Download Throughout paper, take off half a mark for incorrect or missing units

UNIVERSITY OF KWAZULU-NATAL
FOUNDATION PHYSICS SUPPLEMENTARY EXAM
30th NOVEMBER 2009
UNIVERSITY OF KWAZULU-NATAL
SECOND SEMESTER SUPPLEMENTARY MEMORANDUM: 30th November 2009
Subject, Course and Code: Physics Foundation PHYS099 and PHYS199
DURATION: 3 HOURS
TOTAL MARKS:180
Student number: __________________________________
Seat number: _______________
INTERNAL EXAMINERS:
Ms B. Adams, Ms W. Dlamini, Mrs S. Halstead, Mr R. Webber and Mr E Zhandire
EXTERNAL EXAMINER Dr V Couling
ANSWER ALL QUESTIONS ON THIS PAPER.
IN THEIR OWN INTERESTS STUDENTS ARE REQUESTED TO WRITE LEGIBLY.
THIS PAPER CONSISTS OF 14 PAGES. PLEASE SEE THAT YOU HAVE THEM ALL.
Throughout paper, take off half a mark for incorrect or missing units in the
answer to each section.
PROPERTIES OF MATTER
Question 1 (8 marks)
Give your answers to the following questions with the correct number of significant figures
or decimal places.
a.
16.02 litres + 8.3649 dm3 + 27.9 cm3 =
(3)
= 16.02 dm3 + 8.3649 dm3 + 0.0279 dm3 = 24.41 dm3
b.
An engineer finds that a metal has a volume of 0.15 cm3 per gram.
i) What is the density of the metal?
d=
1
g.cm-3 = 6.7 g.cm-3
0.15
ii) What is the mass of a 17.8 cm3 volume of this metal?
How many shares of 0.15 cm3 are there in 17.8 cm3?
Each share will have a mass of 1 gram
Therefore mass =
17.8
= 120 g
0.15
OR m = d x V = 6.7 x 17.8 = 119.26 g = 119
(OR 120 g if you did not round off earlier)
Page 1 of 14
(2)
(3)
UNIVERSITY OF KWAZULU-NATAL
30th NOVEMBER 2009
FOUNDATION PHYSICS SUPPLEMENTARY EXAM
velocity (m/s)
MECHANICS
NOTES:
1.
Take the magnitude of acceleration due to gravity as 9.8 ms-2
2.
Show the reference frame that you use for each situation.
Question 2 (9 marks)
The velocity-time graph below shows the motion of an object over 12 seconds.
10.0
D
8.0
C
B
6.0
E
4.0
A
2.0
F
tim e (s )
0.0
0
2
4
6
8
10
G
12
-2.0
-4.0
The object has a mass of 10 kg and started off travelling north. Refer to the letters on the
graph when answering the following questions.
NB Initially traveling NORTH
a.
Over what section was the object travelling fastest?
D
b.
In what direction is it traveling, over:
i)
section B?
ii)
section E?
iii)
section G?
north
north
south
(1)
(3)
c.
Over what section does the object first remain stationary? F
(1)
d.
What section has maximum acceleration/deceleration?
(1)
e.
What is the resultant force on the object in section D? Show reasons.
(3)
Object has constant velocityT so by Newton I (or show calc with Newton II)T the
resultant force is ZeroT
Page 2 of 14
E
UNIVERSITY OF KWAZULU-NATAL
30th NOVEMBER 2009
FOUNDATION PHYSICS SUPPLEMENTARY EXAM
Question 3 (10 marks)
Zipho throws a ball vertically upward with an initial speed of 15.0 ms-1 to Ntombi who is
leaning over a balcony that is 8.0 m above Zipho’s hand. She catches the ball on its way
down. Calculate:
a.
The velocity of the ball when she catches it. (6)
(take UP as +, –½ if not stated)
v is +
v is –
v = – 8.3 ms-1 (or 8.3 ms-1down) NB Must be negative square root
b.
The time taken to reach Ntombi.
v = u + at = – 8.26 + 15.0 + (–9.8)t
a is –9.8 ms-2 for
whole journey
Zipho
= (15.0)2 +2(–9.8)(8.0) = 68.2
(NB signs must be correct according to ref frame)
∆x or s = + 8.0 m
Ntombi
v2 = u2 + 2as
(4)
(NB signs must be correct according to ref frame)
t = 2.4 s
Question 4
(6 marks)
Sinah stands on a building so she is 11.0 m above the ground. She sees her boyfriend
walking towards the building at a steady pace of 1.6 ms-1. She drops her cellphone to him
when he is 3.5 m from the building. If he keeps the same pace, will he be able to catch the
phone without stretching his arms?
