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Thursday, February 25, 2021

Cosmology HW answers

 

1      C

2      A

M3.(a)      (i)      Spectral line moved to longer wavelength position (allow ‘to red end of spectrum’)

B1




(ii)     Mention of Doppler effect

B1

Expansion of universe / Big Bang

B1

Wavelength increased (or frequency decreased)

B1

Successive ‘peaks’ of wave emitted at increasing distance
from Earth [allow ‘wave stretched’]

B1

Wavelength observed on Earth increases compared with
source stationary

B1

max 4

the use of physics terms is accurate, the answer is fluent / well argued with few errors in spelling, punctuation and grammar

B2

award for 2+
the use of physics terms is accurate, but the answer lacks coherence or the spelling, punctuation and grammar are poor

B1

award for 1
the use of physics terms is inaccurate, and the answer is disjointed with significant errors in spelling, punctuation max 2 and grammar

B0

max 2




(b)     (i)      use of c =
Δ
f = 3 × 108 /(561 × 10–9) – 3 × 108 / (540 × 10–9)

C1

= (5.348 – 5.556) × 1014 [= 2.08 × 1013 Hz]

M1

[explicit subtraction or to 3+ s.f. required for A mark]

A1




(ii)     Δf / f = v / c

V = c * Δf / f [or Δλ / λ] = 3 × 108 × 2.08 × 1013 / 5.556 × 1014

C1

[ecf from bi]

M1

= 1.12 × 107 m / s

A1




(iii)     [d = v / H]
conversion to km / s

C1

d = 11.2 × 103 / 65 = [172 Mpc] [ecf from bii]

B1

= 172 × 106 × 3 × 1016 = 5.17 × 1024 m

A1

[16]

 

M4.(a)

 

P moving away, Q towards (1)

λ increasing and decreasing respectively (1)

(4)




(b)     (i)      









 (1)



(max 8)

[12]

 

Wednesday, February 24, 2021

Momentum and Collisions

 

 

1.       A truck of mass 2kg travels at 8 m/s towards a stationary truck of mass 6kg. After the collision they stuck together and move off in the same direction. What is their common velocity?

a.       16kgms-1 = (2+6 kg)v so 8v = 16 and v =2 ms-1

2.       A car of mass 2000kg, traveling at 10m/s, has a head on collision with a small sports car of mass 500kg. If both cars stop dead on colliding, what was the velocity of the sports car before the collision? Momentum after = 0 so (2000 x 10) + (500 x v) so 500v = 20000 and v = 40 ms-1

3.       A man wearing a bullet proof vest stands on roller skates. His total mass is 80kg. A bullet of mass 20g is fired at him at a speed of 400 m/s. If the bullet is stopped by the vest and falls to the ground, what velocity will the man move at? Momentum before = (0.02kg x 400) + 0 = 8k ms-1

a.       Momentum after = 0 + (80kg x v) = 8 kg ms-1  so 80v = 8 so v = 0.1 ms-1 after being hit

How does your answer compare with “the movies”? They aren’t real. (but then you knew that)

Wednesday, February 10, 2021

momentum questions


1.       A bowling ball has a mass of 10kg and a velocity of 15ms-1. Calculate its momentum.                                150 kgms-1
2.       A speed skater has a velocity of 15ms-1 and a mass of 65 kg. Calculate her momentum.                             975 kgms-1
3.       A bullet of mass 0.068 kg traveling horizontally at a speed of 150 ms-1. Calculate its momentum. 10.2 kgms-1
4.       The world record for bowling at cricket is 80 ms-1. A cricket ball has a mass of 0.17kg. Calculate its momentum.                13.6 kgms-1
5.       Calculate the momentum of a car of mass 1000kg travelling a velocity of 20ms-1                                         2 x 10 4 kgms-1
6.       A jumbo jet has a mass of 30 000 kg. Calculate its momentum at cruising speed of 300ms-1                     9 x 10 6 kgms-1
7.       A Eurostar train has a mass of 752t. (1t = 1000kg).  Calculate its momentum at top speed of 93 ms-1 6.99x 107kgms-1
8.       An oil tanker has a mass of 100 x106 kg and is travelling at 5ms-1. Calculate its momentum.                    5 x 108 kgms-1
9.       The Earth has a mass of 5.9742 × 1024 kg. Its orbital velocity is 29.8 kms-1. Calculate its momentum. 1.73 x1029 kgms-1
10.   Jupiter has a mass of 1.8986×1027 kg and an orbital velocity of 13.06 kms-1. Calculate its momentum. 2.48 x1031 kgms-1
11.   N2 has an atomic mass of 4.65173 x 10-26 kg. Calculate its momentum if it has a velocity 500 ms-1 2.33 x 10-23 kgms-1
12.   The electron mass = 9.10938188 × 10-31 kilograms. Calculate the momentum of an electron travelling at 0.1c (c = 3 x108 ms-1) 2.73 x10-23 kgms-1


Tuesday, February 09, 2021

Materials assessed Homework

 

3.4 Materials MS

 

M1.          (a)    The candidate’s writing should be legible and the spelling, punctuation and grammar should be sufficiently accurate for the meaning to be clear. The candidate’s answer will be assessed holistically. The answer will be assigned to one of three levels according to the following criteria.

