H2 Physic 2009 A Level Solutions

Suggested solutions to H2 2009 Paper 1

1. Ans: D Reasoning: % uncertainty = (0.00005 / 0.00160) x 100% = 3 % (1 sf) 2. Ans: C Reasoning: Electron accelerates across potential difference of 1000 V. Hence, energy = qV = 1.6 x 10-19 x 1000 = 1.6 x 10-16 J 3. Ans: C Reasoning: Gradient of a velocity-time graph gives acceleration. Initially constant gradient, subsequently gradient increases and then decreases to zero. 4. Ans: D Reasoning:

-1

-1

At 3 s, taking upwards as positive:

19.6 ( 9.81)(3) 9.83 m s

9.81 m s(option C is incorrect as the velocity should be downwards, hence negative)

At 5 s, taking upwards as positive:

1

v u at

a

v u at

= + = + − = −= −

= + = -1

-1

9.6 ( 9.81)(5) 29.45 m s

9.81 m sa

+ − = −= −

5. Ans: C Reasoning:

-1

-1

2 2 -1

40cos 45 28.28 m s

40sin45 ( 9.81)(5) 20.77 m s

35 m s

ox x

oy

x y

v u

v

v v v

= = =

= + − = −

= + =

6. Ans: D Reasoning: Weight is the only force and it is providing for the centripetal force during the orbit.

7. Ans: A Reasoning: Using COM:1 (0.100 ) 4 ( 0.050 ) 1 ( ) 4 ( ) -------------(1)

Using relative velocity of approach = relative velocity of seperation:

0.100 ( 0.050 )

p He

p He He p

He p

u c u c u v u v

u u v v

c c v v

+ − = +

− = −

− − = −

0.15 ------------(2)

sub (2) into (1):1 (0.100 ) 4 ( 0.050 ) 1 ( ) 4 (0.15 )

0.100 5 0.6

0.140

He p

p p

p

p

v c v

u c u c u v u c v

c v c

v c

= +

+ − = + +

− = +

= −

0.15 0.140 0.010Hev c c c= − =

8. Ans: C Reasoning: In equilibrium, the resultant force on the object must be zero, hence option A, B and D are correct. Ans C is not correct as the torque by vertical forces is anticlockwise (direction) while the torque by horizontal forces is clockwise. Torque is a vector, hence direction matters. 9. Ans: B Reasoning: G-field – negligible effect as acceleration due to gravity is only 9.81 m s-2 E-field – for the particle to experience a force upwards towards the region of higher potential, the particle has to be negatively charged B-field – using FLHR, B-field points downwards, force experience is out of the page, hence current must be directed to the left. Hence, the particle must be negatively charged as current flows in the direction of positive charges. 10. Ans: D Reasoning: Archimedes’ Principle states that the upthrust experienced by an object partially or entirely immersed in a fluid is equal to the weight of the fluid displaced by the object.

( )

(34000 22.0)(930)(9.81)1030(9.81)

(34000 22.0)(930)(9.81)1030(9.81) 36500

18.5 m

sea oil

oil oil

V g mg

V g

V

Vh

A

ρρ

==

×=

=

×=×

=

11

earth

2GPE

When is decreased, P falls to lower plate decreasesHence GPE becomes more negative i.e. decreases.

pGM m

RV R

= −

Electric force must be directed upwards to balance weight. Hence E field must be directed upwards since charge is positive. Thus lower plate is at a high potential and upper plate is at a low potential. As charge moves down, V ve EPE q V ve∆ = + ∆ = ∆ = + . Hence EPE increases Thus Option B

F

Weight

Upthrust

E Field

+V

0 V

12 Along level road, driving force is used to overcome friction i.e.

friction

1

(200)(20) 4000 W

For every 8 m travelled along the slope, the car gains 1 m in elevation.

