CCC Yenching College C. K.
Ng
AL Physics MC Answers
Year:1994
Question Number:4,5,10,16,22,27,37,38,39
1994MC(4)
Impulse = change of momentum
Change of
Final momentum
momentum
Original momentum
The momenta are moving in different directions, so we need to do a vectorial subtraction.
Impulse = 0.5 20 2 + 30 2 = 18 kg m s-1
1994MC(5)
To execute circular motion, at the top, the centripetal force required is provided by the weight
mv2/R = mg … … … . . (1)
v = gR
Let v’ be the speed at the bottom.
By conservation of energy, mv’ 2/2 = mv2/2 + mg(2R) … … … … … . (2)
From (1) and (2), v' = 5 gR
1994MC(10)
a b
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CCC Yenching College C. K. Ng
(1) Equal mass, so (a) has a larger MI.
(2) Period is proportional to I , so (a) has a longer period.
(3) They are released at the same height. KE at the lowest point = loss in gravitational PE = mass x
g x distance from axis of rotation to center of mass. Their centers of mass are both situated at
their centers.
Same PE loss, so same KE.
1994MC(16)
Beat frequency = f1 – f2
From the information, we know
fX - fY = 3 or fY - fX = 3 (we do not know which one is higher)
fX - fZ = 1 or fZ - fX = 1
(1) Not sure although it is possible
(2) Not sure although it is possible
(3) X is the highest frequency so only fX - fY = 3 and fX - fZ = 1 are possible. The
order is fX > fY > fZ.
1994MC (22)
Fl EA
Young’s modulus E = , so the applied force F = e . Compare with Hooke’s law, we identify
Ae l
EA
k= .
l
l
A wire is cut into two and arranged side by side. Effectively, l → and A → 2 A , so k is four
2
times larger.
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CCC Yenching College C. K. Ng
1994MC(27)
A B
A=B=C=D=1C
C D
To put the four charges together, we consider a procedure like this
(1) At the beginning, there is no charge, so no energy is needed to put A to the
top-left corner.
(2) At the presence of A, B is brought to the top-right corner,
1 1
energy required = qB (potential due to A) = 1( ( )=
4πε 0 1 4πε 0
(3) At the presence of A and B, C is brought to the bottom-left corner,
1 1
energy required = qC (potential due to A and B) = 1( + )
4πε o 4πε 0 ( 2 )
(4) At the presence of A, B and C, D is brought to the bottom-right corner,
2 1
energy required = qD (potential due to A, B and C) = 1( + )
4πε o 4πε 0 ( 2 )
Total energy stored in the system = sum of the above all energies.
4 2 1
= + = (4 + 2 )
4πε 0 4πε 0 2 4πε 0
In general, if there are charge q1, q2, q3, … …..q
n and their separation are r12, r13, r23, … …, then the total
energy stored in them is
1 q1q 2 q1 q3 qq
( + + ......... + 1 3 + .............)
4πε 0 r12 r13 r13
Any two of them “meet” one time only.
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CCC Yenching College C. K. Ng
1994MC(37)
The discrete points will match the solid line when each f is decreased by a fixed amount or V is
increased by a fixed amount.
hf -φ = eV
A. The intensity does not affect the stopping potential
B. "A fixed zero error" à each data differs the true value by a fixed amount.
C. "read the wrong scale on his voltmeter so that his readings always double the actual readings" à
Actual reading Wrong data
2V 4V
3V 6V
4V 8V
The wrong data is always larger than the true reading, but the difference is NOT a
constant.
D. "wrong polarity of the d.c. supply"à the electrons will not be stopped, so no stopping voltage V
will be found.
E. If V is plotted against wavelength, the graph will not be a straight line.
hc
− Φ = eV (a straight line must have the form y = mx + c)
λ
1994 MC (38)
X
En = −
n2
3
First excited state (n = 2) to the ground state (n = 1), hf = E2 – E1 = X.
4
5
Drop from n = 3 to n = 2, hf’= E 3 – E2 = X
36
f' 5
= f’ = 0.19 f
f 27
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CCC Yenching College C. K. Ng
1994MC(39)
a b c
Y
1. The diode is assumed ideal, i.e.
resistance of the diode = 0 (perfect conductor) when the diode is forward-biased.
resistance of the diode = infinity (perfect insulator) when the diode is backward-biased.
2. When two resistors are connected in series, the larger the resistance, the higher the p.d.
e.g.
100 Ω
12 V
When R = 100000000 Ω, V ≈ 12V
When R = 0.000000001 Ω, V ≈ 0V
3.
Referring to the resistor-diode circuit, the diode is forward –biased when X is positive w.r.t. Y (a
to b), The resistance of the diode is much smaller than that of the resistor, so
p. d across the diode ≈ 0
When X is negative w.r. t. Y (b to c), the diode is backward-biased, the resistance of the diode is
much larger than that of the resistor, so
p .d . across the diode ≈ external applying a. c
So, the p.d. across the diode is
a b c
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CCC Yenching College C. K. Ng
1994(AS) MC (17)
Power input = 12 V x 0.5A = 6 W
Useful power output = 10 N x 0.4 ms-1 = 4 W
Power loss due to internal resistance of the armature = I2 R = 2 W
(0.5)2R = 2 R = 8 Ω
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