This blog contains answers to exercises set for students. While every effort is made to ensure that the information posted is correct, mistakes may occur from time to time.
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Wednesday, May 24, 2017
Tuesday, May 23, 2017
y12 Young's slits questions
3. (a) (i) constant phase difference
(allow 1 mark for same phase difference or same frequency/wavelength) B2
(allow 1 mark for same phase difference or same frequency/wavelength) B2
(ii) path difference = λ/2 B1
(b) (i) evidence
shown that fringe width x = 8.0 mm B1
a = λD/x = 6.4 × 10–7 × 1.5/8.0 × 10–3 = 1.2 × 10–4 m C1
(give 2 marks for using x = 4.0 mm giving a = 2.4 × 10–4 m) A1
a = λD/x = 6.4 × 10–7 × 1.5/8.0 × 10–3 = 1.2 × 10–4 m C1
(give 2 marks for using x = 4.0 mm giving a = 2.4 × 10–4 m) A1
(ii) maximum intensity when y = 0 AND
minima at +4 and –4 B1
correct repeat distance, i.e. 8.0 mm with at least 2 full cycles drawn B1
correct repeat distance, i.e. 8.0 mm with at least 2 full cycles drawn B1
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Wednesday, May 17, 2017
Tuesday, May 09, 2017
Wednesday, May 03, 2017
Thursday, April 27, 2017
Materials questions
1. (a) P.E.
at top = 80 × 9.8(1) × 150 = 118 000 (J) (1)
K.E. at bottom and at top = 0 (1)
Elastic P.E. at top = 0, at bottom = P.E. at top for ecf = 118 000 J (1) 3
K.E. at bottom and at top = 0 (1)
Elastic P.E. at top = 0, at bottom = P.E. at top for ecf = 118 000 J (1) 3
(b) 24
N m–1 × 100 m = 2400 N 1
(c) elastic
P.E. is area under F-x graph (1)
graph is a straight line so energy is area of triangle (1)
elastic P.E. = ½ × kx × x = (½kx2) (1) 2
graph is a straight line so energy is area of triangle (1)
elastic P.E. = ½ × kx × x = (½kx2) (1) 2
(d) loss
of P.E. = 100 × 9.8(1) × 150 = 147 000 J (1)
gain of elastic P.E. = ½ × 26.7 × 1052 = 147 000 J (1) 2
gain of elastic P.E. = ½ × 26.7 × 1052 = 147 000 J (1) 2
(e) idea
that a given (unit) extension for a shorter rope requires a greater force 1
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2. (a) (i) speed
= d / t C1
= 24 / 55
= 0.436 (m s–1) allow 0.44 A1
do
not allow one sf
(ii) kinetic energy = ½ m v2 C1
= 0.5 ´ 20 ´ (0.436)2
= 1.9 (J) note ecf from (a)(i) A1
(iii) potential energy = mg h C1
= 20 ´ 9.8 ´ 4
= 784 (J) A1
penalise the use
of g = 10
(b) (i) power = energy / time or work done / time C1
= (15 ´ 784) / 55
note ecf from
(a)(iii)
= 214 (W) A1
(ii) needs to supply children with kinetic
energy B1
air
resistance B1
friction
in the bearings of the rollers / belt B1
total
mass of children gives an average mass of greater than 20 kg B1
Max
B2
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1. (a) (i) Stress = force / area C1
force = stress ´ area
= 180 ´ 106 ´ 1.5 ´
10–4
= 27000 (N) A1
(ii) Y M = stress / strain C1
=
180 ´ 106 / 1.2 ´ 10–3 or using the gradient C1
=
1.5 ´ 1011 N m–2 A1
(b) brittle
elastic/
graph shown up to elastic limit
obeys
Hooke’s law / force α extension / stress α strain
no
plastic region B3
MAX
3
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4. cast iron: brittle
brittle explained as having no plastic region
elastic
elastic explained as returning to original length when
the load is removed / linear graph / Hooke’s law obeyed
or equivalent words MAX 3
brittle explained as having no plastic region
elastic
elastic explained as returning to original length when
the load is removed / linear graph / Hooke’s law obeyed
or equivalent words MAX 3
copper: ductile
ductile explained as can be formed into a wire
initially elastic
plastic where it stretches more and more with little
increase in stress
plastic explained as does not return to its original length
when the load is removed
reference to necking at the end MAX 3
ductile explained as can be formed into a wire
initially elastic
plastic where it stretches more and more with little
increase in stress
plastic explained as does not return to its original length
when the load is removed
reference to necking at the end MAX 3
polythene: easy to deform / deformed with a small force
plastic
ductile
polymeric MAX 2
plastic
ductile
polymeric MAX 2
MAX
8
QWC: spelling, punctuation and grammar B1
organisation and logic B1
organisation and logic B1
[10]
Tuesday, April 25, 2017
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