 ```lOMoARcPSD|17314538
AP Physics 1: Algebra-Based (High School - USA)
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jonnalagadda (saj2436) – Work &amp; CoE – ayida – (4355-3)
This print-out should have 40 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
001 10.0 points
The Joule and the kilowatt-hour are both
units of energy.
19.2 kW &middot; h is equivalent to how many
Joules?
1
Explanation:
Let : F = 709 N ,
d = 9 m , and
t = 14 s .
The power expended is
Correct answer: 6.912 &times; 107 J.
W
Fd
=
t
t
(709 N) (9 m)
=
14 s
= 455.786 W .
P =
Explanation:
1 W = 1 J/s
1000 W 3600 s
&middot;
kW
1h
7
= 6.912 &times; 10 W &middot; s
19.2 kW &middot; h = (19.2 kW &middot; h) &middot;
= 6.912 &times; 107 J .
002 10.0 points
A cheerleader lifts his 59.9 kg partner straight
up off the ground a distance of 0.557 m before
releasing her.
The acceleration of gravity is 9.8 m/s2 .
If he does this 27 times, how much work has
he done?
Explanation:
The work done in lifting the cheerleader
once is
004 10.0 points
You leave your 75 W portable color TV on for
7 hours each day and you do not pay attention
to the cost of electricity.
If the dorm (or your parents) charged you
for your electricity use and the cost was
\$0.1 /kW &middot; h, what would be your monthly
(30 day) bill?
Explanation:
Let :
P = 75 W and
t = 7 h/day ,
W1 = m g h
= (59.9 kg)(9.8 m/s2 )(0.557 m)
= 326.97 J .
The energy consumed in each day is
The work required to lift her n = 27 times is
= (75 W) (7 h/day) &middot;
W = n W1 = (27)(326.97 J) = 8828.19 J.
= 0.525 kW &middot; h/day.
W =Pt
kW
1000 W
In 30 days, you would use
003 10.0 points
A student weighing 709 N climbs at constant
speed to the top of an 9 m vertical rope in
14 s.
What is the average power expended by the
student to overcome gravity?
(30 day) (0.525 kW &middot; h/day) = 15.75 kW &middot; h,
which would cost you
(15.75 kW &middot; h) (\$0.1/kW &middot; h) = \$1.575 .
lOMoARcPSD|17314538
jonnalagadda (saj2436) – Work &amp; CoE – ayida – (4355-3)
005 10.0 points
A 105 kg physics professor has fallen into
the Grand Canyon. Luckily, he managed
to grab a branch and is now hanging 87 m
below the rim. A student (majoring in linguistics and physics) decides to perform a
rescue/experiment using a nearby horse. After lowering a rope to her fallen hero and
attaching the other end to the horse, the student measures how long it takes for the horse
to pull the fallen physicist to the rim of the
Grand Canyon.
The acceleration of gravity is 9.8 m/s2 .
If the horse’s output power is truly 1 horsepower (746 W), and no energy is lost to friction, how long should the process take?
Explanation:
The work against gravity which is necessary
to raise the professor to the rim of the canyon
is given by
W = m g ∆h
= (105 kg) (9.8 m/s2 ) (87 m)
= 89523 J .
If the horse’s power output is one horsepower,
then the the time for the horse to provide the
necessary work is given by
tup =
5. heat
Explanation:
007 10.0 points
A Joule is a measure of
1. density.
2. distance.
3. volume.
4. momentum.
5. energy. correct
Explanation:
008 10.0 points
Shawn and his bike have a total mass of
58.7 kg. Shawn rides his bike 0.77 km in
13.9 min at a constant velocity.
The acceleration of gravity is 9.8 m/s2 .
What is Shawn’s kinetic energy?
Explanation:
Shawn’s velocity is
v=
W
Poutput
89523 J
=
746 W
= 120.004 s .
006 10.0 points
Potential energy and kinetic energy are forms
of what kind of energy?
1. chemical
2
d
.
t
Thus his kinetic energy is
1
K = mv 2
2
0.77 km 1000 m 1 min 2
1
&middot;
&middot;
= (58.7 kg)
2
13.9 min 1 km
60 s
= 25.0184 J .
3. mechanical correct
009 10.0 points
Tim, with mass 71.6 kg, climbs a gymnasium
rope a distance of 2 m.
