2010 Prelims H2 P1 and P2 Guide

8/4/2019 2010 Prelims H2 P1 and P2 Guide

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P1 answers

P2 guide

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13 Q14 Q15 Q16 Q17 Q18 Q19 Q20

A C B D D A D B C A B A C D D D D A B C

Q21 Q22 Q23 Q24 Q25 Q26 Q27 Q28 Q29 Q30 Q31 Q32 Q33 Q34 Q35 Q36 Q37 Q38 Q39 Q40

A A B B D B C B A D B D C B D C D B C A

1(a)(i)3 V d

6

π = × , where d = 22 cm

=∴V 5.6 x 10-3 m3

1(a)(ii) 2

2

1= mv KE , where v = 10 m s

-1(accept values from 8 m s

-1to 12 m s

-1) and

mass = 60 kg (accept values from 40 kg to 100 kg)

=KE ∴ 3000 J (accept values from 1900 J to 7200 J);

(Note must show simple workings where necessary.)

(iii) Density is between 800 to 1100 kg m-3

1(b)(i) The values are close to each other, -35 kg m± ,

but very far from the expected value which should be less than density ofwater (1000 kg m3).Therefore, set B is precise but inaccurate.

1(b)(ii) mass of beaker is included in the measurement of mass orweighing balance has large positive zero errorNote that you need to explain why value is larger than expected and not simplysay weighing balance has zero error

2a(i) Graph is a straight line through the origin fl a is directly proportional x

(must pass through origin for directly proportional)Graph is a straight line with negative gradientfl a is in the opposite direction of x

The graph shows that the motion is bounded within ± 50 mm.

Hence it satisfies the defn of SHM which states that it is a to and fro motionwhere the acceleration is directly proportional to i ts displacement from theequilm pt and it acts in the opp direction to the displacement from the equilibriumpoint.

(ii) ω2 = 250 rad s-1

Τ = 2π/ωΤ = 0.40 s

(iii) Cosine curve

correct amplitude of 0.050 m and with 22

1complete waves

(b)w =

T

π 2=

2

2π =p rad s-1

Apply x = x0 sin wt where x0 = 50 mm and x = 25 mm

Time interval when shuttle remains open ,

fl 25 = 50 sin2

2π t

t = 6

1s fl t = 0.167s

Note that the graphs for SHM is sinusoidal not a linear function hence cannotuse proportionality.

Additional qn: Determine the time interval the shuttle is opened if the shuttle isopened when the bob moves from the 700 mm mark to 650 mm mark.



3 (a) force on magnet/ balance is downwards since reading increasesso by Newton’s third law force on the wire is upwards

(b) Using Fleming’s Left hand rule, it can be deduced that pole P is a north pole(c)

(d) Magnetic force on balance provides additional force recorded on balanceF = B I L sin 90°and F = mg2.3 x 10-3 x 9.8 = B x 2.6 x 4.4 x 10-2 B = 0.20 T

4ai Draw a suitable circuit diagram for this investigation.

A

V

aii

I

V

9V supply

Ammeter

Filamentlamp

Voltmeter

Rheostat

aiii As I and V increases, rate of atomic vibration increases Number of free electrons remains the same, hence resistance increases

Hence the graph shows that the resistance increases as V increases(Note that quite a significant number of candidates mentioned that 1/gradient isthe resistance; that is incorrect. R is defined as the ratio of V to I where V is thepd across the conductor and I, the current flowing through the conductor.)

aivKnow that R =

l

A

ρ ; or mention that R depends on ρ

Therefore ρ must have been unique as R is unique at each temperature.

5(a)(i) Vs is (stopping potential). Electrons with max KE cannot reach the anodehence cannot be collected as even they do not have enough energy toovercome the electrostatic repulsive force between cathode and anode.

(ii) Electrons are emitted with a range of KE, hence when anode is make lessnegative with respect to cathode, some electrons may be able to overcome theelectrostatic repulsive force and reach the anode

(iii) Saturation not achieved immediately once V is +ve because the electrons are

emitted from the metal surface randomly in different directions. (Hence with higher V, the path of more electrons may be altered so that it is able to reach the anode due to the increased in the magnitude of the electric force.)(Saturation current is achieved at V 1 when all emitted electrons are collected .)

5(b)(i) φ = 5.32 x 1.6 x 10-19

= 8.51 x 10-19

J

hc/ λ > 8.51 x 10-19

J

λ< 2.336 x 10−7 m

Hence λ = 2.33 x 10−7 m(ii) p

=h/ λ

p

=2.85 x 10-27

N s





(c)(i)

V QW ×=

= 20 x 1 x 108

= 2 x 109 J

Hence the order of magnitude is to 109.

(ii)VI P = or

t

W P =

=108 x 20 0009

3

2 10

10

x −=

= 2 x 1012 W = 2 x 1012 W

Hence the order of magnitude is to 1012.

(iii)

dr

V d E −=

07.0

1080 3×= E

= 1.14 x 106 V m-1

Since the electric field at the top of the tower is greater than the order of 105 (afew hundred thousands V m -1), a return stroke is likely to occur.

(d) (i) The electrical discharge results in heating up of the atmosphere around thelightning channel.The sudden rise in pressure causes the air around it to expand rapidly resultingin an explosion of the air.

(ii) (Assuming time taken by light is negligible)

Distance (m) = 330 (m s -1) x t (s)

= 330t m

= t 1000

330km t =

8

10

(3 10 330)

9.9 10

x d

x

−

=3

t km

(e)(i)

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ =

0

1log10120 I

I

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ =

0

2log108.124 I

I

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −⎟⎟ ⎠

⎞⎜⎜⎝

⎛ =−

0

1

0

2 log10log101208.124 I

I

I

I I1 = 1.0 W m

-2

andI2 = 100.48 W m-2 found separately

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ =

1

2log108.4 I

I

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ =

1

2log48.0 I

I

1

248.010 I

I =

2

1

r I α

2

22

2

11 r I r I =

48.0

2

2

2

1

1

2 10==r

r

I

I

48.0

2

2

2

102

=r

15.12 =r km





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2010 Prelims H2 P1 and P2 Guide