C A R I B B E A N
E X A M I N A T I O N S
02138020/CAPE/KMS 2024
C O U N C I L
CARIBBEAN ADVANCED PROFICIENCY EXAMINATIONS®
PHYSICS
UNIT 1 – Paper 02
KEY AND MARK SCHEME
MAY/JUNE 2024
Question 1
Specific Objectives: Module 1 - 3.1, 3.2,3.3, 3.5, 3.6, 3.7
KC
(a)
(i)
Displacement is the distance travelled by an
object (1 mk) in a specified direction (1 mk)
OR
2
Distance between two points measured along a
straight line (1 mk) in a specified direction.
(1 mk)
(ii)
Acceleration is the rate of change of velocity
of an object (1 mk)
NB.
Mark
not
given
for
definitions
or any other like them



the
following
rate of change of velocity with time
rate of change of velocity per time
rate of change of velocity over a period
of time.
1
UK
XS
KC
(b)
UK
XS
Graph of velocity, v, versus time, t.
Scale S (1 mk)
No mark off if scale is broken
All points accurately plotted P (1 mk)
Smooth curve C (good symmetry and no straight lines from t=2s to t=6s)(1 mk)
3
KC
(c)
(i)
UK
XS
Gradient = acceleration (1mk)
1
For drawing the appropriate tangent at 1.5 s
(1 mk)
𝑣 −𝑣1
𝑠𝑙𝑜𝑝𝑒 (𝑎) = 2
𝑡2 −𝑡1
13.7 −5.2
slope(a) =
2.7−0.4
8.5 𝑚𝑠−1
=
2
(1mk) for substitution
2.3𝑠
SOI
= 3.7 𝑚 𝑠−2
(1 mk)
(Accept range 3.0 ms-2 to 4.4 ms-2)
(ii)
The
area
under
the
curve
is
equal
1
to
the
displacement.
No. of squares = 158
one square ≡ 0.5 m
Total area = 158 x 0.5 = 79 m
1
OR
Any other method of finding area below graph
Appropriate
mk)
(d)
(i)
method
with
reasonable
Given U = 30 ms-1, θ = 25o
𝑢𝑦 = 𝑢 𝑠𝑖𝑛 𝜃
(1 mk)
-1
= 30 m s (sin 25) (1 mk)
= 12.68 m s-1
(1 mk)
(ii)
answer(1
𝑢𝑥 = 𝑢 𝑐𝑜𝑠 𝜃
(1 mk)
= 30 m s-1 (cos 25) (1 mk)
= 27.19 m s-1
(1 mk)
3
3
KC
(iii)
UK
𝑣𝑦 = 𝑢𝑦 – 𝑔𝑡
but Vy = 0 (1 mk)
and uy =12.68ms-1 and a=-g (correct
convention)(1 mk)
𝑡 =
𝑢𝑦
𝑔
=
12.68 𝑚 𝑠−1
9.81 𝑚𝑠−2
= 1.29 s ignore sign (1 mk)
(ECF/FT value from d(i)
OR
𝑡=
𝑢𝑠𝑖𝑛𝜃
(𝟏 𝐦𝐤)
𝑔
𝑡=
30 sin 25
(𝟏 𝐦𝐤)
9.81
3
𝑡 = 1.29 𝑠 𝟏 𝐦𝐤
OR
Consider the return path from maximum height
to ground
𝑣𝑦 = 𝑢𝑦 – 𝑔𝑡
but Vy = 12.68ms-1
(1 mk)
and uy = 0 and a = g(correct convention)(1 mk)
𝑡 =
𝑢𝑦
𝑔
=
12.68 𝑚 𝑠−1
9.81 𝑚𝑠−2
= 1.29 s ignore sign (1 mk)
(ECF/FT value from d(i)
XS
KC
(iv)
UK
Maximum height, H
𝐻𝑦 =
𝑉2 sin2 𝜃
2𝑔
𝐻𝑦 =
=
𝑈2𝑦
(1 mk)
2𝑔
12.68 𝑚 𝑠−1
2
2× 9.81 𝑚 𝑠−2
(1 mk)
𝐻𝑦 = 8.19 𝑚 (1 mk)
(ECF/FT value from d(i)
OR
𝐻𝑦 = 𝑢𝑦 𝑡 – ½ 𝑔 𝑡2
but uy = 12.68 ms-1; (1 mk)
and g = − 9.81 ms−2 ; (correct convention) (1 mk)
𝐻𝑦 = 12.68 𝑚 𝑠−1 × 1.29 𝑠 − 0.5 9.81𝑚 𝑠−2 × 1.292 𝑠2
𝐻𝑦 = 8.19 𝑚 ignore sign (1 mk)
(ECF/FT value from d(i)
OR
Consider the return path from maximum height
to ground
𝐻𝑦 = 𝑢𝑦 𝑡 + ½ 𝑔𝑡2
uy = 0; (1 mk)
a=g; (correct convention)
𝐻𝑦 =
(1 mk)
(0 𝑚 𝑠−1 × 1.29 𝑠) + [0.5(9.81 𝑚 𝑠−2 × 1.292 𝑠2 )]
𝐻𝑦 = 8.19 𝑚 ignore sign (1 mk)
(ECF/FT value from d(i)
OR
𝑣2𝑦 = 𝑢2𝑦 + 2𝑔𝐻𝑦
𝑣2𝑦 − 𝑢2𝑦
𝐻𝑦 =
2𝑔
but Vy = 0 (1 mk)
and uy =12.68ms-1 and a=-g (correct convention)
(1 mk)
02 − 12.682
𝐻𝑦 =
−2 × 9.81
3
XS
KC
𝐻𝑦 = 8.19 𝑚 ignore sign (1 mk)
(ECF/FT value from d(i)
OR
Consider the return path from maximum height
to ground
𝑣2𝑦 = 𝑢2𝑦 + 2𝑔𝐻𝑦
𝐻𝑦 =
𝑣2𝑦 − 𝑢2𝑦
2𝑔
but Vy = 12.68ms-1
(1 mk)
and uy = 0 and a = g (correct convention)
(1 mk)
𝐻𝑦 =
12.682 − 02
2 × 9.81
𝐻𝑦 = 8.19 𝑚 ignore sign (1 mk)
(ECF/FT value from d(i)
(e)
1
𝑆 = 𝑢𝑡 + 𝑎𝑡2
2
𝑋 = 𝑢𝑡𝑐𝑜𝑠𝜃
1
Eqn 1
(1 mk)
Eqn 2
(1 mk)
2
𝑌 = 𝑢𝑡𝑠𝑖𝑛𝜃 − 𝑔𝑡
2
If g not indicated as negative mark not given
Making t the subject (1 mk) using Eqn 1
𝑋
𝑡=
𝑢𝑐𝑜𝑠𝜃
7
Substitute t in eqn. 2
𝑌 = 𝑢𝑠𝑖𝑛𝜃
𝑋
𝑢𝑐𝑜𝑠𝜃
–½𝑔
𝑋
𝑢𝑐𝑜𝑠𝜃
𝑌 = 𝑋𝑡𝑎𝑛𝜃 𝟏 𝐦𝐤 –
2
(1 mk)
𝑔
𝑋2 (𝟏 𝐦𝐤)
2𝑢2 cos2 𝜃
This equation has the standard form of a
parabola (Y =
Ax - Bx2) (1 mk) conclusion
must be
drawn from previous step in
derivation
UK
XS
KC
UK
XS
10
15
5
OR
𝑌 = 4.66 × 10−1 𝑋 – 6.63 × 10−3 𝑋2 (1 mk)
Total 30 marks