THE INCREDIBLE COLLEGE

A-LEVEL MOCK EXAMINATION

PHYSICS PAPER I

SOLUTIONS

1. (a)(i)= 2 f½

= 2 

= 4.19 rad s-1½

centripetal force= mr2½

= 0.2  0.3  4.192

= 1.05 N½

(ii)correct diagram1

correct labels1

(iii)Resolving the tension into two components,

T sin =mg½

T cos =F½

T2 = F2 + (mg)2½

T2 = 1.052 + (2)2

T = 2.26 N ½

(b)(i)New moment of inertia= mr12½

= 0.2 0.22

= 0.008 kg m2½

1.(b)(Continued)

(ii)When the string is pulled in, the force acts through the axis of rotation.

Thus, there is no torque applied to the bob and its angular momentum is unchanged.

I11 = I22½

mr121 = mr222

0.018 4.19 = 0.0082½

2 = 9.43 rad s-11

(iii)Change in k.e.

= I222 I112½

= mr2222 mr1212

=  0.008  9.432 0.018  4.192

= 0.356  0.158

= 0.198 J½

Work is done by the hand to pull the string downwards and k.e. of the bob increases.1

(iv) From (a)(iii), tan  =

tan  = ½

angle of depression is decreased.½

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2.(a)Applying Hooke's law to the car in equilibrium,

Fs = kx½

where Fs is the force exerted by each string on the car.

Since the car is in equilibrium,

4 Fs = mg½

4 kx = mg

4  k  (0.1) = 1200  10½

k = 30 000 Nm-1½

2.(Continued)

(b)(i)When the car is pressed downwards, the force exerted by each spring becomes

Fs' = k(e + x)½

So the net force in the direction of increasing x(i.e. downward)

mg  4  k(e + x) = ma½

4 ke  4 ke  4 kx = ma

ma =  4 kx

a  - x½

The motion is simple harmonic.½

(ii)From (b)(i), 2 = ½

T = ½

= 0.63 s1

(c)(i)=

= 8.94 rad s-1½

T = 0.70 s ½

(ii)The load remains in contact with the car.

N  0½

mg  mA'2 0½

g  A'2 0

A 0.125 m½

 maximum A is 0.125 m.½

(d)

Without dampers :Cosine curve1

With dampers : Amplitude decreases with time exponentially.1

Periods becomes longer1

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3.(a)(i)

1

(ii)=2l

v=f0

f0= = ½

=

=550Hz½

(iii)v=

f0=½

T=4l2f02 = 4(0.3)2(550)2(3103)½

=326.7 N1

(iv)They have different qualities of sound. ½

Although the sound of the same notes from two instruments have the same fundamental frequencies, they contain different combination of harmonics. ½

(b)(i)Let I be the intensity of noise produced by one machine and n be the maximum number of machine.

For ONE machine,

∴95 = 10 log 1

For n machines,

10 log 1001

10 log n + 10 log 100

10 log n + 95  100

10 log n 5

n 3.16

∴Maximum number of machines is 3.1

3.(b)(Continued)

(ii)For one machine,

95 = 10 log

I = 109.5I01

For background noise,

90 = 10 log

I = 109I01

Total intensity level=10 log []1

=10 log []

=100.2 dB > 100 dB½

So the noise level exceeds the noise limit of 100 dB.½

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4.(a)Since separation between the whistles < distance of observer from the whistles, we can use the equation to determine the distance between the successive minima. 1

Hence, 0.85=

=0.034 m

f=1

(b)Light is emitted by a source as a result of electron transition within individual atoms of the source. These transitions occur randomly and each gives to a short burst of light that lasts about s. 1

The two separate light sources are incoherent since they do not have a constant phase relationship. (Interference occurs only if the sources are coherent.) 1

(c)(i)Fringe separation 1

By , slit separation = 1

4.(c)(Continued)

(ii)Extra optical path caused by the thin transparent sheet is

m1

½

Therefore, the whole interference pattern will shift through 10 complete fringe separations to the side with the plate. 1

The original central position of the interference pattern is now occupied by the 10th order maximum. ½

(iii)The wavelength of laser beam will be reduced in water.½

By , the fringe separation will be reduced too.½

(iv)By , .1

Therefore, and½

there are bright lines on the screen.½

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5.(a)(i)For the larger sphere,

1

For the smaller sphere,

1

(ii)By conservation of charge, 1

After the re-distribution of charge, the spheres have the same electric potential.1

Therefore, 1

1

(b)(i)(1)Maximum e.m.f. = 1

(2)To adjust the maximum p.d. across the potentiometer wire AB so that the balance point can lie in the mid-range of the wire. 1

