Instructor: Hacker Course: Physics 121 Sample Exam 4 Solutions (Newton’s Laws)

Transcription

Instructor: Hacker
Course: Physics 121
Sample Exam 4 Solutions (Newton’s Laws)
Print your name neatly. If you forget to write your name, or if the grader can’t read your
writing, you can lose up to 100 points. Answer all the questions that you can.
This exam will consist of 21 multiple-choice problems. You may not use calculators or
other electronic devices on this exam. The use of such a device will be regarded as an
attempt to cheat, and will be pursued accordingly. All diagrams and figures on this exam
are rough sketches: they are not generally drawn to scale.
No partial credit will be given for these problems. However, you can miss one of the 21
problems without penalty. Your grade will be based on your best 20 problems. You will
not receive extra credit for getting all 21 right.
Your grade on the exam will be based entirely on the answers that you circle on this
sheet. If you have no answer or a wrong answer there, the grader will not look at the
page with the problem to see if the right answer appears there. Illegible or ambiguous
answers will be graded as wrong. You are responsible for copying your answers clearly,
correctly, and in the right place.
Although there is no partial credit on this exam, you must show your work in the space
provided on the exam. There is additional scratch paper at the end of the exam: do
not use it unless you have filled all the scratch space provided on the page with the
problem. If you answer a difficult problem without doing any written work, the grader
will assume that you got the answer by guessing or by copying from someone else, and
will not give you credit for the problem even though you’ve indicated the correct solution
on the answer sheet.
Circle your answers here. Do not detach this sheet from the test.
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Physics 121 sample exam 4 solns
Copyright ©Wayne Hacker 2010. All rights reserved. 2
Introduction to Newton’s laws
Problem 1. You are piloting your spaceship through interstellar space where there is
no gravity and no air, when your timing belt breaks and your engine abruptly quits.
This occurs at time T , when you are travelling at speed V . Which of the graphs below
describes your motion starting at time T ?
v
v
(a)
6
a
*(b)
6
V
a
(c)
6
(d)
6
V
T
-
t
T
-
t
T
-
t
T
-
t
Solution: There is no gravity, no friction, and no air resistance in this situation; so we
assume that once the engine has stopped, there are no forces acting on the spaceship.
Thus it keeps going at the same velocity forever. Graph (b) shows that. Graph (a) shows
a changing speed; and graphs (c) and (d) show nonzero acceleration.
Problem 2. (Similarity Problem) Consider the following two experiments. In the
first experiment you push a block along a level surface at a constant velocity V . To
overcome the friction between the block and the surface, you must apply a constant
horizontal force Fpush,1 . In the second experiment you push the block at a constant speed
2V across the same surface. Assuming that the friction is the same as it was the first
experiment, which of the following choices best describes the new push force Fpush,2 in
terms of the first push force?
*(a)A constant force Fpush,1
(b) A constant force 2Fpush,1
(c) A force that increases from Fpush,1 to 2Fpush,1
(d) A force of zero, since the block does not accelerate.
Solution: For the block to move at constant speed, by Newton’s 1st law the net force
on it must be zero. Doing a force balance in the horizontal direction we find: 0 = Fnet =
Fpush,1 − f ⇒ Fpush,1 = f , where f is the frictional force. Thus when the block is
moving at speed V , the pushing force Fpush,1 is equal to the frictional force. When the
table is moving at 2V , the frictional force is the same; so the pushing force must again
be Fpush,1 . This is a subtle problem because students generally choose (b). The key to
understanding this problem is to realize that that the force that it takes to move the
block at a constant speed is different that the force it takes to accelerate it from rest to
a certain speed. Also the time that it would take. These other issues have to do with
the concepts of work and power, concepts that we have not yet discussed.
Problem 3. Your physics instructor asks you to help him push an atomic bomb into
the classroom. The bomb weighs 8000 N. You push on it with a horizontal force of 400
N. Which of the following statements is true?
(a)
If the atomic bomb moves across the floor, then you feel it pushing back on
you with a force of less than 400 N.
*(b) You feel the atomic bomb pushing back on you with a force of 400 N, whether
it moves or not.
(c)
If the atomic bomb does not move, you feel it pushing back on you with a force
of 8000 N.
(d) You feel the atomic bomb pushing back on you with a force of 8000 N, whether
it moves or not.
Solution: This is Newton’s third law. If you’re pushing on the bomb with a force of
400 N, then the bomb is pushing on you with an equal force.
Problem 4. You come into your physics lab and find
three books stacked on a table, as shown at right. What
is the net force on the middle book?
(a) 10 N downward
(b) 20 N upward
(c) 15 N downward
*(d) 0 N
Solution: The book is motionless, so it’s not accelerating in any direction. Hence the net force on it is zero.
10 N
15 N
20 N
Applications of Newton’s laws
Problem 5. A bucket weighing 500 N is lifted with
an upward acceleration of 2 m/s2 by a uniform chain
weighing 200 N. What is the tension in the middle
link of the chain? For ease of calculation, assume that
g = 10 m/s2 .
(a) 600 N
*(b) 720 N
(c) 840 N
(d) 960 N
u
u
u
wb
Tmid
1
w
2 c
1
w
2 c
6
u
1
w
2 c
?
wb ?
Solution: Sketch a picture of the entire system; then a free-body diagram of the system
consisting of the middle link. The middle link supports the weight of the bucket wb plus
half the weight of the chain, 12 wc . Since the system is accelerating upward, it experiences
a net upward force. We will use Newton’s second law:
wc
mc Tmid − wb −
= Fnet = msystem a = mb +
a
2
2
wc
mc
⇒ Tmid = wb +
+ mb +
a
2
2
wc (wb + wc /2)a
w=mg
−−−→
= wb +
+
2
g
wc g + a
200 N 10 m/s2 + 2 m/s2
= 720 N
= wb +
= 500 N +
2
g
2
10 m/s2
Problem 6. A 5-kg block is initially at rest on a frictionless horizontal surface. It is
pulled with with a constant horizontal force of 10 N. How long must it be pulled before
its speed is 15 m/s?
(a) 2.5 s
(b) 5.0 s
*(c) 7.5 s
(d) 10.0 s
(e) None of these
Solution: Start by using Newton’s 2nd law for 1-d motion to determine a. Once a is
known, this becomes a kinematics problem and we can apply one of our 3 fundamental
equations.
F
10 kg m/s2
÷m
=
= 2 m/s2 .
F = ma −−→ a =
m
5 kg
We now know the acceleration a; so write down what information you are given, what
you want, and which fundamental equation you need to use.