(6)
Sinah
a is +9.8 ms-2 for phone
Phone ∆x or s = + 11.0
Boy
Boy v is 1.6 ms-1
Boy ∆x or s = 3.5 m
a = 0 (const v)
NB DO NOT MIX DATA FOR PHONE & BOY
Vertical Motion - Sinah’s phone:
take down as + (– 12 not stated)
Horizontal motion – Boyfriend
Take direction of motion as +
s = ut + 12 at 2 T
11 = 0 + 12 9.8t2 T
t = 1.5sT
s = ut + 12 at 2
3.5 = 1.6t + 1/2 (0)t2T
t = 1.75s = 1.8sT
Which is longer than 1.5 s
OOPS he does not get there in time. T
Page 3 of 14
UNIVERSITY OF KWAZULU-NATAL
30th NOVEMBER 2009
FOUNDATION PHYSICS SUPPLEMENTARY EXAM
Question 5 (12 marks)
a) A student gets off a bus at university and walks 105 m on a
bearing of 45°, then 87 m on a bearing of 292° and finally 220
m due East. The displacement vectors are represented on the
diagram below. How far and in what direction do they need to
walk to get back to the bus at the end of the day?
(8)
It may help to sketch the vectors individually – BE CAREFUL to get
the correct angles
B = 87 m
y
105 m
87 m
450
220 m
x
2920
y
A =105 m
y
y
By =87 sin 220 m
Ay =105 sin 450 m
450
220
C = 220 m
0
45
0
Ax =105 cos 45 m
Bx = –87 cos 220
m
x
x
x
2920
x-components (east)
105cos45o =74 m
y-components (west)
105sin45o =74 m
87sin22o =33 m
C
87cos22o = –- 81 m
220 m
Resultant
213 m
107 m
A
B
0
To find the journey back to the bus, we are looking for the equilibrant – the opposite of
resultant – so we make the vectors head-to-tail and come back to the start.
s2 = (213)2 + (107)2 therefore
If you want marks for θ you MUST show θ on a diagram.
107
tan θ =
= 1.99
213
θ = 26.7
N
s = 238 m
213 m
θ
107 m
s
B̂
o
b) A pilot wishes to fly due North from Airport A to Airport B. The wind
is blowing at 30 kmhr -1 due west. The small plane averages a velocity
of 200 kmhr-1. What direction should the pilot take in order to reach
Airport B? Use a sketch to help solve.
(4)
The RESULTANT is what must happen overall – the plane must RESULT
IN GOING NORTH
30
NOT tan sin θ =
= 0.15
θ = 8.60
200
The plane must fly bearing 8.60
OR 8.60 East of North
You are not asked to find the resultant speed
Page 4 of 14
Vp= 200 kmhr-1
N
B̂
Resultant – how
plane needs to end up
the student must walk 238 m bearing of 2430 OR 26.70south of west OR 63.3owest of south
θ
vw=30 kmhr-1
UNIVERSITY OF KWAZULU-NATAL
FOUNDATION PHYSICS SUPPLEMENTARY EXAM
30th NOVEMBER 2009
Question 6: (16 marks)
The drawing shows a truck, with mass 2000 kg, towing a
caravan. While the truck accelerates from rest to a maximum
speed of 22.5 ms-1 in 25.0 seconds, the engine of the truck
exerts an average force on the road of 4000 N, and the average
frictional force from the air is 1000 N.
a)
Calculate the mass of the caravan.
(7)
Take right as + direction The car has to accelerate itself AND the caravan
a=
∆v 22.5
T= 0.900 ms-2T
=
∆t 25.0
1000 N
4000 N
F res = 4000 N – 1000 N = 3000 N right. TT
F res = ma, therefore 3000 N = mtotal x 0.900 T
so m total = 3333 kgT (3330 to 3 sig figs)
Therefore caravan mass = 3330 – 2000 = 1330 kgT
b)
When stopping, the maximum braking force on the truck is 9 000 N. Calculate how far
the truck will take to stop from maximum speed when it is NOT towing the caravan?
(6)
We still have friction but NO driving force from the engine,
1000 N
we now also have a braking force
9000 N
F res = –9000 + (–1000) = –10000N T
−10000
T= –5.000 ms-2 T
From F = ma we get: a =
2000
The initial speed is 22.5 ms-1and final speed is 0, v 2 = u 2 + 2as
so
c)
0 = 22.52 + 2 x (– 5.00) s TT
so s = 50.6 metresT
When the car is towing the caravan, will the stopping distance increase, decrease or stay
the same. Give a brief explanation in terms of physics.
