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```SURNAME: ..........................................................
STUDENT NUMBER:
INITIALS: ...........................................
Signature
UNIVERSITY OF CAPE TOWN
DEPARTMENT OF PHYSICS
PHY1012F
Physics A for Engineers
FINAL EXAMINATION
Thursday 06 June 2019
Time: 120 min + 10 min reading time
Internal Examiner:
External Examiner:
Total marks: 72
Full marks: 70
Dr S.M. Wheaton
Dr G. Bosman
INSTRUCTIONS:
1. Complete your details at the top of the page.
2. Answer all questions in the spaces provided on the question paper neatly and legibly.
3. Use blank sheets on question paper for all calculations and/or rough work.
4. A list of constants and formulae are available on the last two (2) pages.
5. Prescribed calculators are allowed.
7. Hand in your question paper at the end of the test.
2
Question 1
a. The number 8 ball in a game of pool moves on a horizontal frictionless table. The graphs
below indicate ๐ฃ๐ฅ and ๐ฃ๐ฆ components of the ballโs velocity. The ball started from the origin.
i.
In which direction (angle) is the ball moving at t = 2 s? (Give your answer as an angle
from the x-axis.)
[2]
ii.
How far from the origin is the ball at t = 4 s?
[4]
PHY1012F Exam 2019-drg
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Question 1โฆcontinued
b. The number 8 ball of mass ๐ = 0.250 kg now strikes the cushion of the pool table at an
angle of ๐1 = 60.0° at a speed of ๐ฃ1 = 27.0 m sโ1 . It bounces off at an angle of
๐2 = 71.0° and a speed of ๐ฃ2 = 10.0 m sโ1.
i.
What is the magnitude of the change in momentum of the ball?
[4]
ii.
In which direction, with respect to the positive ๐ฅ-axis does the change in momentum
vector point?
[1]
PHY1012F Exam 2019-drg
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Question 2
a. Give an example of a particle that has constant kinetic energy but is accelerating. Can a nonaccelerating particle (constant mass) have a changing kinetic energy? If so, give an example.
[3]
b. Calculate the horizontal force ๐นโ required such that ๐1 = 9.00 kg does not slide up or down
along the wedge ๐2 = 150 kg. All surfaces are frictionless and ๐ = 20°.
[4]
PHY1012F Exam 2019-drg
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Question 2โฆcontinued
c. A pendulum of length ๐ฟ = 1.0 m with bob of mass ๐ = 1.0 kg is released from rest at an
angle of ๐ = 30° from the vertical. When the pendulum reaches the vertical position, the
bob strikes a mass ๐ = 3.0 kg that is resting on a frictionless table that has a height
โ = 0.85 m.
i.
Calculate the velocity ๐ฃโ of the bob and the box just after they collide elastically.
[5]
PHY1012F Exam 2019-drg
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Question 2โฆcontinued
ii.
Determine how far away from the bottom edge of the table, โ๐ฅ, the box will strike
the floor.
[3]
PHY1012F Exam 2019-drg
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Question 3
a. Your bicycle tires have a radius of 0.33 m. It takes you 14.2 minutes to ride 14 times around
a circular track of radius 73 m at a constant speed.
i.
What is the angular velocity (in rad/s) of the bicycle around the track?
[3]
ii.
What is the angular velocity (in rad/s) of a tire around its axis?
[4]
PHY1012F Exam 2019-drg
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Question 3โฆcontinued
b. A solid sphere of radius ๐ and mass ๐ is placed at a height โ0 on an inclined plane of slope
๐. When released, it rolls without slipping to the bottom of the incline. Next a cylinder of
the same mass and radius is released on the same incline. If both the cylinder and the
15
sphere reach the same speed at the bottom of the incline, show that โ = โ0 .
[5]
14
PHY1012F Exam 2019-drg
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Question 4
a. An experiment measures the temperature of a 250 g substance while steadily applying heat
to it. The figure below shows the result of the experiment.
i.
Determine the specific heat capacity (c) of the solid phase.
[3]
ii.
Determine the specific heat capacity (c) of the liquid phase.
[3]
iii.
Determine the heat of fusion (Lf).
[2]
iv.
Determine the heat of vapourisation (Lv).
[2]
PHY1012F Exam 2019-drg
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Question 4โฆcontinued
b. A 200 g piece of copper at temperature 450 K and a 100 g piece of aluminium at
temperature 200 K are dropped into an insulated bucket containing water at 280 K. If the
thermal equilibrium temperature of the mixture is 282.6 K, what was the mass of the water
in the bucket?
[4]
๐๐ด๐ = 900 J/kg K
๐๐ถ๐ข = 385 J/kg K
๐๐ป2 ๐ = 4190 J/kg K
PHY1012F Exam 2019-drg
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Question 4โฆcontinued
c. A monatomic gas follows the process 1 โ 2 โ 3 shown below.
[1 atm = 101325 Pa]
i.
How much heat is needed for process 1 โ 2?
[4]
ii.
How much heat is needed for process 2 โ 3?
[2]
PHY1012F Exam 2019-drg
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Question 5
a. What is the order of magnitude of the horizontal force exerted on a passenger car travelling
at the speed limit (120 km/h) on a freeway when a large truck passes close by going in the
opposite direction, also at the freeway speed limit?
