Download Midterm #2 Practice

University of the Fraser Valley
Physics 100
Midterm #2 Practice
Name: __________________________
Directions: Fill in the scantron form with the following information:
1. ID number (student number)
2. Name at top of form
3. Name bubbled into the columns labelled "C1C2C3C4"
4. Test Version Number (shown below); bubble in spaced labelled "version"on the form
Version # 0
1. A steel ball of mass 1.0 kg rolls down an inclined track where the the friction is 2.0 N. The pull
of gravity on the ball supplies a force of 5.0 N along the track. The acceleration along the track
is
A. 7.0 m/s2 down
B. 5.0 m/s2 down
C. 3.0 m/s2 up
D. 3.0 m/s2 down
E. 2.0 m/s2 up
Solution:
Fnet = Fapplied - Ff
a = (Fapplied - Ff)÷m
a = (5.0 - 2.0) ÷ 1.0
a = 3.0 m/s2
2. A locomotive pulling a freight car accelerates at 0.50 m/s2 as shown in the diagram.
What is the tension in the coupling linking the locomotive and car? (Ignore friction.)
A. 5 000 N
B. 25 000 N
C. 30 000 N
D. 390 000 N
Solution:
From the diagram, the net force on the 10 000 kg cart
is due to the tension.
T = ma
T = 10 000 (0.5) = 5 000 N
3. If the tension in the line joining the two masses shown below is 12 N, what is the mass, m1?
(Ignore surface friction.)
A.
B.
C.
D.
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1.1 kg
1.4 kg
2.0 kg
10 kg
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Solution:
10a = 10g – 12
a = 8.6 m/s2
m1a = 12
m1 = 1.39 kg
4. A locomotive is pulling five freight cars along a level stretch of railway track. If the train's
speed is increasing compare the tensions in the coupling connecting the locomotive to the first
car, and the coupling connecting the last two cars. Explain any similarities or diffrerences in
point form.
(3.00 marks)
Solution:
• The tension in the couplings near the locomotive is larger than in the coupling between the last
two cars. The first coupling must provide the force to accelerate all remaining cars, whereas
the last coupling provides the force to accelerate only the last car.
• If the cars have identical mass and frictional forces, the tension in the last coupling would be
1/5 the tension in the first coupling.
(3 marks)
5. Ships are often built on ground that slopes down to the water for easy launching. Normally the
ship is launched before most of its interior structures are completed; this work is done once the
ship is afloat. Is this done because the additional weight would cause the ship to slide down the
launch prematurely? Explain.
(3.00 marks)
Solution:
• The mass of the boat does not effect its acceleration when being launched.
• The additional weight would not cause the boat to slide down prematurely.
• Fnet = mgsinø - µmgcosø
• so 0 = mgsinø - µmgcosø which gives:
• µ = tan ø. So if the angle of the launch is not too great and the material is able to provide the
proper coefficient of friction, the boat will not slide.
(3.00 marks)
6. The inertia of an object is more closely related to its:
A. density
B. mass
C. position
D. shape
E. volume
Solution:
Newton's First Law also known as the law of inertia
7. What is the net force acting on a hang Glider moving at constant velocity in level flight?
A. Zero
B. Equal to gravity.
C. Directed upwards.
D. Directed downwards.
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Solution:
If an object is moving at constant velocity in a linear direction, then the acceleration is zero and
therefore have no net force. All the forces acting on the object will be balanced.
8. A net force F acts on an object of mass m, causing it to accelerate at 4.0 m/s2. If the same net
force F acts on an object of mass 2m, its acceleration will be
A. 1.0 m/s2
B. 2.0 m/s2
C. 4.0 m/s2
D. 8.0 m/s2
Solution:
Fnet = (4.0)m
and Fnet = (a)2m
4.0m÷(2m) = a
a = 2.0 m/s2
9. In the year 2010 an astronaut makes a routine visit to the moon where g is 1.6 N/kg. The
astronaut has a mass of 50 kg on earth where g = 10 N/kg. What force of gravity would the
astronaut feel on the moon's surface?
