2010 H2 Physics Paper 9646/03 suggested solutions Updated 12 Aug 2011 (Noted changes in yellow for students’ copy )
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(a) Latent heat of vaporization is the quantity of heat required to change a substance from liquid to vapour without a change of temperature. (b) (i)
(ii)
The mass m is determined for 2 different values of P so that the heat loss to the surroundings can be determined through calculations. Let H be the heat loss to the surroundings in 5 minutes. E
= m lv + H
14.1 lv + H 1000 8 .2 300(95) = lv + H 1000 300(140) =
- (1) - (2)
(1) – (2),
5. 9 lv 1000 l v = 2.28 × 106 l v = 2.3 × 106 J kg-1 l v = 2.3 × 103 J g-1
300( 45) =
2
(a) (i)
(ii)
The centripetal force is the net force on the ball, and with the rod vertical and the ball above C, the net force is given by T + Weight. Since T is given to be 2 mg, centripetal force would be greater than 2 mg. At top, T + mg = Fc Fc = 2 mg + mg Fc = 3 mg
(iii)
At bottom, T - mg = Fc T = 3 mg + mg T = 4 mg
(b) (i)
Fc = ma 3 mg = mrω2
ω2 =
3g 0.72
ω = 6.39 = 6.4 rad s-1
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(ii)
(c) (i)
(ii)
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(a) (i) (ii)
(b) (i)
v=rω = 0.72 × 6.39 = 4.60 = 4.6 m s-1 The gravitational potential energy is reduced but the kinetic energy is constant. Consequently work must be done to maintain constant KE. Work done = loss in GPE = 0.24 × 9.81 × 0.72 × 2 = 3.39 = 3.4 J A region of space where a force is experienced Strength of field is represented by the closeness of the lines of force. A strong field is represented by lines that are very close together, whilst a weak field is represented by lines that are far apart. Mass
(ii)
Positive Charge
(iii)
A North Pole
(c)
+ v'' < v
B FE
v=
FB
E
S1
–
E B
v' > v
S2
The electric field and magnetic field should be directed such that the electric force exerted on the moving charge is in the opposite direction of the magnetic force acting on the same charge. This setup can be used for velocity selection, where the selected velocity given will be determined by the ratio of E/B, and the charges with this particular velocity will pass through the field undeflected when the magnetic force and the electric force are of the same magnitude. This will allow the charged particle to experience no net force and exit the field on the other side through a narrow opening without any deflection. Any other velocities of the charged particles will hit the screen S2 and be blocked. Note: Diagram must be included to illustrate 1. Direction of Electric Field 2. Direction of Magnetic Field 3. Path of charged particle. 2
4 (a) Switch A Open Open Closed closed
Switch B Open Closed Open Closed
Resistance between X and Y/Ω 12 10 6 6
Deduction R3 + (R1+R2) =12 …..(1) R2 + R1 =10 …….(2) R1 = 6 ……(3) R1 = 6 ……(3)
From (1) & (2), R3 = 12 -10 = 2 Ω From (3), R1 = 6 Ω From (2) & (3), R2 =10 -6 = 4 Ω R2
(b)
R1 X
Z
R3 1 1 1 1 3 + = + = R2 R3 4 2 4 4 1 RXZ = + 6 = 7 = 7.3 Ω 3 3 5 (a) Work function energy of a surface is the minimum amount of energy that has to be given to an electron to release it from the surface of the metal. (b) φ = 2.5 eV = 4.0 x 10-19J Energy of incident photon = hc/λ = 3.68 x 10-19 J Since energy of incident photon less than the work function φ, no electrons would be emitted. (c) Intensity of a radiation at a surface determines the rate of arrival of photons at the surface, and not the energy of an individual photon. The energy of a photon depends on the frequency of the radiation only. Therefore increasing the intensity increases the rate of arrival of the photons at the surface but not the energy of the photon. Since the energy of the individual photon is not increased there will be no electrons emitted from the surface
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6 (a) Force is proportional to the rate of change of momentum of an object (which is free to move). Direction of force is acting in the direction of the change in momentum. (b) (i) Area of triangle above the line drawn.
(ii) work done W = (ave force) x (extension of the wire) ( or = area under F vs x graph) W = ½ Fx Since F = kx Hence, W= ½ Fx = ½ (kx)x = ½ kx2 By conservation of energy, Energy stored in the spring E = W = ½ kx2
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(c) (i) From graph, length of extended spring with load of 4.0 N = 9.2 cm Total length = 9.2 + 0.8 =10.0 cm (ii) 1. Change in gravitational potential energy = mgh = (40) (0.80 x 10-2)(9.81) = 3.2 x 10-2 J = 0.032 J 2. Method 1 (Easiest): Find the difference in area enclosed by F-x graph as it extended from 9.2 cm due to 4 N to 10 cm with an extension of 0.8 cm Shaded area = 3.6x10-2 J
Method 2: From graph, when F = 2.5 N, x = 2.0 cm
k=
F 2.5 = = 1.25 Ncm -1 = 125 Nm -1 x 2.0
Change in elastic potential energy of the spring
1 1 k(x 2 ) 2 − k(x1 ) 2 2 2 1 = (125)(4.02 − 3.22 )x(10−2 ) 2 2 =
= Note:
3.6x10-2 J
(x2)2 – (x1)2 ≠ (x2 - x1)2
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(iii) The W.D to cause the additional extension of 0.80 cm =(increase in elastic p.e.) – ( loss in g.p.e.) = 0.036 – 0.032 = 0.004 J = 4.0 x 10-3 J (d) (i) Total energy of oscillation of the mass = W.D. to cause the additional extension = 4.0 x 10-3 J (ii) 1. Etotal = ½m(vmax)2
1 4.0 4.0x10−3 = ( )(v max ) 2 2 9.81 v max = 0.14 m s -1 2.
