Physics 9702/22 May/June 2019: question paper and mark scheme
Cambridge AS & A Level Physics (9702), Paper 2 AS Level Structured Questions, May/June 2019. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
(a) The diameter d of a cylinder is measured as 0.0125 m ± 1.6%.
Calculate the absolute uncertainty in this measurement.
absolute uncertainty = …… m [1]
(b) The cylinder in (a) stands on a horizontal surface. The pressure p exerted on the surface by the cylinder is given by
4W p= . πd 2
The measured weight W of the cylinder is 0.38 N ± 2.8%.
(i) Calculate the pressure p.
p = …… N m−2 [1]
(ii) Determine the absolute uncertainty in the value of p.
absolute uncertainty = …… N m−2 [2]
[Total: 4]
(a) State Newton’s second law of motion.
……[1]
(b) A car of mass 850 kg tows a trailer in a straight line along a horizontal road, as shown in
Fig. 2.1.
car trailer tow-bar mass 850 kg
horizontal road
Fig. 2.1
The car and the trailer are connected by a horizontal tow-bar.
The variation with time t of the velocity v of the car for a part of its journey is shown in Fig. 2.2.
15
14 v / m s –1
13
12
11
10
9
8
0 5 10 15 20 25 t /s
Fig. 2.2
(i) Calculate the distance travelled by the car from time t = 0 to t = 10 s.
distance = …… m [2]
(ii) At time t = 10 s, the resistive force acting on the car due to air resistance and friction is
510 N. The tension in the tow-bar is 440 N.
For the car at time t = 10 s:
1. use Fig. 2.2 to calculate the acceleration
acceleration = …… m s−2 [2]
2. use your answer to calculate the resultant force acting on the car
resultant force = …… N [1]
3. show that a horizontal force of 1300 N is exerted on the car by its engine
[1]
4. determine the useful output power of the engine.
output power = …… W [2]
(c) A short time later, the car in (b) is travelling at a constant speed and the tension in the tow-bar is 480 N.
The tow-bar is a solid metal rod that obeys Hooke’s law. Some data for the tow-bar are listed below.
Young modulus of metal = 2.2 × 1011 Pa
original length of tow-bar = 0.48 m
cross-sectional area of tow-bar = 3.0 × 10−4 m2
Determine the extension of the tow-bar.
extension = …… m [3]
(d) The driver of the car in (b) sees a pedestrian standing directly ahead in the distance. The driver operates the horn of the car from time t = 15 s to t = 17 s. The frequency of the sound heard by the pedestrian is 480 Hz. The speed of the sound in the air is 340 m s−1.
Use Fig. 2.2 to calculate the frequency of the sound emitted by the horn.
frequency = …… Hz [2]
[Total: 14]
Mark scheme
2(a) (resultant) force proportional/equal to/is rate of change of momentum B1
2(b)(i) distance = area under graph or s = ½ (u + v) t C1
= ½ × (9 + 13) × 10
or
s = ut + ½at 2
= (9 × 10) + (½ × 0.40 × 102)
or
s = vt – ½at 2
= (13 × 10) – (½ × 0.40 × 102)
or
v 2 = u 2 + 2as
132 = 92 + (2 × 0.40 × s)
distance = 110 m A1
2(b)(ii) 1. a = gradient or a = (v – u) / t or a = ∆v / (∆)t C1
e.g. a = (14 – 9) / 12.5 or (13 – 9) / 10
a = 0.40 m s–2 A1
2. resultant force = 850 × 0.40 A1
= 340 N
3. (F =) 510 + 440 + 340 = 1300 (N) A1
4. P = Fv C1
= 1300 × 13 A1
= 1.7 × 104 W
2(c) E = σ/ε C1
E = (F / A) / (∆L / L) or E = FL / A∆L C1
∆L = (480 × 0.48) / (3.0 × 10–4 × 2.2 × 1011) A1
= 3.5 × 10–6 m
2(d) fo = fs v / (v – vs) C1
480 = fs × 340 / (340 – 14)
fs = 460 Hz A1
(a) State what is meant by the centre of gravity of a body.
……[1]
(b) A uniform square sign with sides of length 0.68 m is fixed at its corner points A and B to a wall.
The sign is also supported by a wire CD, as shown in Fig. 3.1.
D wire
54 N
35°
B C sign
E wall 0.68 m
W
A
0.68 m
Fig. 3.1 (not to scale)
The sign has weight W and centre of gravity at point E. The sign is held in a vertical plane with side BC horizontal. The wire is at an angle of 35° to side BC. The tension in the wire is
54 N.
The force exerted on the sign at B is only in the vertical direction.
(i) Calculate the vertical component of the tension in the wire.
vertical component of tension = …… N [1]
(ii) Explain why the force on the sign at B does not have a moment about point A.
……[1]
(iii) By taking moments about point A, show that the weight W of the sign is 150 N.
[2]
(iv) Calculate the total vertical force exerted by the wall on the sign at points A and B.
total vertical force = …… N [1]
(c) The sign in (b) is held together by nuts and bolts. One of the nuts falls vertically from rest through a distance of 4.8 m to the pavement below. The nut lands on the pavement with a speed of 9.2 m s−1.
Determine, for the nut falling from the sign to the pavement, the ratio
change in gravitational potential energy
. final kinetic energy
ratio = …… [4]
[Total: 10]
Mark scheme
3(a) the point where (all) the weight (of the body) is taken to act B1
3(b)(i) vertical component = 54 sin 35° A1
= 31 N
3(b)(ii) the (line of action of the) force (at B) passes through (point) A B1
or
the (line of action of the) force (at B) has zero (perpendicular) distance from (point) A
3(b)(iii) 54 sin 35° × 0.68 or 54 cos 35° × 0.68 or W × 0.34 C1
54 sin 35° × 0.68 + 54 cos 35° × 0.68 = W × 0.34 so W = 150 (N) A1
3(b)(iv) total vertical force = 150 – 31 A1
= 120 N
3(c) (∆)E = mg(∆)h C1
E = ½mv 2 C1
ratio = (m × 9.81 × 4.8) / (½ × m × 9.22) or (9.81 × 4.8) / (½ × 9.22) C1
= 1.1 A1
(a) For a progressive water wave, state what is meant by:
(i) displacement
……[1]
(ii) amplitude.