If car is travelling at 20 m s along slope, then car must be gaining elevation

P Fv

−

===

1

friction

at 2.5 m s

Power required up slope Increase in GPE per unit time

(800)(9.81)(2.5) 4000 23.6 kW

P

−

= += +=

Thus Option D

13

( )( ) ( )

( )

engines Drag

2engines

3 3

3-3 3

3

3engines

3 3

1

Power

Power

2 36 10 12

2 36 1041.67 W m s

12

Power

36 10 41.67

9.52 m s

F v

kv v

k

k

kv

v

v −

= ×

= ×

× × =

× ×= =

=

× =

=

Thus Option D

14 There are only 3 forces acting on the child:

1) Weight: directed downwards. 2) Normal Contact Force: directed upwards to balance weight. 3) Friction which provides the centripetal force since child undergoes

circular motion. Centripetal force is a resultant force not a new force. Thus Option A.

15

( ) ( )6

1

2 6.38 10

24 60 60

463.9 m s

V rωπ

−

=

= ×× ×

=

Thus Option D

16

( ) ( )( )2

2

2

Consider gravitational force on mass 2

2

3

29

Gravitational force on mass is the same in magnitude, opposite in direction.

20.22 Option A

9

M

G M M

x

GMx

M

k

=

=

∴ = =

17

'

GPE of satellite (1)

Total Energy of satellite (2)

For satellite at distance ,

GPE 3.2 MJ

Total Energy 1.6 MJ2

For satellite at distance 2 ,

Total Energy2

GMr

GMr

rGM

rGM

r

rGM

r

= −

= −

= − = −

= − = −

= −

( ) 2 2

0.80 MJ

Thus Option B

GMr

= −

= −

18 12 2 rad s

Thus Option C

f nω π π −= =

19 For gravitational force

where is gravitational potential energy.

For electric force

where is the electric potential energy.

Hence we can expect for the case here, d

(True for

g

E

p

dUF U

dr

dF

dr

EF

dr

= −

Φ= − Φ

= −

' 2

conservative forces)

since (actually this is the spring force and it is conservative)

is of the form

Hence Option A

p

F kx

E k x

= −

20 One molecule of water contains 2 hydrogen atoms.

One mole of water contain NA molecules of water and thus 2NA of H atoms Thus Option D.

21. Ans: B Reasoning: When the gas expands, the work done on the system is negative. When the gas cool, there is a drop in temperature. Since internal energy is proportional to temperature, the change in internal energy is negative. No mention of heat supplied to the system, therefore q = 0 22. Ans: C Reasoning: Since the volume of air is trapped in a tube with one end opened and one end closed, the difference in length is: ¼ = L1 ¾ = L2

Difference in lengths, L2- L1= ½ Therefore wavelength of sound = 2(L2- L1 ) 23. Ans: A Reasoning: Using equation, x = xo sin ωt -0.5 = 1 sin ωt ωt = /6 Therefore, angle between P and Q is 30º 24. Ans: D Reasoning: Using equation, x = D/a Equivalence of terms to be used in context of question: x = v/f ; D=H, a= d Therefore, v/f = H/d => = (dv)/(fH)

25. Ans:A Reasoning: Since the electric field is pointing away from the positive source charges, the equivalence of the electric field must be pointing towards A.

26. Ans: B Reasoning: Given, P = I2 R and p = I2 r; where r is the internal resistance of battery By conservation of energy, e.m.f. = I (R+r) = (P + p)/I 27. Ans: C Reasoning: Resistance of component is given as R = V/I At Current Ix, Rx =Vx/Ix At Current Iy, Ry =Vy/Iy Therefore, change in resistance is Rx – Ry = Vx/Ix - Vy/Iy 28. Ans: C Reasoning: For one row of resistors (in series), the effective resistance is NR

1 1 1 1 1 1...

1 1

1 1

total

total

total

total

R NR NR NR NR NR

NR NR

R R

R R

= + + + +

=

=

=

29. Ans: A Reasoning: Effective resistance across voltmeter = (6000-1 + 2000-1)-1 = 1500

By potential divider method, Vvoltmeter = 1500

121500 20005.14 V

= ×+

=

30. Ans: C Reasoning: By using Fleming’s left hand rule, the particle must be positively charged such that the force acting on the motion of the particle is always perpendicular to the path. (path is clockwise direction). Since the radius of the path is decreasing, it is noted that the speed of the particle must reduce. For the moving charge, F=Bqv This force provides for centripetal force of F = mv2/r Therefore, Bqv = mv2/r B = mv/(qr) Since magnetic field B is constant (mass of particle and charge of particle also do not change), if the radius of the path is decreasing, the speed of the particle must reduce in proportion.