The acceleration of gravity is 9.8 m/s2 .
How much potential energy does Tim gain?
4. electromagnetic
2. nuclear
lOMoARcPSD|17314538
jonnalagadda (saj2436) – Work &amp; CoE – ayida – (4355-3)
Explanation:
Let : m = 71.6 kg ,
h = 2 m , and
g = 9.8 m/s2 .
Potential energy is
U = mgh
= (71.6 kg) (9.8 m/s2 ) (2 m)
= 1403.36 J .
010 (part 1 of 3) 10.0 points
A 40.0 kg child is in a swing that is attached
to ropes 2.30 m long.
The acceleration of gravity is 9.81 m/s2 .
Find the gravitational potential energy associated with the child relative to the child’s
lowest position under the following conditions:
a) when the ropes are horizontal.
Explanation:
Given:
θ = 36.0◦
Solution:
The child is a distance ℓ cos θ from the attachment point, so h2 = ℓ − ℓ cos θ and
Ug = mgh2
= mgℓ(1 − cos θ)
= (40 kg)(9.81 m/s2 )(2.3 m) &middot; (1 − cos 36◦ )
= 172.366 J
012 (part 3 of 3) 10.0 points
c) at the bottom of the circular arc.
Explanation:
Solution: The child is at the lowest point,
so h3 = 0, and
Ug = mgh3
= (40 kg)(9.81 m/s2 )(0 m)
=0J
Explanation:
Basic Concept:
013
Ug = mgh
Given:
ℓ = 2.30 m
m = 40.0 kg
g = 9.81 m/s2
Solution:
The child is at a height h1 = ℓ from the
lowest point, so
Ug = mgh1
= (40 kg)(9.81 m/s2 )(2.3 m)
= 902.52 J
011 (part 2 of 3) 10.0 points
b) when the ropes make a 36.0◦ angle with
the vertical.
3
10.0 points
A block of mass m slides on a horizontal
frictionless table with an initial speed v0 . It
then compresses a spring of force constant k
and is brought to rest.
v
k
m
m
&micro;=0
How much is the spring compressed x from
its natural length?
mk
g
r
m
correct
2. x = v0
k
v2
3. x = 0
2g
k
4. x = v0
gm
1. x = v0
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jonnalagadda (saj2436) – Work &amp; CoE – ayida – (4355-3)
r
Anyone who checks to see if the units are
correct should get this problem correct.
014 10.0 points
A bead slides without friction around a loopthe-loop. The bead is released from a height
of 11.6 m from the bottom of the loop-theloop which has a radius 3 m.
The acceleration of gravity is 9.8 m/s2 .
From conservation of energy, we have
K i + Ui = K f + Uf
m v2
+ m g (2 R)
0 + mgh =
2
v 2 = 2 g (h − 2 R) .
Therefore
p
2 g (h − 2 R)
r
h
i
= 2 (9.8 m/s2 ) 11.6 m − 2 (3 m)
v=
= 10.4766 m/s .
015 (part 1 of 3) 10.0 points
A block starts at rest and slides down a frictionless track. It leaves the track horizontally,
flies through the air, and subsequently strikes
the ground.
570 g
v
b b b b
b b
2.4 m
1
1
m v02 = Ei = Ef = k x2 , or
2
2
m 2
2
x =
v , therefore
k r0
m
.
x = v0
k
Let : R = 3 m and
h = 11.6 m .
4.9 m
k
m
mg
6. x = v0
k
m
7. x = v0
kg
s
k
8. x = v0
mg
r
mg
9. x = v0
k
2
v
10. x = 0
2m
Explanation:
Total energy is conserved (no friction). The
spring is compressed by a distance x from its
natural length, so
5. x = v0
3m
b
b
b
Explanation:
Explanation:
b
b
What is the speed of the ball when it leaves
the track? The acceleration of gravity is
9.81 m/s2 .
What is its speed at point A ?
b
x
A
11.6 m
4
Let : g = −9.81 m/s2 ,
m = 570 g , and
h1 = 2.5 m .
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jonnalagadda (saj2436) – Work &amp; CoE – ayida – (4355-3)
h1
m
017 (part 3 of 3) 10.0 points
What is the speed of the block when it hits
the ground?
v
b b b b
b b
b
b
h2
h
5
b
b
b
Explanation:
b
x
Choose the point where the block leaves the
track as the origin of the coordinate system.
