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Physics 3210
Week 6 clicker questions
In central-force motion, the areal velocity is a constant 
What is the period t of an elliptical orbit with area A?
A.
B.
t A

t
A
2

C. t  A
D.
t
2
A
Area=A
2
Kepler’s First Law states that planets
move in elliptical orbits with the sun at
one focus.
Why is the sun at one focus of the orbit?
Because...
A. Otherwise the planets wouldn’t all be in the same orbital plane.
B. In two-body central-force motion one mass is always at the focus on
the orbit.
C. In two-body central-force motion the center of mass is always at the
focus of the orbit, and the center of mass position is approximately
given by the position of the sun.
D. The moon is at the other focus.
Sun
A system of n particles is described by the masses and
positions of each particle: m , r
M   m

What is the position of the center of mass R of the system?
1
A. R 
r

M 
1
B. R 
m

M 
C. R   m r

D.
R   Mr

E.
1
R   mr
M 
Consider particles  and b, and the internal force fb=the
force on particle  due to particle b.
How does Newton’s third law relate the internal forces?
A.
fb  fb
B.
fb  fb
C.
fb  fb
D.
fb  fb  0
E.
fb  fb  1
fb

b
Suppose a system of particles experiences only internal forces
(no external forces).
What can you say about the momentum of the system?
A. Nothing – having no external forces doesn’t determine the
momentum of the system.
B. The momentum of each particle is constant.
C. The momentum of some (but not all) of the particles is
constant.
D. The momentum of the center of mass of the system is constant.
Physics 3210
Wednesday clicker questions
A chain with length b and mass density r is initially attached
at both ends. One end of the chain is cut. After this falling end
has moved down a distance x, what is the mass of the right
(falling) side of the chain?
r
A. m right   b  x 
2
r
B. m right   b  x 
2
C. m right  r  b  x 
D.
m right  r  b  x 
E.
mright  rb
x
A chain with length b and mass density r is initially attached
at both ends. One end of the chain is cut; the other end of the
chain experiences a tension T. As the chain falls, what is the
total external force on the chain (in the x direction)?
A.
FT
B.
F  Mg
C.
F  Mg  T
D.
F  Mg  T
x
A chain with length b and mass density r is initially attached
at both ends. The height is measured by x (increasing down)
and the potential energy is zero at the attachment point. What
is the total potential energy of the chain?
A.
U  rgb
B.
U  rgb
rgb 2
C. U 
2
2
r
gb
D. U 
2 2
rgb
E. U 
4
x
A system of n particles is described by the masses and
positions of each particle, relative to the center of mass: m , r
What can you say about the quantity
A.
 m r  0

B.
 
 
 m r  0

D.

 m r  0

C.
 m r
 
 m r  the position of the CM

 
 
?
Physics 3210
Friday clicker questions
Consider particles  and b; the internal force fb=the force on
particle  due to particle b; the position vector rb= the vector
from particle  to particle b. If the force between the two
particles is central, what can you say about rbfb?
A. Nothing.
B.
rb  fb  0
C.
rb  fb  0
D.
rb  fb  0

rb
b
A system of n particles is described by the masses and positions
of each particle, relative to the center of mass: m , r
The squared velocity of each particle is therefore
v 2  r2  2r  R  R 2  v2  2r  R  V 2
What is the total kinetic energy of the system?
1
1
2 1
T   m v   m v V  MV 2
2 
2 
2
1
1
2
2
B. T   m v  m v V  MV
2 
2

1
1
C. T   m v2  MV 2
2 
2
1
2

T

m
v

V
D.


 
2 
A.
The work done in unwrapping a rope from a cylinder is given by
R
x 


W   rg  x  R sin    dx
 R 

0
What is the total work done?
 2

A. W  rgR 
 cos  1
 2

2



B. W  rgR 2
 2  cos  1


2



C. W  rgR 2 
 cos 
 2

2


cos 
2

 1
D. W  rgR 
R
 2

What is the total kinetic energy of the rope-cylinder system?
1
1
2 2
2 2
T

r
R
2



R


MR



A.
2
2
1
1
2
2
B. T  rR  2    R  MR
2
2
C.
1
T   m  M  R2
2
1
2 2
D. T   m  M  R 
2