High Level (Good to excellent): 5 or 6 marks

The information conveyed by the answer is clearly organised, logical and coherent, using appropriate specialist vocabulary correctly. The form and style of writing is appropriate to answer the question.

Candidate must suggest

·        drawing a graph of F vs ΔL (or vice versa)

·        AND that k is in some way linked to the gradient

·        AND use of a suitable named instrument to measure or determine extension

·        AND 1 further means of reducing uncertainty: repeats / minimum 8 different readings / use of vernier scale / check values of mass with balance / parallax elimination with set square, pointer in contact with scale, mirror.

For 6 marks:

must also give suitable range at least up to 10N but not beyond 20N (accept ‘up to 20N’ / ‘not beyond 20N’)

AND minimum 8 different readings OR parallax elimination must be included

AND repeats must be included

AND correctly explains how k is obtained from their graph.

Intermediate Level (Modest to adequate): 3 or 4 marks

The information conveyed by the answer may be less well organised and not fully coherent. There is less use of specialist vocabulary, or specialist vocabulary may be used incorrectly. The form and style of writing is less appropriate.

Candidate must suggest:

·        to measure / determine extension OR initial and final length

·        AND to use F = k ΔL or k = F / ΔL
OR drawing a graph of F vs ΔL (or vice versa)

·        AND use of suitable instrument to measure extension
OR 1 means of reducing uncertainty:
repeats / use of vernier scale / check values of mass with balance / parallax elimination with set square,    pointer in contact with scale,
mirror / minimum 8 different readings / graphical approach

For 4 marks, uncertainty comment AND instrument required

Low Level (Poor to limited): 1 or 2 marks

The information conveyed by the answer is poorly organised and may not be relevant or coherent. There is little correct use of specialist vocabulary. The form and style of writing may be only partly appropriate.

Any relevant statement from the marking points above

For 2 marks: must mention minimum two points including:

·        to measure / determine extension OR initial and final length

6




(b)     (i)     (k = 2 × 85 = 170 (N m–1) )

(ΔL = F / k =) 15 / 170 ( or 7.5 / 85 )

= 0.088  (m) (0.0882)

2



(ii)     (k = ½ × 85 = 42.5 )

(ΔL = F / k = ) 15/42.5 ( or 2 x 15/85)

= 0.35  (m) (0.3529)

2




(iii)     (W = ½ FΔL or ½ k ΔL2)

= ½ × 15 × 0.0882 ( or 2 x ½ × 7.5 × 0.0882)  ecf 5bi

= 0.66  (J) (0.6615) ecf 5bi

2



(iv)     (series) greater  ecf for answer ‘less’ or ‘same’ where candidates
incorrect answers to bi and bii support this.

extension is more (in series) and the force is the same
(in both situations)

AND quotes Energy stored = ( ½ )Fs or ½ FΔL OR energy proportional to
extension

3

[15]

M2.          (a)     tensile stress: (stretching) force (applied) per unit
cross-sectional area (1)
tensile strain: extension (produced) per unit length (1)

2

 

(b)     Hooke’s law (or stress  strain) obeyed up to point A (1)
A is limit of proportionality  (1)
elastic limit between A and region B (1)
region C shows plastic behaviour or wire is ductile  (1)
region B to C wire will not regain original length  (1)
beyond region C necking occurs (and wire breaks) (1)

max 5

QWC

[7]

 

M3.          (a)    

Suitable scale on both axes (eg not going up in 3s) and > ½ space used

≥ points correct (within half a small square)

line is straight up to at least stress = 2.5 × 108 and curve
is smooth beyond straight section

3




(b)     understanding that E = gradient (= Δyx)

allow y/x if line passes through origin

= 1.05 × 1011 (Pa) (allow 0.90 to 1.1) ecf from their line in (a)
if answer outside this range and uses a
y value ≥ 2

when values used from table;

•        two marks can be scored only if candidates line passes
through them

•        one mark only can be scored if these points are not on their line

2








[8]