Download ExamView - SEMESTER 1 PRACTICE TEST.tst

THE ACTUAL TEST WILL BE MULTIPLE CHOICE ONLY AND NOT ALL OF THE TOPICS ARE
COVERED COMPLETELY ON THIS SHEET,GOOD LUCK. MR N
semester 1 practice test
Multiple Choice
Identify the choice that best completes the statement or answers the question.
____
1. 1 micrometer (1 µm) equals:
a. 10–3 m b. 10–6 m c. 103 m d. 106 m
____
2. What is the number of significant digits in the measurement 34.000 m?
a. 2 b. 3 c. 4 d. 5
____
3. A car is moving with a uniform speed of 15.0 m/s along a straight path. What is the distance covered by the
car in 12.0 minutes?
a. 1.02 × 10–3 km b. 1.80 × 10–1 km c. 8.00 × 10–5 km d. 1.08 × 101 km
____
4. Which of the following is a pair of vector quantities?
a. Speed — Distance b. Velocity — Distance c. Velocity — Displacement d. Speed — Displacement
2
Name: ________________________
____
____
ID: A
5. A man starts his car from rest and accelerates at 1 m/s2 for 2 seconds. He then continues at a constant velocity
for 10 seconds until he sees a tree blocking the road and applies brakes. The car, decelerating at 1 m/s2, finally
comes to rest. Which of the following graphs represents the motion correctly?
a.
b.
c.
d.
6. Which of the following system of forces provides the block the highest net force?
a.
c.
11 N
71 N
b.
153 N
d.
405 N
____
227 N
22.7 N
403 N
7. The path of a projectile through space is called its:
a. equilibrant b. torque c. range d. trajectory
2
15.3 N
Name: ________________________
ID: A
____
8. A stone is thrown horizontally from the top of a 25.00-m cliff. The stone lands at a distance of 40.00 m from
the edge of the cliff. What is the initial horizontal velocity of the stone?
a. 2.260 m/s b. 15.60 m/s c. 17.70 m/s d. 22.05 m/s
____
9. A ball is thrown horizontally at 10.0 m/s from the top of a hill 50.0 m high. How far from the base of the cliff
would the ball hit the ground?
a. 23.6 m b. 26.4 m c. 31.9 m d. 45.0 m
____ 10. A player kicks a football at an angle of 30.0° above the horizontal. The football has an initial velocity of 20.0
m/s. Find the horizontal component of the velocity and the maximum height attained by the football.
a. 10.0 m/s, 17.6 m b. 17.3 m/s, 5.10 m c. 25.1 m/s, 7.40 m d. 30.3 m/s, 9.50 m
____ 11. A player kicks a football with an initial velocity of 3.00 m/s at an angle of 60.0° above the horizontal. What is
the horizontal distance traveled by the football?
a. 0.312 m b. 0.397 m c. 0.673 m d. 0.795 m
____ 12. The movement of an object or a point mass at a constant speed around a circle that has a fixed radius is called
uniform:
a. circular motion b. parabolic motion c. elliptical motion d. rotational motion
____ 13. A sprinter runs at a speed of 3.00 m/s on a circular track that has a radius of 40.00 m. Find the centripetal
acceleration of the sprinter.
a. 0.225 m/s2 b. 4.44 m/s2 c. 0.750 m/s2 d. 0.0750 m/s2
____ 14. A 0.50-kg ball is attached to a string of 0.50 m and swung in a horizontal circle with a velocity of 1.0 m/s.
Find the centripetal force of the ball.
a. 0.50 N b. 1.0 N c. 2.0 N d. 2.5 N
____ 15. To melt 4 kg of a solid, 106 J of heat is required. Which of the following expressions gives the heat of fusion
of the solid?
4 × 106
10 6
25 × 106
J ⋅ kg c.
J/kg d.
J/kg
a. 4 × 10 6 J ⋅ kg b.