31. Ans: C Reasoning: The current in coil X and coil Y flow in opposite direction to each other. For a flat circular coil, the magnetic flux density at the centre of the coil is given by

rNI

B o

2µ

= . The magnitude of magnetic field is inversely proportional to the radius

of the coil and the magnitude of the current. At P, the flux density produced by the current in coil Y is in the same direction as the flux density produced by the current in coil X. At Q, the flux density produced by the current in coil Y is in the opposite direction to the flux density produced by the current in coil X. 32. Ans: A Reasoning: Since Lenz’s law applies, the effect of the iron rod must be to oppose the cause. This must reduce the current when the iron rod is inserted.

33. Ans: B Reasoning: Since almost none of the flux escapes from such a system, it is the flux that is constant around the whole ring. 34. Ans: D Reasoning: w = 2/T = 2/(2.5 x 10-3) = 800 35. Ans: B Reasoning: Po = Io

2R = (2)2(200) = 800 W <P> = Po/2 = 400 W 36. Ans: A Reasoning: Emax = hf – y-intercept = = work function of metal A smaller work function means a small value of y-intercept. 37. Ans: C Reasoning: An electron colliding with an atom can transfer part or all of its kinetic energy to the atom to excite the atom. E = -3.4 – (-13.6) = 10.2 eV 38. Ans: B Reasoning: The other 3 options were not correct. In ideal intrinsic semiconductor, there is no extra holes or electrons created.

39. Ans: D Reasoning: B.E = E = (m)c2 B.E per nucleon = (m)c2/y

40. Ans: C Reasoning: Assume that after same period of decay, t for all the nuclides, the number of moles (proportional to activity, A) remained for all the nuclides are given in the table. A = Aoe-t

Ao = A/e-t

Suggested solutions to 2009 H2 Paper 2 Q1(a) Scalar quantity is a physical quantity with magnitude only and vector quantity is a

physical quantity that has both magnitude and direction. (b) Acceleration of an object is defined as the rate of change of its velocity with

respect to time. (c)(i) v = vB – vA

= 2 218 18+ = 25. 5 m s-1 (ii) a = v / t = 25.5 / 4.4 = 5.80 m s-2

(iii)

- resultant force pointing towards the centre of the circle - resultant force must be perpendicular to the curve at P

(iv) As the resultant force is perpendicular to the motion, there is no work done by

this force, hence there is no change in kinetic energy. The force only cause a change in direction of the car, hence velocity changes

which results in an acceleration.

- vA

vB v

P

F

Q2 (a) 2 2

2

1

2

0 2( 9.81)(1.8) taking upwards to be +ve

5.94 m s (3sf)

y y

y

y

v u as

u

u −

= +

= + −

=

(b)

( )

( )

Average force,

0.45 5.2 6.6 taking pointing to the left to be positive

0.22

f i

pF

tm v

tm v v

t

∆=∆

∆=∆

−=

∆ − − =

24.1 N (3 sf)

Force calculated is a positive value i.e. force acting on ball is directed towards the left.

=

(c)

The relative speed of approach between the ball and the wall before collision is not equal to the relative speed of separation between the ball and the wall after collision. Hence collision is inelastic. Alternatively, The KE of the wall before and after collision is zero. The KE of the ball after collision is lesser than the KE of the ball before collision. Hence the total KE of the system i.e. ball + wall, is lower after collision. Thus KE is not conserved i.e. collision is inelastic.

(d) The time taken for the ball to fall a vertical distance of 1.8 m is equal to the

time taken by the ball on its flight up. Since the horizontal speed has decreased after collision, the horizontalndistance covered by the ball after collision will be lesser compared to the horizontal distance covered by the ball before collision. Hence ball will not rebound to pint on ground where it was launched.

Q3 (a) There is uneven distribution of mass in the boat.

(b) Vector sum or resultant of the forces was zero. Vector sum of the moments or torque about the same axis was zero.