vmax = ωxo 0.14 = ω(0.80 x 10 -2)
ω = 2πf =
0.14 0.80x10−2
f = 2.8 Hz (e) A spring with mass would increase the effective mass of the oscillating system, hence the frequency of the oscillation would decrease. Use F = ma = - kx (spring) a = - (k/m)x Use a = -ω2x = - (2πf)2x
Additional information to find relationship between mass and frequency
Thus (k/m) = (2πf)2
f=
1 k 2π m
Note: frequency of a loaded spring, f =
1 k 2π m
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(NOT in syllabus)
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(a)
Progressive: A progressive wave is the movement of a disturbance from a source which transfers energy from the source to places around it. Transverse: A transverse wave is a wave where the particles of the medium oscillate in a direction at right angles to the direction of its propagation/ direction of energy transfer.
(b)
For a polarised transverse wave, the oscillations are in one direction only, in the plane normal to the direction of energy transfer whereas for unpolarised wave, the oscillations are in many directions.
(c)
Amplitude = A Intensity =I I ∝ A2
(d) Angle θ 180° 90° 60°
amplitude A 0 A cos 60 = A/2
intensity I 0 I/4
(e) (i) The Principle of Superposition states that when two or more waves meet at a point, the resultant displacement at that point is the vector sum of the displacements produced at that point by each of the wave separately. (ii) When progressive sound waves travel from the source to the other end of the pipe, they are reflected from the open end of the pipe. The superposition of the incident and the reflected waves set up a stationary sound wave. The open end of the pipe is always an antinode and the node to node distance is half a wavelength. (f) (i) N
A
A
33.5 cm (ii) λ = 2 x 0.67 = 1.34 m v=fλ = 250 x 1.34 = 335 m s-1 (g) The antinodes would be outside the pipe. Since the value in (f)(ii) is an underestimate, this imply that the distance from antinode to antinode is an underestimate. The effective length of the air column is therefore greater than the length of the pipe. (Or The air just outside the open ends of the pipe is set into vibration and the antinodes of a stationary wave occur a distance c (called the end-correction) beyond the ends.)
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8 (a)(i)
(ii)
From momentum conservation, 0 = (A – 4) (-v) + 4V A−4 Hence V = v 4
Initial k.e. of α particle = ½ 4u V2, where u is the atomic mass unit. Initial k.e. of daughter nucleus = ½ (A – 4)u v2 Therefore the ratio initial ke of α particle to initial ke of daughter =
1 ( 4uV 2 ) 2
1 ( A − 4) uv 2 2 (b)(i)
=
4(
A−4 4
)2 v 2
( A − 4) v
2
=
A−4 4
=
1 A −1 4
using ΔE = Δm c2 = (211.9459 – 207.9374 – 4.0015 )(1.66 x 10-27)(3.00 x 108)2 = 1.0458 x 10-12 J hence energy in MeV = 1.0458 x 10-12 J / (1.6 x 10-19 x 106) = 6.54 MeV
(ii) From (b)(i) (initial ke of α particle + initial ke of daughter nucleus) = 6.54 MeV
and from (a)(ii) initial k.e. of α particle = (¼(212) – 1) (initial k.e. of daughter nucleus) = 52 (initial k.e. of daughter nucleus) Hence initial k.e of α particle + 1/52 (initial ke of α particle) = 6.54 MeV Therefore initial k.e. of α particle = 6.54 ( 52/53) = 6.42 MeV (c)(i) Gamma ray photon is emitted during the decay. (ii)
During the decay, momentum of the system is conserved which implies that the momentum of the individual particles involved in the decay has to be considered, i.e. besides the momentum of the thallium and α particle. The momentum of the gamma ray photon emitted has to be taken into account. Hence the momentum of the thallium is unlikely to be in opposite direction to the momentum of the α particle, as the direction of the gamma ray photon emitted need to be taken into account when determining the direction of both the thalium and α particle.
(d)(i) (ii)
T½ = ln2 / λ = 3648 = 3650 s 2 hours (or 7200s ) is roughly equal to the two half life of bismuth ( 3700s ) which explains why there is approximately ¼N existing after 2 hours. During the two hour period 3/4 N of bismuth would have decayed into thallium. There should be 3/4N Thallium after two hours. But thallium, the daughter nucleus of bismuth, by itself is radioactive. It can undergo further nuclear decay. Hence, the number of thallium will be less than 3/4N. Also because the half life of thallium is much smaller than that of bismuth, the number of thallium after two hours is not just less than 3/4N but much less than 3/4 N.
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