Download 2009-YJC-PH-H2-P1-Prelim-soln

Yishun Junior College
2009 H2 Physics Prelim P1 (Solutions)
2009 Yishun JC Preliminary Examination
H2 Physics Paper 1 Suggested Solution
1.
1
A
2
B
3
D
4
B
5
A
6
B
7
A
8
A
9
A
10
C
11
B
12
D
13
A
14
B
15
B
16
A
17
B
18
D
19
B
20
C
21
B
22
A
23
C
24
B
25
A
26
D
27
B
28
B
29
D
30
A
31
A
32
C
33
B
34
D
35
B
36
B
37
D
38
C
39
C
40
B
d
 VT  0.3520 V
L
VA d L VT
0.3 0.001 0.3






 0.214
VA
d
L
VT
23.0 0.980 1.5
VA 
VA  0.08 V
VA  0.35  0.08  V
2.
By adding a resistor in series with the driver cell, the potential difference across
resistance wire will become smaller. Hence the balance length will increase and this
reduces the percentage uncertainty of VT.
3.
Speed of the sphere decreases with time.
Velocity is positive as downwards is positive.
Therefore change in velocity is negative.
4.
Initial velocity of balloon is resultant of the leftward velocity of skateboarder and the
upward velocity during the throw.
u
uy
ux
After the balloon leaves the hand of the skateboarder, the trajectory is influenced by
the weight of the balloon and air resistance to the balloon’s motion.
5.
VI = (mc)/t + h. If rate of heat loss to surrounding, h, is to be accounted for, c is
supposed to be smaller. Hence, experimental value of c will be overestimated if heat
loss is not accounted for.
6.
From ideal gas equation, pV = nRT  pV  T. From graph, (pV)Y < (pV)Z < (pV)X.
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Yishun Junior College
7.
2009 H2 Physics Prelim P1 (Solutions)
To create rotational equilibrium, the sum of moments exerted by the weight of each
material about the knife-edge must be equal. Another way is to use visual inspection
to estimate the location of the centre of gravity (CG) for the composite rod. The
knife-edge must be below the CG for rotational equilibrium.
For C and D, the CG lies in the middle of the rod so knife-edge is placed correctly
below the CG.
For B, the CG is off-centre to the right since material B is concentrated to the right.
For A, the CG is also off-centre to the right since material B is concentrated to the
right, but the knife-edge is at the centre. Rotational equilibrium cannot be
established.
8.
Forces acting on the plank are in equilibrium. Hence, they must be able to form a
closed triangle. Also, the three forces must pass through a concurrent point, which is
Q in this case.
S
R
T
W
T
F
Q
F
P
W
9.
The largest force exerted by Q on P is the same magnitude as the force exerted by P
on Q (by Newton’s 3rd law of motion). Such force can be determined from the
product of the mass of Q and the acceleration given (by Newton’s 2nd law of motion).
For A:
For B:
For C:
For D:
10.
FQP = 3 ma
FQP = 2 ma
FQP = 1.5 ma
FQP = 2 ma
Mass is the property of a body which resists change in motion. Magnitude of speed
is irrelevant for consideration. Hence in the order of decreasing mass, the list is:
aircraft, car, block, alpha particle.
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Yishun Junior College
11.
2009 H2 Physics Prelim P1 (Solutions)
Horizontal forces acting on man:
v
Fon man by cord
Fon man by ground
Fon trolley by cord
Fon trolley by ground
Work done is positive if the applied force and displacement are in the same direction.
12.
1 2
kx 2
kx  mgh  h 
2
2mg
13.
At top : mg  N 
mv 2
r
In contact N  0 
mv 2
 mg  v  gr
r
mv 2
r
N always  0 ; weightlessness is felt when N = 0
At bottom : N  mg 
14.
15.
16.
Speed and magnitude of acceleration are constant but direction of velocity and
acceleration are changing. Velocity and hence momentum are tangential to the path.
Acceleration and hence force (rate of change of momentum) are perpendicular to the
path (towards centre).
1
Trend depicting g  2 (Note: since r is measured from the surface of the Earth, g
r
does not start from a very large value).
1 2 GMm
2GM
.
mv 
v 
2
r
r
Larger mass of Sun results in stronger gravitational field, hence making it harder for
hydrogen molecules to escape.
17.
Period of wave, T = 4 x 4.0 ms = 16 ms
Hence, frequency, f = 1/T = 63 Hz
18.
If the new vertical sensitivity is applied to the original waveform, amplitude of trace
would have reduced by half on the c.r.o. Since the same trace is obtained, the new
intensity corresponds to twice of the original wave amplitude.
As intensity  (amplitude)2, the new intensity should be four times greater.
19.
Since the target’s motion is simple harmonic, it will be the slowest in regions 1 and 5
and fastest in region 3.
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Yishun Junior College
2009 H2 Physics Prelim P1 (Solutions)
20.
Partial vacuum implies less damping on the pendulum. Thus maximum amplitude of
forced oscillation would be higher. Frequency at maximum amplitude increases
slightly due to lesser damping.
21.
Using  
22.
Using d sin   n , for the same order (n = 1) and wavelength , d is inversely
proportional to  (for small angles). (OR diffraction effects become poorer with larger
grating spacing)
23.
E
24.
W  F .s  eE s cos 60 o . Work done is positive because the external force (acting
ax
, the fringe separation x would be largest if both wavelength and slit-toD
screen distance are large and slit-to-slit distance is small.
2
E  r' 
E
     r' 2
 4.0 m
2
E'  r 
E 4
4o r
Q