(c)

Take moments about position of rope 1, T2(2.00) = 15000(0.75) T2 = 5625 N T1 = 15000 – 5625 = 9375 N

Q4 ai) Diffraction- Diffraction is the phenomenon of bending of waves when they travel through a small opening or when they pass round a small obstacle. Since diffraction grating has slits that resemble small openings, waves get diffracted as they pass through the slits. (need to answer to the context of question) aii) Coherence- Two waves are considered as coherent when they have constant phase difference between them. (Note: For light waves, it is impossible to get coherent light waves unless they start from the same source. This is that there are very frequent irregularities in the wave created by any light source, and these irregularities cannot possibly be matched in the light from even two apparently identical but separate laser beam sources. ) aiii) Superposition- The Principle of Superposition states that when two waves of the same kind meet at a point in space, the resultant displacement at that point is the vector sum of the displacements that the two waves would separately produce at that point. bi) Using equation dsin = n d = 1x10-3/500 = 2 x 10-6 m n = 2 = 650 nm (largest wavelength to find maximum angle) d sin = n 2 x 10-6 sin = 2 (650 x 10-9) = 40.5º

bii) For minimum angle, = 350 nm d sin = n 2 x 10-6 sin = 3 (350 x 10-9) = 31.7º biii) A problem with viewing the second or third order maxima of the white light with this diffraction grating is that there is overlapping of orders such that it is difficult to identify clearly the correct orders. Q5 a) Using equation,

1 22

19 19

15 2

4

1.6 10 1.6 1057.6 N

4 (2.0 10 )

o

o

Q QF

r

F

πε

πε

− −

−

=

× × ×= =×

b) The shaded area of the graph represents the work done by an external force in bringing particle P1 from infinity to a distance of 2.0 x 10-15 m away from P2. c) When fusion occurs (assuming exorthermic), the product has lesser mass than the reactants. Therefore, energy available is due to the increase in mass defect (final total mass is less than initial total masses). This energy can be referred as the binding energy released when the two particles fuse in the form of kinetic energy of the product or electromagnetic radiation (e.g. gamma radiation). (Note that in nuclear physics, energy includes E= mc2, i.e. mass-energy conservation) (you should not answer to energy transformation before fusion occurs)

Q6 (a) Since field and the coil are at right angles, Magnetic flux linkage = NBA = (15) ( 0.018) (0.3)(0.24) = 0.01944 = 0.019 Wb

(b) (i) Due to the decreasing magnetic field from t = 2.0 s to 6.0 s, there is a decrease in magnetic flux linkage to the coil. Hence there is induced e.m.f and according to Faraday’s law, the induced e.m.f is directly proportional to the rate of change of magnetic flux linkage.

(ii) ddt

ε Φ= − = - ( )d NBA dB

NAdt dt

= −

magnitude of induced e.m.f = |dB

NAdt

− |

= (15) (0.3)(0.24) 0.018

( )4.0

= 4.86 x 10-3 = 4.9 x 10-3 V (2 s.f) (iii)

t/s

4.9 x 10-3

2.0 4.0 6.0 8.0

E/V

Q7 (a) At volume V1 = 2.0 x 10-3 m3, Pressure P1 = 1.0 x 105 Pa, P1V1 = 200 Pa m3 At volume V2 = 8.0 x 10-4 m3, Pressure P2 = 4.0 x 105 Pa P2V2 = 320 Pa m3

Considering the ideal gas equation PV = nRT. Since R is a constant, and the mass of the gas is constant, n (number of moles of gas) is constant therefore PV is directly proportional to temperature, T. When gas is compressed to V2 = 8.0 x 10-4 m3, P2V2 > P1V1, hence T2 >T1 temperature at the reduced volume would be greater. Temperature of the air increases.

(b) pV cγ = apply lg on both sides of equation;

lg lg lglg lg lg

p V c

p V c

γγ

+ == − +

A graph of lg p versus lg V can be plotted. If the relationship is true, a straight line graph will be obtained with lg c as the vertical intercept and γ− as the gradient.