vi = 0 m/s



v = 15 m/s
f
Given:
Want: t
Which Equation: Fundamental equation (1)
2

a
=
2
m/s



ti = 0
Solving for t in equation 1: t =
15 m/s
vf − vi
=
= 7.5 s.
a
2 m/s2
Problem 7. The acceleration due to gravity at the surface of Neptune’s moon Triton is
approximately 2.4 m/s2 . If a rock weighs 24 N on Triton, what is its mass? Round your
answer to the nearest kilogram.
*(a)
(c)
(e)
10 kg
1/10 kg
None of these
Solution: w = mgT
(b) 20 kg
(d) 1/20 kg
÷g
T
−−−→
m=
24 N
w
= 10 kg
=
gT
2.4 m/s2
Problem 8. A block is hanging from a spring scale that is attached
to the ceiling of an elevator. Under which of the following circumstances will the reading on the scale be less than the weight of the
block?
(a) The elevator is moving downward and slowing down.
*(b) The elevator is moving upward and slowing down.
(c) The elevator is moving downward at constant speed.
(d) The elevator is moving upward at constant speed.
Tu
6
u
wb ?
Solution: Sketch a free-body diagram for the block, as at right. There are two forces
acting on the block: the weight wb pulling down, and the tension in the scale Tu pulling
upward. If Tu < wb , then the net force is downward; so the block is experiencing downward acceleration. This can happen if the elevator is moving upward and slowing down,
or if it is moving downward and speeding up. Only the former option is in the answers.
Problem 9. An object is hung from a spring scale attached to the ceiling of an elevator.
The scale reads 100 N when the elevator is sitting still. What is the reading on the scale
if the elevator is moving downward with a constant speed of 10 m/s? Approximate g as
10 m/s2 .
(a) 0 N
(b) 50 N
*(c) 100 N
(d) 200 N
(e) None of these
Solution: Since the elevator is moving at constant speed, the acceleration a is 0 m/s2 .
From Newton’s 2nd law it follows that the net force is just the force due to gravity, which
was present when the elevator was sitting still. Thus the reading on the scale is 100 N.
Problem 10. A rubber band has an unstretched length of 8 cm and a force constant of
4 N/cm. What is the force required to stretch it to a length of 12 cm?
(a) 3 N
*(b) 16 N
(c) 32 N
(d) 48 N
Solution: The rubber band is being stretched by ∆x = 12 cm − 8 cm = 4 cm. The
force required to stretch it is F = k∆x = (4 N/cm)(4 cm) = 16 N.
Problem 11. A block with weight w = 100
N is suspended by two thin ropes, labelled A
and B. Rope A makes an angle of θA = 30◦
with the horizontal; rope B makes an angle of
θB = 60◦ with the horizontal. The tension in
rope A is TA ; the tension in rope B is TB . Find
TB ; round your answer to the nearest newton.
√
(a) 50 N
*(b) 50 √3 N
(c) 110 N
(d) 110 3 N
(e) None of these
Z θA
Z
Z
θB
Z
Z TA
Z
Z
TB Z
Z
Z
w
Solution: The block is not moving, so the net force on it must be zero. The net force
in the horizontal direction is the sum of the horizontal components of the two tensions:
TA cos θA pulling leftward; and TB cos θB pulling rightward. The net force in the vertical
direction is the sum of the weight w, pulling downward; and the vertical components of
the two tensions, TA sin θA and TB sin θB , pulling upward. Thus we get two equations
that we can solve for the two variables TA and TB :
TA cos θA = TB cos θB
TA sin θA + TB sin θB = w
We want to solve for TB , so we will begin by solving the first equation for TA , then
substitute into the second equation.
TA cos θA = TB cos θB
substitute
−−−−−→
÷ cos θ
A
−−−−→
TA =
TB cos θB
cos θA
TB cos θB sin θA
+ TB sin θB = w
cos θA
× cos θ
A
−−−−→