While on the ramp,
K b = Ut
1
m vx2 = −m g h1
2
vx2 = −2 g h1
p
vx = −2 g h1
q
= −2 (−9.81 m/s2 ) (2.5 m)
Let :
h = 4.9 m .
Now choose ground level as the origin. Energy conservation gives us
K f = Ui
1
m vf2 = −m g h
2
p
vf = −2 g h
q
= −2 (−9.81 m/s2 ) (4.9 m)
= 9.805 m/s .
= 7.00357 m/s .
Alternate Solution:
016 (part 2 of 3) 10.0 points
What horizontal distance does the block
travel in the air?
Explanation:
h2 = −2.4 m .
Let :
With the point of launch as the origin,
h2 =
t=
1 2
gt
2
s
2 h2
.
g
Thus
x = vx t = vx
s
2 h2
g
s
= (7.00357 m/s)
= 4.89898 m .
2 (−2.4 m)
−9.81 m/s2
p
−2 g h2
q
= −2 (−9.81 m/s2 ) (2.4 m)
vy =
= 6.86207 m/s , so
q
vf = vx2 + vy2
q
= (7.00357 m/s)2 + (6.86207 m/s)2
= 9.805 m/s .
018 (part 1 of 3) 10.0 points
A block is pushed against the spring with
spring constant 9.1 kN/m (located on the lefthand side of the track) and compresses the
spring a distance 4.5 cm from its equilibrium
position (as shown in the figure below).
The block starts at rest, is accelerated by
the compressed spring, and slides across a frictionless track (as shown in the figure below).
It leaves the track horizontally, flies through
the air, and subsequently strikes the ground.
The acceleration of gravity is 9.81 m/s2 .
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6
9.1 kN/m
v
522 g
2.4 m
b b b b
b b
4.5 cm
9.81 m/s2
Solution: Using Eqs. 1, 2, and 3, we have
b
b
b
b
b
1
1
m vx2 = k d2
2
2
(4)
k d2
m
r
k d2
vx =
s m
vx2 =
b
x
What is the speed v of the block when it
leaves the track?
=
Explanation:
(5)
(6)
(9100 N/m) (0.045 m)2
0.522 kg
= 5.94153 m/s .
Let : g = 9.81 m/s2 ,
m = 0.522 kg ,
k = 9100 N/m ,
vx = v , and
d = 0.045 m .
019 (part 2 of 3) 10.0 points
What is the horizontal distance x the block
travels in the air?
Explanation:
At ∆y = h = −2.4 m (below the jump off
height),
5.94 m/s
b b b b
b b
b
b
h
d
g
m
k
Basic Concepts:
chanical Energy
b
b
b
b
4.16 m
Conservation of Me-
Ui = Uf + K f ,
(1)
since vi = 0 m/s.
K=
1
m v2
2
(2)
1
(3)
Us = k d 2 .
2
Choosing the point where the block leaves the
track as the origin of the coordinate system,
∆x = vx ∆t
1
∆y = − g ∆t2 ,
2
since axi = 0 m/s2 and vyi = 0 m/s.
1
h = − g t2
s2
−2 h
.
t=
g
(7)
(9)
Since x = vx t , therefore using Eq. 6 and 8,
we have
x = vx t
s
r
k d2
−2 h
=
m
g
s
2 k d2 h
= −
mg
2 (9100 N/m)
= −
0.522 kg
(0.045 m)2 (−2.4 m)
&times;
9.81 m/s2
= 4.15608 m .
(10)
1/2
020 (part 3 of 3) 10.0 points
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7
What is the total speed of the block when it
hits the ground?
Explanation:
Explanation:
Basic Concept:
v=
K f = Ui ,
since vi = 0 m/s and hf = 0 m.
1
Solution: Using m vy2 = m g h , we have
2
p
vy = −2 g h
(11)
q
= −2 (9.81 m/s2 ) (−2.4 m)
= 6.86207 m/s , so
q
(12)
vf = vx2 + vy2
q
= (5.94153 m/s)2 + (6.86207 m/s)2
= 9.07688 m/s .
