Download 1. (a) - PLK Vicwood KT Chong Sixth Form College

00’ AL Physics/Structural Questions/Marking/P.1
PLK VICWOOD K.T. CHONG SIXTH FORM COLLEGE
00’ AL Physics : Structural Questions
Marking Scheme
1.
(a)
2
road sign
5m
ramp
2
30 m
(b) u cos t = 30
0 = u sin - gt
(u sin)2 - 2g(5) = 0
1 mark for at least one equation correct
2 marks if all equations correct
On solving, (i) u = 31.6 ms-1
(ii)  = 18.4
(iii) t = 1 s
(c)
F = ma
2
1
1
1
(or Fs =
1
1
mu 2  mu12 )
2
2
5
1
-8000 = 1000 a
a = - 8 ms-2
And
v2 = u2 + 2as
31.62 = u12 + 2(-8)(39)
u1 = 40.3 ms-1
1
(d) The car in fact took a parabolic path due to gravity.
The actual angle of projection should be greater than 9.5, consequently the above relation
would give an over-estimated value of u.
2.
2
1
1
2
(a)
tension
reaction on m1
pan
m1
weight of pan
3
3
weight of m1
action on pan
(b) (Rest or) uniform motion
1
1
00’ AL Physics/Structural Questions/Marking/P.2
(c)
&
(m + M)g - T = (m + M)a
T - Mg = Ma
a =
&
T = Mg + Ma
= 1.55 N
v2 = u2 + 2as
v2 = 2(0.32)(0.36)
v = 0.48 ms-1
(e) At equilibrium,
For s.h.m.,
mg = kx0
0.01  10 = 4 x0
x0 = 0.025 m below O
 =
1
1
3
1
1
2
1
1
k
m  2M
4
0.01  2  0.15
= 3.6 rads-1
=
3.
1
mg
m  2M
= 0.32 ms-2
On solving,
(d) By
(or mg = (m + 2M)a )
1
1
4
(a) Ensure that the full range of discharging current and therefore discharging time is obtained. 1
1
(b) - uncertainty in starting the stop watch
- uncertainty in the readings of current
- uncertainty in the readings of time
- uncertainty due to the tolerance of the resistor
- uncertainty due to the resistance of the microammeter
2
2
1
1
ANY TWO @1
(c) Reduce the random error in the readings taken.
(d)
V = IR
= 50  10-6  100  103
= 5 V (minimum value)
1
Therefore the e.m.f. of battery is 6 V.
1
2
-t
(e) (i)
ln I = -9.4  10-2 t - 9.9 (compare with I = I 0 e RC )
1
= -9.4  10-2
RC
C = 106  10-6 F
slope =
(ii) Time constant RC = 100  103  106  10-6
= 10.6 s << 60 s
Time charging time is much greater than the time constant so the capacitor can be
considered as fully charged.
(iii)
y-intercept less negative (larger initial current)
or steeper slope (more negative)
1
1
1
3
1
1
2
1
1
00’ AL Physics/Structural Questions/Marking/P.3
4.
(a)
1
Nmc 2
3
1
P = c 2
3
PV =
1
3  16  10 5
1.57
= 1750 ms-1
c2 =
1
(b) W = P V
= 105  1.2
= 1.2  105 J (or 120 kJ)
1
1
(c) (i) Initially,
P0V0 = n0RT
After inflating one balloon,
Since
2
for the gas inside the cylinder.
P1V0 = n1RT
in the cylinder
PbVb = nbRT
in the balloon
n0 = n1 + nb
P0V0 = P1V0 + PbVb
2
1
1
PbVb
10 5  1.2
=
V0
0.5
5
= 2.4  10 Pa
P = P0 - P1 =
(ii) After inflating k balloons,
PkV0 = P0V0 - k(PbVb)
PV
Pk = P0  k ( b b ) > 10  105
V0
1
3
1
16  105 - k (2.4  105) > 10  105
k < 2.5
Therefore at most 2 balloons should be inflated.
5.
1
2
(d) Work has to be done against the atmospheric pressure as well as the intermolecular forces
between the gas molecules, therefore by the conservation of energy heat is transferred from the
surroundings into the cylinder.
2
2
(a) voltage follower
1
1
(b) At the beginning when the p.d. between the metal can and the sample increases, more ions 1
migrate to the opposite electrodes per unit time, ionization current increases, p.d. across the
resistor also increases.
When the p.d. is greater than ~20 V, all the ions produced in unit time would be collected by 1
the can or sample.
The p.d. across the resistor is saturated as further increase in applied p.d. would not increase 1
the ionization current.
(c) (i) mass defect = 238.0508 - (234.0436 + 4.0026) = 0.0046 u
energy released = 0.0046  934
= 4.30 (MeV)
70  10 3
10 9
= 70  10-12 A
(ii) Saturated ionization current =
3
1
1
1
2
00’ AL Physics/Structural Questions/Marking/P.4
70  10 12
1.6  10 19
= 4.38  108 (ionizations per second)
number of positive ions produced =
1
(iii) energy needed to produce 4.38  108 ionizations
= 4.38  108  30
= 1.31  104 MeV
activity =
1
1.31  10 4
= 3050 (disintegrations per second)
4.30
1
Assumption: - all  particles emitted have undergone ionization.
or
- nearly all the energy released in the decay goes to the  particle
(iv) No.
- The ionization power of  particles is very weak.
- The  particles emitted do not possess the same amount of energy.
- The penetrating power of  particles is high/range of  particles is long.
6.
2
ANY
TWO
1
3
1
1
1
2
(a)
-Q
+Q
1
1
VS2  VR2
1
= 2.00 2  1.80 2
= 0.87 V (r.m.s.)
1
(b) (i) (I) VAA’ =
1
) &
C
VAA'
1
So
=
VR
CR
(II) VAA’ = I (
VR = IR
1.80
0.87  2  100  10 3  10  10 3
= 3.3  10-10 F
C
C0 =
= 1.1  10-10 Fm-1
3.0
C =
(ii) (I) Zero
(II) L = L
At 100 kHz, L = 2  100  103  10-7 = 0.06 m-1
At 1 MHz, L = 2  1  106  10-7 = 0.6 m-1
1 MHz is better since the order of magnitude of L is comparable to that of the
resistor.
2
1
1
1
3
1
1
1
1
2
00’ AL Physics/Structural Questions/Marking/P.5
7.
(a) (i)
2
2
(ii)
4  10 7  5 1
1
(

