VICTORIA JUNIOR COLLEGE
SUGGESTED SOLUTIONS TO
2009 PHYSICS H2 P2 PRELIM
EXAMS
1(a) Along OA, the ball released at
time t = 0, moves down the slope with
a constant acceleration before impact
with the block
Along AB, the ball hits the block and
is brought momentarily to a stop
before moving up the slope.
Along BC, the ball moves up the slope
with a constant deceleration. At time
t = 2.2 s, the ball momentarily comes
to a stop. It then moves down the slope
with a constant acceleration.
(b)(i) From the graph,
acceleration =gradient of line OA
v−u
=
t
2.4 − 0
=
= 2.0 m s-2
1.2 − 0
(ii) Length of the incline = area under
the graph OA
1
=
(1.20)(2.4) = 1.44 m
2
(iii) Mean force =
(0.70)(2.4 + 1.6)
0.2
= 14 N
m(∆v)
∆t
=
2(a) At C, for circular motion of the
particle, T + mg = mv2/L
T = 0 if the string is just taut.
Hence v = (gL)1/2
(b) Using conservation of energy, as
the particle goes from B to C,
(1/2)mV2 = mg(2L) + (1/2)mv2
Hence V = (5gL)1/2
At this point, the tension in the string is
the greatest.
3(a)(i) The tension in the string will be
larger when the pendulum is
accelerating upwards. This is
equivalent to an increase in the
effective g.
(ii) geff = 9.81 + 1.0
= 10.81 m s-2
T = 2π(L/geff)1/2
= 2π(1.0/10.81)1/2
= 1.91 s
(b)(i) In a space-craft orbiting the
earth, geff is zero.
L
In the formula T = 2π
, the
g eff
period tends to infinity.
This means that after the bob of the
pendulum has been displaced, the
pendulum will not oscillate.
(ii) In a spring-mass system, both the
mass m and the restoring force
constant k are independent of
gravitational acceleration.
When the mass is given a small
vertical displacement downwards and
released, the system will oscillate
simple harmonically with a period
m
given by T = 2π
.
k
Alternative answer:
In the space-craft, elastic PE in the
spring can be converted into KE of the
mass, and vice versa.
When the mass is given a small
vertical displacement downwards and
released, the system will oscillate
simple harmonically with a period
m
.
given by T = 2π
k
(c) The string is most likely to break at
point B, i.e. the lowest point.
1
H2 Physics Papers, Notes at alevelphysics.co
4(a)
N = normal reaction
5(a) Electrical PE of electron at
potential of +4.00 V is U = qV
= (−1.6 × 10-19)(+4.0) = −6.4 × 10-19 J
= - 4.0 eV
N
BIL
mg
indicating that Q is the higher
potential.
(b)
300
Potential energy / J
position
0
Vo
B
−6.4 × 10
-19
first block
(b) Net force = BILcosθ − mgsinθ
= ma
a=
(1.0)(
12
)(0.50) cos 30 0 − (0.020)(9.81) sin 30 0
24
0.020
≈ 5.9 m s-2 up the slope.
(c) When terminal velocity of the rod
is reached, resultant force acting on it
is zero. Hence BILcosθ = mgsinθ
(0.020)(9.81) sin 30 0
I=
(1.0)(0.50) cos 30 0
I = 0.23 A
(c) Refer to the plan view of the
diagram below.
P
second block
gap
(c) Height of potential barrier = Vo
= 4.0 eV.
Transmission coefficient,
T ∝ exp(−2kd)
2m(V0 − E )
where k =

We require that the transmission
coefficient remains unchanged:
T′ = T
∴ A exp(−2k ′d ′) = A exp(−2kd )
where A = constant of proportionality.
∴
exp(−2
2m(V0 − E ′)

d ′) = exp(−2
2m(V0 − E )

d)
∴
−2
2m(V0 − E ′)

