2007 H2 Paper 2

1 (a) (i) s ut at

h gt

g

g

2

2

2

-2

1( )

2

1

2

12.66 (0.740)

2

9.72 m s

Examiner Comments: If you had done 266/0.740, what you obtained is the average speed, not the final speed.

(ii) 1 1100% 0.38%

266

h

h

2 0.005100% 0.68%

0.740

t

t

Examiner Comments: You must leave the answers to 2 and only 2 sig. fig. as instructed by the question.

(b) 2

2

2

0.38% 2 0.68%

1.74%

hg

t

g h t

g h t

1.74% 9.72 0.169g m s-2

-2

-2

9.72 0.169 m s

9.7 0.2 m s

g

Examiner Comments: By convention, uncertainties should be expressed to only 1 sig. fig., hence the final answer for g must be given to 1 decimal place.

(c) 1 Due to human reaction, the timer might have been started or stopped too late.

2 The timer may be miscalibrated.

Examiner Comments: The question asked to explain why the value of t is precise but not accurate. Answers which explained why the value of g might be precise but no accurate (such as due to air resistance) gets no credit.

2 (a) Electric field strength at a point is the force per unit positive charge acting on a small test charge placed at that point.

Examiner Comments: The answers were surprisingly poor with many candidates not referring to a positive charge or not including the term per in their definition.

(b) (i) Leftward

Examiner Comments: Many students got this part wrong. Students must remember that the direction of electric field is defined in terms of positive charges. So the direction of field must be in opposite direction to the direction of force experienced by electrons.

(ii) From Fig 2.2, we obtained the field strength at d = 2 cm to be 2.4 kV m-1. 19 3 16(1.6 10 )(2.4 10 ) 3.84 10F qE N

(c) (i) W qV

If the electric force is assumed to be constant, the work done on the electron can be calculated easily using W=Fd. Hence

16 19(3.84 10 )(0.04) (1.6 10 )

96.0 V

Fd qV

V

V

Examiner Comments: Many students had no idea.

(ii) It is an underestimate.

The actual field strength (and hence electric force) is not constant, but varies as shown in Fig. 2.2. The average field strength (and hence average electric force) is higher than whats assumed in part (c)(i).

Examiner Comments: Most students had no idea.

3 (a) A progressive wave is a propagation of vibrations or oscillations, transferring energy or momentum in the direction of wave propagation.

Examiner Comments: Answers with no reference to vibrations or oscillations as the cause of wave motion are penalised.

(b) (i) 1. The two waves must overlap

2. The two wave sources must be coherent

3. The two waves should have comparable amplitude

Examiner Comments: Since the question already stated the waves produced have the same frequency, answers specifying same frequency/wavelength/speed get zero credit. Since the wave we have in mind is waves on the water surface, polarisation is quite meaningless.

(ii) 1

Number of antinodal lines = 3+1+3 = 7

2

Examiner Comments: The correct answer must be a curve drawn roughly mid way between (any) two antinodal lines.

4 (a) Vertically out of the page (using Flemings LHR)

Examiner Comments:

(b) (i) Since magnetic force provides the required centripetal force

2v

Bqv mr

The -particle had a charge of 2e

(2 ) ( )

2

vB e v mv

r

Ber p

Examiner Comments: This is a show question which demand answers to be very clear. Answers which do not contain the two explanatory sentences are penalised.

(ii) 2 2 221 1 1 (2 ) 2( )

2 2 2K

p Ber BerE mv

m m m

Examiner Comments: Working must be clear.

(c)

Examiner Comments: From (b)(i), one can conclude that at higher momentum, the radius of circular arc is larger (despite the magnetic force being larger). Answers which did not show clearly that the path is straight once the particle leave the magnetic field are penalised.

Straight path

Circular path

5 (a)

Examiner Comments: Must students know the diffusion of electrons is from n-type to p-type

(b) - Due to concentration gradient

- electrons diffuse from the n-type to p-type, and holes from p to n.

