Download SOLUTIONS Aug 2016 exam TFY4102 1) In a perfectly ELASTIC

SOLUTIONS Aug 2016 exam TFY4102
1) In a perfectly ELASTIC collision between two perfectly rigid objects
A) the momentum of each object is conserved.
B) the kinetic energy of each object is conserved.
C) the momentum of the system is conserved but the kinetic energy of the system is not
conserved.
D) both the momentum and the kinetic energy of the system are conserved.
E) the kinetic energy of the system is conserved, but the momentum of the system is not
conserved.
Answer: D
2) In a double-slit experiment, if the slit separation is increased, which of the following
happens to the interference pattern shown on the screen?
A) The minima get closer together.
B) The maxima stay at the same position.
C) The minima and maxima stay at the same position.
D) The minima stay at the same position.
E) The maxima get further apart.
Answer: A
3) Fluid fills the container shown in the figure. At which of the indicated points is the
pressure greatest?
A) A
B) B
C) C
D) D
E) The pressure is the same at
each of the labeled points.
Answer: E
4) Which of the following
statements about Gauss's law
are correct? (There may be more than one correct choice.)
A) Gauss's law is valid only for symmetric charge distributions, such as spheres and
cylinders.
B) If there is no charge inside of a Gaussian surface, the electric field must be zero at
points of that surface.
C) Only charge enclosed within a Gaussian surface can produce an electric field at points
on that surface.
D) If a Gaussian surface is completely inside an electrostatic conductor, the electric field
must always be zero at all points on that surface.
E) The electric flux passing through a Gaussian surface depends only on the amount of
charge inside that surface, not on its size or shape.
Answer: D, E (A is incorrect; it is just easier to calculate the field in this case. B is
violated if the surface is placed inside a uniform field, C is false because there can be a
field from outside charges, but there will be no net flux through the surface from these)
5) An object is executing simple harmonic motion. What is true about the acceleration of
this object? (There may be more than one correct choice.)
A) The acceleration is a maximum when the displacement of the object is a maximum.
B) The acceleration is a maximum when the speed of the object is a maximum.
C) The acceleration is a maximum when the displacement of the object is zero.
D) The acceleration is zero when the speed of the object is a maximum.
E) The acceleration is a maximum when the object is instantaneously at rest.
Answer: A, D, E use the fact that the acceleration is the second time derivative and the
speed is the magnitude of the first derivative of displacement; x(t) = sint or cost
MECHANICS
An inclined plane that makes an angle of 30 degrees with the horizontal has a spring with
spring constant 4500N/m at the bottom. A 2.2 kg block released near the top moves
down the plane and compresses the spring to a maximum of 0.024 m from its relaxed
length (point A).
a) ignoring friction, calculate the distance the lower edge of the block moves from the
instant it is released until it comes to a momentary stop against the compressed spring.
Conservation of energy: initial potential energy = final elastic energy
mgy=1/2 kx2 , where x is the compression amount and y is the height change.


b) what is the momentum of the block when it reaches point A (the point where the block
first met the spring) after the spring pushes the block back up the plane?
There is no friction, so the momentum is the same when it reaches point A going up as it
was going down, and you can use either conservation of energy (PE=KE) or the
acceleration due to gravity along the slope to solve this one. Momentum is a vector, so
you have to specify components to get full credit.


| |
⃗
̂
̂
̂
̂
c) Now include the effects of friction, with µk= 0.10 If the block again compresses the
spring by 0.024m, how far does the front(lower) edge of the block travel from release
until the moment it stops against the compressed spring?
Conservation of energy again, but this time there is a loss(work) term due to gravity.