25
4
4
____ 16. 50 g of ice at 0°C is dropped in a beaker containing 100 g of water at 0°C. What will be the contents of the
beaker after 5 hours? Assume that the room temperature is 0°C.
a. 150 g of water b. 25 g of ice and 125 g of water c. 75 g of ice and 75 g of water d. 50 g of ice and
100 g of water
3
Name: ________________________
ID: A
____ 17. Heat is added to an open container of a liquid. The liquid is brought to its boiling point and half the liquid
boils away. Which of the following graphs shows how the temperature changes with time in this period?
a.
b.
c.
d.
____ 18. A solid is heated at a constant rate until it reaches the vapor state. The temperature of the substance changes
with time as shown in the graph below. Which part(s) of the graph indicate(s) that the substance exists in
solid-liquid and liquid-vapor state?
a. OP b. OP, QR, ST c. PQ, RS d. OP, ST
____ 19. A student walks to class at a velocity of 3 m/s. To avoid walking into a door as it opens, the student slows to a
velocity of 0.5 m/s. Now late for class, the student runs down the corridor at a velocity of 7 m/s. The student
had the least momentum
a. while walking at a velocity of 3 m/s. b. while dodging the opening door. c. immediately after the door
opened. d. while running to class at a velocity of 7 m/s.
____ 20. A rubber ball with a mass of 0.30 kg is dropped onto a steel plate. The ball’s velocity just before impact is 4.5
m/s and just after impact is 4.2 m/s. What is the change in the ball’s momentum?
a. –0.09 kg•m/s b. –2.6 kg•m/s c. –4.0 kg•m/s d. –12 kg•m/s
4
Name: ________________________
ID: A
____ 21. A moderate force will break an egg. However, an egg dropped on the road usually breaks, while one dropped
on the grass usually does not break because for the egg dropped on the grass,
a. the change in momentum is greater. b. the change in momentum is less. c. the time interval for
stopping is greater. d. the time interval for stopping is less.
Problem
F 1 is 57 N, of force ä
F 2 is 57 N, and of ä
F 3 is 57 N. The angles θ 1 and θ 2 are 30°
22. The magnitude of the force ä
ä,F
ä , and F
ä.
each. Use the graphical method to find the resultant of the forces F
1
2
3
A has magnitude 58.0 units and is inclined to the positive axis of x at 42.0°. Vector ä
B has magnitude
23. A vector ä
ä
36.0 units and is inclined to the positive axis of x at 121°. Vector C has magnitude 40.0 units and is inclined
ä.
A, ä
B , and C
to the positive axis of x at 53.0°. Use the graphical method to find the resultant of the vectors ä
24. A 2.7-kg box is released on a horizontal surface with an initial speed of 2.9 m/s. It moves on the surface with
a deceleration of 0.27 m/s2. Calculate the kinetic friction force on the box.
25. The distance between Pluto and the Sun is 39.1 times more than the distance between the Sun and Earth.
Calculate the time taken by Pluto to orbit the Sun in Earth days.
26. Calculate the force of gravitational attraction between two spheres of mass 10.1 kg and 45.4 kg that are 38.5
m apart.
27. A disc of radius 5.70 cm rotates about its axis and a point 1.90 cm from the center of the disc moves 34.5 cm
in 12.2 s. Calculate the angular speed of the disc.
28. A disc rotates at 5.60 rad/s. An angular acceleration acts on the disc to change its angular speed to 16.8 rad/s.
If the disc turns through 12.2 rad during this time, calculate the angular acceleration of the disc.
29. A 6110-kg bus traveling at 20.0 m/s can be stopped in 24.0 s by gently applying the brakes. If the driver
slams on the brakes, the bus stops in 3.90 s. What is the average force exerted on the bus in both these stops?
30. A toy car X of mass 0.200 kg moves along a frictionless surface with a velocity of 0.180 m/s. It collides with
another toy car Y, with a mass of 0.250 kg and a speed of 0.130 m/s in the same direction. After the collision,
toy car X continues to move in the same direction with a velocity of 0.177 m/s. Calculate the speed of toy car
Y after the collision.
