# Physics Paper 3 Questions and Answers with Confidential- Form 3 Term 3 Opener Exams 2023

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:

1. Measure accurately 1 metre of thread and cut it leaving an allowance of 50cm on either end. Mark the centre of the thread
2. Clamp the thread between two retort stands and suspend a 50g mass at the same level 55cm above the bench, as shown below.

3. By moving one end of the retort stands, adjust d to value of 40cm.
4. pull the mass slightly in a direction parallel to xy then release it to oscillate. Record the time t,for 20 oscillations.
5. 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)
6. plot a graph of T2 against d2 (5mrks)
7. Determine the slope of your graph.(3mrks)
8. Given that      T2 = 3d2 + C using your graph determine the values.
m
1. M            (3mrks)
2. 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

1. Determine the number of complete turns of the helical spring.
N=……………………………………………….(1mrk)
2. Measure the external diameter of the spring using the vernier calipers
D=……………………………………m (1mrk)
3. Use the micrometer screw gauge to determine the diameter of the wire of the spring.
D=……………………………………m (1mrk
4. Determine the value of M       N = 0.4D      (2mrk)
dM
5. 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

6. 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)
7. Plot a graph of weight (N) against 1/e (cm-1) (5mrks)
8. Determine the slope (s) of the graph at a mass of 45g (3mrks)
9. 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) 1.5 1.45 1.4 1.53 1.3 1.25 1.2 1.15 1.1 1mk T2(s2) 2.25 2.1 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

1. Row 2(t-value)
2. 1-1 values correct  -  0mk
3. 2-3 values correct  -  1mk
4. 4-5 values correct  -  2mks
5. 6-7 values correct  -  3mks
6. 8-9 values correct  -  4mks
7. slope = ΔT2 = (1.44 –1.96)    (1mrk)
Δd2    (4900-2500)
8.
1. T2 = 3d2 + C
M
3 = – 0.00022 s2/cm2   (1mrk)
M
M =     3             (1mrk)
0.00022
= – 13846.15cm2/s2           (1mrk)
2. C = y – intercept             (1mrk)
Y intercept = 2.55s2
C = 2.55s2   (1mrk)

QUESTION 2

PART A

1. N = 83    (1mrk)
2. D = 0.0116m(4 d.p) ± 0.0002    (1mrk)
3. D = 6.3x10-4m     (1mrk)
4. N = 0.4D       (½ mrk)
dM
m = 0.4D0.4x0.0116           1mrk = 0.0887 4 d.p   (½ mrk)
dN      6.3x10-4x83
5. Lo= 50.0

6.  W(N) 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1mk L (cm) 50 51.2 52.4 53.6 54.8 56 57.2 58.4 59.6 60.8 62 1mk E (cm) 0 1.2 2.4 3.6 4.8 6 7.2 8.4 9.6 10.8 12 1mk 1/e cm-1 0 0.8333 0.4166 0.2777 0.2083 0.1666 0.1389 0.119 0.1042 0.0926 0.0833 1mk
7.
• Scale = 1mk –Appropriate scale
• Axes = 1mk-well labelled
• plots = 2mks   8-11 plots correctly plotted
• 4-7 correctly plotted     1mk
• Smooth curve - 1mrk

8.  slope 5.5 – 3.5 = 2.5cm 1mrk
2.2 –1.4     0.8

9. T= m(5+60)2    1mrk
– 255
T = 0.0887(2.5+60)2   1mrk  = 1.3588N2cm2    1mrk
– 255

Question  A

Question B

#### Download Physics Paper 3 Questions and Answers with Confidential- Form 3 Term 3 Opener Exams 2023.

• ✔ To read offline at any time.
• ✔ To Print at your convenience
• ✔ Share Easily with Friends / Students

### Related items

.
Subscribe now

access all the content at an affordable rate
or
Buy any individual paper or notes as a pdf via MPESA
and get it sent to you via WhatsApp