)
2
0.02 0.07
= 3.6  10-5 T
B =
1
(iii) Hall probe.
(b) (i) Steady d.c. would produce a static magnetic field which would easily mix up with the
earth’s magnetic field.
Even with large d.c. current the B-field produced is weak and is therefore difficult to
detect.
(ii) (I) The CRO trace represents the induced e.m.f. in the search coil and it is
proportional to the peak value of the sinusoidal B-field produced by the wires.
- The search coil detects alternating B-field. Therefore the earth’s field is not
detected.
- Measurable induced e.m.f. can be produced by increasing the frequency even
with relatively small current.
(II) - the length of the wire should be as long as possible ~ 2 m
- the two wires should be well away from any magnetic materials
- set-up should be well away from any stray fields, such as those
from mains socket
- twist the two connecting wires
- adjust the orientation of the coil so that the peak-to-peak value
of the trace on the CRO is maximum
- avoid placing the coil near the ends of the wires
(Accept any other reasonable answers)
1
2
1
1
1
1
2
1
1
1
3
ANY
TWO
@1
2
2
00’ AL Physics/Structural Questions/Marking/P.6
8.
(a) (i)
X
FB
1
FE
1
Y
(ii) Magnetic force acting on a moving electron, FB = Bev
Electrons migrate to Y, making there more negative, an E-field is set up across XY
Electrons also experience the electric force, FE = eE = e/l
If the two forces are equal, migration stops and the p.d. across the two ends becomes
steady.
e
Bev =
l

 = Blv
GMm
mv 2
=
R
R2
(b) (i)

gRE2
= v2
R

v =
1
1
1
3
1
(or g’ = 8.86 ms-2)
1
10  (6.4  10 6 ) 2
6.8  10 6
= 7800 ms1
1
(ii)  = Blv = 30  10-6  20  103  7800
= 4700 V
- the cable is always perpendicular to the B-field
- the magnetic field is uniform over this 20-km cable
- the satellite and the shuttle move with the same speed
(i.e. ensure that v, l and B are mutually perpendicular)
3
1
ANY TWO
@1
(iii) Current direction: upward.
The ions and free electrons in the ionosphere would complete the circuit for electric
power generation.
2
3
1
2
3
00’ AL Physics/Structural Questions/Marking/P.7
9.
(a)
sin 
(656.3, 0.394)
0.4
0.3
(486.1, 0.292)
(434.0, 0.260)
(410.2, 0.246)
(364.6, 0.219)
0.2
360
460
560
660
Appropriate quantities are chosen for plotting the graph
Points correctly plotted
Correct graph
1
1
1
1
0.375  0.225
= slope =
nm-1
d
625  375
= 600 (lines per mm)
1
(b) (i)  = 434.0 nm
(ii) ultra-violet
(c) Electrons transitions from a high energy level to a lower one within an excited atom.
Energy levels are discrete/quantized.
1
5
1
1
1
1
1
1
2
/nm
00’ AL Physics/Structural Questions/Marking/P.8
(d) (i)  = (364 .6 nm)
h
n2
n 4
2
hc
1 1
c
=
 4( 2  )
364 .6 nm
4

n
(i.e. h = K (
1 1
 ))
n2 4
1
 6.626  10 34  3  10 8
1
 4
9
364 .6  10
1.6  10 19
= -13.6 (eV)
(ii) K =
1
1
Negative value means the electron is bounded and the magnitude is the energy of that
level with respect to E.
10. (a) (i) Feeding back a certain portion of the output to the inverting input
(or feeding back part of the output to the input, and they are in antiphase.)
- gain is predictable (more or less independent of the characteristics
of the op-amp)
- stability is higher
- distortion of the output is less, i.e. amplification is more linear
- gain is constant over a wider range/band of frequencies
(ii) Gain will increase.
Rf
Ri
Ans. (I) -100 mV
(II) -15 V (saturation)
1
1
1
4
1
ANY
TWO
@1
2
3
1
1
1
1
2
(iii) Vout =  Vin
00’ AL Physics/Structural Questions/Marking/P.9
(iv)
Vin / V
2
1
0
time
-1
-2
Vout / V
15
7.5
2
0
time
2
-7.5
-15
(b) (i)
Gain =
0.5
Rf
=
Ri
10  10  3
0.5
Rf =
 10 k
10  10  3
= 500 k
(ii) A step-up transformer differs from an amplifier in a way that it does not have a high
impedance input and the op amp draws its power from a separate source.
1
1
2
1
1
2