5
Name: ________________________
ID: A
31. Carol and Bruno drag a box of mass 58.0 kg along a frictionless floor. Carol pushes the box with a force of
11.4 N at an angle of 40.0° downward from the horizontal. Bruno pulls the box from the other side with a
force of 11.0 N at an angle of 40.0° above the horizontal. What is the net work done on the box if the
displacement of the box is 14.5 m?
32. Raul pushes a stalled car with a force of 204 N. If the required force decreases at a constant rate from 204 N
to 44.0 N for a distance of 16.3 m in 16.0 s, calculate the average power required to move the car.
33. Andrew throws a 0.11-kg ball toward Donald, who is standing on a ledge. The ball leaves Andrew’s hands at
a height of 0.24 m and Donald catches it at a height of 0.82 m. Calculate the gravitational potential energy of
the ball relative to the ground before being thrown.
34. A warehouse worker pushed a cart weighing 4.50 kg to the top of an inclined plane. Initially, the cart was
0.670 m above the floor. If the top of the inclined plane is 2.70 m above the floor, calculate the work done by
gravity as the worker pushed the cart to the top of the plane.
35. A 2.90 × 10 −3 -kg bullet is fired with a velocity of 154 m/s toward a 5.44-kg stationary solid block resting on a
surface that has a coefficient of friction 0.215. The bullet emerges with a reduced velocity of 20.2 m/s after
passing through the block. What distance will the block slide before coming to rest? Assume that the block
does not lose any mass.
6
ID: A
semester 1 practice test
Answer Section
MULTIPLE CHOICE
1. ANS: B
10–6 is equivalent to the metric prefix micro (µ). Therefore, 1 micrometer (1 µm) = 10–6 m.
Feedback
A
B
C
D
The metric prefix milli (m) is equivalent to 10e–3 m.
Correct!
The metric prefix kilo (k) is equivalent to 10e+3 m.
The metric prefix mega (M) is equivalent to 10e+6 m.
PTS: 1
KEY: SI units
MSC: 1
NOT: /a/ The metric prefix milli (m) is equivalent to 10e–3 m. /b/ Correct! /c/ The metric prefix kilo (k) is
equivalent to 10e+3 m. /d/ The metric prefix mega (M) is equivalent to 10e+6 m.
2. ANS: D
As per the common rules for counting significant figures: (i) All nonzero digits are significant. (ii) All zeros
on the right of the last nonzero digit in the decimal part are significant.
Feedback
A
B
C
D
All zeros on the right of the last nonzero digit in the decimal part are significant.
The last two zeros are also significant.
The last zero is also significant.
Correct!
PTS: 1
KEY: Significant digits
MSC: 2
NOT: /a/ All zeros on the right of the last nonzero digit in the decimal part are significant. /b/ The last two
zeros are also significant. /c/ The last zero is also significant. /d/ Correct!
3. ANS: D
Distance = speed × time
Substituting the values in the relation for distance, we get, d = (15.0 m/s)(12.0 min)(60 s/1 min) = 10,800 m
= 1.08 × 101 km
Feedback
A
B
C
D
Speed is multiplied with time to calculate the distance. All the quantities should have
units of the same system.
Time must be converted to seconds.
Time should be multiplied with speed, not divided, to calculate the distance.
Correct!
PTS: 1
NAT: B.4
KEY: Average speed
MSC: 2
NOT: /a/ Speed is multiplied with time to calculate the distance. All the quantities should have units of the
same system. /b/ Time must be converted to seconds. /c/ Time should be multiplied with speed, not divided,
to calculate the distance. /d/ Correct!
1
ID: A
4. ANS: C
Velocity and displacement are vector quantities.
Feedback
A
B
C
D
Speed and distance are scalar quantities.
Velocity is a vector, but distance is scalar.
Correct!
Displacement is a vector, but speed is scalar.
PTS: 1
NAT: B.4
KEY: Vectors
MSC: 1
NOT: /a/ Speed and distance are scalar quantities. /b/ Velocity is a vector, but distance is scalar. /c/ Correct!
/d/ Displacement is a vector, but speed is scalar.
5. ANS: B
Feedback
A
B
C
D
This graph does not show a period of constant velocity.