……[1]
(b) Two coherent waves X and Y meet at a point and superpose. The phase difference between the waves at the point is 180°. Wave X has an amplitude of 1.2 cm and intensity I. Wave Y has an amplitude of 3.6 cm.
Calculate, in terms of I, the resultant intensity at the meeting point.
intensity = …… [2]
(c) (i) Monochromatic light is incident on a diffraction grating. Describe the diffraction of the light waves as they pass through the grating.
……[2]
(ii) A parallel beam of light consists of two wavelengths 540 nm and 630 nm. The light is incident normally on a diffraction grating. Third-order diffraction maxima are produced for each of the two wavelengths. No higher orders are produced for either wavelength.
Determine the smallest possible line spacing d of the diffraction grating.
d = …… m [3]
(iii) The beam of light in (c)(ii) is replaced by a beam of blue light incident on the same diffraction grating.
State and explain whether a third-order diffraction maximum is produced for this blue light.
……[2]
[Total: 11]
Mark scheme
4(a)(i) distance (in a specified direction of particle/point on wave) from the equilibrium position B1
4(a)(ii) the maximum distance (of particle/point on wave) from the equilibrium position B1
or
the maximum displacement (of particle/point on wave)
4(b) I ∝ A2 C1
IR / I = (3.6 – 1.2)2 / (1.2)2 A1
resultant intensity = 4.0I
4(c)(i) as wave(s) pass through the slit(s) B1
wave(s) spread (into geometric shadow) B1
4(c)(ii) nλ = d sin θ C1
3λ = d sin 90° or 3λ = d C1
d = 3 × 630 × 10–9 A1
= 1.9 × 10–6 m
4(c)(iii) wavelength of blue light is shorter (than 540 nm/630 nm/wavelengths of original light) M1
(so) third order diffraction maximum is produced A1
(a) State Kirchhoff’s second law.
……[2]
(b) A battery of electromotive force (e.m.f.) 5.6 V and internal resistance r is connected to two external resistors, as shown in Fig. 5.1.
5.6 V r
V
90
18
Fig. 5.1
The reading on the voltmeter is 4.8 V.
(i) Calculate:
1. the combined resistance of the two resistors connected in parallel
combined resistance = …… Ω [2]
2. the current in the battery.
current = …… A [2]
(ii) Show that the internal resistance r is 2.5 Ω.
[2]
(iii) Determine the ratio
power dissipated by internal resistance r
. total power produced by battery
ratio = …… [3]
(c) The battery in (b) is now connected to a battery of e.m.f. 7.2 V and internal resistance 3.5 Ω.
The new circuit is shown in Fig. 5.2.
5.6 V 2.5
7.2 V 3.5
Fig. 5.2
Determine the current in the circuit.
current = …… A [2]
[Total: 13]
Mark scheme
5(a) sum of e.m.f.(s) = sum of p.d.(s) M1
around a loop/around a closed circuit A1
5(b)(i) 1. 1 / R = 1 / R1 + 1 / R2 C1
1 / R = 1 / 90 + 1 / 18
R = 15 Ω A1
2. I = V/R C1
I = 4.8 / 15 or I = 4.8 / 90 + 4.8 / 18 A1
I = 0.32 A
5(b)(ii) E = V + Ir C1
or
E = I(R + r)
5.6 = 4.8 + 0.32 r so r = 2.5 (Ω) A1
or
5.6 = 0.32 × (15 + r) so r = 2.5 (Ω)
5(b)(iii) P = EI or P = VI or P = I2R or P = V2 / R C1
ratio = (0.322 × 2.5) / (5.6 × 0.32) or 0.256 / 1.792 C1
= 0.14 A1
5(c) 7.2 – 5.6 – 2.5I – 3.5I = 0 C1
I = 0.27 A A1
(a) State what is meant by a field line (line of force) in an electric field.
……[1]
(b) An electric field has two different regions X and Y. The field strength in X is less than that in Y.
Describe a difference between the pattern of field lines (lines of force) in X and in Y.
……[1]
(c) A particle P has a mass of 0.15 u and a charge of −1e, where e is the elementary charge.
(i) Particle P and an α-particle are in the same uniform electric field. Calculate the ratio
magnitude of acceleration of particle P
. magnitude of acceleration of α-particle
ratio = …… [3]
(ii) Particle P is a hadron composed of only two quarks. One of them is a down (d) quark.
By considering charge, determine a possible type (flavour) of the other quark.
Explain your working.
……[3]
[Total: 8]
Mark scheme
6(a) path/direction in which a (free) positive charge will move B1
6(b) (lines) closer together in Y/further apart in X B1
6(c)(i) a = Eq / m C1
or
F = Eq and F = ma
ratio = (1e / 0.15 u) × (4 u / 2e) or 1 / 0.15 × 4 / 2 C1
ratio = 13 A1
6(c)(ii) down quark charge is –(1 / 3)e C1
– (1 / 3)e + q = –1e so q = –(2 / 3)e A1
(–(2 / 3)e is) anti-up / u (quark) (allow charm or top antiquark) B1
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Physics 9702/22 for May/June 2019 is out of 60 marks, with 1 hour 15 minutes to answer it, across 6 questions.
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