 540 N = 106 620 N = 1.1 × 105 N

maximum stress = A

Tmax 400 MPa

area of cable A =maxT

5

6

1.1 10 N

400 10 Pa

= 2.75 × 10−4 m2

cross-sectional area of 1 strand = r2 = (1 × 10−3 m)2 = 3.1 × 10−6 m2

number of strands =4

6

2.75 10

3.1 10

≈ 89

(Using raw numbers without rounding gives approximately 85.)

(c) Multi-strand cable is safer, as damage to a single strand is less likely to cause catastrophic failure.

Q 67 (a) Mean diameter = 1.026 mm = 1.026 × 10−3 m

cross-sectional area A = 8.3 × 10−7 m2

(b) See Table 4.1.

Load/g Length/mm Extension/ mm

Strain Load/N Stress/ 107 Pa

0 200 0 0.0 0.00 0

20 220 20 0.1 0.20 0.24

40 240 40 0.2 0.39 0.48

60 260 60 0.3 0.59 0.71

80 280 80 0.4 0.78 0.95

100 320 120 0.6 0.98 1.19

120 380 180 0.9 1.18 1.42

140 480 280 1.4 1.37 1.66

160 snapped

Table 4.1 Answer to Question 67(b)

12




Chapter 6

SUR

(c) See Figure 4.1.

Figure 4.1 Answer to Question 67(c)

(d) Modulus E = gradient of straight-line part = 2.37 × 106 Pa

(e) (i) Uncertainty in diameter Δd = ±0.03 mm (or 0.04 mm)

(ii) Percentage uncertainty in diameter =

0261

030

.

.× 100% = 3%

uncertainty in area ΔA = 6%

(f) Marked length l = 200 mm; percentage uncertainty is 0.5%

(g) When the load was 80 g, the extension was 80 mm. The actual uncertainty in extension is ±2 mm, so

percentage uncertainty =

80

2× 100 = 2.5%

(h) The sum of the percentage uncertainties in all quantities is 10% to 1 significant figure, so the result should be quoted as E = (6.0 ± 0.6) × 105 Pa.

Q 68 (a) Image is at distance f from lens:

f = P

1 =

D.218

1= 0.0549 m = 0.055 m = 55 mm

(b) (i) Time for infrared pulse to reach and return from object, t = 5.2 × 10−9 s 2 × distance = speed of infrared × time 2 × distance = (3.00 × 108 m s−1) × (5.2 × 10−9 s) =1.56 m distance = 0.78 m = 78 cm.

(ii) Film must be at distance of image, v:

v

1=

f

1−

u

1=

m.05490

1−

m.780

1= 16.93 m−1

v = 0.059 m = 5.9 cm

(c) f = 0.0549 m, v = 0.063 m,

u

1=

f

1−

v

1 =

m.05490

1−

m.0630

1 = 2.34 m−1

u =1342

1m.

= 0.43 m = 43 cm

13




Chapter 6

SUR

Q 69 (a) The intercept on either axis is equal to 1/f.

From x-axis,

f

1= 10.7 m−1, f = 0.093 m

From y-axis,

f

1= 11.3 m−1, f = 0.088 m

So f = (0.091 ± 0.03) m

(b) (i) Measurements have only been made to two significant figures – to the nearest centimetre. They should certainly have been made to the nearest half centimetre, and perhaps to the nearest millimetre.

(ii) There are only four measurements. Eight or so would have been better.

(iii) The measurements do not cover a great enough range of values. In particular, v only varies between 0.11 m and 0.16 m. The experimenter should have used smaller values of u, as this would have given bigger values of v.

Q 70 (a) Diffraction effects mean that features smaller than the wavelength cannot be resolved.

= f

v =

1

6

1600ms

15 10 Hz

= 1.067 × 10−4 m = 0.11 mm

(b) Time delay = 14 s

s = 2

vt=

1 61600 ms 14 10 s

2

= 0.0112 m = 11.2 mm

(c) (i) Ultrasound reflected from blood travelling towards the probe will be shifted to a higher frequency because of the relative motion between the blood cells and the probe.

(ii) When blood is travelling away from the probe, the ultrasound will be shifted to a lower frequency.

Q 71 (a) Time delay between front and back surface of lens

t = 10 s − 4.5 s = 5.5 s

s = 2

vt=

1 61640 ms 5.5 10 s

2

= 4.51 × 10−3 m = 4.51 mm

(b) Time delay between back of lens and retina = 31 s − 10 s = 21 s

s = 2

vt=

1 61530 ms 21 10 s

2

= 0.0161 m = 16.1 mm

(c) The ultrasound is transmitted into the cornea with very little reflection, so a coupling gel must have been used that matches the material in the cornea.

(d) The lens fitted would be too powerful, so that it would bring the rays to a focus in front of the retina and the person would be short-sighted.

(e) Power = f

1=

m. 310523

1

= 42.6 D

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ANSWERS TO QUESTIONS ON THE WHOLE CHAPTER · Salters Horners AS/A level Physics 1 Additional Sheet...

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