Big Body Benz
Big Body Benz
Big Body Benz
* 0 7 6 4 3 5 8 0 9 2 *
PHYSICS 9702/22
Paper 2 AS Level Structured Questions February/March 2020
1 hour 15 minutes
INSTRUCTIONS
● Answer all questions.
● Use a black or dark blue pen. You may use an HB pencil for any diagrams or graphs.
● Write your name, centre number and candidate number in the boxes at the top of the page.
● Write your answer to each question in the space provided.
● Do not use an erasable pen or correction fluid.
● Do not write on any bar codes.
● You may use a calculator.
● You should show all your working and use appropriate units.
INFORMATION
● The total mark for this paper is 60.
● The number of marks for each question or part question is shown in brackets [ ].
DC (LK/SW) 180016/4
© UCLES 2020 [Turn over
2
Data
Formulae
1
uniformly accelerated motion s = ut + 2 at 2
v 2 = u 2 + 2as
Gm
gravitational potential φ = −
r
1 Nm 2
pressure of an ideal gas p = 3 〈c 〉
V
simple harmonic motion a = − ω 2x
Q
electric potential V =
4πε0r
capacitors in parallel C = C1 + C2 + . . .
1
energy of charged capacitor W = 2 QV
resistors in series R = R1 + R2 + . . .
BI
Hall voltage VH =
ntq
0.693
decay constant λ =
t 1
2
current
1. ...............................................................................................................................................
voltage
2. ...............................................................................................................................................
[2]
(b) The acceleration of free fall g may be determined from an oscillating pendulum using the
equation
4π2l
g=
T2
where l is the length of the pendulum and T is the period of oscillation.
l = 1.50 m ± 2%
and T = 2.48 s ± 3%.
9.63
g = ................................................ m s–2 [1]
8
percentage uncertainty = ..................................................... % [2]
(iii) Use your answers in (b)(i) and (b)(ii) to determine the absolute uncertainty of the
calculated value of g.
0.08 × 9.6
= 0.8 m s-2
0.8
absolute uncertainty = ................................................ m s–2 [1]
[Total: 6]
BLANK PAGE
(a) The dolphin emits a sound as it swims directly towards a stationary submerged diver. The
frequency of the sound heard by the diver is 9560 Hz. The speed of sound in the water is
1510 m s–1.
Determine the frequency, to three significant figures, of the sound emitted by the dolphin.
9560 = f × 1510 / (1510 – 4.50)
f = 9530 Hz
9530
frequency = .................................................... Hz [2]
(b) The dolphin strikes the bottom of a floating ball so that the ball rises vertically upwards from
the surface of the water, as illustrated in Fig. 2.1.
path of
ball height of
ball above
ball surface
surface of water
speed 5.6 m s–1
Fig. 2.1
The ball leaves the water surface with speed 5.6 m s–1.
(i) Calculate the maximum height reached by the ball above the surface of the water.
v2 =u2 +2as
h = 5.62 / (2 × 9.81)
= 1.6 m
1.6
height = ..................................................... m [2]
(ii) The ball leaves the water at time t = 0 and reaches its maximum height at time t = T.
On Fig. 2.2, sketch a graph to show the variation of the speed of the ball with time t from
t = 0 to t = T. Numerical values are not required.
speed
0
0 time t T
Fig. 2.2
[1]
Use your answer in (b)(i) to calculate the change in gravitational potential energy of the
ball as it rises from the surface of the water to its maximum height.
7.1
change in gravitational potential energy = ...................................................... J [2]
(iv) State and explain the variation in the magnitude of the acceleration of the ball as it falls
back towards the surface of the water if air resistance is not negligible.
The air resistance will increase thus the acceleration will decrease
...........................................................................................................................................
...........................................................................................................................................
...........................................................................................................................................
...........................................................................................................................................
..................................................................................................................................... [2]
[Total: 9]
force × displacement
...................................................................................................................................................
...................................................................................................................................................
............................................................................................................................................. [1]
(b) A skier is pulled along horizontal ground by a wire attached to a kite, as shown in Fig. 3.1.
wire
kite
speed 4.4 m s–1
140 N
skier 30° ground
horizontal
The skier moves in a straight line along the ground with a constant speed of 4.4 m s–1. The
wire is at an angle of 30° to the horizontal. The tension in the wire is 140 N.
(i) Calculate the work done by the tension to move the skier for a time of 30 s.
d = 4.4 × 30
work =140*cos30 × 4.4 × 30
= 1.6 x 10^4 J
1.6 × 10^4
work done = ...................................................... J [3]
(ii) The weight of the skier is 860 N. The vertical component of the tension in the wire and
the weight of the skier combine so that the skier exerts a downward pressure on the
ground of 2400 Pa.