(
)
d ′ = −2
(
2m(V0 − E )

d
)
∴ (V0 − E ′) d ′ = (V0 − E ) d
We know that Vo = 4.0 eV and
E = 1.0 eV and new d′ = 1.10d.
∴
(4.0 − E ′) (1.10)d = (4.0 − 1.0) d
v
(
)
(
)
∴ 1.10 2 (4.0 − E ′) = 3
∴ E′ = 1.52 eV.
Q
End Q is at the higher potential.
Fleming’s Right Hand Rule shows the
second finger pointing from P to Q,
(d) No. of electrons that hit barrier per
N I 4.0 × 10 −6
second =
= =
t
e 1.6 × 10 −19
= 2.5 × 1013 s-1.
2
H2 Physics Papers, Notes at alevelphysics.co
∴ no. of electrons that successfully
tunnel through per second
= 1.83 × 10-12 × 2.5 × 1013
= 45.8
B
B
6(a)(i) R = Ae T ⇒ ln R = ln A +
T
0
0
Temperatures θ = 50 C and 80 C
correspond to T = 50 + 273 =323K and
353 K respectively.
From graph, R = 110 Ω at 50 0C and
R = 50 Ω at 80 0C respectively.
B
Hence ln 110 = ln A +
…..(1)
323
B
…..(2)
and ln 50 = ln A +
353
(1) – (2) gives
B
B
ln 110 − ln 50 =
−
323 353
Hence 0.78845 = B(2.63 x 10-4)
B ≈ 3.0 x 103 K
(d)(i)
In (1), A ≈ 1.03 x 10-2 Ω
(d)(ii) V =
(ii) A more reliable method to
determine the values of A and B would
be to plot a graph of ln R against 1/T
B
using the equation ln R = ln A + .
T
A and B would be more reliably
determined from calculations involving
the y-intercept and the gradient of the
line plotted.
B
T
−2
(b) R = Ae ⇒ R = (1.03x10 )e
R ≈ 35.8 Ω
3.0 x103
95 + 273
6.0 V
40.0 Ω
Ideal
voltmeter
At 30.0 0C, the resistance of X is
approximately 188 Ω.
By the potential divider principle,
voltage across the R = 40.0 Ω resistor
is
R
V =
E
R + RX
40.0
V =
(6.0) ≈ 1.05 V
40.0 + 188
R
E
R + RX
As temperature rises, the resistance RX
decreases.
According to the equation above, the
voltmeter reading should increase.
(d)(iii) The voltmeter could be
replaced by a buzzer.
When the temperature is low, the p.d.
across the buzzer is low and it does not
sound. When the temperature rises
beyond a certain level, the p.d. across
the buzzer rises, causing it to sound.
(e) E g (T ) = E g (0) −
(c)
I
V
V
αT 2
T +β
From the graph for X,
when T = 900 K, Eg (T) ≈ 0.40 eV.
when T = 400 K, Eg(T) ≈ 0.62 eV
α 900 2
….(1)
0.40 = 0.744 −
900 + β
α 400 2
…..(2)
400 + β
In (1), α 900 2 = 0.344(900 + β ) ….(3)
0.62 = 0.744 −
3
H2 Physics Papers, Notes at alevelphysics.co
In (2), α 400 2 = 0.124(400 + β ) ….(4)
9 2 0.344(900 + β )
(3)
gives 2 =
0.124(400 + β )
(4)
4
10.0(400 + β ) = 5.504(900 + β )
4000 + 10 β = 4953.6 + 5.504 β
β ≈ 212 K
0.344(900 + 212)
900 2
-4
α ≈ 4.72 x 10 eV K-1
In (3), α =
(f) When temperature rises, more
electrons have sufficient energy to go
from the valence band to the
conduction band in the semiconductor. These electrons in the
conduction band and the holes left
behind in the valence band contribute
to the increased conductivity of the
semi-conductor.
Additionally, the jump from valence to
conduction band by electrons is made
easier at higher temperatures by the
energy bandgap decreasing in value.
******** END ********
4
H2 Physics Papers, Notes at alevelphysics.co