021 (part 1 of 2) 10.0 points
While skiing in Jackson, Wyoming, your
friend Ben (of mass 72.9 kg) started his descent down the bunny run, 20.6 m above the
bottom of the run.
If he started at rest and converted all of
his gravitational potential energy into kinetic
energy, what is Ben’s kinetic energy at the
bottom of the bunny run?
m = 72.9 kg ,
g = 9.8 m/s2 ,
h = 20.6 m .
2K
s m
2 (14717.1 J)
72.9 kg
= 20.0938 m/s .
023 (part 1 of 4) 10.0 points
A 162 g particle is released from rest at point
A along the diameter on the inside of a frictionless, hemispherical bowl of radius 47.5 cm.
The acceleration of gravity is 9.8 m/s2 .
A
R
C
B
2R/3
Calculate its gravitational potential energy
at point A relative to point B.
Explanation:
The potential energy at A is
Explanation:
Let :
=
r
and
Kbottom = Utop
= mgh
= (72.9 kg) (9.8 m/s2 ) (20.6 m)
= 14717.1 J .
022 (part 2 of 2) 10.0 points
What is his final velocity?
UA = m g hA
= (162 g)(0.001 kg/g)(9.8 m/s2 )
&times; (47.5 cm)(0.01 m/cm)
= 0.75411 J .
024 (part 2 of 4) 10.0 points
Calculate its kinetic energy at point B.
Explanation:
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jonnalagadda (saj2436) – Work &amp; CoE – ayida – (4355-3)
The kinetic energy at B is equal to the
potential energy at A, since all the energy
went to kinetic energy.
K B = UA
= 0.75411 J .
025 (part 3 of 4) 10.0 points
Calculate its kinetic energy at point C at
2R
.
height
3
Basic Concepts: Conservation of Mechanical Energy
Ui = K f
since vi = 0 m/s and hf = 0 m.
1
K = mv 2
2
Ug = mgh
hi = ℓ − ℓ cos θ
Given:
ℓ = 31.0 m
θ = 36.0◦
g = 9.81 m/s2
Explanation:
The kinetic energy at C is
KC = m g (hA − hC )
2R
= mg R −
3
UA
=
3
0.75411 J
=
3
= 0.25137 J .
Solution:
1
mghi = mvf2
2
2
vf = 2ghi
= 2gℓ(1 − cos θ)
= 2(9.81 m/s2 )(31 m)(1 − cos 36◦ )
= 116.16 m2 /s2
so that
vf =
026 (part 4 of 4) 10.0 points
Calculate its potential energy at point C.
q
116.16 m2 /s2
= 10.7777 m/s
028 (part 2 of 2) 10.0 points
b) pushes off with a speed of 6.00 m/s?
Explanation:
The potential energy at C is
UC = UA − K C
= 0.75411 J − 0.25137 J
= 0.50274 J .
027 (part 1 of 2) 10.0 points
Tarzan swings on a 31.0 m long vine initially
inclined at an angle of 36.0 ◦ with the vertical.
The acceleration of gravity if 9.81 m/s2 .
What is his speed at the bottom of the
swing if he
a) starts from rest?
Explanation:
8
Explanation:
Basic Concept:
Ui + K i = K f
since hf = 0 m.
Given:
vi = 6.00 m/s
Solution:
1 2
1
mvf = mghi + mvi2
2
2
vf2 = 2gℓ(1 − cos θ) + vi2
= 2(9.81 m/s2 )(31 m)
&middot; (1 − cos 36◦ ) + (6 m/s)2
= 152.16 m2 /s2
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jonnalagadda (saj2436) – Work &amp; CoE – ayida – (4355-3)
so that
A block of mass m slids up a ramp making
an angle θ with the horizontal. The block has
1
an initial KE of
m v02 as it starts up the
2
ramp and travels a distance L along the ramp
before coming to momentary rest.
q
vf = 152.16 m2 /s2
= 12.3353 m/s
029
10.0 points
A 200 kg block is released at a 4.2 m height
as shown. The track is frictionless. The block
travels down the track, hits a spring of force
constant k = 1623 N/m .
The acceleration of gravity is 9.8 m/s2 .
200 kg
4.2 m
1623 N/m
x
Determine the compression of the spring x
from its equilibrium position before coming to
rest momentarily.