TB cos θB sin θA + TB sin θB cos θA = w cos θA
factor
−−−→ TB (cos θB sin θA + sin θB cos θA ) = w cos θA
w cos θA
divide
−−−→ TB =
cos θB sin θA + sin θB cos θA
(100 N) cos 30◦
=
cos 60◦ sin 30◦ + sin 60◦ cos 30◦
√
= 50 3 N
Problem 12. In your physics lab, you find the
apparatus shown at right. A glider with a weight of
m1 is resting on a horizontal air-track with the air
off. A light string runs horizontally from the glider,
over a massless frictionless pulley, and down to a
hanging block with a weight of m2 . The coefficient
of static friction between the glider and the air-track
is µs , and the coefficient of kinetic friction between
the glider and the air-track is µk . The coefficients of
friction are assumed known. After the glider-stringblock system have been released to move freely, and
before the hanging block has hit the floor, what is
the tension T in the string?
m1
(a)
(1 + µk )g
m1 + m2
m1 m2
(1 − µk )g
(c)
m1 + m2
m1
µs > µk
m2
m1 m2
*(b)
(1 + µk )g
m1 + m2
m1 + m2
(d)
(1 + µk )g
m1 m2
Solution: First, you should determine a condition between m1 and m2 that assures
us that the system moves when the glider-block system is released. The maximal static
friction of the glider on the air-track is fk = µs N1 = µs m1 g. That is, we need m2 g >
µs m1 g ⇒ m2 > µs m1 . This is our condition for motion.
Assuming the system will move, we can determine the forces on it. We will first analyze
the forces on the two separate subsystems: mass m1 and m2 , and then we will connect
them by examining the combined system: glider, massless string, and hanging-mass. We
will connect the forces on the systems by observing that because the string is taunt and
assumed not to stretch, that the velocity and the acceleration of the two masses must be
the same.
The glider m1 : If we take the x1 -axis for m1 to point in the
direction of motion and the y1 -axis to point upward, then applying
Newton’s second law we arrive at the system of equations:
N1
6
T1
Fnet1,x = T1 − fk = m1 a1,x ,
Fnet1,y = N1 − m1 g = m1 a1,y = 0 ,
f
u- k
m1 g
?
where we’ve used the fact that the glider does not move in the
vertical direct, so ay,1 = 0 (see free-body diagram).
free-body diagram
for mass m1
From the second equation we have
N1 = m1 g
⇒
fk = µk N1 = µk m1 g .
Upon substituting this equation into the first equation we have
T1 = fk + m1 a1,x = µk m1 g + m1 a1,x .
The hanging mass m2 : Since the motion of m2 is pure vertical, we only need a y2 axis to describe the motion. We take the y2 -axis pointing downward in the direction of
motion. Applying Newton’s second law to mass m2 , we arrive at the system of equations:
Fnet1,y = m2 g − T2 = m2 a2,y
⇒
T2 = m2 g − m2 a2,y .
By hypothesis, the pulley is massless and frictionless, so it only serves to redirect the
forces. Thus, T1 = T2 and we have
T1 = T = µk m1 g + m1 a1,x ,
T2 = T = m2 g − m2 a2,y .
Equating these equations through T and using the fact that since the glider and hanging
mass are connected by a string, they move as one and therefore have the same magnitude
of acceleration (i.e., a1,x = a2,y = a) gives the following equation for a
µk m1 g + m1 a = T = m2 g − m2 a
re-arraging terms
−−−−−−−−−→ (m1 + m2 )a = (m2 − µk m1 )g
m2 − µk m1
÷ (m1 +m2 )
g
−−−−−−→ a =
m1 + m2
We can now use a to determine T .
T = m2 g − m2 a
m 2 − µk m 1
= m2 g − m2
g
m1 + m2
(m1 m2 + m22 ) − (m22 − µk m1 m2 )
=
g
m1 + m2
m1 m2
=
(1 + µk )g
m1 + m2
As a check on our formulas that we found for acceleration and tension, let’s ask what
happens if the mass of the glider goes to zero (i.e., m1 → 0+ )? Then a = g and T = 0
