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Variable Mass
Break Up

Before the split, momentum
is P = MV
• M total mass
• V center of mass velocity

After the split, the sum of
momentum is conserved.
• P = m1v1 + m2v2


Center of mass velocity
remains the same.
The kinetic energy is not
conserved.
v1
M
V
V
v2
Explosions


A 325 kg booster rocket and
732 kg satellite coast at 5.22
km/s.
Explosive bolts cause a
separation in the direction of
motion.

• K = (1/2)(1057 kg)(5.22 x
103 m/s)2 = 14.4 GJ

Find the kinetic energy
before separation, and the
energy of the explosion.
Kinetic energy after
separation
• K1 = 16.4 GJ
• K2 = 0.592 GJ
• Satellite moves at 6.69 km/s
• Booster moves at 1.91 km/s

Kinetic energy before
separation is (1/2)MV2

The difference is the energy
of the explosion.
• Kint = 2.6 GJ
Change in Momentum

The law of action was redefined to use momentum.
 dp
F
dt

The change can be due to change in velocity or a
change in mass
 dmv
F
dt
Infinitessimal Change

In a short time the following happens:
• The mass goes from m to m + Dm
• The velocity goes from v to v + Dv
• The mass added or removed had a velocity u compared to
the object

Momentum is conserved
(m  Dm)v  m(v  Dv)  Dm(v  u )
mDv  uDm
Thrust

If there is no external force the force to be applied
must be proportional to the time rate of change in
mass.
m
Dv
Dm
u
Dt
Dt
m
dv
dm
u
dt
dt
The force -u(dm/dt) is the thrust
Water Force

Thrust can be used to find
the force of a stream of
water.

A hose provides a flow of 4.4
kg/s at a speed of 20. m/s.

The momentum loss is
dp
dm
u
dt
dt
• (20. m/s)(4.4 kg/s) = 88 N

The momentum loss is the
force.
Rocket Speed


Rockets decrease their
mass, so we usually write
the mass change as a
positive quantity.
The equation can be
integrated to get the
relationship between the
mass and increased velocity.
dv
dm
m  u
dt
dt
dm
dv  u
m
dm
vi
mi
m
v f  vi  u (ln m f  ln mi )
vf
dv  u 
mf
 mi
v f  vi  u ln 
m
 f




Applied Force

If there is an external force
that must equal the time rate
of change in momentum.
dv
dm
Fext  m  (u  v)
dt
dt

Force is needed to maintain
the speed.
Heavy Water

Water is poured into a
beaker from a height of 2 m
at a rate of 4 g/s, into a
beaker with a 100. g mass.

There is extra momentum
from the falling water.
v  2 gh
  m/t

What does the scale read
when the water is at 200. ml
in the beaker (1 ml is 1 g)?

Answer: 302 g
dp
dm
v
  2 gh
dt
dt

This is about 0.024 N or an
equivalent mass of 2.4 g.