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83’ AL Physics/Structural Questions/Marking/P.1
PLK VICWOOD K.T. CHONG SIXTH FORM COLLEGE
83’ AL Physics: Structural Questions
Marking Scheme
1.
(a)
Force
time
0
Award marks for
(i) single polarity (note that
is acceptable).
(ii) increase/decrease in magnitude of force.
(iii) sinusoidal type shape (need not be exact sinusoidal
since an exact ‘Hooke’s Law’ sponge was not
specified. However, we would expect
½
½
rather
½
½
than
, for example).
(iv) symmetry about ‘highest point’.
(b)
Force
0
time
Award marks for
(i) same polarity as in (a).
(ii) constant magnitude of force (any decrease would
not be significant for a ball thrown in the air,
and should not be credited).
2
(c) Experiments required are:
(i) To show that a constant force produces a constant
acceleration (e.g. form 4/5 experiments using
trolleys, runways and ticker tape vibrators).
2
(ii) To show that acceleration due to gravity is
constant (e.g. using a falling mass and a
ticker tape vibrator, or stroboscopic photographs
of falling bodies).
( Experiments (ii) and (i) together suggest that the
gravitational force is constant, which would obviously
not be the case for a ‘sponge’ theory. )
2
83’ AL Physics/Structural Questions/Marking/P.2
2.
(a) Using Bernoulli’s principle.
1
Difference in pressure =  1.29  (90 2  55 2 ) Pa
2
= 3270 Pa
1
 Lifting force = 3270  200 N
= 6.5  105 N
1
At the point of taking off, wt. of plane = lifting force
1
6.5  10
kg
10
= 6.5  104 kg
1
 Mass of plane =
5
(b) out of syllabus
3.
(a) Student A
Agree.
No intermolecular attraction; no potential energy.
Internal energy is the total kinetic energy.
Student B
Disagree.
There will be intermolecular potential energy which
will contribute to the internal energy.
1
1
Student C
Agree.
The internal energy of a gas can be increased by the
process of work.
(Must give correct reason to get mark(s))
(b) Work done on piston (converted into internal energy)
Ei = pressure  volume change
1
1
= (101  10 3 )  m(

)
0.60 958
Energy supplied to evaporate water,
Es = m(2.3  106)
E1
101  10 3
=
 100 % = 7.3%
Es
0.60  (2.3  10 6 )
4.
(a) The maximum collect current is given by
6
6
I~
R~
= 600 
R
10  10 3
1
1
1
1
1
1
83’ AL Physics/Structural Questions/Marking/P.3
(b) Thermistor should be in position X.
1
Reason: To make the lamp glow
collector current must increase
 base current must increase
 p.d. across Y component must increase
thermistor
X
6V
resistor
Y
(c)
thermistor
4.7 k
The full 6 V supply p.d. is
divided between X and Y.
If the resistance of the thermistor decrease, the
proportion of this p.d. across
Y (the resistor) will rise.
1
The lamp starts glowing for a p.d.
across R of 1.1 V (given).
1

6V
R
1.1 V
1



R
1.1
=
R  4.7
6
6 R = 1.1 R + (4.7  1.1)
4.9 R = 1.1  4.7
1.1  4.7
R =
4.9
= 1.05 k
1
1
Choose R = 1 k, say.
5.
Lamp A comes on before Lamp B.
1
When a constant voltage V is applied across an inductance L
the rate of increase of current through the coil is V/L i.e.
the current will always take a finite time to reach its final
value. In the resistor the current change is virtually
instantaneous.
1
(N.B. Candidates need not mention induced back e.m.f. In
fact Nuffield suggests that this term leads to confusion
and should be avoided. However, credit should be given
for correct explanations in terms of back e.m.f.)
(b) The vector diagram is as follows:
Vy1
VL = IL
1

Vy2
VR = IR
83’ AL Physics/Structural Questions/Marking/P.4
tan  =
L =
L
R
R tan 
…… (1)

1
From the CRO traces
phase difference between Vy1 and Vy2 = 3 ms
1
Since period = 20 ms
3
3
=
 2 = 0.94 rad. =
20
10
1
sub. in (1),
R
tan 

500  tan(0.94 )
L =
100 
~ 2.18 H
L =
1
(c)
y1
y2
For full marks candidates should show
L
)
R
(ii) /2 phase difference i.e. 2.5 cm (not 1.5 cm as
in first case).
(iii) Y2 leading Y1
(iv) Y2 inverted (as compared with original)
(v) Y1 same polarity (as original)
(i) amplitude ratio approx. 1.4 (=
½
1
½
½
½
83’ AL Physics/Structural Questions/Marking/P.5
( Note that the exact positions of the waveforms on the CRO screen will
depend on the triggering level and slope. Candidates need not draw
waveforms in positions shown above (the combined pattern can be
shifted all to the left or to the right). The relative positions,
however, must be as indicated above.)
6.
out of syllabus
7.
(a) (i)
1066
MW
0.323
= 3300 MW
Power generated =
1
(ii) Power lost to the atmosphere
= 3300  (0.18 + 0.007) MW
= 617 MW
1
(b) Power delivered to cooling water
= 3300  0.49 MW
= 1617 MW
1
 48  1000  4200   = 1617  106 where  : rise in temperature
1617  10 6
C
48  1000  4200
= 8 C
i.e.  =
1
1
(c) Mass difference = (235.0409 + 1.0086) (140.9141 + 91.9250 + 3  1.0086) u
= 0.1846 u = 3.064  10-28 kg
Using E = mc2,
energy released by one nuclide of U235
= 0.1846  1.66  10-27  (3  108)2 J
= 2.76  10-11 J
No. of nuclides undergone fission =
1
3300  10 6
2.76  10 11
= 1.2  1020
(d)
1
1.2  10 20
 0.235 kg
6  10 23
= 4.7  10-5 kg
Mass of U235 need in 1 sec =
1
Mass required in 10 years
= 4.7  10-5  365  24  3600  10 kg
= 14,800 kg ~ 15,000 kg
1
1