[4]
PHY1012F Exam 2019-drg
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Question 5โฆcontinued
b. A 5.00 kg air-filled, sealed, rigid float tank that has a volume of 1.00 m3 is pulled 50.4 m
down to the seafloor in order to assist in lifting a sunken object. A diver standing on the
seafloor cranks a winch to pull the tank down.
i.
How much work is required to pull the tank down? (You may safely ignore the mass
in the air tank and the mass of the winch rope.)
[7]
ii.
What is the average power required if it takes 10 min to winch down the tank? [3]
PHY1012F Exam 2019-drg
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Physical constants and useful formulae
Speed of light
Acceleration due to gravity (Earth)
Gas constant
Density of air
Stefan-Boltzmann constant
๏ณ
๐
๐(๐) = ๐๐ + ๐๐๐ ๐ + ๐๐ ๐๐
๐
๐
๐(๐) = ๐๐ + ๐๐๐ ๐ + ๐๐ ๐๐
๐
โโ๐๐๐ = ๐๐
โโ
๐ญ
๐=
๐๐ฝ
๐๐
๐๐ = ๐๐
3.00 × 108 ๐/๐
9.80 ๐/๐  2
6.022 × 1023 /๐๐๐
8.314 ๐ฝ/๐๐๐ ๐พ
1.225 ๐๐/๐3
5.67 × 10โ8 ๐/๐2 ๐พ 4
c
g
NA
R
๐๐๐๐
๐๐ (๐) = ๐๐๐ + ๐๐ ๐
๐๐ (๐) = ๐๐๐ + ๐๐ ๐
๐๐ (๐๐๐) = ๐๐ ๐ต
๐ถ=
๐๐
๐๐
๐ = ๐๐ฝ
๐๐
๐๐ =
= ๐๐ ๐
๐
๐๐ = ๐๐ ๐ต
๐ป=
๐๐๐ ๐๐
=
๐
๐
๐๐ = ๐ถ๐
๐
๐ฝ๐ = ๐ฝ๐ + ๐๐ โ๐ + ๐ถ(โ๐)๐
๐
๐๐ = ๐๐ + ๐ถโ๐
๐๐๐ = ๐๐๐ + ๐๐ถโ๐ฝ
PHY1012F Exam 2019-drg
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โโ =
๐ญ
โโ๐ = ๐
โโ๐
๐
โโ
๐๐
๐๐
๐๐
โโ = ๐ฑโ
โ๐
โโ (๐)๐๐
๐ฑโ = โซ ๐ญ
๐๐
๐๐
โโ๐ ๐๐
๐พ=โซ ๐ญ
โโ โ โ๐
โโ = ๐ญโ๐ ๐๐๐ ๐ฝ
๐พ=๐ญ
๐ท=
๐๐
๐ผ๐๐ =
๐
๐(โ๐)๐
๐
๐ญ๐ = โ
โ๐ฌ๐๐๐ = โ๐ฒ + โ๐ผ + โ๐ฌ๐๐ = ๐พ๐๐๐
๐พ
โ๐
๐๐ผ
๐๐
๐พ๐ต๐ช = โ๐ฒ + โ๐ผ
๐ฒ๐ + ๐ผ๐ + ๐พ๐๐๐ = ๐ฒ๐ + ๐ผ๐ + โ๐ฌ๐๐
โ๐โ = โ๐โ × โ๐ญโ
๐ = ๐๐ญ ๐๐๐ ๐ = ๐๐ญ๐ = ๐๐ญ
๐ฐ = โ ๐๐ ๐๐๐ = โซ ๐๐ ๐๐
๐๐๐๐ = ๐ฐ๐ถ
๐
โโ = ๐
โโ × ๐
โโ
๐ณ
โโ = ๐ฐ๐
โโโโ
๐ณ
โโ๐๐๐ =
๐
๐ฌ(๐๐๐๐ ๐๐๐๐๐๐๐๐) = ๐ฒ๐๐๐ + ๐ผ๐ฎ =
๐ฌ(๐๐๐๐๐๐๐) = ๐ฒ๐๐๐ + ๐ฒ๐๐ + ๐ผ๐ฎ =
๐=
๐ญ
๐จ
โโ
๐๐ณ
๐๐
๐ ๐
๐ฐ๐ + ๐ด๐๐๐๐
๐
๐ ๐ ๐
๐ฐ๐ + ๐ด๐๐๐๐ + ๐ด๐๐๐๐
๐
๐
๐ = ๐๐ + ๐๐๐
๐ญ๐ฉ = ๐๐ ๐ฝ๐ ๐
๐๐ ๐จ๐ = ๐๐ ๐จ๐ = ๐ธ
๐
๐
๐๐ + ๐๐๐๐ + ๐๐๐๐ = ๐๐ + ๐๐๐๐ + ๐๐๐๐ = ๐๐๐๐๐๐๐๐
๐
๐
๐ธ = ๐ด๐โ๐ป
๐ธ = ๐๐ช๐ฝ โ๐ป
๐ธ = ±๐ด๐ณ๐
๐ธ = ๐๐ช๐ท โ๐ป
๐๐ธ ๐๐จ
=
โ๐ป
๐๐
๐ณ
๐ธ = ±๐ด๐ณ๐
๐พ = โ๐๐น๐ป๐๐(๐ฝ๐ โ๐ฝ๐ )
๐๐ธ
= ๐๐๐จ๐ป๐
๐๐
PHY1012F Exam 2019-drg
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