Explanation:
L
v0
How much work did kinetic friction do on
the block between its starting point and the
point it came to momentary rest?
v02
correct
1. m g L sin θ −
2
2. Zero, since friction is perpendicular to
mg
3. m g L tan θ
4. m g (sin θ − cos θ)
5. m g L
Let : m = 200 kg ,
k = 1623 N/m ,
h = 4.2 m , and
g = 9.8 m/s2 .
From conservation of energy
∆E = Wf
or
6. m g L sin θ
1
7.
m v02 , since the block comes to rest:
2
W = ∆KE
8. m g L cos θ
Explanation:
k x2
− mgh = 0.
2
L
θ
Hence,
x=
=
r
2mgh
s k
2 (200 kg) (9.8 m/s2 ) (4.2 m)
(1623 N/m)
= 3.18499 m .
030
9
10.0 points
L sin θ
If we take the gravitational potential energy
to be zero at the point where the block has its
initial speed, then from E = K + Ug
Ei =
1
m v02 + 0 and
2
Ef = 0 + m g L sin θ .
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jonnalagadda (saj2436) – Work &amp; CoE – ayida – (4355-3)
The work done by friction, a nonconservative force, is thus
v02
Wf = Ef − Ei = m g L sin θ −
.
2
031 10.0 points
A 16.9 kg block is dragged over a rough, horizontal surface by a constant force of 81.6 N
acting at an angle of 29 ◦ above the horizontal. The block is displaced 95.4 m, and the
coefficient of kinetic friction is 0.176.
The acceleration of gravity is 9.8 m/s2 .
81.
6N
◦
29
10
= F sx cos θ
= (81.6 N) (95.4 m) cos 29◦
= 6808.6 J .
To find the frictional force, Ff riction = &micro; N ,
we need to find N from vertical force balance.
Note that N is in the same direction as the
y component of F and opposite the force of
gravity. Thus
F sin θ + N = m g
so that
N = m g − F sin θ .
Thus the friction force is
16.9 kg
~ f riction = −&micro; N ı̂ = −&micro; (m g − F sin θ)ı̂ .
F
&micro; = 0.176
The work done by friction is then
If the block was originally at rest, determine
its final speed.
Explanation:
Let : m = 16.9 kg ,
F = 81.6 N ,
sx = 95.4 m ,
&micro; = 0.176 , and
g = 9.8 m/s2 .
~ f riction &middot; ~s = −|f&micro; | |s|
W&micro; = F
= −&micro; (m g − f&micro; sin θ) sx
h
= −(0.176) (16.9 kg) (9.8 m/s2 )
i
−(81.6 N) sin 29◦ (95.4 m)
= −2116.59 J .
The net work done on the block is equal to
the change in kinetic energy, so
1
m v 2 − 0 = WF + W&micro;
2
Consider the force diagram
N
F
θ
fk
r
2 [WF + W&micro; ]
v h m
i
u
u 2 (6808.6 J) + (−2116.59 J)
t
=
(16.9 kg)
v=
= 23.5641 m/s .
mg
~ &middot; ~s, where ~s is the distance
Work is W = F
traveled. In this problem ~s = 5ı̂ is only in the
x direction.
⇒ WF = Fx sx
032 10.0 points
The spring has a constant of 10 N/m and
the frictional surface is 0.4 m long with a
coefficient of friction &micro; = 1.56 . The 2 kg
block depresses the spring by 29 cm, then is
lOMoARcPSD|17314538
jonnalagadda (saj2436) – Work &amp; CoE – ayida – (4355-3)
released. The first drop is 1 m and the second
is 2 m.
The acceleration of gravity is 9.8 m/s2 .
11
the spring, then it gains kinetic energy by
dropping a distance h1 , then it loses kinetic
energy by doing work against friction.
k
m
x
Us + Ug − Wf r = Kf =
1
m v2
2
Since
h1
L
2 (Us + Ug − Wf r )
m
2 (0.4205 J + 19.6 J − 12.2304 J)
=
2 kg
2 2
= 7.7901 m /s
v2 =
&micro;
h2
s
How far from the bottom of the cliff does it
land?
then
v=
Explanation:
Basic concepts The potential energy of a
spring is
Kinematics gives the time for the mass to drop
a distance h2
1 2
gt
2
2 h2
t2 =
g
s
2 h2
t=
g
s
2 (2 m)
=
(9.8 m/s2 )
= 0.638877 s .