as would be expected since the hanging mass would then be in free fall. Notice that we
cannot let m2 go to zero because of the constraint m2 > µs m1 > 0.
Applications of uniform circular motion
Centripetal force in the horizontal plane
Problem 13. A car goes around a banked curve on an icy highway. What is the source
of the centripetal force?
(a) Kinetic energy
*(c) Normal force
(e) None of these
(b) The car’s engine
(d) There is no centripetal force
Solution: The highway is described as icy, which suggests that there’s no friction. The
curve is banked, so there’s a normal force directed toward the inside of the curve. Kinetic
energy is not a force at all; and since the car is going around a curve, there must be some
centripetal force.
Problem 14. A car is moving down a straight and level road. The driver comes to a
stop sign, at which he slows down but does not stop completely, then speeds up again.
What is the source of the centripetal force?
(a) Kinetic friction
(c) Normal force
(e) None of these
(b)
*(d)
Air resistance
There is no centripetal force
Solution: The car is moving in a straight line. Since it is not moving in a curve, there
is no centripetal force.
Copyright ©Wayne Hacker 2010. All rights reserved.10
Problem 15. (Similarity Problem) An object with a mass of M is on a frictionless
horizontal surface. The object is attached to a horizontal string with length R whose
other end is attached to a pivot fixed in the surface; the object is then set to moving in
a horizontal circle at a constant speed V . Under these circumstances, the tension in the
string is T . If the speed of the object is doubled to a constant 2V , and the length of the
string is increased to 2R, what will be the tension in the string?
√
2T
(a) T
(b)
*(c) 2T
(d) 4T
(e) None of these
Solution: The tension in the string is the sole source of centripetal acceleration; so the
tension equals M arad . When the speed of the object is V and the radius is R, the tension
is T = M V 2 /R. When the speed and the radius are both doubled, it becomes
Tnew
4M V 2
M (2V )2
=
= 2T
=
2R
2R
Centripetal force in the vertical plane
Problem 16. A block with mass M is whirled on the end of a thin rigid rod that is
attached to the axle of a motor so that it moves at a constant speed in a vertical circle
with radius R. At the top of the circle, the tension in the rod is twice the weight of the
block. What is the speed of the block?
√
√
*(b) √3gR
(a) √2gR
(c) 2 gR
(d) 4 gR
(e) None of these
Solution: At the top of the circle, two forces are pulling straight down on the block:
gravity, pulling downward with the weight of the block W = M g; and tension in the
rod at the top of the circle Ttop , pulling downward with twice the weight of the block.
The sum of these two is the centripetal force: Frad = W + Ttop = 3W = 3M g. Since
Frad = M arad , it follows that the centripetal acceleration is 3g. Hence
arad = 3g =
v2
R
⇒
v=
p
3gR
Force of gravitational attraction
Problem 17. A space probe has a mass of 10 kg. It is sent to the surface of Planet X,
whose radius is 2 × 106 m. On the planet’s surface, the probe’s weight is 100 N. What is
the mass of Planet X? For ease of calculation, assume that G = 23 × 10−10 N m2 /kg2 .
(a) 23 × 1023 kg
(c) 3 × 1023 kg
(b)
*(d)
2
3
× 1023 kg
6 × 1023 kg
Solution: The weight of an object is the force of gravity on it. We know the gravitational
force, the mass of the probe mp , and the radius of the planet; we want to know the mass