Correct!
The car starts from rest instead of at constant speed and the slopes of the different parts
of the graph are incorrect.
The car moves at a constant velocity for 12 seconds.
PTS: 1
NAT: B.4
KEY: Velocity-time graph
MSC: 3
NOT: /A/ This graph does not show a period of constant velocity. /B/ Correct! /C/ The car starts from rest
instead of at constant speed and the slopes of the different parts of the graph are incorrect./D/ The car moves
at a constant velocity for 12 seconds.
6. ANS: A
Feedback
A
B
C
D
Correct!
The difference between the forces is not the maximum.
The net force is the difference between the two forces.
This system does not give the highest net force.
PTS: 1
NAT: B.4
KEY: Force
MSC: 1
NOT: /A/ Correct! /B/ The difference between the forces is not the maximum./C/ The net force is the
difference between the two forces. /D/ This system does not give the highest net force.
2
ID: A
7. ANS: D
A projectile is any object that has been given an initial thrust and moves through air. Its path through space is
called its trajectory.
Feedback
A
B
C
D
Equilibrant is a type of force.
Torque is the product of the force and length of the lever arm.
Range is the horizontal distance traveled by a projectile.
Correct!
PTS: 1
NAT: B.4
KEY: Projectile, Trajectory
MSC: 1
NOT: /a/ Equilibrant is a type of force. /b/ Torque is the product of the force and length of the lever arm. /c/
Range is the horizontal distance traveled by a projectile. /d/ Correct!
8. ANS: C
Solve the equation for the time the stone is in the air. Then solve the equation for the initial horizontal
velocity by substituting the values of time and the range.
Feedback
A
B
C
D
The time period is 2.260 s. Divide the range by the time period to get the initial
horizontal velocity.
Divide the range by the period, instead of multiplying the range by the period.
Correct!
You calculated the vertical velocity instead of the horizontal velocity.
PTS: 1
NAT: B.4
KEY: Projectiles launched horizontally
MSC: 3
NOT: /a/ The time period is 2.260 s. Divide the range by the time period to get the initial horizontal velocity.
/b/ Divide the range by the period, instead of multiplying the range by the period. /c/ Correct! /d/ You
calculated the vertical velocity instead of the horizontal velocity.
9. ANS: C
Use the equation for the y-position, and solve the equation for the time the stone is in the air. Multiply the
time period with the horizontal velocity to obtain the horizontal distance (range).
Feedback
A
B
C
D
Use the equation for the y-position to find the time period.
Multiply the period with the horizontal velocity.
Correct!
The magnitude of the acceleration due to gravity is 9.80 m/s^2.
PTS: 1
NAT: B.4
KEY: Projectiles launched horizontally
MSC: 3
NOT: /a/ Use the equation for the y-position to find the time period. /b/ Multiply the period with the
horizontal velocity. /c/ Correct! /d/ The magnitude of the acceleration due to gravity is 9.80 m/s^2.
3
ID: A
10. ANS: B
Write expressions for the vertical and horizontal velocity components. Solve the velocity equation for the
time of maximum height. Substitute this time into the equation for vertical position to find the height.
Feedback
A
B
C
D
You calculated the vertical component instead of the horizontal component.
Correct!
The actual velocity is always more than its horizontal component.
Can the horizontal component of the velocity be more than the actual velocity?
PTS: 1
NAT: B.4
KEY: Maximum range
MSC: 3
NOT: /a/You calculated the vertical component instead of the horizontal component. /b/ Correct! /c/ The
actual velocity is always more than its horizontal component. /d/ Can the horizontal component of the
velocity be more than the actual velocity?
11. ANS: D
Solve the vertical-position equation for the time at the end of the flight, when y = 0. Substitute that value of
time into the equation for horizontal distance to get the range.
Feedback
A
B
C
D
Solve the vertical-position equation for the time at the end of the flight when y = 0.
To find the range, multiply the horizontal component of velocity with the time obtained
when y = 0.
The vertical component of velocity is the product of the velocity and the sine of the
angle the football makes with the horizontal.
Correct!