5.(b)(Continued)

(ii)The e.m.f. of the cell E, 1

When K is closed, the terminal p.d. across the 20  resistor is where r is the internal resistance of the cell. 1

Since 1

Therefore, 1

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6.(a)Lenz’s law states that the induced e.m.f. and induced current oppose the original change of magnetic flux. 1

Faraday’s law of electromagnetic induction states when the magnetic flux linkage passing through an area enclosed by a loop changes, an e.m.f. is induced which equals minus the rate of change of the flux linkage. 1

(b)(i)At the beginning, the rod XY is accelerated by F.½

By Lenz’s law, a magnetic deflection force FB, opposite to F, is induced on the rod XY. ½

FB increases with the speed of the rod XY.½

When , the rod XY attains a terminal speed.½

(ii)The induced current flows from X to Y.1

(iii)(1)1

(2)Equivalent resistance of the circuit, 1

Induced e.m.f., 1

(3)Terminal speed, 1

(iv)Mechanical power =1

Power loss =1

Mechanical power developed by the applied force equals to the total power loss through the resistors since energy should be conserved. 1

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7.(a)(i)I2R2=VBE ½

I2= =

=8.23 A½

(ii)Vin=I1R1 + I2R21

I1=

=

=10.6 A1

(iii)I1=I2 + IB1

IB=I1 I2

=(10.6  8.23) A

=2.37 A1

(b)(i)IL=IB ½

=100(2.37106)

=0.237 mA½

(ii)Vout=VCC ICRL½

=6  (0.237103)(3.5103)

=5.17 V½

(c)Yes, the output voltage is so high that it will be easily saturated when the input ½

voltage is lower.

Moreover, a voltage that is too close to either VCC or zero will have a large distortion½

when it is used as a linear amplifier.

(d)Maximum output swing when Vout = VCC/21

∴VCC=ICRL + Vout

6=(0.237103)RL + 3

RL=12.68 k1

7.(Continued)

(e)

quiescent Vout at 3V½

1800 out of phase with input signal½

peak to peak voltage of output signal = 76.08 V1

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8.(a)(i)K.E. gained=P.E. loss

½ mv2=mgh

v=

Before impact,v1= = 4 ms1 (downwards)1

After impact,v2= = 2 ms1 (upwards)1

(ii)F=rate of change of momentum

=

=

=150 N½

Total resultant force on frame

=impulsive force + weight of block

=150 + 510

=200 N½

(iii)Stress= =

=6.67108 Nm-21

8.(a)(Continued)

(iv)E= =

e= = ½

=6.06103 m½

(v)From (a)(iv),

e= =

e=

=½

=(6.06103)

=1.52103 m½

(b)(i)F=½

=½

(ii)At equilibrium separation r0, F = 0,½

=0

r0= =

=2.8701010m½

(iii)k = when r = r0½

=  ()½

∴k= []

=0.122 Nm1½

8.(b)(iii)(Continued)

∴E= =

=4.26108 Pa½

(iv)The atomic interaction starts to break down after the interatomic attractive force has reached its maximum value. ½

 = 0½

 = 0

r = 3.181010 m 1

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9.(a)K=eV1

=(1.61019)(100103)

=1.61014J1

(b)I= =

=6103A 1

P=IV = (6103)(100103)

=600 W1

(c)No. The energy supplied to the electrons from the cathode is very small. The energy of the thermionic energy at the cathode can be assumed to be zero. 1

Their gain in K.E. is due to the accelerating electric field between the electrodes with a p.d. of 100 kV. 1

(d)Kmax=hmax = h1

0=h = (6.631034)

=1.241011 m1

9.(Continued)

(e)

(f)- High melting point : it can prevent the target from melting

- High atomic number : it is more efficient to decelerate the electron to give X-rays

- Good conductor of heat: it can transfer heat to the cooling system more efficiently.

(ANY TWO, each 1 mark)2

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10.(a)(i)Gain in p.e. = Loss in k.e.½

½

1

(ii)By (a)(i), nuclear radius of a gold atom m1

1

The density of the nucleus of a gold atom

1

(b)(i)1

1

(ii)Let V be the volume of the blood inside the dog.

Assume that atoms are evenly distributed in the blood of the dog.1

Activity of the sample = 1

1

1

10.(b)(Continued)

(iii)Archaeological dating for the estimation of the age of rocks and fossils

Radiotherapy for the treatment of cancer

Tracers for the detection of an underground leak

Any TWO reasonable answers1

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