1. No category

 dx / dt 
 dr 
v
 dy / dt ;
dt
dz / dt 
Velocity
Momentum


p  mv ;
Impulse




dp  [dp] A  [dp]B  [dp]C ...
Force
 [dp ] A
FA 
dt
Weight


Fg  m g
Total Momentum

 

ptot  p1  p2  p3 ...
System’s Center Of Mass
Motion of COM

rCM



m1 r1  m2 r2  m3 r3 ...

M





M vCM  m1 v1  m2 v2  m3v3 ...  ptot
Internal versus external interactions.
No external interactions  ptot conserved
Inertial frames of reference:
Not accelerated (not rotating or any other kind of accel…)
C5A.1 Consider a rocket in deep space whose
empty mass is 5200 kg that can carry 52,000 kg
of propellant. If the rocket engine can eject 1300
kg/s of propellant from its nozzle at a speed of
vg=3300 m/s relative to the rocket, and if the
rocket starts at rest, what is its approximate final
speed?
A.
B.
C.
D.
102 m/s;
104 m/s;
106 m;
108 m.
C5A.1 Consider a rocket in deep space whose empty mass is
5200 kg that can carry 52,000 kg of propellant. If the rocket
engine can eject 1300 kg/s of propellant from its nozzle at a
speed of vg=3300 m/s relative to the rocket, and if the rocket
starts at rest, what is its approximate final speed?
Can we assume that the gas is all expelled at speed vd relative to the
ground, so we can equate the momentum of the gas to the one
gained by the rocket (implies the rocket is moving much slower than
vd)?
Then the rocket speed would be:
v 
mg v g
M
 10v g
Doesn’t work! This rocket will move at speeds comparable to vg, so
we really have to take into account the varying velocity and mass.
C5A.1 Consider a rocket in deep space whose empty mass is
5200 kg that can carry 52,000 kg of propellant. If the rocket
engine can eject 1300 kg/s of propellant from its nozzle at a
speed of vg=3300 m/s relative to the rocket, and if the rocket
starts at rest, what is its approximate final speed?
Consider momentum conservation in ejection of dm at speed vg during time dt
m(t ) v(t )  m(t  dt ) v(t  dt )  dm(v g  v(t  dt ))
before
after
m(t  dt )  m(t )  dm 0  v(t )dm  m(t )dv  dm(v g  v(t  dt ))
 
v(t  dt )  v(t )  dv 
dm v g  m(t )dv
M
dm
dm
dv  v g
 v f  v g 
 v g ln(11)  v g 2.4
m(t )
m(t )
M m
C5A.1 Consider a rocket in deep space whose
empty mass is 5200 kg that can carry 52,000 kg
of propellant. If the rocket engine can eject 1300
kg/s of propellant from its nozzle at a speed of
vg=3300 m/s relative to the rocket, and if the
rocket starts at rest, what is its approximate final
speed?
A.
B.
C.
D.
102 m/s;
104 m/s;
106 m;
108 m.
Energy
E  K1  K 2  K3  ...  V (r12 )  V (r13 )  V (r23 )  ...;
Kinetic energy K  1 mv 2 ;
2
Potential energy
gravitational V ( z )  mgz
C6T.2 How does the kinetic energy, Kfast, of
a car traveling at 50 mi/h compare with the
kinetic energy, Kslow, of an identical car
traveling at 25 mi/h? (The consequences of
a collision are in rough proportion to the
energy involved.)
A.
B.
C.
D.
Cars are identical so Kfast = Kslow;
Kfast ≈ 1.5 Kslow;
Kfast ≈ 2 Kslow;
Kfast ≈ 4 Kslow.
C6T.3 A person throws three identical rocks off a
cliff of height h with exactly the same speed v0
each time. Rock A is thrown almost vertically
upward, rock B is thrown horizontally, and rock C
is thrown almost vertically downward. Which rock
hits the ground with the greatest speed? (Ignore
air friction).
A.
B.
C.
D.
Rock A;
Rock B;
Rock C;
All rocks hit with
the same
speed.
C6T.4 A Imagine that we know from experiments that
when the object moves from point A to point B, the
potential energy of its gravitational interaction with the
earth increases by 24 J, and that when the object moves
from point B to point C, the potential energy decreases
by 18 J. If we define the system’s reference separation
to be when the object is at point C, what is the value of
the system’s potential energy when the object is at point
A?
A.
B.
C.
D.
E.
+6 J;
-6 J;
-52 J;
+52 J;
0 J.
C6T.5 In a coordinate system where the z axis is
vertical, we choose the gravitational potential
energy of a 4-kg rock interacting with the earth to
be zero when z=-5 m. The formula for the
potential energy as a function of z is thus
V(z)=mgz+C. What is the (approximate) value of
C?
A.
B.
C.
D.
E.
-50 J;
+50 J;
-200 J;
+200 J;
0 J.
C6T.7 Consider a rock interacting gravitationally with the
earth. Imagine that we define the interaction potential
energy to be zero if the rock is at ground level. A person
standing at the bottom of a well throws the rock vertically
upward from 20 m below ground level. The rock makes it
all the way up to 1 m below ground level before falling
back into the well. The total energy of the rock earth
system is:
A.
B.
C.
D.
E.
F.
Negative;
Zero;
Positive (in this particular case);
Positive because energy is always positive;
The answer depends on the rock’s mass;
The answer depends on the rock’s initial speed.