PTS: 1
NAT: B.4
KEY: Maximum range
MSC: 2
NOT: /a/ Solve the vertical-position equation for the time at the end of the flight when y = 0. /b/ To find the
range, multiply the horizontal component of velocity with the time obtained when y = 0. /c/ The vertical
component of velocity is the product of the velocity and the sine of the angle the football makes with the
horizontal. /d/ Correct!
12. ANS: A
Uniform circular motion is the movement of an object or point mass at a constant speed around a circle with a
fixed radius.
Feedback
A
B
C
D
Correct!
A parabola is not a circle.
An elliptical orbit does not have a fixed radius.
A rotating object moves around its own vertical axis.
PTS: 1
NAT: B.4
KEY: Uniform circular motion
MSC: 1
NOT: /a/ Correct! /b/ A parabola is not a circle. /c/ An elliptical orbit does not have a fixed radius. /d/ A
rotating object moves around its own vertical axis.
4
ID: A
13. ANS: A
Substitute the values of velocity and radius in the equation for centripetal acceleration.
Feedback
A
B
C
D
Correct!
Divide the square of the velocity by the radius to calculate centripetal acceleration.
Square the velocity in the formula for centripetal acceleration.
Divide the square of the velocity by the radius.
PTS: 1
NAT: B.4
KEY: Centripetal acceleration
MSC: 2
NOT: /a/ Correct! /b/ Divide the square of the velocity by the radius to calculate centripetal acceleration. /c/
Square the velocity in the formula for centripetal acceleration. /d/ Divide the square of the velocity by the
radius.
14. ANS: B
mv 2
.
Use Newton’s second law, F =
r
Feedback
A
B
C
D
Did you include the value of radius in the formula?
Correct!
Use Newton's second law.
Multiply the mass with the square of the velocity and divide it by the radius.
PTS: 1
NAT: B.4
KEY: Centripetal force
MSC: 2
NOT: /a/ Did you include the value of radius in the formula? /b/ Correct! /c/ Use Newton's second law. /d/
Multiply the mass with the square of the velocity and divide it by the radius.
15. ANS: C
Feedback
A
B
C
D
Instead of multiplying the heat required and the mass of the solid melted, divide the heat
required by the mass of the solid melted.
Instead of dividing the heat required by the melting point temperature, divide it by the
mass of the solid melted.
Correct!
Instead of multiplying the melting point temperature and the ratio of heat and mass of
the solid, divide the heat required by the mass of the solid melted.
PTS: 1
KEY: Heat of fusion
MSC: 1
NOT: /a/ Instead of multiplying the heat required and the mass of the solid melted, divide the heat required
by the mass of the solid melted. /b/ Instead of dividing the heat required by the melting point temperature,
divide it by the mass of the solid melted. /c/ Correct! /d/ Instead of multiplying the melting point temperature
and the ratio of heat and mass of the solid, divide the heat required by the mass of the solid melted.
5
ID: A
16. ANS: D
Feedback
A
B
C
D
For the ice to melt completely, there must be a source of energy.
For half the ice to melt, there must be a source of heat.
For some water to freeze into ice, it must lose heat to the other contents of the beaker.
Correct!
PTS: 1
KEY: Change of state
MSC: 1
NOT: /a/ For the ice to melt completely, there must be a source of energy. /b/ For half the ice to melt, there
must be a source of heat. /c/ For some water to freeze into ice, it must lose heat to the other contents of the
beaker. /d/ Correct!
17. ANS: B
Feedback
A
B
C
D
This graph does not show the change of state.
Correct!
If heat is supplied at a constant rate, the temperature should not increase so sharply.
The temperature at the end of the graph should not rise if some of the liquid has yet not
evaporated.
PTS: 1
KEY: Heat of vaporization
MSC: 2
NOT: /a/ This graph does not show the change of state. /b/ Correct! /c/ If heat is supplied at a constant rate,
the temperature should not increase so sharply. /d/ The temperature at the end of the graph should not rise if
some of the liquid has yet not evaporated.
