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Past Test Papers FSKS113
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, FSKS113Geskeduleerde toets 1 / Scheduled test 1 Maart / March 2019
____________________________ ___________ ___________________
Van / Surname) Voorletters / Initials Studente no. / Student nr.
Tyd 1 uur / Time 1 hour
Konstantes en Vergelykings Constants and Equations
Gravitasieversnelling 9,8 m/s2 Acceleration due to gravity
Gravitasiekonstante 6,67 x 10-11 N m2 kg-2 Gravitational constant
v = v0 + at
Bewegingsvergelykings met Equations of motion with constant
Δx = v0t + ½at2
konstante versnelling (vir x en y) acceleration (for x and y)
v2 = v02 + 2a Δx
AFDELING A / SECTION A 16 punte/marks
(1) 4
(2) 1
(3) 3
(4) 3
(5) 5
(6) 5
(7) 1
(8) 4
AFDELING B / SECTION B 24 punte/marks
Beantwoord die volgende vrae in die spasies. / Answer the following questions in the space below.
1. Die maksimum spoed van ‘n leeu is 81 km/h. 1. The maximum speed of a lion is 81 km/h.
(a) Wat is die verste afstand wat hy in 20 sekondes kan (a) What is the furthest distance he can cover in 20
hardloop? (3) seconds? (3)
(b) Definieer die begrip verplasing. (2) (b) Define the concept displacement. (2)
(c) Gebruik die definisie in (b) om te verduidelik waarom (c) Use the definition in (b) to explain why the lion’s
die leeu se verplasing moontlik kleiner sal wees as displacement will probably be smaller than the distance
die afstand bereken in (a). (2) calculated in (a). (2)
(a) 81 km/h = 22.5 m/s
Δx = v. Δt (a = 0 m/s2)
= (22.5)(20)
= 450 m
(b) Verplasing is ’n vektor met rigting van die beginposisie na die eindposisie, en met grootte die reglynige
afstand van die beginposisie tot die eindposisie.
Displacement is a vector quantity with magnitude the straight line distance from the starting point to the
end point, and direction from starting point to the end point.
1/4
,(c) As die leeu nie reglyning hardloop nie, sal sy afstand (die werklike lengte wat hy aflê) groter as die
verplasing wees. / ʼn Reguit lyn is altyd die kortste afstand tussen 2 punte.
If the lion does not run in a straight line, the distance that he covers will be larger than the
displacement. / A straight line is always the shortest distance between 2 points.
2/4
, 2. ‘n Motoris wat aanvanklik teen 25 m/s ry, sien ‘n 2. A motorist who initially drives at 25,0 m/s sees a
staande trok 64,0 m voor hom in die pad. Hy rem standing truck 64,0 m in the road in front of him. He
onmiddelik met konstante versnelling van −5 m/s2. brakes immediately with constant acceleration of
Bepaal of hy sal stop voordat hy met die trok bots.(4) −5m/s2. Determine whether he will stop before colliding
with the truck. (4)
v2 = v02 + 2a Δx
𝑣 2 − 𝑣02
∆𝑥 =
2𝑎
0−(25)2
= 2(−5)
= + 62.5 m
< 64.0 m
Sal betyds stop / Will stop in time
3. ⃗ dat die boks
In die skets hieronder, voorkom krag 𝑇 3. ⃗ prevents the box (mass
In the sketch below, force 𝑇
(massa 50,0 kg) teen die wrywinglose skuinsvlak afgly. 50,0 kg) from sliding down the frictionless inclined
Bereken die grootte van krag 𝑇⃗. (3) plane. Calculate the magnitude of force 𝑇 ⃗ . (3)
T
FN
Fgx
Fgy
ΣFx=0
T = Fgx
= mg sin 400
= (50)(9,8)(0,643)
=314,97 N
3/4
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