Download chapter 7 blm answer key

CHAPTER 7 BLM ANSWER KEY
BLM 7-1: Section 7.1 Review/
Reinforcement
The speed of the cart at point B was 4.6 m/s.
W  33 J  ( 3.0 N)(4.0 m)
W  21 J
Answers
1.
2.
(a)
kinetic
W  E k
(b)
work-energy theorem
Ek  12 mvB 2
(c)
isolated system
vB 
(d)
decreases, increases, constant
2 E k
m
vB 
2(21 J)
2.0 kg
The boy did 919 J of work on the wagon.
vB  4.6
m
s
The speed of the cart at point C was 0 m/s.
W  21 J  (3.0 N)  7.0 m 
W 0 J
The question asks only for the work done by the
applied force, so the force of friction is not needed for
the calculation. In the equation for work used below,
F represents the magnitude of the force and d
represents the magnitude of the displacement.
W  Ek
v0
4. (a)
W  F d cos
The speed of the skier at the 60 m height (point B)
was 34 m/s.
ETA  ETB
W   60.0 N  20.0 m  cos 40.0
3.
m
s
W  919.25 J
EgA  EkA  EgB  EkB
W  919 J
mg hA  12 mvA2  mg hB  12 mvB2
vB  2 g  hA  hB 
The speed of the cart at point A was 5.7 m/s.
m

vB  2  9.81 2  120 m  60 m 
s


m
vB  34
s
W  12 (5 m)(6 N)  (3 m)(6 N)
W  33 J
W  Ek
Ek  12 mvA 2
2Ek
vA 
m
m
vA  5.7
s
(b)
The speed of the skier at the take-off point
(point C) was 42 m/s.
vB  2 g  hA  hB 
m

vB  2  9.81 2  120 m  30 m 
s 

m
vB  42
s
CHAPTER 7 BLM ANSWER KEY
(c)
The skier is moving at half of the take-off speed at
a height of approximately 98 m. Half of the takeoff speed is 21 m/s. Since the point is not known,
label the position “21.”
EgA  EkA  Eg21  Ek21
2
mg hA  12 mvA2  mg h21  12 mv21
2
g hA  12 v21
h21 
g
h21 
 9.81  120 m    21 
m
s2
1
2
m
s
2
9.81 sm2
Ek A  Eg B  Ek B  Eg A  Eg B  Ek B
1
2
mA v 2  mB g hB  12 mB v 2  mA g hA  0 J  0 J
 12 mA  12 mB  v 2  mA g hA  mB g hB
v 
v 
2  mA g hA  mB g hB 
mA  mB


BLM 7-2: Conservation of Total
Energy/Problem Solving
3. (a)
Answers
1. (a)
Ee  12 kx 2
N

Ee  12  70.0  (3.5 m) 2
m


2
ET  4.3  10 J
mv 2  F d
2 F d
m
v
(2)(225 N)(1.20 m)
0.30 kg
v  42
(b)
When the mass is at x = 0, Ee = 0 and
Ek  12 mvmax 2 . The maximum velocity is 17 m/s.
1
2
Eg  Ek at the arrow's highest point.
(b)
mg h  12 mv 2
v2
2g
 42.426 

m 2
s
2 9.81 sm2

h  92 m
2.
When the 75.0 kg box hits the floor, both boxes will
possess the same speed, assuming that the string does
not stretch:
5.4 m/s.
2(428.75 J)
3.0 kg
vmax  17
The arrow rises to a height of 92 m.
h 
mvmax 2  428.75 J
vmax 
m
s
h 
2
At x = 3.5 m, all of the energy is elastic potential
energy. So, the total energy of the system is
4.3  102 J.
Ek  W
v
m


 (3.0 m)  (25.0 kg)  9.81  (3.0 m) 


s 
s 
2
m
s
Arrow’s velocity as it leaves the bow:
42 m/s
1
2
m
75.0 kg  25.0 kg
v  5.4
h21  98 m


2 (75.0 kg)  9.81
m
s
The total energy of the system at any point x is
given as follows, where A is the amplitude of
the system.
1
2
mkA2  12 mv 2  12 kx 2
v
k 2
( A  x2 )
m
At 0 m displacement
v
70.0 mN
(3.52  0.02 )
3.0 kg
v  17
m
s
CHAPTER 7 BLM ANSWER KEY
Ee  12 kx 2
At 2.0 m displacement
v
v  14
m
s
ET  Ek  Ee  428.75 J
At 3.5 m displacement
v
v0
4.
N
m
70.0
(3.52  3.52 )
3.0 kg
m
s
The ± sign above indicates that the mass could be
moving to the right or to the left at this instant.
(c)
The acceleration when the displacement is 0.0 m is
F
a
m
 kx
a
m
70
a
70
N
m
N
m
( 2.0 m)
3.0 kg
m
a  47 2
s
d h  vh t  vi t cos
5.64 m  vi t cos 45
5.64 m
(0.7071) t
Equation 1
(3.5 m)
a
3.0 kg
m
a  82 2
s
N
m
The ± sign above indicates that the mass could be
accelerating to the right or to the left at this instant.
The results for part (b) can be used to give
Ek  12 mv 2
m