A. 31 N
B. 31 kg
C. 80 N
D. 80 kg
E. 500 N
Solution:
F = mg
F = 50 x 1.6
F = 80 N
10. If frictional forces are considered minimal, what is the tension in the string system shown
below?
A. 1.7 N
B. 17 N
C. 25 N
D. 76 N
Solution:
a = mg÷mtotal
a =(2.6 x 9.8)÷7.8
a = 3.27 m/s2
T = mg - ma
T = m(g - a)
T = 16.9 N
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11. An elevator weighing 12 000 N is accelerating upwards. The tension in the cable is 20 000 N
and the frictional resistance to motion is 5 000 N. The unbalanced force on the elevator is
A. 37 000 N up
B. 27 000 N up
C. 13 000 N up
D. 8 000 N up
E. 3 000 N up
Solution:
Free Body Diagram:
Net Force = Tension - Fg - Ff
Net Force = 20 000 - 12 000 - 5 000
Net Force = 3000 N upwards
12. An 80 kg man stands on a spring scale while accelerating upwards in an elevator. If the scale
indicates that his "weight" is 980 N, what is the magnitude of the acceleration of the elevator in
meters per second squared?
A. 2.5
B. 13
C. 12.5
D. 0.0080
Solution:
Using a Free Body Diagram:
980 - mg = ma
980 - 80 x 9.81 = 80 a
a = 2.44 m/s2
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13. A 2.0 kg frictionless puck resting on a level table is attached to a mass by a nylon thread over a
frictionless pulley. This mass is hanging over the edge of the table and is also 2.0 kg. The mass
is released and pulls the puck along the table. The acceleration of the puck is:
A. zero
B. 5.0 m/s2
C. 10.5 m/s2
D. 2.0 x 10 m/s2
Solution:
Puck
Mass
Fnet = T
Fnet = mg - T
2a=T
2a=2g-T
Substitute and solve to get: 4 a = 2 g
So a = 0.5 g
a = 4.9 m/s2
14. A student exerts a 120 N horizontal force on a 25 kg carton of apples, causing it to accelerate
over level ground at 1.8 m/s2.
Find the coefficient of friction between the carton and the ground.
A. 0.31
B. 0.38
C. 0.49
D. 0.67
Solution:
Fnet = Fapplied - Ffriction
F
µ= f
mg
Ffriction = Fapplied - Fnet
Ffriction = 120 - (25 x 1.8)
Ffriction = 75 N
where Ff = µmg, so
µ = 75/(25 x 9.8)
µ = 0.31
15. NASA has finally finished a space program that can send a person to Mars. The mass of the
astronaut is measured to be 75 kg.
(6.00 marks)
a) What is the weight of the astronaut on the Earth?
b) What is the weight of the astronaut on Mars where g = 3.8 m/s2?
c) What is the value of g on top of Mt. Everest if the astronaut weighs 683 N?
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Solution:
a) W = mg
W = (75kg)(9.80 m/s2)
W = 7.4x102 N
b) W = mg
W = (75kg)(9.80 m/s2)
W = 2.9x102 N
c) g=W÷m
g = 683N÷75kg
g = 9.1 m/s2
(6.00 marks)
16. In the diagram below, the table is frictionless. Calculate the acceleration of the 2.6 kg weight.
(4.00 marks)
Solution:
• From the force diagram (2 marks)
Fnet = Fg − FT + FT
Fnet = Fg = (2.6 kg)(9.8 sm2 ) = 25.48 N
• To find the acceleration: (2 marks)
Fnet = mtotal a
25.48 N = ( 7.8kg )( a )
a = 3.27 sm2
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Use the following diagram to answer the next 1 questions.
Two masses are connected by a light cord passing over frictionless pulleys as shown in the
diagram below.
17. A block is then added to the system.
Adding the block decreases the acceleration of the system. Identify and explain two reasons for
this decrease.
(4.00 marks)
Solution:
Since the system mass has increased, the acceleration must decrease (a = Fnet ÷ m).
Since there is friction, the net force will decrease and the acceleration will be smaller yet.
