INSTRUCTIONS TO CANDIDATES
- Answer all questions in the spaces provided in the question paper.
- You are supposed to spend the first 15minutes of the 2 ¼ hours allowed for the paper reading the whole paper carefully.
- Marks are awarded for clear record of the observations actually made, their suitability and accuracy.
- Candidates are advised to record their observations as soon as are made.
- Mathematical tables and calculators may be used.
QUESTION 1
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) - plot a graph of T2 against d2 (5mrks)
- Determine the slope of your graph.(3mrks)
- Given thatT2 = 3d2 + C using your graph determine the values.
m- M (3mrks)
- C (2mrks)
QUESTION TWO
PART A
You are provided with the following :
- Vernier caliper
- 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 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) 1/e (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)
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 | 1.35 | 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 |
(7mrks)
Row 2(t-value)
values correct -0mk
2-3 values correct 1mk
4-5 values correct 2mks
6-7 values correct 3mks
8-9 values correct 4mks
g) slope = ΔT2 = (1.44−1.96) 1mrk
Δd2 (4900−2500)
h)( i ) T2 = 3d2 +C
M
3 = −0.00022 s2/cm2 1mrk
M
M = 3 1mrk
0.00022
=−13846.15cm2/s2 1mrk
(ii) C=y – intercept 1mrk
Y intercept=2.55s2
C=2.55s2 1mrk
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
-
Mass (g) 0 10 20 30 40 50 60 70 80 90 100 Weight (N) 0 0.1 0.2 0.3 0,4 0.5 0.6 0.7 0.8 0.9 1.0 ✔1mk Reading (L)cm 50 51.2 52.4 53.6 54.8 56.0 57.2 58.4 59.6 60.8 62.0 ✔1mk Extension 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 0.41666 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
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