QUESTION ONE
You are provided with the following apparatus
- A metre rule
- A cotton or a silk thread
- A stop watch or stop clock
- A 50g mass
- Four pieces of wood
Proceed as follows:
- Measure accurately 1 metre of thread and cut it leaving an allowance of 50cm on either end. Mark the centre of the thread
- Clamp the thread between two retort stands and suspend a 50g mass at the same level 55cm above the bench, as shown below.
- By moving one end of the retort stands, adjust d to value of 40cm.
- pull the mass slightly in a direction parallel to xy then release it to oscillate. Record the time t,for 20 oscillations.
- Repeat the procedure (c) and (d) with other values of d, increasing at intervals of 5cm and complete the table below. Where T is the periodic time.
d(cm) 40 45 50 55 60 65 70 75 85 t(s) T(s) T2(s2) d2(cm2)
(7mrks) - plot a graph of T2 against d2 (5mrks)
- Determine the slope of your graph.(3mrks)
- Given that T2 = 3d2 + C using your graph determine the values.
m- M (3mrks)
- C (2mrks)
QUESTION TWO
PART A
You are provided with the following :
- Vernier calipers
- Micrometer screw gauge
- Masses;two 10g,20g,50g and 100g
- A helical spring
- Metre rule or half metre rule
Proceed as follows
- Determine the number of complete turns of the helical spring.
N=……………………………………………….(1mrk) - Measure the external diameter of the spring using the vernier calipers
D=……………………………………m (1mrk) - Use the micrometer screw gauge to determine the diameter of the wire of the spring.
D=……………………………………m (1mrk - Determine the value of M N = 0.4D (2mrk)
dM - Suspend the helical spring vertically alongside the clamped half metre rule as shown in figure 1below.Determine the length Lo,of the spring before loading it.
Lo=………………………cm - load the spring with a mass of 20g and determine the new reading on the metre.(L) record this in the table below. Calculate the extension e=L– Lo due to the mass of 20g and record the value in the table given below. Repeat step f for other masses and complete the table.
Mass(g) 0 10 20 30 40 50 60 70 80 90 100 Weight(N) Reading(L)cm Extension e(cm) Ye (cm-1) - Plot a graph of weight (N) against 1/e (cm-1) (5mrks)
- Determine the slope (s) of the graph at a mass of 45g (3mrks)
- Given that M = – 255T
(S+60)2
Determine the value of T where S is the slope at 45g (3mrks)
PHYSICS PRACTICAL CONFIDENTIAL
Each candidate should be provided with the following apparatus.
Question 1
- Two retort stands , 2 clamps and 2 boss heads.
- A metre rule.
- A cotton thread (110cm long).
- A stop watch or stop clock.
- 4 pieces of wood (3cm x3cmx1cm).
- A 50g mass.
Question 2
- Vernier calipers (To be shared).
- Micrometer screw gauge (To be shared).
- Masses two 10g, 20g, 50g and 100g.
- A helical spring.
- Metre rule or half metre rule.
MARKING SCHEME
Table
d(cm) | 40 | 45 | 50 | 55 | 60 | 65 | 70 | 75 | 85 | |
t(s) | 30 | 29 | 28 | 27 | 26 | 25 | 24 | 23 | 22 | 4mks |
T(s) | 1.50 | 1.45 | 1.40 | 1.53 | 1.30 | 1.25 | 1.20 | 1.15 | 1.10 | 1mk |
T2(s2) | 2.25 | 2.10 | 1.96 | 1.82 | 1.69 | 1.56 | 1.44 | 1.32 | 1.21 | 1mk |
d2(cm2) | 1600 | 2025 | 2500 | 3025 | 3600 | 4225 | 4900 | 5625 | 6400 | 1mk |
- Row 2(t-value)
- 1-1 values correct - 0mk
- 2-3 values correct - 1mk
- 4-5 values correct - 2mks
- 6-7 values correct - 3mks
- 8-9 values correct - 4mks
- slope = ΔT2 = (1.44 –1.96) (1mrk)
Δd2 (4900-2500) -
- T2 = 3d2 + C
M
3 = – 0.00022 s2/cm2 (1mrk)
M
M = 3 (1mrk)
0.00022
= – 13846.15cm2/s2 (1mrk) - C = y – intercept (1mrk)
Y intercept = 2.55s2
C = 2.55s2 (1mrk)
- T2 = 3d2 + C
QUESTION 2
PART A
- N = 83 (1mrk)
- D = 0.0116m(4 d.p) ± 0.0002 (1mrk)
- D = 6.3x10-4m (1mrk)
- N = 0.4D (½ mrk)
dM
m = 0.4D = 0.4x0.0116 1mrk = 0.0887 4 d.p (½ mrk)
dN 6.3x10-4x83 - Lo= 50.0
W(N) 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1mk L (cm) 50 51.2 52.4 53.6 54.8 56.0 57.2 58.4 59.6 60.8 62.0 1mk E (cm) 0 1.2 2.4 3.6 4.8 6.0 7.2 8.4 9.6 10.8 12.0 1mk 1/e cm-1 0 0.8333 0.4166 0.2777 0.2083 0.1666 0.1389 0.1190 0.1042 0.0926 0.0833 1mk -
- Scale = 1mk –Appropriate scale
- Axes = 1mk-well labelled
- plots = 2mks 8-11 plots correctly plotted
- 4-7 correctly plotted 1mk
- Smooth curve - 1mrk
- slope 5.5 – 3.5 = 2 = 2.5cm 1mrk
2.2 –1.4 0.8 - T= m(5+60)2 1mrk
– 255
T = 0.0887(2.5+60)2 1mrk = 1.3588N2cm2 1mrk
– 255
Question A
Question B
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