(3)
There will be more mass/inertia with the caravanT,
so with same braking force we have smaller acceleration (deceleration) T
therefore stopping distance increasesT
Page 5 of 14
UNIVERSITY OF KWAZULU-NATAL
FOUNDATION PHYSICS SUPPLEMENTARY EXAM
30th NOVEMBER 2009
Question 7: (10 marks)
The diagram below shows a woman at the airport pulling her 20 kg suitcase at a constant
speed along a horizontal surface, using a strap that makes an angle θ above the horizontal.
She pulls on the strap with a 35 N force and the frictional force on the suitcase is 20 N.
FNormal
Ty = 35 sin θ N
T = 35 N
θ
Ffriction = 20 N
Tx = 35 cos θ N
Fg = 20 x 9.8 N
a)
What angle does the strap make with the horizontal?
(3)
Constant velocity therefore Fresultant = 0
Instead of T or 35 N we use the horizontal and vertical components Tx and Ty
Notice that there are now 2 forces vertically upwards
Consider horizontal forces: (Take forward as positive)
FRes x = 0
this means: Tx + Ffriction = 0
20
cosθ =
= 0.5714
35
θ = 55o
therefore
35 cos θ +(–20) = 0
b) What normal force does the ground exert on the suitcase?
Consider vertical forces: (Take UP as positive)
(4)
FRes y = 0
FN + Ty + Fg = 0
therefore: FN + 35 sin θ – 20 x 9.8 = 0
FN = 196 – 35sin55o =
167 N (or 170N)
c) If the strap breaks, what would be the magnitude and direction of the acceleration of
the suitcase?
(3)
The only horizontal force will be Ffriction or –20N
F = ma
a=
−20
= −1.0ms −2
20
Page 6 of 14
UNIVERSITY OF KWAZULU-NATAL
FOUNDATION PHYSICS SUPPLEMENTARY EXAM
30th NOVEMBER 2009
ELECTROSTATICS
NOTE: Take the electrostatic constant in air to be k = 9.00 x 109 Nm2C-2 and the charge on
an electron to be Qe = 1.602x10-16 C
Question 8
a)
n=
(26 marks)
In walking across a carpet, you acquire a net negative charge of 50 µC. How
many excess electrons do you have?
(2)
50x10−6
= 3.1x1011 electrons
−16
1.602x10
b)
The diagram below shows four charges located at the corners of a square.
q1 = - 10µC
q2 = - 10µC
F12
F32
F42
q4 = + 5.0µC
q3 = + 5.0µC
i)
Use the diagram to show F12, F32 and F42.
(3)
ii)
What is the magnitude of the force F12?
(3)
F12 =
iii)
kQ1Q2 9 x10 9 x(10 x10 −6 ) 2
=
= 90 N
r2
(0.10) 2
What is the distance separating q2 and q4?
(2)
r2 = (0.10)2 + (0.10)2
r = 0.14 m
iv)
F42 =
Determine the magnitude of the force F42.
kQ 2Q 4 9.00x109 × (10x10−6 ) × (5.0 × 10−6 )
=
= 23N
r2
(0.14) 2
Page 7 of 14
(3)
UNIVERSITY OF KWAZULU-NATAL
c)
FOUNDATION PHYSICS SUPPLEMENTARY EXAM
30th NOVEMBER 2009
Charge Q acts as a point charge to create an electric field. Its strength, measured a
distance of 30 cm away, is 40 NC-1. What would be the electric field strength
30 cm away from charge 2Q ?
i)
Enew = 2 E1 = 80 NC-1 (in the same direction)
90 cm away from charge 0.5Q?
ii)
E new =
iii)
1
2
2
3
=
1
x 40 = 2.2 NC-1 (in the original direction)
18
(4)
Calculate the force on a 40 nC charge 30 cm away from charge Q.
F = E1 x q = 40 x 40x10-9 = 1.60x10-6 N
d)
(2)
(in direction of field)
(3)
A proton is accelerated between two points where the potential difference is10 V.
What is the work done on the proton? Qp= + 1.602 x 10-19 C.
(4)
W = q X Vab = 1.602 X 10-19 X 10
= 1.6 X 10-18 J
Page 8 of 14
UNIVERSITY OF KWAZULU-NATAL
FOUNDATION PHYSICS SUPPLEMENTARY EXAM
30th NOVEMBER 2009
ELECTRIC CIRCUITS
Question 9
(10 marks)
Questions a, b and c refer to the circuits below, which contain identical batteries and bulbs.
The internal resistance of the batteries can be ignored.
A
E
C
F
B
D
a.
If bulb C is unscrewed, the brightness of bulb B will ____same___________(2)
b.
If bulb E is unscrewed, the brightness of bulb D will ____increase________ (2)
c.
If bulb E is unscrewed, the brightness of bulb F will ____ decrease ______ (2)
Questions d and e refer to the circuit on the right. A battery is
connected to two bulbs, an ammeter, and a voltmeter as shown.