1
k x2
2
1
= (10 N/m) (29 cm)2
2
= 0.4205 J .
h2 =
Gravitational potential energy is
Ug = m g h
= (2 kg) (9.8 m/s2 ) (1 m)
= 19.6 J .
Wf r = &micro; m g L
= (1.56) (2 kg) (9.8 m/s2 ) (0.4 m)
= 12.2304 J .
Thus the mass will land a distance
s = vt
= (2.79108 m/s) (0.638877 s)
= 1.78315 m .
from the bottom of the cliff.
Kinetic energy is
K=
7.7901 m2 /s2
= 2.79108 m/s .
Us =
Work done against friction on a flat surface is
q
1
m v2 .
2
The height of the cliff determines how long it
takes for the mass to reach the bottom.
Energy considerations give us the horizontal velocity as the mass leaves the cliff top.
The mass gets its initial kinetic energy from
033 10.0 points
A solid metal ball and a hollow plastic ball
of the same external radius are released from
rest in a large vacuum chamber.
When each has fallen 1 m, they both have
the same
1. inertia.
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12
What average force is exerted on the car?
2. momentum.
3. speed. correct
Explanation:
4. kinetic energy.
Let :
5. change in potential energy.
Explanation:
All of the quantities mentioned above except terminal velocity depend upon mass.
Note: If this experiment had not taken
place in a vacuum chamber, none of these
034 10.0 points
A child bounces a 59 g superball on the sidewalk. The velocity change of the superball is
from 20 m/s downward to 16 m/s upward.
1
If the contact time with the sidewalk is
800
s, what is the magnitude of the average force
exerted on the superball by the sidewalk?
Momentum is defined as
∆p = m &middot; ∆v
= (972 kg) (30.5556 m/s − 0 m/s)
= 29700 kg &middot; m/s .
Let : t = 1.09 s .
Momentum and impulse are related by
∆p = F t ,
so
F =
Explanation:
Let :
m = 59 g = 0.059 kg ,
vu = 16 m/s ,
vd = 20 m/s , and
∆t = 0.00125 s
Choose the upward direction as positive.
The impulse is
I = F ∆t = ∆P
= m vu − m (−vd )
= m (vu + vd )
m (vu + vd )
F =
∆t
(59 g) (16 m/s + 20 m/s)
=
0.00125 s
= 1699.2 N .
035 10.0 points
A(n) 972 kg drag race car accelerates from
rest to 110 km/h in 1.09 s.
m = 972 kg and
v = 110 km/h = 30.5556 m/s .
29700 kg &middot; m/s
∆p
=
= 27247.7 N .
t
1.09 s
036 10.0 points
A(n) 680 N man stands in the middle of a
frozen pond of radius 12.1 m. He is unable to get to the other side because of a
lack of friction between his shoes and the
ice. To overcome this difficulty, he throws his
1.3 kg physics textbook horizontally toward
the north shore, at a speed of 8 m/s.
The acceleration of gravity is 9.81 m/s2 .
How long does it take him to reach the
south shore?
Explanation:
Let : Wm
r
mb
vb′
= 680 N ,
= 12.1 m ,
= 1.3 kg , and
= 8 m/s .
lOMoARcPSD|17314538
jonnalagadda (saj2436) – Work &amp; CoE – ayida – (4355-3)
13
The mass of the man is
3. decrease because of conservation of momentum. correct
Wm
mm =
.
g
4. increase because of conservation of momentum.
From conservation of momentum,
′
mm vm + mb vb = mm vm
+ mb vb′
′
0 = mm vm
+ mb vb′
5. remain the same because the raindrops
are falling perpendicular to the direction of
cart’s motion.
mb ′
v
mm b
g mb ′
=−
v
Wm b
9.81 m/s2 (1.3 kg)
(8 m/s)
=−
680 N
= −0.150035 m/s .
′
vm
=−
Explanation:
This is an inelastic collision in the direction
along which the cart is rolling. Only momentum ~p along that direction is conserved.
Because the raindrops fall vertically, they do
not carry momentum horizontally. Assume
∆m of rain water accumulates on the cart:
pi = pf
m v = (m + ∆m) v ′ .
