F5 Physics First Term Exam (14-15) Marking Scheme
1.(a)Applying v =fλ, the wavelength of the wave is
λ = v/f = 340/3000 ≈ 0.113 m (1M+1A)
(b)(i)Since the sound waves come from a single source (signal generator), they are in phase and of the same frequency, i.e. they are coherent. (1A)
Where the sound waves meet in phase, constructive interference occurs and a loud sound is heard;(1A)
where the sound waves meet exactly out of phase, destructive interference occurs and a soft sound is heard. (1A)
(ii)Loud sound (1A)
(iii)The sound waves travel different distances and the amplitudes are not the same when they interfere, as a result, the cancellation is not complete. (1A)
The microphone has a finite size. (1A)
The sound waves coming from S2 and S3cannot cancel each other out completely upon destructive interference. (1A)
(c)When the frequency of the waves is increased to30 000 Hz, their wavelength ismuch smaller than the width of slit S1and the degree of diffraction is very small. Thus the waves cannot reach slits S2 and S3. (1M for +1M for less diffraction)
(d)S3R = = 302.6549 cm = 3.027 m
The path difference is 3.027-3 = 0.027 m and the wavelength is 0.027/5 = 0.0054 m (1M)
The frequency is 340/0.0054 = 62960 Hz. (1A)
(e)Two bright fringes would be observed on the screen. (1A)
2.(a)(i)a sin = m (1M)
1.67 x 10-6 sin = 2 x590 x 10-9
= 45.0o(1A)
(ii)a sin = n
sin 1=>n / a 1(1M for sin 1)
n a/ = 1.67 x 10-6 / 590 x 10-9 = 2.83(1A for n a/ or 2.83)
Thus n must be smaller than 2.83, 3rd order fringes cannot be seen. (1M)
(b)a sin = m
1.67 x 10-6 sin 42.10 = 2 (1M)
= 5.60 x 10-7 m(1A)
(c)(i)a sin = m
1.67 x 10-6 sin 13.50 = 1 (1M)
= 3.90 x 10-7 m (violet)(1A) (1A for violet)
(ii)2 = 3V(1M)
2 = 3 (3.9 x 10-7)(1M)
= 5.85 x 10-7 m(1A)
3.(a)Single frequency or single wavelength(1A)
(b)Subsidiary maxima peaks further away from center. (1M)
Central maximum wider and symmetrical. (1M)
(c) Cause fluorescent material to glow. (1M)
(d)Wavelength of light is of order 10-7m which is small compared with ordinary objects.(1A)
Wavelength of sound is comparable to the length of daily life obstacles.(1M)
4.(a)(i) = c/f = 3 x 108 / 9.4 x 109
= 3.19 x 10-2 m
(ii)Path difference = 6 - 5 = = 3.19 x 10-2 m (1A for 6 - 5 = , 1A)
(b)Maximum is detected at the position shown.
It is because the path difference is ,
So constructive interference occurs.
(c)s = D/a
0.11 = 3.19 x 10-2 (0.42) / a(1M)
a = 0.122 m (1A)
(d)When the frequency is doubled, the wavelength is halved. (1M)
Path difference is still integral multiple of wavelength. (1M)
So constructive interference still occurs and a maximum is detected as before. (1A)
5.(a)
/ 1A for 2 correct
2A for 3 correct
(b) The left hand side is positive. / 1A
(c)Along the vertical direction:
T cos30 = W = mg
T = mg / cos 30o = 0.005 x 9.81 / cos 30o = 0.0566 N / 1M
1A
(d)Along the horizontal direction:
Fe = T sin30 = 0.0566 x sin30o= 0.0283 N / 1M+1A
(e) QE = 0.0283
E = 0.0283 / 20 x 10-6 = 1415 N C-1 / 1M
1A
6.(a)The electron is retarded by a constant electric force in the downward direction. Applying F = qE and Newton’s law of motion, the deceleration a of the electron is given by
(1M+1A)
(b)This is a uniformly accelerated motion. Applying, the distance s travelled by the electron before it comes to rest momentarily is
(1M + 1A)
(c)Applying v = u + at, the time elapsed t in travelling the distance s is
(1M + 1A)
(d)Since a proton carries positive charge (1A), it will be accelerated upwards by a constant electric force in the electric field (1A). It will not be retarded by the field.
MC1-5A A B C C6-10 A D B C B11-15 B D B D A
16-20 B B B D A21-25 C A C A C26-28 C D C
Explanations to selected mc
2.Only electrons can move, protons cannot move.
3.Force on A = force on B because of action and reaction.
9.After contact, neutralization occurs and the net charge is 8Q – 2Q = 6Q. Because of charge sharing, each sphere will contain 3Q, so the force between them is proportional to (3Q)(3Q) = 9Q2.
17.Let the speed of sound be v
v = 300 m s–1