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YSSIIC
CSS 221111
I) CONCEPTUAL QUESTIONS
1) A rock is dropped at the same instant that a ball, at the same initial height, is thrown
horizontally. Which one will have greater velocity when it reaches the ground?
Explain using 2-D equations.
2) Can the equations of kinematics be used in a situation where the acceleration varies with the
time? NO!! They can be used ONLY when acceleration is CONSTANT.
Can they be used when the acceleration is zero?
YES!! Zero is a constant.
3) The Gravitational Force exerted by the Earth on the Moon holds the Moon in a circular orbit.
Does Gravity Force do?
(c) is the correct. Since Fg on the Moon acts toward the earth to provide ac inward. The Moon’s
displacement at any moment is tangent to the circle, parallel to velocity and perpendicular to Fg.
So the angle between Fg and the displacement is 90° (cos90°=0), and the work done by Fg on the
moon is therefore ZERO! W = Fdcos90°
4) Why the Total Energy of a system can be either positive or negative, whereas the Kinetic Energy
is always positive?
Total Energy is: K + U. U can be negative, so the sum can also be NEGATIVE!
II) Straightforward problems
1) A satellite of mass m is moving in a circular orbit around the Earth.
Applying Newton’s 2nd Law:
∑F = Fg = m aR = mv2/r
Fg = G(MEm)/r2
Taking (1) = (2):
(1)
(2)
G(MEm)/r2 = m v2/r (3)  v2 = GME/r
Since r = RE +h 
GM E
RE  h
v
2) A hockey puck on a frozen pond is given initial speed vi
A) Sketch a free-body diagram
B) Use Newton’s 2nd Law (
y  axis :
 F  ma
å
F = ma with x and y directions) to show that:
y
 0 FN  FG  0  FN  FG  mg
y
 0 FN  FG  0  FN  FG  mg
y  axis :
 F  ma
III) Intermediate problems
1) A glider of mass m is free to slide along a horizontal air track.
A) Find an equation that establishes the conservation of energy of the system spring-mass.
K i + U si = K f + U sf Þ
kx 2 = mv 2 Þ v =
0 + 12 kx 2 =
1
2
mv 2 + 0 Þ
kx 2
k
= x
m
m
B) Show that the impulse
I = mv = mx
k
mx m k
m x km
=
=
= x km
m
m
m m
2) The bead is released from a height h = 4R.
Etop = E A Þ U top + K top = U A + K A Þ mgh + 0 = mgy A + 12 mv A2 Þ
mg (4 R) = mg (2 R) + 12 mv A2 Þ 4 gR = 2 gR + 12 v A2 Þ
1
2
3)
v A2 = 4 gR - 2 gR = 2 gR Þ v A2 = 4 gR Þ v A = 2 gR
Atwood Machine.
A)
B)
ΣFy = T – m1g = m1 a
ΣFy = T – m2g = – m2 a
Then:
T = m1g + m1 a
(4)
Equating: (3) = (4) and Solving for a
m1g + m1 a = m2g – m2 a 
m1 a + m2 a = m2g – m1g 
a (m1 + m2) = (m2 – m1)g 
D)
(2)
(3)
T = m2g – m2 a
C)
(1)
a
 m2  m1  g
 m1  m2 
(5)
Substituting (5) into (3):T = m1g + m1 a (3) 
  m  m1  
m m g  m1m1 g
T  m1 g  m1  2
g   T  m1 g  1 2

m1  m2
  m1  m2  
T
 mm 
m1m1 g  m1m2 g  m1m2 g  m1m1 g
 T  2 1 2  g
m1  m2
 m1  m2 
IV) Challenging problems
1) As shown in the figure.
A) Write an equation for conservation of the mechanical energy of the system Bob-Earth.
(K + U )bottom = (K + U )top
1
2
m vB2 + 0 = 0 + m g (2l ) Þ vB2 = 4 g l Þ vB = 2 g l
B) Write an equation for conservation of linear momentum of the system Bullet-Bob.
v
v
pi = p f Þ mv = MvB + m Þ mv - m = MvB Þ
2
2
2M 2 g l
v
4M
m = MvB since vB = 2 g l Þ v =
=
2
m
m
gl
2) A 4 kg particle is subject to a total force that varies with the position
A) Using the Work-Kinetic Energy Theorem find the speed of the particle at x = 5 m
Wnet  K = Kf − Ki = ½mv2 − ½mvi2 = ½mv2 − 0
Wnet  Area Under the curve =
v
B h 53

J  7.5 J  12 mv 2 
2
2
2(7.5)
m / s  1.94m / s
4
B) Using the Work-Kinetic Energy Theorem find the speed of the particle at x = 12 m
Wnet  K = Kf − Ki = ½mv2 − ½mvi2 = ½mv2 − 0
B h
h
 B  h   B  b 
2
2
53
2
Wnet 
J  5  3J   3  1.5  J  7.5 J  15 J  4.5 J  27 J 
2
2
2(27)
27 J  12 mv 2  v 
m / s  3.67m / s
4
Wnet  Area Under the curve =
3) Let us “weigh” the Earth.
A) The mass of the Earth ME can be written as:
M Em
ME m
gRE2
Fg = G 2
& Fg = mg Þ G
= m g Þ ME =
RE
RE2
G
B) The Average Density (E) of the Earth is given by:
gRE2
3g RE2
ME
3g
G
E 



3
3
4 RE 4 GRE 4 GRE
VE
3