π
<-

## practice

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Catégorie :Category: mViewer GX Creator Lua TI-Nspire
Auteur Author: codper3
Type : Classeur 3.6
Page(s) : 7
Taille Size: 211.44 Ko KB
Mis en ligne Uploaded: 15/09/2021 - 19:22:22
Mis à jour Updated: 15/09/2021 - 19:22:43
Visibilité Visibility: Archive publique

### Description

Reading School Page 1 of 7
Mark schemes
D
1.


A
2.


3. C


D
4.


5. A


C
6.


D
7.


B
8.


D
9.


D
10.


(a) (i) elastic potential energy and gravitational potential energy ✓
11.
For elastic pe allow “pe due to tension”, or “strain energy” etc.
1

(ii) elastic pe → kinetic energy → gravitational pe
→ kinetic energy → elastic pe ✓✓
[or pe→ke→pe→ke→pe is ✓ only]
[or elastic pe → kinetic energy → gravitational pe is ✓ only]
If kinetic energy is not mentioned, no marks.
Types of potential energy must be identified for full credit.
2

(b) (i) period = 0.80 s ✓
during one oscillation there are two energy transfer cycles
(or elastic pe→ke→gravitational pe→ke→elastic pe in 1 cycle)
or there are two potential energy maxima per complete oscillation ✓
Mark sequentially.
2

Reading School Page 2 of 7
(ii) sinusoidal curve of period 0.80 s ✓
– cosine curve starting at t = 0 continuing to t = 1.2s ✓
For 1st mark allow ECF from T value given in (i).
2

(c) (i) use of T = gives 0.80 = ✓

∴ = 22 (21.6) ✓ N m–1 ✓

Unit mark is independent: insist on N m–1.
Allow ECF from wrong T value from (i): use of 0.40s gives
86.4 (N m–1).
3

(ii) maximum ke = ( ½ mvmax2) = 2.0 × 10–2 gives

vmax2 = ✓ (= 0.114 m2s–2) and vmax = 0.338 (m s–1) ✓

vmax = 2πfA gives A = ✓

and A = 4.3(0) × 10–2 m ✓ i.e. about 40 mm

[or maximum ke = (½ mvmax2) = ½ m (2πfA)2 ✓
½ × 0.35 × 4π2 × 1.252 × A2 = 2.0 × 10–2 ✓

∴ A2 = ✓ ( = 1.85 × 10–3 )

and A = 4.3(0) × 10–2 m ✓ i.e. about 40 mm ]
[or maximum ke = maximum pe = 2.0 × 10–2 (J)
maximum pe = ½ k A2 ✓
∴ 2.0 × 10–2 = ½ × 21.6 × A2 ✓

from which A2 = ✓ ( = 1.85 × 10–3 )

and A = 4.3(0) × 10–2 m ✓ i.e. about 40 mm ]
First two schemes include recognition that f = 1 / T
i.e. f = 1 / 0.80 = 1.25 (Hz).
Allow ECF from wrong T value from (i) – 0.40s
gives A = 2.15 × 10–2m but mark to max 3.
Allow ECF from wrong k value from (i) –86.4Nm–1 gives
A = 2.15 × 10–2m but mark to max 3.
4


Reading School Page 3 of 7
(a) 2.2 s
12.
c.a.o.

B1
1

(b) exactly two reasonable sine wave cycles drawn

B1

displacement = 10 cm when time = 0

B1

time = 2.2 s after one cycle

B1
4

peaks decrease to approximately 7.0 cm after two cycles
or 8.4 cm after one cycle

B1

award two marks if half-cycle confused with full cycle
but otherwise correct

(c) (i) the period would be decreased

B1

(ii) there would be less damping/more oscillations before
the pendulum comes to rest

B1
2


(a) (i) a normal reaction shown and labelled on either diagram
13. B1

a frictional force correctly shown and labelled on either diagram (may be
outward on second diagram)

deduct 1 mark for each wrong force (condone poor friction / reaction)
B1
(2)

Reading School Page 4 of 7
(ii) friction (between surface and wheel / tyre)
B1

(normal) reaction (at the surface)
B1

horizontal component of either force / component towards the centre
B1

sum of horizontal components
B1
(4)

(b) use of mg = mv2 / r or g = v2r, centripetal force = mv2 / r
C1

correct substitution v2 = 9.8 × 5.2
C1

7.1 m s–1
A1
(3)


Reading School Page 5 of 7
(a)
14.
vmax=2π × 2.0 × 2.5 × 10–2
vmax= 0.314 m s–1 ✔
(use of Ek = ½mv2)
54 × 10–3 = ½m × (0.314)2
m = 1.1 (kg) ✔

2.0 × 2π=√(k/1.1 ) ✔
(k = (4π)2 × 1.1)
k = 173 (172.8) ✔(N m–1)
Can

OR

5.4 × 10–3 = ½ k (2.5 × 10–2)2 ✔
k = 173 (172.8) N m–1 ✔
If either of these methods used can then find mass from frequency
formula or from kinetic energy

OR

54 × 10–3 = ½ F × 2.5 × 10–2
F = 4.32
4.32 = k × 2.5 × 10–2
k = 173 (N m–1)
Accept 170 and 172.8 to 174
1
1
1
1

(b) )
same mass so

frequency = 3.5 (3.46) (Hz) ✔
Allow CE from (a) for k or m
...

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