against the electric force) is in the same direction as the vertical displacement.
25.
RA 
 3t 
2t
2


3
2
, RB 
, RC 
.
6t
2t
3t
Hence, maximum resistance can be established across face P and its opposite face.
26.
Resistance of thermistor decreases as temperature increases.
27.
The next lower wavelength corresponds to a higher frequency of light which is only
slightly higher than f2.
This is because energy states of a hydrogen atom are not equally spaced.
VS
B
P2
P1
f
f1
28.
f2 f3
For fixed p.d. across PQ, maximum power occurs when equivalent R is smallest.
Net RA = 4r, RB = r/4, RC = r, RD = r
29.
Wb = T m2 = N m –1 A –1 m2 = (N m) A –1 = J A –1
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Yishun Junior College
2009 H2 Physics Prelim P1 (Solutions)
30.
To find force on Y:
Using right hand grip rule, B field on Y due to current in X is out of the plane.
Using Fleming’s left hand rule, with B acting out of plane and current flowing left, an
upward force is created.
Using N3L, force on X is equal in magnitude but opposite in direction.
31.
Since there is no change in magnetic flux for the rod when it moves, the induced
e.m.f. is zero.
32.
 
d
which is the negative of the gradient of flux linkage-time graph
dt
Since the gradient is a negative constant, induced ε is a positive constant.
33.
Vo
1
  V 2   Vo 2 .
For full sinusoidal wave, Vrms 
2
2
For half-wave rectified version given,
V2 
34.
P
Vp
Vs
V
1 Vo 2  Vo 2


. Thus Vrms  o  0.35Vo .
4  2 
8
8
V2
 Vs  50  2.0  10 V
R
Np
2000

 Vp 
 10  400 V
Ns
50
35.
KEmax = 3.5  10–19 – 3.1  10–19 = 0.4  10–19 J = 0.25 eV.
Thus stopping potential is 0.25 V.
36.
Shortest wavelength  largest photon energy  largest difference between excited
state and ground state. Highest possible excited state is closest and lower than
energy W for atom B.
37.
Doping does not change the energy gap of an intrinsic semiconductor. By introducing
dopant atoms, a new energy level is found within the energy gap. Majority charge
carriers are easily created because smaller energy difference is found between either
a) donor level and conduction band, or b) acceptor level and valence band.
38.
An important aspect of lasing action is to create more members at the excited state
than the ground state. Thus, it is more important for members at the excited state to
stay relatively longer than those at the ground state. Atoms at the ground state will
stay at this stable state until they are excited by the energy supplied from the pump.
39.
 is unique and constant for a particular radioactive source.
40.
Using C  Co e t ,
100
X
 e  ( 8 )    0.347 ; then
 e 0.347 ( 6 )  X  200 s 1
1600
1600
~ End of Paper 1 Solution~
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