(4 marks)
18. A professional golfer can hit a 0.15 kg golf ball so that it leaves the club with a speed of 46 m/s.
Immediately after hitting the ball, his golf club is measured to have a velocity of 4 m/s. If the
club has a mass of 0.6 kg, calculate the velocity of the club just before hitting the ball.
(3.00 marks)
Solution:
∆p=0
mcvc = mbvb + mcvc'
vc = (mbvb + mcvc') ÷ mc
vc =(0.15 x 46 + 0.6 x 4) ÷ 0.6
vc = 15.5 m/s
(3.00 marks)
19. The front of an automobile is designed to crumple in a collision in order to reduce the injury to
the occupants. Discuss briefly the physics of how this design feature improves safety for the
occupants.
(4.00 marks)
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Solution:
The crumpling of the automobile decreases the acceleration experienced by the occupants by
increasing the distance to stop and/or increasing the time taken to stop.
Ek = F d
increasing d, decreases F
change in momnetum = F t
increasing t, decreases F
(4 marks)
20. A 200 gram baseball comes into home plate at a velocity of 9.0 m/s and is batted toward the
pitcher at 12 m/s. The total change in momentum of the ball is:
A. 6.0 kg m/s
B. 42 grams m/s
C. 600 N m/s
D. 4.2 N•s
Solution:
∆p = pfinal - pinitial
∆p = m(vf - vi) remember to include directions
∆p = 4.2 kgm/s
21. A 2.0 kg ball moving at 12 m/s hits a 5.0 kg ball moving at 6.2 m/s in the other direction. The
5.0 kg ball continues in the same direction at 1.3 m/s. Calculate the speed of the 2.0 kg ball
immediately after impact.
A. 0.25 m/s
B. 3.5 m/s
C. 24 m/s
D. 31 m/s
Solution:
∆p=0
pf = pi (Up is positive)
m5v'5 + m2v'2 = m5v5 - m2v2
v'2 = (m5v5 - m2v2 - m5v'5) ÷ m2
v'2 = (5 x 6.2 - 2 x 12 - 5 x 1.3) ÷ 2
v'2 = 0.25 m/s
22. Which of the following is not a vector?
A. energy
B. momentum
Solution:
Energy is a scalar
C. impulse
D. force
23. When the speed of a car is tripled, its kinetic energy changes by a factor of:
A. 9
B. 4
C. 1
D. 1/4
Solution:
Since velocity is tripled and Ek is proportional to v2, then Ek changes by a factor of 9
24. A certain object has a kinetic energy of Ek. If it's mass is doubled and its velocity is tripled,
what effect do these changes make to the object's kinetic energy?
A. 2.5 Ek
B. 9 Ek
C. 10 Ek
D. 18 Ek
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Solution:
Ek is proportional to mass and v2, so Eknew is 2 x 9 Ek = 18 Ek
25.
Calculate the potential energy lost if the 10 kg block shown above tips over on to its side.
A. 7.8 J
B. 11 J
C. 15 J
D. 29 J
Solution:
center of mass of block at initial position = 0.15 m above table
center of mass of block at final position = 0.04 m above table
∆Ep = mg∆h
∆Ep = 10.78 J
26. What is the kinetic energy of a 0.850 kg sea gull flying at 18 m/s and 20 m above the ocean?
A. 8.3 J
B. 140 J
C. 170 J
D. 300 J
Solution:
Ek = 0.5 mv2
Ek = 0.5 x 0.85 x 182
Ek = 137.7 J
Ek = 140 J
27. The graph below shows the distance an object moved when a varying force was applied to it.
What is the work done in moving the object from 0 m to 20 m?
A. 2.5 x 101 J
B. 5.0 x 103 J
Solution:
W = area under graph
W = 1/2 Fd
W = 0.5 x 500 x 20 = 5.0 x 103 J
C. 1.0 x 104 J
D. 4.9 x 104 J
28. A 200 N force acts on a 10.0 kg block and moves it 5.0 m in 2.5 seconds. How much work is
being done?