(c) (i) From Fig 7.2, when V = 1.00 x 10-3 m3, P = 2.65 x 105 Pa, lg (2.65 x 105) = 5.42 (ii) When V = 1.00 x 10-3 , lg P = lg (1.00 x 10-3 ) = -3.0

(iii) Gradient = 5.40 5.06

2.97 ( 2.75)−

− − −= -1.545 = -1.55 = -γ

γ = 1.55 using ( -2.97, 5.40), y = mx + c 5.40 = (-1.545) (-2.97) + c vertical intercept, c = 0.81135 = lg c lg c = 0.81135 c = 6.476 = 6.48

(iv) _ _ _ _

3.85_ _ _ _

Volume of gas before compressionVolume of gas after compression

=

3

3 3

2.00 103.85

_ _ _ _

_ _ _ _ 0.519 10

Volume of gas after compression

Volume of gas after compression m

−

−

× =

= ×

By extending the graph, and checking for value of T when V = 0.519 x 10-3 m3 From the graph, T = 535 K

(v) From the first law of thermodynamics U Q W∆ = + . During the explosion, expansion of the gas occurs rapidly so W is negative at a very short period of time and there is very little time for heat exchange. Change in internal energy of the gas therefore becomes negative rapidly as gas expands rapidly, and hence temperature of the gas falls rapidly.

Paper 3 Q1 (a) Simple harmonic motion (S.H.M.) is defined as the oscillatory motion of a

particle whose acceleration a is always directed towards a fixed point and is directly proportional to its displacement x from that fixed point.

(b) (i) (1) w = 2f = 2(620) = 3896 rad s-1

(2) a = - w2xo = - (3896)2(0.21 x 10-3) = - 3200 m s-1

(ii) A straight line with negative gradient, cutting the x-axis at the origin

and with ends of the line at coordinates (0.21, - 3200 m s-1) and (-0.21, 3200 m s-1).

(c) If the frequency produced by the loudspeaker is equal to the natural

frequency of the vibration of the cone of the loudspeaker, resonance occurs. There is maximum transfer of energy and the cone vibrates with maximum amplitude. This is not desirable .

Q2 (a) (i)

4.5 (2.5)1.8 (2 sf)

V IRR

R

==

= Ω

(ii)

( )( )

8

23

3

Resistance of 1 turn of solenoid

1.6 10 0.088

0.30 10

4.979 10 Resistance of solenoid 1.8

lAρ

π

−

−

−

=

×=

×

= × Ω=

3

no. of turns resistance of each turn 1.8

1.8no. of turns

4.979 10 361.4 360 turns (2 sf)

−

Ω× = Ω

=×

=≈

(iii) 360

No. of turns per unit length is0.120

3000 turns per metre (3 sf)

=

=

(b) (i) 1 Tesla is defined as the magnetic flux density of a uniform magnetic field when a wire of length 1m, carrying a current of 1A, placed perpendicular to the field, experiences a force of 1N in a direction at right angles to both the field and the current.

(ii)

( ) ( )o

7

3

4 10 3000 2.5

9.42 10 T (3 sf)

B nIµ

π −

−

=

= ×

= ×

(c) (i) 7 o

7 1

7 1

4.0 10 cos30

3.46 10 m s

3.5 10 m s (2 sf)

xU−

−

= ×

= ×

≈ ×

(ii) 7 o

7 1

4.0 10 sin30

2.0 10 m s (2 sf)

yU−

= ×

= ×

(d) When the moving charge enters the magnetic field, it experience a

magnetic force of magnitude BF qvB= . This magnetic force acts in a direction that is perpendicular to the field and the charge’s velocity. Since the force and hence accelerations acts perpendicularly to the charge’s velocity, the charge will undergo circular motion with the magnetic force providing the centripetal force. Hence,

2

2

Bmv

Frmv

qvBr

mvr

qB

=

=

=

(e) Electron on entering the solenoid will travel in a helical path.

Radius of helical path,

31 7

19 3

(9.11 10 )(2.0 10 )

(1.6 10 )(9.42 10 ) 0.01208 1.2 cm (2 sf)

ymvr

qB−

− −

=

× ×=× ×

=≈

Since radius of helical path is smaller than radius of solenoid, electron will not collide with wall of solenoid.

Q3 (a) Using De-broglie’s equation:

(b) For this question, it would be useful to memorise the equation relating Kinetic

energy to momentum. However, it is also very easy to derive the formula yourself.

