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FND Solutions Semester 1, 2008 Page 1
THE UNIVERSITY OF SYDNEY
FACULTIES OF ARTS, EDUCATION & SOCIAL WORK,
ENGINEERING AND SCIENCE
PHYS 1002 PHYSICS 1 (FUNDAMENTALS)
SOLUTIONS
JUNE 2008
Time allowed: THREE Hours
MARKS FOR QUESTIONS ARE AS INDICATED
TOTAL: 90 MARKS
INSTRUCTIONS
**- All questions are to be answered.
- Use a separate answer book for section A and section B.
- All answers should include explanations in terms of physical principles.**
DATA
Density of fresh water at 20 °C and 1 atm ρ = 1.00 × 103 kg.m-
Density of sea water at 20 °C and 1 atm ρ = 1.024 × 103 kg.m-
Atmospheric air pressure P = 1.00 × 105 Pa
Free fall acceleration at earth's surface g = 9.81 m.s-^2 Speed of light in vacuum c = 3.00 × 108 m.s-^1
FND Question 1
A cork is held underwater, completely immersed, and then released.
(a) Draw a labelled diagram to show the force(s) acting on the cork immediately after release. Your diagram should clearly identify each force.
(b) Is the cork in static equilibrium immediately after release? Explain your reasoning.
(c) Consider the force(s) on the cork after it has floated to the surface. Explain how these forces differ from those in part (a).
(5 marks)
Solution
(a)
(2 marks)
(b) No it is not in static equilibrium. The buoyant force exceeds the weight and so the cork moves (accelerates) upwards towards the surface. (1 mark)
(c)
FND Question 2
The diagram shows the vectors representing the gravitational force Fg
ur , acceleration , and velocity v of a satellite in a circular orbit around the Earth. Your friend (not a physics student) gives her opinions about the diagram. Briefly comment on the physics of each opinion.
a
r r
(a) “The gravitational force is pulling the satellite towards the Earth and it is also accelerating towards the Earth. Therefore the satellite cannot remain in its circular orbit.” (b) “An astronaut in the satellite can not be weightless because the force of gravity is pulling him towards the earth.” (c) “Satellites move around the Earth. But the dish for my satellite television service points at the same location in the sky. Surely, it should move to follow the satellite in its orbit.” (d) “If the Earth suddenly disappeared, the satellite would fly off into space moving at a constant velocity”
Solution (a) The gravitation force is pulling the satellite towards the earth and the satellite is accelerating towards the Earth as a result. But the tangential velocity means that the satellite is continually falling around the Earth. (2 marks)
(b) The astronaut is accelerating towards the Earth with the same acceleration as the local value of g. Therefore he experiences apparent weightlessness. (1 mark)
(c) A communication satellite such as used for satellite television is in geosynchronous orbit and has a rotation period of 24 hours, the same as that of the Earth. It therefore moves with the same angular speed as a location on Earth and appears to remain at the same position in the sky. (1 mark)
(d) This is a true statement. Without the Earth’s gravity there would be no net force on the satellite and it would move off into space at constant velocity. (1 mark)
FND Question 3
A roller coaster goes down a slope and around a vertical loop. Does a passenger in the roller coaster experience a greater apparent weight at the top of the loop or the bottom? Explain your answer in terms of physical principles. Include free-body diagrams for the passenger at the top and bottom of the loop.
(5 marks)
Solution Since the roller coaster is travelling in a circle, there must be a net acceleration
towards the centre to provide the circular motion, so there must be a net force, , also acting towards the centre of rotation. The magnitude of this force depends on the speed of rotation of the roller coaster.
F net
ur
This net force is the vector sum of two forces. Firstly, there is the force of the due to
the passenger’s weight, W
uur , which is constant in magnitude and always acts vertically
downwards. Secondly, there is the normal force, N
uur , acting on the passenger from his contact with his seat. This acts normal to his seat and in the direction of the centre of rotation (provided the rotation speed of sufficient). This is his apparent weight.
The vector equation can be written as: F net = N + W
ur uur uur .
