Clicker Quiz
 Assume that the rock is launched with an angle of θ = 45°.
With what angle with respect to the horizontal does the
rock strike the ground in front of the castle?
v0 = 14.2 m/s
v f = 18.5 m/s
a)
b)
c)
d)
e)
March 21, 2008
Physics for Scientists&Engineers 1
25.4°
37.9°
45.0°
57.1°
65.2°
1
Clicker Quiz Solution
v0 = 14.2 m/s
v f = 18.5 m/s
vx v0 cos"
cos ! =
=
vf
vf
$ v0 cos" '
#1 $ (14.2 m/s ) cos 45° '
! = cos &
= cos &
)
)(
%
v
18.5
m/s
%
(
f
#1
! = 57.1°
March 21, 2008
Physics for Scientists&Engineers 1
2
Clicker Quiz


A ball with mass m is thrown vertically into the air with an initial speed v.
Which of the following equations correctly describes the maximum height h of
the ball?
v
2g
• A)
h=
• B)
h=
• C)
h = 2mv / g
• D)
mv 2
h=
g
• E)
v2
h=
2g
March 21, 2008
g
1 2
v
2
Physics for Scientists&Engineers 1
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Clicker Quiz


A ball with mass m is thrown vertically into the air with an initial speed v.
Which of the following equations correctly describes the maximum height h of
the ball?
v
2g
• A)
h=
• B)
h=
• C)
h = 2mv / g
• D)
mv 2
h=
g
• E)
v2
h=
2g
March 21, 2008
g
1 2
v
2
1 2
mv = mgh
2
1 2
v = gh
2
v2
h=
2g
Physics for Scientists&Engineers 1
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Clicker Quiz
 A curling stone with mass 19.96 kg is given an initial
velocity of 2.46 m/s. The stone slides on the ice
with a coefficient of kinetic friction of 0.0109.
How far does the stone slide before it stops?
•
•
•
•
•
a) 18.7 m
b) 28.3 m
c) 34.1 m
d) 39.2 m
e) 44.5 m
March 21, 2008
Physics for Scientists&Engineers 1
5
Clicker Quiz Solution
 A curling stone with mass 19.96 kg is given an initial
velocity of 2.46 m/s. The stone slides on the ice
with a coefficient of kinetic friction of 0.0109.
How far does the stone slide before it stops?
•
•
•
•
•
a) 18.7 m
b) 28.3 m
c) 34.1 m
d) 39.2 m
e) 44.5 m
March 21, 2008
!K = W
1 2
mv = µ k mgx
2
2.46 m )
1 v
1
(
x=
=
= 28.3 m
2
2 µ k g 2 ( 0.0109 ) 9.81 m/s
2
2
(
Physics for Scientists&Engineers 1
)
6
Clicker Quiz
 A ballistic pendulum is used to measure the speed of a
bullet shot from a gun. The mass of the bullet is 50.0 g and
the mass of the block is 20.0 kg. When the bullet strikes
the block, the bullet/block system rises a vertical distance
of 5.00 cm. What was the speed of the bullet as it struck
the block?
• a) 397 m/s.
• b) 426 m/s.
• c) 457 m/s.
• d) 479 m/s.
• e) 503 m/s.
March 21, 2008
Physics for Scientists&Engineers 1
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Clicker Quiz Solution
 A ballistic pendulum is used to measure the speed of a
bullet shot from a gun. The mass of the bullet is 50.0 g and
the mass of the block is 20.0 kg. When the bullet strikes
the block, the bullet/block system rises a vertical distance
of 5.00 cm. What was the speed of the bullet as it struck
the block?
m 2v2
1
(m + M )gh = 2
(m + M )
• a) 397 m/s.
2
2gh(m
+
M
)
2
• b) 426 m/s.
v =
m2
• c) 457 m/s.
2gh(m + M )2
v=
• d) 479 m/s.
m2
• e) 503 m/s.
2 ( 9.81 m/s 2 ) ( 0.05 m ) (0.05 kg+20.0 kg)2
v=
( 0.05 kg )2
v = 397 m/s
March 21, 2008
Physics for Scientists&Engineers 1
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Clicker Quiz

Assume we have two identical cars each with mass m. One is parked. The other
car is moving with a speed of v and collides with the parked car head-on. The
two cars stick together and slide a distance d. What is the coefficient of
sliding friction between the tires and the road?
• A)
µk =
2mv
g
• B)
µk =
v
2mg
• C)
µ k = 2mgv 2
• D)
µk =
• E)
v2
µk =
8gd
March 21, 2008
v
2gd 2
Physics for Scientists&Engineers 1
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Clicker Quiz Solution