The time to travel the 12.1 m to shore is
∆x
′ |
|vm
12.1 m
=
0.150035 m/s
Therefore
t=
m
v
m + ∆m
v′ &lt; v .
v′ =
= 80.6477 s .
The speed of the cart will decrease because
of conservation of momentum.
037 10.0 points
An open cart on a level surface rolls without
frictional loss through a vertical downpour of
rain, as shown below.
As the cart rolls, an appreciable amount of
rain water accumulates in the cart.
rain
038 10.0 points
Two ice skaters approach each other at right
angles. Skater A has a mass of 75 kg and
travels in the +x direction at 2.58 m/s. Skater
B has a mass of 38.2 kg and is moving in the
+y direction at 1.68 m/s. They collide and
cling together.
Find the final speed of the couple.
rain water
v
cart
Explanation:
From conservation of momentum ∆p = 0
mA vA ı̂ + mB vB ̂ = (mA + mB ) vf
The speed of the cart will
1. decrease because of conservation of mechanic energy.
2. increase because of conservation of mechanic energy.
Therefore
p
(mA vA )2 + (mB vB )2
vf =
mA + mB
p
(193.5 kg m/s)2 + (64.176 kg m/s)2
=
75 kg + 38.2 kg
= 1.80092 m/s
lOMoARcPSD|17314538
jonnalagadda (saj2436) – Work &amp; CoE – ayida – (4355-3)
039 10.0 points
A 1.12 kg skateboard is coasting along the
pavement at a speed of 6.33 m/s when a 0.660
kg cat drops from a tree vertically downward
onto the skateboard.
What is the speed of the skateboard-cat
combination?
Let : m1
m2
mf
v1
v2
p1
p2
p1 p2
Explanation:
Basic Concept: All of the motion related
to the momentum is horizontal:
px
m1~v1,i = (m1 + m2 )~vf
py
since v2,i = 0 m/s.
Given:
m1 = 1.12 kg
vi,1 = 6.33 m/s
m2 = 0.660 kg
θ
π−θ
m1
v1
14
= 7 kg ,
= 16 kg ,
= m1 + m2 = 23 kg ,
= 9.5 m/s ,
= 10 m/s ,
= m1 v1 = 66.5 kg m/s ,
= m2 v2 = 160 kg m/s ,
= m1 v1 m2 v2
= 21280 kg2 m2 /s2 ,
= p1 + p2 cos θ
= 217.783 kg m/s ,
= p2 sin θ
= 52.0909 kg m/s ,
= 19◦ , and
= 161◦ .
vf
mf
φ
θ
Solution:
m1 vi,1
vf =
m1 + m2
(1.12 kg)(6.33 m/s)
=
1.12 kg + 0.66 kg
= 3.98292 m/s
v2
m2
The final momentum is
pf = (m1 + m2 ) vf .
(1)
Momentum is conserved
040 10.0 points
A(n) 7 kg object moving with a speed of
9.5 m/s collides with a(n) 16 kg object moving
with a velocity of 10 m/s in a direction 19 ◦
from the initial direction of motion of the 7 kg
object.
7 kg
9.5 m/s
19◦
10 m
/s
16 kg
What is the speed of the two objects after
the collision if they remain stuck together?
~p1 + ~p2 = ~pf .
Using the law of cosines, we have
p21 + p22 − 2 p1 p2 cos(π − θ) = p2f .
Solving for vf , we have
q
p21 + p22 − 2 p1 p2 cos(π − θ)
vf =
m1 + m2
1
=
(7 kg) + (16 kg)
h
&times; (4422.25 kg2 m2 /s2 )2
+ (25600 kg2 m2 /s2 )2
2
Explanation:
(2)
2
2
◦
− (21280 kg m /s ) cos(161 )
= 9.73592 m/s .
i1/2
lOMoARcPSD|17314538
jonnalagadda (saj2436) – Work &amp; CoE – ayida – (4355-3)
Alternate Solution: Since
p2f = p2x + p2y ,
we have
vf =
q
p2x + p2y
m + m2
1
1
=
(7 kg) + (16 kg)
h
&times; (47429.4 kg2 m2 /s2 )
i1/2
+ (2713.46 kg2 m2 /s2 )
= 9.73592 m/s .