- electron-hole pair recombination occur near the junction, resulting in a

- positively charged donor ions on the n-side and negatively charged acceptor ions on the p-side

Examiner Comments: Many students lose marks because of lack of detail. Both (not either) the electrons and holes, n and p sides, donor and acceptor ions, must be explained. Reasons for the migration and recombination must be mentioned.

(c)

(To increase the width of the depletion region, the pn junction must be reverse biased.

Examiner Comments: Answers which did not use proper circuit symbols (for the cell) were penalised.

6 (a) 14 4 17 17 2 8 1N He O H

Hydrogen nucleus

(b) (i) 2 2

8 2

13

( ) ( )

[(17.004507 1.008142 ) (14.007525 4.003860 )](3 10 )

1.9 10 J

f im c m m c

u u u u

(ii) There is an increase in rest-mass energy after the reaction. This suggests that the additional energy must have come from the kinetic energy of the bombarding -particles.

Examiner Comments: Many students did not realise the rest-mass energy increased.

(iii) The principle of conservation of momentum dictates that the oxygen and hydrogen nuclei cannot be stationary after the reaction.

The incoming KE of the -particles is equal to the gain in rest-mass energy plus the KE of the oxygen and hydrogen nuclei.

Examiner Comments: Many students could not make the link that the KE of the product nuclei must have come from the KE of the incoming -particle.

7 (a) (i) Work done by a force is the product of the force with displacement in the direction of the force.

For the specific situation in this question, the increase in electric potential energy can be expressed by work done by electric force

pE F r

With some algebraic manipulation

pE

F Gr

Examiner Comments: A proper definition of work done must be presented. Stating distance as distance perpendicular to direction of force is of course wrong.

(ii) For r < 2.8 x 10-10 m, G is negative, so force is repulsive.

For r > 2.8 x 10-10 m, G is positive, so force is attractive.

Examiner Comments: Many students have their concepts messed up. Recall that

dUF

dx , so

it is the energy gradient (and not energy) that relates to the force. Also realize that the negative sign is a reminder that the direction of force is opposite to direction of increasing energy. Hence the force is always directed towards decreasing Ep.

(iii)1 G=0, so F=0.

(iii)2 19

10

[( 1.5) ( 10.0)] (1.60 10 )1.70 N

[8.0 0.0] 10F G

Examiner Comments: Despite the hint from (a)(i), many students still did not realize the force is to be determined from the energy gradient. Many students made careless mistakes e.g. omitting 10-10 and e.

(b) 1 It is repulsive.

2 It is very short-range. (It is very strong at small r, but drops rapidly as r increases).

Writer Comments: The graph in Fig. 7.2 was the result of two forces at play: (1) the repulsive force between two positively charged nuclei when they get too close and (2) the attractive force between two oppositely charged ions when they get too close.

By matching the shape of a 8

1y

x graph to the graph in Fig. 7.2, we know that this term is

related to the part of the graph where Ep was positive and decreasing. Hence this term is related to the repulsive force (and not the attractive one)

(c) (i) 1. rmin = 2.35 x 10-10 m

rmax = 4.25 x 10-10 m

(Answers are obtained by reading off the graph where Ep = -6.0 eV. At the extreme ends of the vibration, kinetic energy is zero. Since total energy is always equal to the sum of KE and PE, at the extreme positions, PE must be -6.0 eV)

2. From graph, at r = 3.5 x 10-10 m, Ep = -7.1 eV

6.0 7.1

1.1 eV

TE PE KE

KE

KE

(ii) If the motion were simple harmonic, the Ep graph should be symmetrical about the equilibrium position (at r = 2.8 x 10-10 m).

Examiner Comments: Answers should be based energy considerations since the graph in Fig 7.3 is an energy graph. Answers based on acceleration, or just regurgitation of the definition of shm get no credit.

(d) Heating causes the total energy (and maximum PE) of each ion to increase.

The ions will then be vibrating between smaller minimum r, and larger maximum r.

Because the restoring force is not symmetrical, the oscillating ions spend more time at larger separations, resulting thermal expansion.

Examiner Comments: Answers which merely state an increase KE without any reference to changes in PE get zero credit.