Download 1. Use the following information to answer the next question. An

1. Use the following information to answer the next question.
An explosion of a stationary object produces three fragments that move away from each other in a horizontal plane.
The speed, vC, of fragment C is
a.
36.9 m/s
b.
48.6 m/s
c.
51.7 m/s
d.
148 m/s
Answer: A
2. A tennis ball with a mass of 110 g is traveling 18.5 m/s east. It is struck by a racquet that applies a force of
950 N west. The ball and the racquet are in contact for 3.20 ms. The change in momentum of the tennis ball is
a.
c.
2.04 kg • m/s, west
2.04 × 103 kg • m/s, west
b.
d.
3.04 kg • m/s, west
3.04 × 103 kg • m/s, west
Answer: B
Use the following information to answer she next three questions.
At the starting gate of a downhill ski race, a skier pushes with his arms and legs to get as fast a start as possible. The skier
leaves the starting gate with an initial spud of 1.60 m/s.
The race course is made up of both sleep and flat sections. At the end of the first flat section. the vertical distance from the
skier to the finish line is 80.0 m and the skier has a speed of 20.0 m/s. At the end of the steepest section, the skier reaches his
maximum speed of 33.3 m/s (120 km/h). The mass of the skier is 68.2 kg.
3. The magnitude of the impulse on the skier as he leaves the starting gate is
a.
1.09 x 102 N-s
b.
1.36 x 103 N-s
c.
2.27 x 103 N-s
d.
8.18 x 103 N-s
Answer: A
4. If there were no friction on the skier. the minimum vertical distance from the starting gate that the skier would
have to descend in order to reach his maximum speed would be
a.
1.70 m
b.
20.3 m
c.
56.4 m
d.
734 m
Answer:
Use the following information to answer she next three questions.
At the starting gate of a downhill ski race, a skier pushes with his arms and legs to get as fast a start as possible. The skier
leaves the starting gate with an initial spud of 1.60 m/s.
The race course is made up of both sleep and flat sections. At the end of the first flat section. the vertical distance from the
skier to the finish line is 80.0 m and the skier has a speed of 20.0 m/s. At the end of the steepest section, the skier reaches his
maximum speed of 33.3 m/s (120 km/h). The mass of the skier is 68.2 kg.
5. At the end of the first flat section, the mechanical energy of the skier, with respect to the finish line is
a.
13.6 kJ
b.
53.5 kJ
c.
54.2 kJ
d.
67.2 kJ
Answer: D
6. Use the following information to answer the next question.
When the switch in the circuit above is closed, the electric potential drop across the 5 Ω resistor is V5. The electric
potential drop across the 10 Ω resistor is
a.
12V5
b.
13V5
c.
2 V5
d.
3 V5
Answer: C
7. Rutherford's planetary model of the atom was an improvement over previous models because it was able to
explain the
a.
b.
c.
d.
existence of atoms
line spectra of hydrogen
existence of energy levels within atoms
scattering of alpha particles incident on a thin gold foil
Answer: D
8. Use the following information to answer the next question.
Vector Fields
I Gravitational
II Electrical
III Magnetic
Which of the vector fields above are produced by a charged particle at rest?
a.
I, II, and III
b.
I and II only
c.
I and III only
d.
II and III only
Answer: B
9. Use the following information to answer the next question (%PointValue% marks)
A wheelchair lift on a bus is used to move a wheelchair and its occupant from ground level to the level of the floor
of the bus. The wheelchair lift is powered by an electric motor. The efficiency of an electric motor in a wheelchair
lift is determined by using the formula
Develop a series of instructions for a maintenance team to follow in order to determine the efficiency of the electric
motor in an occupied wheelchair lift.
In your instructions,
• identify the energy conversions involved in the operation of a wheelchair lift. Start with the energy
input to the electric motor, and end with the output energy when the wheelchair and its occupant are on
the bus
• design a procedure that the maintenance team could use to determine the input power of an electric
motor.
• include a labelled circuit diagram
• design a procedure that the maintenance team could use to determine the output power of the electric
motor as it powers the occupied wheelchair lift
• describe the calculations, including necessary formulas, required to determine the efficiency of this
electric motor
Marks will be awarded for the physics used to solve this problem and for the effective communication of your
response.
For students to achieve the standard of excellence with their responses, they needed to demonstrate
internal
consistency. This was evidenced in the discussion of energy conversions in which the final energy form
was
gravitational potential energy, in the procedure for the output power in which the variables to be
measured were mass, height moved, and time taken, and in the citing of the equation p = mgh/t. These
students often stated that the lift had kinetic energy as it rose, but since there was no kinetic energy once
the lift and occupant were at the top, they did not need to make any measurements of speed. (A student
could have decided to measure the mechanical energy at the midpoint but would then have needed to
provide steps to measure all the variables associated with the mechanical energy.) The other main area
that evidenced internal consistency was in the circuit diagram, design for input power, and the electrical
power formula.
Students at this level provided a schematic circuit diagram with both an ammeter and a voltmeter properly
placed. They provided a procedure which clearly stated that the readings on the meters were required and
they stated the formula P = IV.
Student responses that received a score of 5 were well written, addressing the full scope of the question
and showing significant application of physics principles to a real-life situation. One common weakness
of these responses was that they completely missed the idea that the motor is continually doing work to
the system. These students addressed the idea by saying things like “at the start, the motor does work on
the lift and occupant which produces kinetic energy (energy of motion) and potential energy (because it is
above the bottom level) but then gave the memorized “all the kinetic energy is converted into potential
energy once the occupant is on the bus.” This internal inconsistency was not significant enough to drop
the response to a score of 4.