of the planet mX . We can use the formula for gravitational force and solve for mX .
Gmp mX
wr2
·(r2 /Gmp )
−
−
−
−
−
−
→
m
=
X
r2
Gmp
6
2
(100 N)(2 × 10 m)
= 6 × 1023 kg
mX =
(2/3 × 10−10 N m2 /kg2 )(10 kg)
Fg = w =
Orbits
Problem 18. A satellite has a mass of 6 kg. It is in circular orbit around a planet
with a mass of 6 × 1024 kg. The radius of the planet is 106 m; the radius of the orbit
is 108 m. What is the satellite’s orbital period? For ease of calculation, assume that
G = 32 × 10−10 N m2 /kg2 .
(a) π × 102 s
(c) π × 1014 s
(e) None of these
*(b)
(d)
π × 105 s
π × 1017 s
Solution: The formula for the orbital period is
2πr3/2
T =√
GM
where T is the period, r is the radius of the orbit, G is the gravitational constant, and
M is the mass of the primary (in this case, the planet). The mass of the satellite and
the radius of the planet don’t enter into the formula at all, and have nothing to do with
this problem. Hence:
2π(108 m)3/2
2πr3/2
=
T =√
1/2
GM
( 23 × 10−10 N m2 /kg2 )(6 × 1024 kg)
We’ll look at the numbers first, then check the units and make sure that they work.
T =
2π × 1012
2π × 1012
=
= π × 105
[4 × 1014 ]1/2
2 × 107
Now, we need to check the units. In the numerator, the units are m3/2 . In the denominator, remember that 1 N = 1 kg m/s2 . Hence the units of the denominator are:
N m2 kg
·
1
kg2
1/2
kg2 m3
=
kg2 s2
1/2
=
m3/2
s
Thus the units of the whole formula (numerator over denominator) are:
m3/2
=s
m3/2 /s
Our solution is: T = π × 105 s.
Applications of Newton’s laws to rotating bodies
Moment of inertia
Problem 19. A system consists of a massless turntable with a lead block with mass m1
attached at a distance of r1 from the center. A second system has the same moment
of inertia, but uses a block with mass m2 = 2m1 . At what distance r2 is this block
attached?
(a) r1 /4√
(b) r1 /2
*(c) r1 / 2
(d) r1
(e) None of these
Solution: We assume that the blocks can be treated as point masses.
I1 = m1 r12
and I2 = I1 = m2 r22 = (2m1 )r22
⇒
r12 = 2r22
⇒
r1
r2 = √
2
Torque
Problem 20. A massless horizontal pole projects 2 m from the side
of a building. It is held up by a cable running from the side of
the building to the end of the pole, making an angle of 30◦ to the
horizontal. A sign weighing w = 30 N hangs from the pole 1.5 m out
from the building. What is the tension in the cable? Round your
answer to two significant figures.
(a) 30 N
(c) 40 N
(e) None of these
HH
HH
(b) 35 N
*(d) 45 N
HH
H
w
Solution: Since the system is not moving, the torques produced by the weight and by
the tension must match. Let T be the tension.
T rT sin θT = wrw
⇒
T =
wrw
(30 N)(1.5 m)
=
= 45 N
rT sin θ
(2 m) sin 30◦
Problem 21. A pendulum consists of a lead sinker with weight W
attached to the end of a thin string with length L; the other end of
the string is attached to a hook in the ceiling. The pendulum is pulled
back and released. When the string makes an angle of θ with the
vertical, it has a tension of T . At that point, what is the net torque
about the hook in the ceiling?
(a) 0
*(c) W L sin θ
(e) None of these
(b) (T + W ) sin θ
(d) (T − W )L cos θ
A
A
θA
L
A
A
A
Au
θ
W?
Solution: The tension is pulling straight away from the hook, so it makes no contribution to the torque: r = 0. The weight W is pulling straight downward, at an angle of
180◦ − θ to the string and at a distance of L from the hook. Hence
τ = W L sin(180◦ − θ) = W L sin θ

Instructor: Hacker Course: Physics 121 Sample Exam 4 Solutions (Newton’s Laws)

Transcription

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