B THERMODYNAMICS
A sample of monatomic ideal gas is in a thermally insulated cylinder fitted with a piston
that can change the cylinder volume. Initially, the cylinder volume is V1= 2.00 x 10-3 m3,
and the temperature is 290 K. The piston then moves until the cylinder volume is 0.200
V1. If the compression is quasi-static and the entropy change is zero, How much work is
done on the gas?
Q=0; PV= NkT and the change in thermal energy = Work = NCv∆T
Being monatomic, each gas particle has three degrees of freedom, so CV=2/3 kB= 2.07 x
10-23J/K
C. ELECTROSTATICS
(b)


    2 cos( ) 
Ex   2 sin( )  and E y   1

2

2 0
 0



where   30.
Hence at (3 m, 1 m, 0)
 (90.0 109 C/m 2 )

Ex  
sin(30)   2.54 103 N/C and
12
2
2
2(8.85

10
C
/N

m
)


9
2
9
 (130.0 10 C/m )  (90.0 10 C/m 2 )cos(30 ) 
3
Ey  
  2.94 10 N/C,
12
2
2
2(8.85

10
C
/N

m
)


D. WAVES
You are blindfolded and given one end of a rope that has two sections joined end to end;
the two sections has different linear mass densities, but identical radii and lengths. One
end is attached to a pole, but you do not know if it is fixed or sliding.
You are able to detect the time and direction of rope displacements that result from
inputs that you make with your hand.
a) How can you determine whether you are holding the less or more dense rope?
There will be reflections from both the interface between the ropes, and from the end.
The first pulse to come back will be from the interface; if the reflected pulse is in the
same direction as the input, you are holding the heavier rope (no phase change upon
reflection).
b) How can you determine if the far end of the rope is fixed or sliding?
You have to wait for a second pulse to come back – one that has gone all the way to the
far end and back. The pulse that is transmitted through the interface and reaches the far
end will always be in the direction of the input. If the end of the second rope is tied in
place, it will be inverted. If the far end of the rope is free, the second pulse that returns
will be in the same direction as the input.
c) Find an expression for the ratio of the linear mass densities of the ropes
We can estimate the relative densities by measuring the times for the different pulses to
return. If the input pulse is at t=0, and the first pulse that returns (from the interface) is
arrives at t1 and the second pulse (from the far end) arrives at t2, we can write these times
as
√
√
√
√
√
⇒
(
)
E) CIRCUITS
The battery has been connected to
the circuit in the figure for several
minutes.
R1=R2=5.0 Ω
R3= 4.0 Ω R4= 6.0 Ω
R7 = 1.0 Ω
R6 =R8= 0.50 Ω
There was a misprint and the value
of R5 was not given; either use
symbol or assume value – no credit
was lost by either approach.
a) Determine the current through each resistor
Let I n be the current through resistor Rn . Since the circuit has been closed a long time,
the capacitors will be fully charged and no additional current will flow to or from them.
In particular, this means no current will flow across R6 or the 30.0 F capacitor. Current
can only flow through that parallel arrangement involving the capacitors by flowing
through R7 and R8 , which puts them in series with R1 and R2 . This drastically simplifies
the equivalent resistance of the circuit to
1
 1
1 
Req  R1  R2  
   R7  R8
 R3  R4 R5 
This allows us to determine the current supplied by the battery:
I batt 
Vbatt
Req
Note that this
must also be the current through resistors R1 , R2 , R7 , and R8 .
b) Determine the charge on either plate of each capacitor.
The potential difference across the 30.0 F capacitor must be the same as the potential
difference across R7 , because they are in parallel. Thus, we can write
q30.0
F
 C30.0 FV30.0
F
 C30.0 F I R7 R7 .
Similarly, the potential difference across the 20.0 F capacitor must be the same as the
potential difference across the series combination of R7 and R8 , because those paths are
in parallel. Thus
q20.0
F
 C20.0 FV20.0
F
 C20.0 F I R7 (R7  R8 )
Collecting these results, and assuming R5 = 10 ohms: I battery  I1  I 2  I 7  I8
I3  I 4  I5  3.03 A, and I 6  0. The charge on the 30 F capacitor is 1.82 104 C. The charge
on the 20 F capacitor is also 1.82 104 C. (you did not need to provide a numerical
answer)
c) Which capacitor charges more quickly?The 30 µf capacitor will have a lower voltage
across it during charging due to the parallel resistor, and the capacitance is larger, so it
will take longer to charge up.