At the bottom of the loop the normal force (acting upwards towards the centre of rotation) is opposite to the weight force (acting downwards), so must be larger than the weight force to provide the required centripetal force. At the top of the loop the normal force and the weight force are both pointing downwards towards the centre of rotation, so the normal force is smaller than the weight force, even zero, depending on the speed. If the speed is insufficient the normal force (and his apparent weight) can be negative (he needs to be held in his seat.
FND Question 4
Position A Position B
Suppose you are trying to cut a branch off a tree using a pair of garden shears. Assume the blades are equally sharp at all positions along the blades.
Will it be easier to cut through the branch in Position A or Position B? Explain your answer in terms of physical principles. (5 marks)
Solution It is easiest to cut the branch in Position A. (1 mark)
Assuming the system is close to equilibrium, torque applied by the hands is the same in magnitude as torque at the position of the branch. (1 mark)
Assuming the hands are in the same positions and the same force is applied by the
hands in both cases then the torque τ exerted on the branch is the same in both
Position A and Position B. (1 mark)
If F is the "cutting" force exerted on the branch by the blade and d is the distance of the branch to the pivot ( or axis) then the torque exerted on the branch is given by
τ = F d ( or^ F R sin θ ) where^ θ^ ≈^ 90°^ (1 mark)
Because d (or R) is smaller at Position A, the "cutting" force F must be larger in Position A so it cuts the branch more easily. (1 mark)
Note: 1 mark for each italicised point mentioned or implied
OR
Note: Students might cite Archimedes' lever rule instead of using the concept of torque explicitly. This is correct but less deep but can still attract full marks if complete. See below.
It is easiest to cut the branch in Position A. (1 mark)
Assuming the system is close to equilibrium, Archimedes' lever rule applies (1 mark)
Archimedes' lever rule states that for a lever (such as the blade) Fh Rh = Fb Rb (
mark)
where Fh is the force applied by hands and Fb is the ("cutting") force applied to the
branch and Rh is the distance between hands and axis and Rb is the distance from the branch to the axis. Fh and Rh are the same in both cases. (1 mark)
However, in Position A, R b is smaller than at Position B ... the cutting force Fb is
larger in position A, so the branch is easier to cut. (1 mark)
Note: 1 mark for each italicised point mentioned or implied
FND Question 6 = REG Question 6 = ADV Question 6
A string is attached to a wall. A lecturer holding the unattached end pulls with a constant force (parallel to the string) while moving his hand up and down (perpendicular to the string) to create a pulse travelling towards the wall.
The lecturer now wants to produce a pulse that takes a longer time to reach the wall. Four ways of doing this are considered below.
(a) Choose the best answer from the following and give a brief justification. i. He should move his hand up and down more quickly. ii. He should move his hand up and down more slowly. iii. Neither (i) nor (ii) will change the time for the pulse to each the wall.
(b) Choose the best answer from the following and give a brief justification. i. He should use a heavier string of the same length, under the same tension. ii. He should use a lighter string of the same length, under the same tension. iii. Neither (i) nor (ii) will change the time for the pulse to each the wall.
(c) Choose the best answer from the following and give a brief justification. i. He should displace his hand a greater distance up and down but at the same speed. ii. He should displace his hand a lesser distance up and down but at the same speed. iii. Neither (i) nor (ii) will change the time for the pulse to each the wall.
(d) Choose the best answer from the following and give a brief justification. i. He should pull the string harder to increase the tension in the string. ii. He should pull the string less hard to decrease the tension in the string. iii. Neither (i) or (ii) will change the time for the pulse to each the wall.
The lecturer now detaches the string from the wall and invites one of the students to hold that end of the string and keep it fixed. The lecturer and student both pull the string with the same force as the lecturer did before. The lecturer then moves his hands up and down in the same way as he did originally.
