2014 Prelim H2 Physics Paper 2 Solutions

1a Conditions for equations to be valid: Constant acceleration Motion in a straight line

1b(i) vR=√(v x2+vy 2)

= (10.02 + 3.02) = 10.44 10 m s-1

= tan-1 () = 73.3 73

1b(ii) 1. Initial velocity of sandbag is vR with angle to horizontal.Path is symmetrical and parabolic

1bii 2. Taking downwards as positive,Sbag = -10.0 (4.0) + (9.81)(4.0)2

= 38.48 mSballoon = - 10.0 (4.0) = - 40.0mDistance apart = 40.0 + 38.48 = 78.48 78 m

1

vR

vR

2ai There is zero net force acting on the signboard.There is zero net torque about any axis acting on the signboard.

ii

G is at the intersection of lines of action of T1 and T2

iii Weight = T1 cos 40 + T2 cos 50= 300 cos 40 + 252 cos 50= 391.8= 390 N

b Net force acting on the board immediately after released = 20 N

Initial net acceleration = F/m = 20/(391.8/9.81) = 0.501 0.50 m s-2 [allow ecf from (aiii)]

2

T1

T2signboard

4050

G

3ai

aii

3bi

When the rock is momentarily at rest, KE is zero and total energy = GPE(ai) Hence, total energy of rock is the intersection of line AB and GPE graph(aii) At distance R, KE is zero. At distance RE, KE = total energy – GPE

(bi) For a rock to escape to infinity with minimum speed, total energy at infinity is zero.

3bii Possible explanation includes Moon would attract the rock potential at Moon(’s surface) is

o not zeroo less than 0o lower than at infinity

moon is nearer; less gain in GPE from Earth to moon

Thus, escape speed would be lower.

3

Correct point P

Correct curve with: KE = 0 at R and KE + PE = ET

Roughly correct energy at RE

KE + PE = ET

Roughly correct shape between RE and R

Kineticenergy

Correct point Q

X P

X QR

total energy for rock in (a)

EnergyA

B

r0 RE

4a DiffractionDiagram (if drawn) must show light source (e.g. light, laser), object causing diffraction (e.g. single slit) and means of observation (e.g. screen)For observable diffraction, the size of the slit should be of the same order as that of the wavelength.OR Description of setup must mention the above items.

It was expected that the light on the screen would be the size of the slit but the width of the band of light seen on the screen is greater than the slit width. This demonstrates the spreading effect on a wave as it passes through an aperature.

OR

2-Source InterferenceDiagram (if drawn) must show light source (e.g. light / laser), object causing interference (i.e. double slit) and means of observation (e.g. screen)

Light source must be monochromatic and coherentOR Description of setup must mention the above items.

Instead of two bright fringes of the screen, a fringe pattern containing light and dark fringes is observed.

OR

Difraction gratingDiagram (if drawn) must show light source (e.g. light / laser), diffraction grating and means of observation (e.g. screen)

Light source must be monochromatic and coherentOR Description of setup must mention the above items.

A pattern of bright narrow lines/spots with dark regions between them are observed on the screen. The spacing between the lines/spots are not equal.

Note: drawing of wavefronts / ripples is not accepted as there are multiple planes for light waves.

4b 1. For a given metal, there is a threshold frequency below which no emission of photoelectrons occurs regardless of the intensity of radiation.

2. Emission of photoelectrons occurs with no observable time lag at frequencies of greater than threshold frequency, even at very low intensity.

3. Increasing the intensity of the radiation has no effect on the maximum energy of the electrons

4. The rate at which electrons are emitted increases proportionally with the intensity of the radiation.

5. The maximum kinetic energy (not kinetic energy) of the emitted photoelectron is dependent on / increases linearly (not proportional) with the frequency of the incident radiation (above threshold frequency), even with low intensity.

Any 3 from the above.

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5 (a) One ohm is the resistance of a conductor in which the current is 1 ampere when a potential difference of 1 volt is applied across it.OROne ohm is the resistance of a conductor in which the ratio of potential difference across it to the current flowing through it is 1 volt per ampere.

(b) Power dissipated in r = Total Power – Power dissipated in LoadI2 r = ε I – PL

(0.10)2 r = (2.5)(0.10) – 0.23r = 2.0 Ω

(c) Single cell

Resistance of Load, R1 ¿0.230.102 =23 Ω

Two cells5.0 = 0.15R + 0.15 x 4.0Resistance of Load, R2 = 29 Ω

As an ohmic device has constant resistance, since the load’s resistance has changed/ increased, the load can be deduced as a non-ohmic device.

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6a One tesla is the magnetic flux density which causes a force per unit length of one newton per metre on a straight wire carrying a current of one ampere and is at right angles to the direction of the magnetic field.