Assume we have two identical cars each with mass m. One is parked. The other
car is moving with a speed of v and collides with the parked car head-on. The
two cars stick together and slide a distance d. What is the coefficient of
sliding friction between the tires and the road?
• A)
• B)
µk =
µk =
2mv
g
v
2mg
• C)
µ k = 2mgv 2
• D)
µk =
• E)
v2
µk =
8gd
March 21, 2008
v
2gd 2
vf =
mv v
=
2m 2
1
v2
2
( 2m ) v f = m
2
4
v2
!K = W = m = Fk d = µ k ( 2mg ) d
4
v2
= 2 µ k gd
4
v2
µk =
8gd
Physics for Scientists&Engineers 1
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Clicker Quiz
 The period of rotation of the Earth is 24 hours. At this
angular velocity, the centripetal acceleration at the surface
of the Earth is small compared with the acceleration of
gravity. What would the period of rotation of the Earth be
if the magnitude of the centripetal acceleration at the
surface of the Earth at the equator due to the rotation of
the Earth were equal to the magnitude of the acceleration
of gravity?
• A) 0.043 hours
• B) 0.340 hours
• C) 0.841 hours
• D) 1.41 hours
• E) 12.0 hours
March 21, 2008
Physics for Scientists&Engineers 1
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Clicker Quiz Solution
 The period of rotation of the Earth is 24 hours. At this
angular velocity, the centripetal acceleration at the surface
of the Earth is small compared with the acceleration of
gravity. What would the period of rotation of the Earth be
if the magnitude of the centripetal acceleration at the
surface of the Earth at the equator due to the rotation of
the Earth we equal to the magnitude of the acceleration of
gravity?
2
2
a
=
!
r
=
!
Rearth = g
• A) 0.043 hours c
g
9.81 m/s 2
• B) 0.340 hours
!=
=
= 0.00124098 rad/s
6
Rearth
6.37 "10 m
• C) 0.841 hours
2#
• D) 1.41 hours T = 2# =
= 5063.08 s = 1.41 hr
! 0.00124098 rad/s
• E) 12.0 hours
March 21, 2008
Physics for Scientists&Engineers 1
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Clicker Quiz
 In the previous roller coaster problem, what is the speed with which the
roller coaster car has to enter at the bottom of the loop of 5.00 m
radius in order to achieve the feeling of weightlessness at the top?
• a) 7.00 m/s
Hint: Use energy conservation
• b) 12.1 m/s
• c) 13.5 m/s
• d) 15.6 m/s
• e) 21.4 m/s
March 21, 2008
Physics for Scientists&Engineers 1
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Clicker Quiz Solution
 In the previous roller coaster problem, what is the speed with which the
roller coaster car has to enter at the bottom of the loop of 5.00 m
radius in order to achieve the feeling of weightlessness at the top?
• a) 7.00 m/s
Hint: Use energy conservation
• b) 12.1 m/s
• c) 13.5 m/s
K0 = K + U
• d) 15.6 m/s
U = mgh = mg ( 2R ) = 2mRg
• e) 21.4 m/s
1 2
1 2
mvbottom = 2mRg + mvtop
2
2
2
vtop
= Rg
1 2
1
mvbottom = 2mRg + mRg
2
2
(
)
vbottom = 5Rg = 5 ( 5.00 m ) 9.81 m/s 2 = 15.6 m/s
March 21, 2008
Physics for Scientists&Engineers 1
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Clicker Quiz
 We want to design a curve in the road with a radius of 110.0 m such that
a car can travel around the curve without friction. The car enters the
curve with a speed of 31.3 m/s. What is the required bank angle?
• a) 1.25°
• b) 15.2°
• c) 28.2°
• d) 37.5°
• e) 42.2°
March 21, 2008
Physics for Scientists&Engineers 1
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Clicker Quiz Solution
 We want to design a curve in the road with a radius of 110.0 m such that
a car can travel around the curve without friction. The car enters the
curve with a speed of 31.3 m/s. What is the required bank angle?
• a) 1.25°
• b) 15.2°
Rg(sin ! + µ s cos! )
• c) 28.2°
v=
cos! " µ s sin !
• d) 37.5°
take µ s = 0
• e) 42.2°
v=
Rg sin !
= Rg tan !
cos!
v2
( 31.3 m/s )
tan ! =
=
= 0.90787
2
Rg (110.0 m ) 9.81 m/s
2
(
)
! = 42.2°
Could also start with components of the forces in the x-direction
v2
mg sin ! = m cos!
R
March 21, 2008
Physics for Scientists&Engineers 1
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Clicker Quiz Solution
 Consider a uniform rod with length L and mass M as shown below. What
is the moment of inertia of this rod if rotated about one end?
•
•
•
•
•
A) (1/3)ML2
B) (1/5)ML2
C) (1/6)ML2
D) (1/10)ML2
E) (1/12)ML2
March 21, 2008
Physics for Scientists&Engineers 1
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Clicker Quiz Solution
 Consider a uniform rod with length L and mass M as shown below. What
is the moment of inertia of this rod if rotated about one end?
•
•
•
•
•
A) (1/3)ML2
B) (1/5)ML2
C) (1/6)ML2
D) (1/10)ML2
E) (1/12)ML2
I=
V = AL
dV = Adx
r! = x
M
M
2
I=
x
Adx
=
AL "L
L
I=
March 21, 2008
M 2
r! dV
V V"
L
M L3
"0 x dx = L 3
2
1
ML2
3
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