Ek  12 (3.0 kg) 13.8744

s

Ek  2.9  102 J
m

0  1.30 m  vi t (sin 45 )  12  9.81 2  t 2
s 

2
Equation 2
Substitute vi from Equation 1 into Equation 2.
5.64 m
m

t (0.7071)  12  9.81 2  t 2
(0.7071)t
s


m 2
6.94 m  4.905 2 t
s
0  1.30 m 
t 
The acceleration when the displacement is 3.5 m is
70
The angle the gun makes with the desktop is 45°. The
height of the end of the gun is 1.30 m. When the ball
hits the ground, d v  0.
d v  h  vv t  12 g t 2
The acceleration when the displacement is 2.0 m is
a
If a projectile starts from rest from a point above the
ground, then at time t, its horizontal and vertical
components of position will be as shown below. The
gun’s spring constant is 6.0 N/m.
vi 
(0.0 m)
3.0 kg
m
a0 2
s
(d)
N

2
Ee  12  70.0
 (2.0 m)
m


Ee  140 J
70.0 mN
(3.52  2.02 )
3.0 kg
6.94 m
4.905 sm2
t  1.19 s
Substitute t = 1.19 s into Equation 1 to find vi, the
initial velocity of the ball.
5.64 m
(0.7071)1.19 s
m
vi  6.70
s
vi 
The kinetic energy of the ball as it leaves the gun is
equal to the elastic potential energy of the gun’s
spring. Solve for the spring constant.
CHAPTER 7 BLM ANSWER KEY
Ee  Ek
1
2
Edme Mariotte
kx  mv
2
1
2
(1620–1684)
2
mv 2
x2
0.00056 kg(6.70
k
(0.065 m) 2
N
k  6.0
m
k
m 2
s
)
BLM 7-3: The Law of Conservation of
Momentum/Science Inquiry
Answers
The following is some information on the scientists
selected.
French physicist Mariotte demonstrated the conclusions of John Wallis on the
conservation of momentum by conducting a series of experiments on
pendulums allowed to strike each other. The following Internet iste provides
more information on Mariotte. www.newadvent.org/cathen/09671a.htm
Isaac Newton
(1642–1727)
The law of conservation of momentum follows directly from Newton’s second
and third laws. Newton described the momentum experiments made by
Huygens, Mariotte, Wren, and Wallis and his own careful experiments in his
book, the Principia. Newton’s experiments consisted of verifying the law of
conservation of momentum by carefully correcting for air resistance of colliding
pendulums. More information about Newton can be found on the following
Internet site.
www.arts.richmond.edu/~rubin/pedagogy/131/
131notes/131notes_130.html
Scientist
Role
John Wallis
(1616–1703)
John Wallis was one of the founders of the Royal Society. He was also one of
the first scientists to suggest the law of the conservation of momentum when,
in 1668, the Royal Society accepted the challenge to investigate the behaviour
of colliding bodies. Wallis is credited with the discovery of the law of
momentum — even though Wren and Huygens sent in similar solutions to the
Royal Society — because Wallis considered perfectly inelastic bodies and the
other two scientists investigated only elastic bodies. Further information on
Wallis can be found on the Internet at
www.maths.tcd.ie/pub/HistMath/People/Wallis/
RouseBall/RB_Wallis.html
Christopher Wren
(1632–1723)
Along with John Wallis and Christiaan Huygens, Christopher Wren produced
similar results related to the conservation of momentum. More details about
Wren’s work can be found on the Internet at
www.arct.cam.ac.uk/~Campbell/phd/wren/science.html
Christiaan Huygens
(1629–1695)
Huygens conducted experiments that demonstrated how momentum (mv) is
conserved in physical interactions. He also showed that mv 2 (2Ek) was
conserved. This was a first step in working out the law of conservation of
energy. More information is posted on the Internet at
www.clas.ufl.edu/users/rhatch/08sr-hygns-bio.html
2. (a)
Speed of the roller coaster at point B: 12.7 m/s
Ek  Eg  Eheat  Ek  Eg
1
2
mvB 2  mg hB  Eheat  12 mvA 2  mg hA
1
2
mvB 2  mg hB  Eheat  12 mvA 2  mg hA
vB 
vB 
2(mg hB  Eheat  12 mvA 2  mg hA )
m
2  (6.1803  10 4

kg  m
s
2
2
)  (1.24  105 J)  (1.1250  10 4 J)  (2.4721  105
kg  m
s
2
2
)

