# 2009-jct-jc2-h2-solutions ```Solution for 2009 JCT (H2)
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Solution Paper 2
(1a) (i)
(ii)
(c)
2
or sum of moments is zero (about any point) when the system is in
equilibrium
1800 &times; 0.35 = F &times; 2.5
F = 250 N (or 252 N)
F + FR = 1800 or correct moments equation about towbar
FR = 1550 N ecf from (i)





There is no change in the kinetic energy. Work done by car is used to
overcome friction, air resistance, heat and sound.

(a) The internal energy is a function of the state of the system. The increase in
internal energy of a system is equal to the sum of the heat absorbed by the
system and the work done on the system.

(b) (i) The temperature remains constant because there is no change in the random
kinetic energy of the substance.

(ii) The substance is undergoing a change of phase (from liquid to vapour)
because there is an increase in the random potential energy and hence an
increase in the separation between the atoms/molecules.

(c)
3
Q=ml
where l is the specific latent heat of vaporization of the substance
[-0.7 &times; 106 – (-7.9 &times; 106)] = 3.2 &times; l
l = 2.25 &times; 106 J kg-1

(a) (i) The field lines are radial and when extrapolated they appear to originate from
the centre of the sphere.

(ii) VAB = VBC
Q
1
1
Q
1
1
(
)
(
)
4o 0.40 0.50
4o 0.50 r
c
rC = 0.67 m

(b) Horizontal component of v remains constant since no force acts in this direction. An
electric force acts on the proton downward. So the vertical component of v
increases uniformly downward with time. Hence the proton describes a parabolic
path in the plane of the paper as shown.

path of
proton
E
4(a) px = 15.0 kg m s-1
px = mx vX
15.0 = mx (3.0)
=&gt; mx = 5.0 kg
(b)
time t /s
0.25
0.30
0.35
0.40
momentum of trolley X, px / kg m s-1
15.0
10.0
5.8
5.0
(c)
(d)
(i)
&lt;ax&gt; =
2.85  1.00
= -13.7 m s-2
0.250  0.385
Velocity of X, vx / m s-1
3.0
2.0
1.2
1.0
(ii)
(e)
sx = area under v-t graph = &frac12; (2.85 + 0.80) (0.40 - 0.25) = 0.27 m
By conservation of momentum,
px + pY = px’ + pY’
15.0 + 2.0(1.5) = 5.0 + pY’
pY’ = 13 kg m s-1 to the right
(f)
min EPE = change in total KE of masses
= &frac12; mx (vx’2 – vx2) + &frac12; mY (vY’2 – vY2)
= &frac12; (5.0) (1.02 – 3.02) + &frac12; (2.0) [(13/2.0)2 – 1.52]
= 20J
Section C
1 (a)
(b)
Correct shape (concentric circles) with increasing separation 
Direction correct 
(i)
Torque on coil due to magnetic force
  F d
 NBIL sin   d


 40  0.60  2.0 103  0.02  sin 90o  0.02


= 1.92 x 10-5 N m

(ii) By having the magnetic field always parallel to the plane of the coil, the
torque on the coil would then be only dependent on the current in the coil.
This ensures that the angle of deflection of the pointer would be proportional
to current. 
1 (c)
(i)
After the magnet is released, it accelerates under gravity and its velocity
increases. When the magnet approaches the coil, the magnetic flux linking
the coil increases. According to Faraday’s law, an induced emf is produced
which increases to a maximum when the N-pole of the magnet reaches the
coil. By Lenz’s law, the direction of the induced emf opposes the increasing
flux. After the S-pole leaves the coil, the magnetic flux linking the coil
reduces and the induced emf is greater in magnitude due to the higher
speed of the magnet. This emf is also in opposite direction compared to
that for incoming magnet.
(ii) 1.
As the magnet moves downwards, the induced emf produced would
generate an induced current in the coil  which flows in such a way as
to produce an induced Magnetic field which exerts an upward magnetic
force on the magnet.
2. Since an induced current is produced in the coil, electrical energy must
be generated in the coil. Loss in p.e. of magnet is converted to gain in
k.e. of magnet plus electrical energy generated in coil. Hence, lesser
amount of k.e. is gained by the magnet. So it is slower than before.
(iii)
1.
e.m.f./V
2.50
1.25
0
-1.25
-2.50
0
t/ms
100
200
300
400
300
400
 for greater magnitude
 for same time interval
2.
e.m.f./V
2.50
1.25
0
-1.25
-2.50
0
t/ms
100
200
 for greater magnitude
 for later time
(d)
(i)
P  IV
P 4200 kW
I 
 382 A
V
11 kV

(ii) For transformer U,
N s Vs
11


 0.040
N p V p 275

2(a)(i)
E
hc

6.63  10 34  3.0  10 8

254  10 9
 7.83  10 19 J
2(a)(ii)
ET
N
 ( )hf
t
t
N P 210  12  10 6
 

t
hf
7.83  10 19
 3.22  1015 s 1
P
2(a)(iii)
N'
)e
t
 2.7  1013  1.6  10 19
I (
 4.32  10 6 A
2(b)(i)
Stopping potential is the potential difference that must be applied
between the cathode and the anode so that the most energetic
photoelectron emitted just fails to reach the anode.

From eVs  h
2(b)(ii)
c


If light of higher wavelength was used, the energy of the photons would
decrease. Since the work function &Oslash; is unchanged (same metal used),
the kinetic energy of the emitted electrons would decrease. Hence the
stopping potential would decrease.

I
2(b)(iii)
1. λ = 254 nm
2. λ &gt; 254 nm
V
0
(c) (i)
(ii)

Since the energy levels are discrete (quantized), electron transitions
between the levels involve absorption or emission of photons of
discrete/specific wavelengths resulting in spectral lines.

1.
Energy of photon =
hc

=
6.63x10 -34 x 3.00 x 10 8
= 4.56 x10-19 J
9
436x10
= 2.85 eV
Since E4 – E2 = 2.85 eV, the transition is from E4 to E2.
2.

The bombarding electron must have a minimum kinetic energy of 13.06
eV in order for the hydrogen atom to be excited from the ground state E1
to E4. Only then can a transition E4 to E2 take place. So the minimum
value of V is 13.06 V.

```