18. ANS: C
Feedback
A
B
C
D
The temperature increases with time in part OP, so the substance must be in one state.
The temperature increases with time in parts OP, QR, and ST, so the substance must be
in one state.
Correct!
The temperature increases with time in parts OP and ST, so the substance must be in
one state.
PTS: 1
KEY: Change of state
MSC: 2
NOT: /a/ The temperature increases with time in part OP, so the substance must be in one state. /b/ The
temperature increases with time in parts OP, QR, and ST, so the substance must be in one state. /c/ Correct!
/d/ The temperature increases with time in parts OP and ST, so the substance must be in one state.
19. ANS: B
PTS: 1
20. ANS: B
PTS: 1
21. ANS: C
PTS: 1
6
ID: A
PROBLEM
22. ANS:
Zero
0
PTS: 1
NAT: B.4
KEY: Sum of vectors
MSC: 3
NOT: Draw the vectors to scale and use the protractor to draw them tip to tail. The closing side of the
polygon in reverse order is the resultant.
23. ANS:
The resultant is 113 units at an angle of 64.4° to the positive axis of x.
PTS: 1
NAT: B.4
KEY: Sum of vectors
MSC: 3
NOT: Draw the vectors to scale and use the protractor to draw them tip to tail. The closing side of the
polygon in reverse order is the resultant. Measure the length of the resultant to get the magnitude of the
vector, and use a protractor to find the direction of the resultant.
24. ANS:
0.73 N
PTS: 1
NAT: B.4
KEY: Friction
MSC: 2
NOT: The only force on the box is the kinetic friction, which should be equal to the product of mass and
acceleration of the box.
25. ANS:
89300 Earth days
89300
PTS: 1
NAT: B.4
KEY: Kepler's laws
MSC: 3
NOT: The square of the period of a planet's orbit around the Sun is proportional to the cube of its mean
distance from the Sun.
26. ANS:
2.06 × 10 −11 N
PTS: 1
MSC: 3
27. ANS:
1.49 rad/s
NAT: B.4
KEY: Gravitational force
NOT: Use the mathematical form of Newton's law of gravitation.
PTS: 1
NAT: B.4
KEY: Angular velocity
MSC: 3
NOT: The angular speed of a rotating body is equal to the linear speed of a point divided by the distance of
the point from the axis of rotation.
7
ID: A
28. ANS:
10.3 rad/s2
PTS: 1
NAT: B.4
KEY: Rotational motion
MSC: 3
NOT: The angular acceleration is equal to the difference between the squares of the final and initial angular
speeds divided by twice the angle turned.
29. ANS:
F gentle braking = 5.09 × 10 3 N
F hard braking = 3.13 × 10 4 N
PTS: 1
NAT: B.4
KEY: Impulse
MSC: 3
NOT: Apply the impulse-momentum theorem to obtain the force needed to stop the vehicle.
30. ANS:
0.132 m/s
PTS: 1
MSC: 3
31. ANS:
2.49 × 10 2 J
KEY: Conservation of momentum, Newton's third law
NOT: Apply Newton's third law and the law of conservation of momentum.
PTS: 1
KEY: Work
MSC: 3
NOT: Work is equal to the product of force and displacement times the cosine of the angle between the force
and the direction of the displacement.
32. ANS:
126 W
PTS: 1
KEY: Power
MSC: 3
NOT: Power is equal to the work done divided by the time taken to do the work.
33. ANS:
PE ground = 0.26 J
PTS: 1
KEY: Gravitational potential energy
MSC: 3
NOT: The gravitational potential energy of an object is equal to the product of its mass, the acceleration due
to gravity, and its height from the reference level.
34. ANS:
–89.5 J
PTS: 1
KEY: Gravitational potential energy
MSC: 3
NOT: The work done by gravity is the weight of the cart times the vertical distance through which the cart is
pushed.
35. ANS:
1.21 × 10 −3 m
PTS: 1
KEY: Collisions, Change in kinetic energy
MSC: 3
NOT: The principle of conservation of momentum can be used to find the final velocity of the block. The
kinetic energy of the block is used in doing work against friction.
8