Determine the total area of the skis in contact with the ground.
p=F/A
F = 860 - 140*sin30
= 790
A= 790 / 2400
= 0.33 m^2
0.33
area = .................................................... m2 [3]
(iii) The wire attached to the kite is uniform. The stress in the wire is 9.6 × 106 Pa.
diameter = .....................................................
4.3 × 10^–3 m [2]
(c) The variation with extension x of the tension F in the wire in (b) is shown in Fig. 3.2.
300
F/N
250
200
150
100
50
0
0 0.20 0.40 0.60 0.80
x / mm
Fig. 3.2
A gust of wind increases the tension in the wire from 140 N to 210 N.
[Total: 12]
..................................................................................................................................... [1]
...........................................................................................................................................
the maximum displacement of a point on a wave
..................................................................................................................................... [1]
(i) Diffraction of the light waves occurs at each slit of the grating. The light waves emerging
from the slits are coherent.
1. diffraction
.............................................................................................................................. [1]
2. coherent.
....................................................................................................................................
phase difference is constant
.............................................................................................................................. [1]
(ii) The wavelength of the laser light is 650 nm. The angle between the third order diffraction
maxima is 68°, as illustrated in Fig. 4.1.
third order
diffraction maximum
laser light
68°
wavelength 650 nm
third order
diffraction diffraction maximum
grating
Calculate the separation d between the centres of adjacent slits of the grating.
nλ = d sinθ
d = 3 × 650 × 10–9 / sin34
d = 3.5 × 10^–6 m
3.5 × 10^–6
d = ..................................................... m [3]
(iii) The red laser light is replaced with blue laser light.
State and explain the change, if any, to the angle between the third order diffraction
maxima.
The wavelength of the blue light is shorter compared to red light thus the angle will decrease
...........................................................................................................................................
...........................................................................................................................................
..................................................................................................................................... [2]
[Total: 9]
...................................................................................................................................................
...................................................................................................................................................
............................................................................................................................................. [1]
(b) A wire has a resistance of 1.8 Ω. The wire has a uniform cross-sectional area of 0.38 mm2 and
is made of metal of resistivity 9.6 × 10–7 Ω m.
1.8 Ω 0.60 Ω
X Y
Fig. 5.1
The battery P has an electromotive force (e.m.f.) of 1.2 V and negligible internal resistance.
(i) Explain, in terms of energy, why the potential difference (p.d.) across resistor X is less
than the e.m.f. of the battery.
...........................................................................................................................................
...........................................................................................................................................
..................................................................................................................................... [1]
(d) Another battery Q of e.m.f. 1.2 V and negligible internal resistance is now connected into the
circuit of Fig. 5.1 to produce the new circuit shown in Fig. 5.2.
1.2 V
Q
1.2 V
1.8 Ω 0.60 Ω
X Y
Fig. 5.2
State whether the addition of battery Q causes the current to decrease, increase or remain
the same in:
(e) The circuit shown in Fig. 5.2 is modified to produce the new circuit shown in Fig. 5.3.
1.2 V
3.6 Ω
1.8 Ω 0.60 Ω
X Y
Fig. 5.3
Calculate:
[Total: 12]
6 A uniform electric field is produced between two parallel metal plates. The electric field strength is
1.4 × 104 N C–1. The potential difference between the plates is 350 V.
(b) A nucleus of mass 8.3 × 10–27 kg is now placed in the electric field. The electric force acting
on the nucleus is 6.7 × 10–15 N.
(i) Calculate the charge on the nucleus in terms of e, where e is the elementary charge.
(iii) Use your answers in (b)(i) and (b)(ii) to determine the number of neutrons in the nucleus.
[Total: 7]
...................................................................................................................................................
............................................................................................................................................. [1]
(i) State the two leptons that are produced by the decay.
...........................................................................................................................................
..................................................................................................................................... [2]
(ii) Part of the energy released by the decay is given to the two leptons.
...........................................................................................................................................
..................................................................................................................................... [2]
[Total: 5]
BLANK PAGE
Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Every
reasonable effort has been made by the publisher (UCLES) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the
publisher will be pleased to make amends at the earliest possible opportunity.
To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced online in the Cambridge
Assessment International Education Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download
at www.cambridgeinternational.org after the live examination series.
Cambridge Assessment International Education is part of the Cambridge Assessment Group. Cambridge Assessment is the brand name of the University of
Cambridge Local Examinations Syndicate (UCLES), which itself is a department of the University of Cambridge.