Student responses that received a score of 4 contained the idea that at least two types of energy are
present at any time (electrical energy in the motor, kinetic energy and gravitational potential energy as the
lift rises or electrical power and gravitational potential energy divided by time) but contained significant
gaps in logic or design so that the reader needed to interpret the response.
Student responses that achieved the acceptable standard on this question were often disjointed in that
they would answer each of the parts without addressing the question: “Develop a series of instructions for
a maintenance team to follow in order to determine the efficiency of the electric motor in an occupied
wheelchair lift.”
Strengths of responses that received a score of 3 included: a series circuit diagram containing a closed
loop, a load, and often a power supply; the citing of the formula P = IV; and a procedure that would allow
for the measurement (or the knowledge) of one of the variables.
Answer:
10. Use the following information to answer the next question. (%PointValue% marks)
An ion with a charge of -2e and a mass of 1.33 × 10-25 kg is initially at rest. A high-voltage source is used to accelerate
it. The ion then enters a magnetic field perpendicularly. In the magnetic field, the ion experiences a centripetal force of
4.41 × 10-14 N and travels in an orbit with a radius of 0.200 m.
Analyze the situation above by
• determining the speed of the ion as it travels in the magnetic field. If you are unable to determine the
speed of the ion, use the hypothetical value of 1.90 × 105 m/s for the rest of the question
• determining the electric potential difference needed to accelerate the ion from rest
• determining the magnetic field strength
Use the following additional information to answer the next part of the question. An electric field is then added to this
region so that the ion passes undeflected through the region.
Analyze the situation in which the ion moves in both fields by
• determining the magnitude of the electric field that would be required to make the ion pass undeflected
through the region
• completing the diagram by
- drawing the parallel plates
- labelling the positively charged plate and the negatively charged plate
- drawing at least three field lines showing the shape and direction of the electric field
Clearly communicate your understanding of the physics principles that you are using to solve this question. You may
communicate this understanding mathematically, graphically, and/or with written statements.
Written Response Question 2 was scored anaholistically and independently by two teachers and
required the student to address four major concepts: the effect of an external magnetic field on a moving
charged object, the conservation of energy in an electric field accelerating a charged object, balanced
forces as the object moves through mutually perpendicular magnetic and electric fields and the electric
field associated with charged parallel plates. Students achieving the standard of excellence provided clear
statements of the physics principles that they were using at each stage of the question. These include
statements such as the unbalanced magnetic force causes the circular deflection (Fm = Fc), energy is
conserved (Eelectric = Ek), and for undeflected motion the forces must be balanced (Fm = Fe). These students
then recorded the appropriate formulas from the data sheets, showed the substitution of the numbers into
the formulas, and stated the result of the calculation to the correct number of significant digits and with
the appropriate units. They showed their understanding of the final major concept by providing diagrams
that included labelled, horizontal parallel plates located above and below the path of the ion (either inside
the magnetic field or just above and below the magnetic field) and added field lines that began at the
positively charged plate and ended with an arrowhead at the negatively charged parallel plate. It was
beyond the scope of the question for students to explain in terms of hand rules or in terms of the direction
of the electrostatic force why they placed their positively and negatively charged plates where they did.
Student responses that received a score of 4 tended to imply the physics principles being used. For
example, rather than stating that energy is conserved they would start with the equation 2 1 2 = Vq mv , or
they would calculate Ek using 2 12 mv and then substitute that value into DE = Vq. They had no difficulty
arriving at the correct answers and tended to have appropriate significant digits and units. These responses
tended to provide electric field diagrams with one of these significant errors: field lines running from the
negatively charged to the positively charged plate, field lines that went from both plates half-way into the
region, or the plate polarity reversed for the correct direction of the electrostatic force.
Students achieving at the acceptable standard tended to calculate speed using circular motion formulas,
calculate the magnetic field intensity using a formula from the data sheet, and begins either the electric
potential difference calculation by calculating the kinetic energy or the electric field diagram by labelling
the polarity of horizontal parallel plates.
Weaknesses of student responses that received a score of 3 include:
• Using point source formulas for electric field intensity in a parallel plate application.
• Sloppy notation in which electric field intensity E and energy E are interchanged.
• Using electric field intensity and energy values interchangeably within a solution.
• Using electromagnetic induction formulas (V = Blv) to calculate the accelerating electric potential
difference.
• Electric field lines that looked like induced magnetic field lines, i.e., lines that circle the charged plates.
17
Student responses that did not meet the acceptable standard tended to calculate the speed of the ion using
a circular motion formula and to make valid or invalid starts to the rest of the question. These responses
often contained vertical parallel plates on either side of the magnetic field. Many formulas, valid and
invalid, were recorded but there was no citing of any physics principles that guided their solutions.
a)
b)
mv 2
Fmag = Fcent =
R
F R
v = cent
m
v=
(4.41 x 10-14 (200)
(1.33 x 10-25
v = 2.58 x 105
m
s
Ee = Ek
1
mv 2
2
mv 2
V=
2q
Vq =
V=
(1.33 x 10-25 )(2.58 x 105 ) 2
2(3.2 x 10-19 )
V = 1.38 x 104 V
d)
c)
β qv = Fcent
β =
Fcent
qv
(4.41 x 10-14 )
(3.2 x 10-19 )(2.58 x 105 )
β = 0.535 T
β =
e)
Answer:
Fe = Fmag
E q = β qv
E = βv
E = (5.35 x 10-1 )(2.57518 x 105 )
E = 1.38 x 105
N
C