(e) Compared with the original travel time for the pulse, choose the best answer from the following options and give a brief justification. i. The travel time does not change because the tension in the rope is the same. ii. The travel time gets longer because the tension in the rope is greater. iii. The travel time gets shorter because the tension in the rope is greater. (5 marks)
Solution
(a) Option (iii) (½ mark) The velocity of the pulse along the string does not depend on frequency. (½ mark)
(b) Option (i). (½ mark) The velocity depends on 1/μ1/2^ where μ is the mass per unit length of the string. Hence greater μ means smaller velocity (and hence longer travel time). (½ mark)
(c) Option (iii) (½ mark) The size of the vertical displacement does not chance the velocity. He also changes the period of the wave but this also does not affect the velocity. (½ mark)
(d) Option (ii). (½ mark) The velocity depends on F1/2^ where F is the tension in the string. Hence smaller F means smaller velocity (and hence longer travel time). (½ mark)
(e) Option (i). (½ mark) The tension in the string will be the same. The action-reaction pair “lecturer-wall” is the same as “lecturer-student”. Flecturer= -Fwall = -Fstudent. Hence the travel time stays the same. (½ mark)
Total (5 marks)
(b)
The surface area of the box is: 2 × 2 =4 m^2 and so the force on the top is: 5 5 5
(2.0045 10 )(4) 8.0182 10 N
= 8.02 10 N (3 sig figs)
F = P A = × = ×
×
(1 mark) (c) The bottom of the box is at a depth of h = 12 m. So the force (up on bottom) is given by: 5 3 0 5 5
1.00 10 (1.024 10 )(9.81)(12) Pa 2.2055 10 Pa = 2.21 10 Pa
p = p + ρ swg h = × + ×
= ×
×
5 5 5
(2.205 10 )(4) 8.8218 10 N
= 8.82 10 N
⇒ F = P A = × = ×
×
(1 mark for approach; 1 for correct answer) (d) Net upward force is given by:
5 5 4
(8.8218 8.0182) 10 N
0.8036 10 8.04 10 N
F upward = Fbottom − Ftop
= − × = × = ×
Note that this result is affected significantly by the precision with which the forces are calculated. (1 mark) Buoyant force
3 4 4
8.036 10 N
8.04 10 N
F B = ρ sw g Vbox
= × × ×
= ×
= ×
(1 mark)
The two forces are expected to be equal because the buoyant force is caused by the difference in forces due to the pressure of the fluid. They are the same force. (1 mark)
(e)
(Use either FB or net force due to pressures but not ‘both’)
The box is in static equilibrium with the upward buoyancy force balanced by the downward weight force + tension. Therefore
4 4
7.44 10 N
T = F B − W
= × −
= ×
(1 mark for method; 1 mark for correct result)
(b) Time taken is given by:
0.010 s 8000
f i f i
v v a t v v t a
(1 mark for method; 1 mark for correct result)
(c) velocity to right taken as positive
(2 marks for graph and scale information)
(d) velocity upwards taken as positive
(2 marks for graph and scale information)
(e) acceleration upwards taken as positive
(2 marks for graph and scale information)
( )
3.0 kg.m.s
0.59 m.s (5.0 0.06)
i f
f
p p M m v
v
(1 mark for method; 1 mark for correct result)
(d) The kinetic energy just after collision is (^1) ( ) 2 (0.5)(5.06)(0.59 ) (^2) 0.88 J 2 f
K = m + M v = =
Or 0.89 if they kept all the sig figures for (^) v (^) f.
So the energy lost is Δ E = 75 − 0.88 = 74 J. (i.e. nearly all of the energy)
(1 mark for method; 1 mark for correct result)
(e) The block will rise until all of the kinetic energy is transformed into potential energy so: 0.88 (^) ( ) (5.06)(9.81)
0.018 m 1.8cm (5.06)(9.81)
K J m M g h
h
h
(1 mark for method; 1 mark for correct result)
FND Question 10
An elephant drags a tray of logs of total weight 10000 N up a uniform slope. The slope is 20 m in length and is inclined at an angle of 30 to the horizontal. The value of the coefficient of kinetic (sliding) friction between the tray of logs and the slope is 0.50. The elephant walks at a constant speed.
o
(a) Draw a free-body diagram for the tray of logs.
(b) What is the net force on the tray of logs.
(c) Calculate the tension in the rope.
(d) When the logs have reached the top of the slope, the rope between the elephant and the logs breaks. If the slope was frictionless , how long would it take for the tray of logs to slide back down to the bottom of the slope. (10 marks)
Solution
(a)
(3 marks)