A. 40 J
B. 510 J
C. 1000 J
D. 1490 J
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Solution:
W = Fd cosø
F is in the direction of the displacement
W = 200 x 5.0
W = 1000 J
29. The graph below shows how the force acting on an object varies with distance.
What is the work done in moving the object from 20 m to 60 m?
A. 50 J
B. 100 J
C. 400 J
D. 900 J
Solution:
Work equals the area under the graph line. The area under the graph line is represented by a
triangle and rectangle from 20-60 m.
Area of a triangle = 0.5 base x height.
Area of rectangle = length x width
Triangle area = 0.5 x 40 m x 10 N
Rectangle area = 40 m x 5 N
W = sum of two areas
W = 200 J + 200 J
W = 400 J
30. What is the minimum work done when a 65 kg student climbs an 8.0 m-high stairway in 12 s?
A. 420 J
B. 520 JN
C. 5 100 J
D. 6 200 J
Solution:
W = Fd, where F = mg
W = mgd
W = 65 x 9.8 x 8
W = 5096 J
31. An 1100 kg car increases its speed from 12 m/s to 25 m/s in 10 s. What is the work done on the
car?
A. 9.3 x 104 J
Solution:
W = ∆Ek
B. 3.5 x 104 J
C. 2.6 x 105 J
D. 3.4 x 105 J
W = 0.5 m (v2final - v2initial)
W = 0.5 x 1100 (252 - 122)
W = 2.6 x 105 J
32. A car moving at 80 km/h skids 50 m with locked brakes. How far will the car skid with locked
brakes if it is travelling at 40 km/h?
A. 13 m
B. 18 m
C. 25 m
D. 35 m
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Solution:
∆Ek = W
∆Ek = Fd
d = ∆Ek÷F
d = 0.5 mv2 ÷ F, so d α v2
If the velocity is decreased by a factor of 2, then distance will be decreased by a factor of 4.
The new distance will be 50÷4 = 12.5 m.
33. Two identical bullets, one with twice the kinetic energy of the other, are fired into a large target.
The faster bullet will penetrate
A. the same distance as the slower one
B. twice as far as the slower arrow
C. four times as far as the slower one
D. more than four times as far as the slower one
Solution:
∆Ek = W
∆Ek = Fd
d = ∆Ek÷F, so d α Ek
If the kinetic energy is increased by a factor of 2, then distance will be increased by a factor of 2.
34. An escalator is able to move 20 passengers of average mass 60 kg. If the full escalator moves
the passengers from the first floor to the second floor in 60 s, as shown. What power is required
by the escalator?
A. 9.8 x 101 W
B. 9.8 x 102 W
Solution:
P=W÷t
P=mgh÷t, were m = 20 x 60 kg
P = 1200 x 9.81 x 5 ÷ 60
P = 981 W
C. 2.0 x 103 W
D. 6.0 x 104 W
35. A machine with a power rating of 3.0 x 105 W must complete a job requiring 1.5 x 106 J of
energy by lifting a 1.0 x 103 kg object. How long will it take to complete the job?
A. 0.2 s
B. 5.0 s
C. 150 s
D. 300 s
Solution:
P=E÷t
t = E÷ P
t = 1.5 x 106 ÷ 3.0 x 105
t = 5.0 s
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36. A motor using 3.7 × 103 W is 81% efficient. This motor is pulling a 450 kg block along a
horizontal surface. If the coefficient of friction is 0.35, what is the speed of the block?
(7.00 marks)
Solution:
W Fd
P=
=
= Fv
t
t
∴
P
=v
F
v=
P × 0.81
µmg
v = 1.94m / s
37. An 80 kg driver in a 900 kg Porsche is travelling at 35 m/s. The driver suspects that a parked
"van" on the side of the road contains "photo-radar". The driver slows down to the maximum
posted limit of 28 m/s. What is the change in kinetic energy of the vehicle?
(3.00 marks)
Solution:
∆Ek = 0.5 m (vf2 - vi2)
where m = total mass of 980 kg
∆Ek = 0.5 x 980 (282 - 352)
∆Ek = - 2.16 x 105 J
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