Using Conservation of energy,

Gain in Kinetic energy = Loss in Potential energy

Q4 (a) The decay equation for Cobalt-60 when it undergoes beta decay is , where represents the resulting daughter nucleus from the reaction. These daughter nucleus REMAIN inside the box, such the number of nuclei inside the box IS CONSTANT. i.e. At any given time,

number of( ) nuclei + number of( ) nuclei = INITIAL number of( ) nuclei The half-life of 60Co should more correctly be defined as the time taken for the number of 60Co to decay to one half of its initial value. (more generally, the half-life of a particular nuclide is defined as the time taken for the number of that particular nuclide to decay to one half of its initial value.) (b) N = (total mass)/(mass of 1 cobalt-60 atom)

= = 4.3159 × 1013 atoms Thus A = λN = N

t½ = N ln 2 / A = (4.3159×1013) ln 2 / 1.8 × 105 Bq =1.66 × 108 s = 5.27 yrs. Alt, N = nNA, where n is number of moles, NA is Avogadro’s number n = M/MM, where MM is the molar mass. For 60Co, MM = 60 g = 0.060 kg n = 4.3×10-12 / 0.060 = 7.167 × 10-11 mol N = nNA = … = 4.3159 × 1013 atoms, same as before. … …

Q5 (a)(i) The gravitational field strength at a point is defined as the (gravitational) force PER unit mass acting on an object placed there. (n.b. it is NOT defined as the force acting ON A UNIT MASS). (ii) Newton’s law of gravitation states that the gravitational force of attraction between 2 point masses is directly proportional to the product of the masses and inversely proportional to the square of the separation between them.

This can be mathematically expressed as .

Hence, Gravitational field strength, (shown)

(b)(i) Density = 2.5268×1017 = 2.53×1017 kg m-3 (3s.f.)

(b)(ii) Disclaimer: Neutron stars are NOT explicitly within the A-level syllabus, and this part of the question is intended to discriminate the better students. Even so, many will probably not score for this part as it represents a somewhat difficult 3 marks to score. Based on the examiner’s report, the answers expected from students require them to 1-recognize what a neutron star is, & 2-recognize that neutron star density is greater towards the core due to compression of the outer layers. This is NOT COMMON KNOWLEDGE, and even within an introductory nuclear physics course in the university it is taught that nuclear matter has an approximately constant density, which might lead one to INCORRECTLY conclude the same of a neutron star. Online sources do confirm variation in the density of neutron stars (see http://en.wikipedia.org/wiki/Neutron_star#Structure; http://www.astro.umd.edu/~miller/nstar.html ). Outright wrong answers revolve around variation of density due to the different densities of rock/soil/magma material etc, as even a typical star will not be comprised of such solid material.

However, knowledge that a typical gaseous star will experience variations in density and pressure due to the inner layers experiencing greater compression due to the outer layers can be used and extended to neutron stars to arrive at the same conclusion

Students are instead invited to visit the website: http://www.universetoday.com/24219/what-is-a-neutron-star/ for a brief introduction to neutron stars as well as simple explanation as to how they form. The insight gained is worth more than a model answer to parrot. (c)(i) The gravitational field strength at the surface of a neutron star,

= 1.2006 × 1012 = 1.20 × 1012 N kg-1 (3 s.f.) (c)(ii)

Centripetal acceleration, a = r ω2 = r = (1.7 × 104) = 1.5218 × 107 = 1.52 × 107 m s-2 (3 s.f.) (c)(iii) The gravitational acceleration determined in (i) is greater than the centripetal acceleration necessary to keep it in orbit at that radius, as calculated in part (ii). This suggests that not only would a particle at the surface NOT leave the surface, but an additional force is required in the OUTWARD direction if the particles continue to orbit at that particular radius. [the resultant motion for any situation where the actual force experienced exceeds the necessary centripetal force will not be circular motion at constant radius but might instead be a inward spiraling motion ] Consider the analogy of a person standing at the equator vs a satellite in orbit • For the satellite, mg’ = mv2/r’ , where g’ is the gravitational field strength at distance

r’ from the centre. • For the person at the equator, (mg – N) = mv2/R, where R is the radius of the planet

and N is the normal contact force in the outward direction acting against mg to make the resultant (mg – N) just sufficient to provide for the centripetal force but not exceed it)