6bi As seen from the increased balance reading, there is a downward force on magnet due to wire carrying current.By Newton’s third law, there is an upward force on wire by magnet.By Fleming’s left hand rule, pole P is a north pole

6bii By Newton’s 2nd law, W – BIL =0 W= BIL2.3 × 10–3 × 9.81 = B × 2.6 × 4.4 × 10–2

B = 0.20 T (g = 10, loses this mark)

6cReading for maximum current = 2.6 x √2 = 3.68 AF = BIL =(0.20)(3.68)(4.4x10-2) =0.032 Nmg = 0.032 N, m = 0.032/9.81 = 3.3 gtotal variation of mass = 2 x 3.3 = 6.6 gOR

m= BLg

I, m I

mmax

mdc=

I max

I dc

mmax

2 .3=2 .6 √2

2.6

mmax = 2.3√2 = 3.3 gtotal variation of mass = 3.3 x 2 = 6.6 g

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7(a)

The cut-off wavelength corresponds to the most energetic photon that can be produced. That happens when all the kinetic energy of an accelerated electron is lost in a single collision/interaction with the target atom in producing one photon.

(b) (i) area = 0.200 m x 0.300 m = 0.0600 m2

Accept a range of 0.0300 m2 ≤ area ≤ 0.120 m2

(ii)

Area of a graph:

no of grains =

0 . 060010−6×10−6

= 6.00×1010

no of photons = 6.00×1010 ×10 = 6.00×1011

energy = 6.00×1011 × 10-15 = 6.00 x 10-4 J

(c) implied probability of arrival proportional to square of amplituderatio = 42 = 16

7

10-6 m

10-6 m

8ai From Fig. 8.1, d = 57.5 m, d/v = 57.5/25.0 = 2.30 sAccept 57.0 m < d < 58.0 m, 2.28 s < d/v < 2.32 s

aii Plot the point corresponding to v = 25.0 m s-1

Draw the line of best fit for all the points (fair scatter of points about the best fit line)

bi Gradient = (3.00-1.40)/(36.0-11.5) = 0.065312 = 0.0653Accept range : 0.062 < gradient < 0.069

bii From d = + Bv, = + B, gradient = 1/(2A)A = 1/(2x0.0653) = 7.652 = 7.66 m s-2 range : 7.3 < A < 8.1

biii Sub. (11.5, 1.40) into the equation = + B1.40 = 0.0653(11.5) + B , B = 0.6489=0.649 s range : 0.61< B < 0.69

biv Since A has the unit of acceleration, is most likely representing the braking distance or the stopping distance where A is the deceleration of the motion. The braking distance depends on the speed and the deceleration of the braking force applied.

B has the unit of time. Bv will represent the thinking distance where B is the thinking time or reaction time of the rider. Before the braking force is applied, the thinking distance depends on the speed and the thinking time.

c Using d = + Bv, d = 50.02/(2x7.656) + 0.6489x50 = 195.7= 196 m Distance from obstacle = 200 – 195.7 = 4.28 m

d Friction between the tyres and track surface will be reduced, reducing the magnitude of the

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v / m s-1

5.0 10.0 15.0 20.0 25.0 30.0 35.0 40.0ˡ ˡ ˡ ˡ ˡ ˡ ˡ1.00

1.40

1.80

2.20

2.60

3.00

3.40

•

•

•

•

•

d/v / s

(11.5,1.40)

(36.0,3.00)

•

motorcycle’s deceleration. This increases the required braking distance.Visibility of motorist’s is reduced, increasing the reaction time and hence the thinking distance. Thus stopping distance has to be increased.

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Suggested Solution

AimTo investigate how the magnetic flux density B between the 2 circular coils varies with the current I in the coils

Procedure

1) Set up the apparatus as shown above.2) Switch on the power source to pass a current through both coils to produce a magnetic

field.3) Record the value of I from the ammeter.4) Place the hall probe between the coils. Record the value of the magnetic flux density B

from the datalogger.5) Repeat steps 2 to 4 to obtain further values of B for different I by changing the resistance of

variable resistor.

6) From B=kIn lg B=n lg I +lg k . Plot a graph of lg B against lg I. The value of n can be

obtained from the gradient of the graph.

Control of variables1) The separation of the 2 coils should be kept constant by using the meter rule to check and

adjust the separation before every reading. 2) The position of the hall probe between the 2 coils should be kept constant by clamping the

hall probe to a fixed position between the coils.

Safety and Accuracy 1) The plane of the hall probe should be perpendicular to the magnetic field within the 2 coils.2) The axes of the 2 coils must coincide to ensure that the magnetic field in the region

between the 2 coils is uniform.3) The coils are to be close to one another to ensure that the magnetic field between the coils

is uniform.4) The position of the hall probe is along the centre axis of the 2 coils and it is equidistant from

the coils.5) The range of the current used should be kept small to reduce heating effect in coils by

using the variable resistor.

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Hall probe

datalogger

Power source A