Ignore internal resistance. If a third bulb is added in parallel to
the two bulbs shown, the
V
d.
voltmeter reading will ____ same_____
(2)
e.
ammeter reading will ____ increase ___
(2)
Page 9 of 14
A
UNIVERSITY OF KWAZULU-NATAL
30th NOVEMBER 2009
FOUNDATION PHYSICS SUPPLEMENTARY EXAM
Question 10 (11 marks)
In the circuit shown calculate:
a)
I4Ω =
b)
the current through each 4.0 Ω resistor,(2)
2
7
0.50Ω
x 0.700 = 0.20 A
4.0Ω
the total resistance of the circuit,
4.0Ω
8.0Ω
(3)
700mA
1
1 1 1 1
7
= + + + =
Req 4 4 4 8 8Ω
Req = 1.1Ω
RT = 0.5 +1.1 = 1.6Ω
c)
4.0Ω
the battery voltage, and
(3)
VT = ITxRT = 0.70 x 1.6 = 1.1 V
d)
the voltage across the 8.0 Ω resistor.
V8 = IxR = 0.1 x 8.0 = 0.80 V
OR
(3)
V8=VT – 0.70x0.50 = 0.75 V
Question 11 (10 Marks)
a) For the circuit on the right, calculate the unknown
current I.
(2)
I + 1.0 = 5.0
I = 4.0 A
I
Loop 1
5.0Ω
X
[3]
c) Use loop 1, as shown, to calculate the resistance R.
24 – 22.5 = 4.0R + (-1.0)5.0
6.5
R=
= 1.6Ω
4.0
Page 10 of 14
Loop 2
3.5Ω
24V
b) Use loop 2, as shown, to calculate voltage X. (4)
X = 5.0 x 3.5 + 1.0 x 5.0
= 22.5 V
= 23 V
5.0A
1.0A
R
node a
(4)
UNIVERSITY OF KWAZULU-NATAL
FOUNDATION PHYSICS SUPPLEMENTARY EXAM
30th NOVEMBER 2009
Question 12 (16marks)
In the circuit on the right, the “lost volts” across the battery is 1.0 V. If a current of 2.5 A
flows in the circuit, calculate:
a)
the internal resistance (r)
(3)
2.5A
Vl = I.r
V
1.0
r= l =
= 0.40Ω
I
2.5
5.2Ω
2.4Ω
emf
b)
the terminal potential difference
(3)
r
Vtpd = IRext = 2.5 x 7.6 = 19 V
c)
the emf of the battery
(3)
emf = I(R + r)
= 2.5 x 8.0
= 20 V
d)
the energy lost in the external circuit in 1 minute, and
W = I2 Rt
V2 t
R
192 (60)
=
7.6
OR W =
= (2.5) 2 (7.6)(60)
= 2.9 ×103 J (or 2850 J)
e)
(4)
OR W = VIt
= 19 × 2.5 × 60
= 2.9 ×103 J
= 2.9 ×103 J
the power loss in the battery
P =I V
OR P = I2 R
= (1.0)( 2.5)
= (2.5)2 ( 0.4)
= 2.5 W
Page 11 of 14
= 2.5 W
(3)
OR
P=
V 2 (1.0) 2
=
= 2.5W
R
0.4
UNIVERSITY OF KWAZULU-NATAL
FOUNDATION PHYSICS SUPPLEMENTARY EXAM
30th NOVEMBER 2009
Question 13
(7 marks)
a) Complete the following diagrams to show what happens to a ray of light passing
through the perspex prisms or lenses:
.
f
O
Page 12 of 14
f
O
(7)
UNIVERSITY OF KWAZULU-NATAL
Question 14
FOUNDATION PHYSICS SUPPLEMENTARY EXAM
30th NOVEMBER 2009
(16 marks)
(5)
F
I
O
F
(5)
1
v
=
1
f
− u1 =
1
6.0
− 21.5 = - 0.2334
v = - 4.28 = - 4.3 cm
(4)
mag =
−v
u
=
4.3
2.5
= 1.72 = 1.7
(2)
Any 2 of:
Virtual
Enlarged
upright
(same side as object)
(can’t be cast on screen)
Page 13 of 14
UNIVERSITY OF KWAZULU-NATAL
Question 15
FOUNDATION PHYSICS SUPPLEMENTARY EXAM
30th NOVEMBER 2009
(13 marks)
(13)
φwater
y
tany = 1.0/1.5
y = 33.7o
θair = 56.3o
npsinθp = nasinθa
sinθp =
1.00 x sin 56.3
= 0.5738
1.45
θp = 35o
sinφw =
1.45 x0.5738
= 0.6255
1.33
φw = 39o
and θp = φp
nwsinφw = npsinφp
Page 14 of 14