(d) Disclaimer: The Bremsstrahlung process is technically NOT in the A-level syllabus, although students are introduced to it as the mechanism responsible for the continuous portion of the X-ray spectrum produced when electrons are accelerated towards a metal target. The characteristic lines are explained using quantum theory, whilst the continuous portion of the X-ray spectrum relies on classical Maxwell’ s Electromagnetic Theory which is well beyond the A-level syllabus. Students cannot be expected to perform calculations but can at least be expected to describe the Bremsstrahlung process of an accelerating charged particle radiating EM radiation. The suggested answer should revolve around the idea that protons are (positively) charged particles, and when subjected to large (gravitational) acceleration of a neutron star, will produce E.M. radiation via the Bremsstrahlung process, as with any accelerating charged particle.

Q6(a) Fig 6.1 shows the velocity – time graph of a ball during its upward motion and the gradient of the tangent at various points on the curve is a measure of the acceleration. The force due to air resistance is proportional to acceleration (Fdrag = ma – mg). From Fig 6.1, it is evident that the gradient of the tangent decreases with decreasing velocity, hence it can be deduced that air resistance varies with speed.

(b) Magnitude of acceleration of free fall can be determined by drawing a tangent at

the point v = 0 as at this point, the ball is at rest instantaneously and the only force acting on it is its weight, hence the acceleration is that of g.

(c)(i) Maximum height reached = area under velocity-time graph = no. of squares x area of 1 square 30(2.5 x 0.25) =18.8 m (ii) Initial energy = KE = ½ m v2 = ½ m (252) At the highest point, Final energy = GPE = mgh = m(9.81)(18.8)

2

2

1 (25 ) (9.81)(18.8)Energy lost 2Ratio 0.4101Initial KE (25 )2

m m

m

−= = =

(d)(i) Draw tangent at v = 10 m s-1

-217.1 2.0Acceleration 13.5 m s

0.30 1.45−= = −−

(ii) At v = 10 m s-1, the ball is travelling upwards, Hence, Fdrag + mg = ma Fdrag = ma – mg = 0.350 (13 – 9.81) = 1.12 N (e)

- smooth and gentler curve with decreasing gradient

- final time must be greater than 2(1.75) = 3.5 s as ball takes a longer time to travel down

- area bounded by the curve above and below the x-axis should be approximately equal as the distance traveled by the ball in the upward motion and downward motion must be equal

mg Fdrag

v

Q7 (a)

(i) One volt is the potential difference between two points in an electrical circuit when one joule of electrical energy is converted to other forms of energy as one coulomb of charge passes from one point to the other.

(ii) The electromotive force (e.m.f.) of a source is the energy converted from other

forms to electrical energy per unit charge delivered round a complete circuit.

The potential difference between two points in an electrical circuit is defined as the electrical energy converted into other forms of energy per unit charge passing from one point to the other.

(b) (i) ××

(ii) (from aii, you are given a hint that you should be using this equation relating energy to potential difference and charge.)

(iii)

×(to 3 sf)

(iv)

R = 6.0 (to 2 s.f.) (c) (i) After connecting the second similar cell, there is now a total emf of 3.0 V and

total internal resistance of 0.5 (to 1 s.f.)

(ii) As current increases, temperature increases and the vibration of the lattice ions increases. Free electrons collide more frequently with the ions. The electrons experience more obstructions and suffer a drop in drift velocity. Resistance thus increases.

(d) (i) The total internal resistance of 0.50 is 104 orders of magnitude smaller than the

resistance of the thermistor which has a range from 1800 and 4000 and hence can be considered negligible.

(ii) 1. At 0°C, RT = 4000

= 2.0 V

2. At 20°C, RT = 1800

= 1.42 V

Note: from (di), we already established that internal resistance can be considered negligible, hence we do not consider it in our working.

(iii) For the fixed resistor to have 1.2 V at 0°C,

R = 6000

For the fixed resistor to have 2.4 V at 20°C,

R = 450

It is not possible to substitute a different fixed resistor in the circuit to achieve this range of potential differences as the required resistance for the fixed resistor varies too significantly.