900.0 kg
vB  12.7
m
s
BLM 7-4: Concept Review/Reinforcement
Answers
1.
The average force exerted is 2.3  103 N, acting in
the opposite direction of your head’s motion. This is a
very large force, and illustrates how dangerous it can
be to bump your head.
F t  p
p  m(vf  vi )
m
 m
p  (4.5 kg)  0
 2.5 
s
 s
kg  m
p  11.25
s
p
F
t
11.25 kgsm
F
0.0050 s
F  2.3  103 N
CHAPTER 7 BLM ANSWER KEY
2.
The momentum of the players before the collision
equals the momentum of the players after the
collision. The defender was moving toward the other
player at a velocity of –3.4 m/s.
mp vp  md vd  0
vd 
vd 
 mp vp
md

vd  3.4
3.

 85 kg (4.0
m
s
)
100 kg
m
s
The initial momentum of the tool and astronaut is
equal to the final momentum. The astronaut is moving
at 0.093 m/s[forward] toward the spaceship.
0  ma va  mt vt
 mt vt
va 
ma
va 
  0.70 kg  (12.0 ms )
90 kg
m
va  0.093 [forward]
s
4.
The law of the conservation of momentum states that
the sum of the momenta of two objects before a
collision is equal to the sum of their momenta after
they collide. As shown below, momentum is
conserved in collisions (A) and (B) because the vector
sum of the momenta of the balls is equal before and
after the collisions.
5.
The total momentum before the collision is equal to
the total momentum after the collision. Sphere B
moves at a velocity of 3.23 m/s[E46.6°S]. Refer to
the diagram and calculations below.
CHAPTER 7 BLM ANSWER KEY
Momentum in the x direction
mvA x  0  mvA x  mvB x
2. (a)
Ek  Eg  Eheat  Ek  Eg
vB x  vA x  vA x
m
m
 (4.20 cos34.0 )
s
s
m
vB x  2.218
s
vB x  5.70
Momentum in the y direction
0  mvA y  mvB y
1
2
mvB 2  mg hB  Eheat  12 mvA 2  mg hA
1
2
mvB 2  mg hB  Eheat  12 mvA 2  mg hA
vB 
vB 
vB y  vA y
m
s
Pythagorean theorem and the tangent function
2
vB  vB x 2  vB y 2

(b)
mechanical
(c)
positive, increases
(d)
conservative
(e)
third
(f)
equal
(g)
impulse
4
3.
5
kg  m
s
0  12 (900.0 kg)(12.7 ms ) 2
10.0 m
F  7.26  103 N
The water going over the falls will increase in
temperature by approximately 2.3°C.
W  Eg
Qgained  W
opposite
5
)  (1.24  10 J)  (1.1250  10 J)  (2.4721  10
F
W  mg h
Answers
2
F d  E k f  E k i
m
s
BLM 7-5: Chapter 7 Test/Assessment
s
2
2
)
The brakes do work on the roller coaster to reduce
its kinetic energy to zero after 10.0 m. The
frictional braking force is in the opposite direction
of the motion and is 7.26  103 N.
2.218
  46.6 clockwise from the x- axis
kg  m
m
s
m
s
  tan 1.0586
4
900.0 kg
2.348
1
1. (a)
2   (6.1803  10
2
W  Ek
2
Positive x- and negative y -components
place the resultant in the fourth quadrant.
tan  
m
(b)
vB y  2.348
2
2(mg hB  Eheat  12 mvA 2  mg hA )
vB  12.7
vB y  4.20sin 34.0
m 
m
2

vB   2.218    2.348 
s 
s

m
vB  3.23
s
vB y
tan  
vB x
Speed of the roller coaster at point B: 12.7 m/s
m

Qgained  (1.0 kg)  9.81 2  (979.0 m)
s 

Qgained  9603.99 J
mcT  Qgained
T 
Qgained
T 
9603.99 J
1.0 kg(4186 kgJ C )
mcw
T  2.3 C

CHAPTER 7 BLM ANSWER KEY
4.
The average force of the racquet on the ball was
1.9  102 N.
F t  p
F
F
m  vf  vi 
t
 0.057 kg  10 ms    13 ms 
7.00  103 s
F  1.9 N
5.
Cars are made with bumpers that retract during a
collision in order to increase the time of the collision,
thus reducing the force of the impact