Continuous Mass Flow
Rockets
8.01
W05D2
Today’s Reading:
MIT 8.01 Course Notes
Chapter 11 Reference Frames
Sections 11.1-11.3
Chapter 12 Momentum and the Flow of Mass
Sections 12.1,12.3
Announcements
Problem Set 4 due Week 6 Tuesday at 9 pm in box outside
26-152
Math Review Week 6 Tuesday 9-11 pm in 26-152.
Add Date Friday Oct 4
Conservation of Momentum:
System
For a fixed choice of system, if there are no external
forces acting on the system then the momentum of the
system is constant is constant.




Fext = 0 ⇒ Δpsystem = 0
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Strategy: Momentum of a System
1. Choose system
2. Identify initial and final states
3. Identify any external forces in order to determine whether
any component of the momentum of the system is
constant or not
i) If there is a non-zero total external force:

dp sys
 total
Fext
=
dt
ii) If the total external force is zero then momentum is
constant


p sys,0 = p sys,f
External Forces and Constancy of
Momentum Vector
The external force may be zero in one direction but not
others
The component of the system momentum is constant in
the direction that the external force is zero
The component of system momentum is not constant in
a direction in which external force is not zero
Modeling: Instantaneous
Interactions
•  Decide whether or not an interaction is instantaneous.
•  External impulse changes the momentum of the
system.
t +Δtcol




I[t , t + Δtcol ] = ∫ Fext dt = (Fext ) ave Δtcol = Δp sys
t
•  If the collision time is approximately zero,
Δtcol  0
then the change in momentum is approximately zero.


Δp system ≅ 0
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Problem Solving Strategy:
Momentum Diagrams
1) Identify all objects in system.
2) Choose a reference frame.
3) Draw a momentum diagram for
i)  state of the system before interaction
ii)  state of system after collision.
Diagram should include
1.  Choice of unit vectors for direction of momentum
2.  Show each mass element. Show the speed and arrow for
direction of velocity of each mass element.
Momentum and External Forces
Momentum law:


Fext = dPsys / dt
For discrete changes:



Fext,ave Δtinteraction = Psys, f (t f ) − Psys,i (ti )
For continuous changes:



Psys (t + Δt) − Psys (t)
Fext = lim
Δt→0
Δt
Recoil
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Relative Speed
The person jumps off the cart with a speed u relative to the
final speed of the cart vc. The speed of the person vp
relative to ground is
v p = u − vc
Table Problem: Recoil
A person of mass mp is standing on a cart of mass mc that is on ice.
Assume that the contact between the cart’s wheels and the ice is
frictionless. The person jumps off the cart with speed u relative to the
final speed of the cart.
a) Does the momentum of the system (person and cart) change when the
person jumps off?
a)  What is the final velocity of the cart as seen by an observer fixed to the
ground?
b)  What is the final velocity of the person as seen by an observer fixed to
the ground?
Continuous Mass Transfer
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Concept Question: Rain Falling Into Cart
Suppose rain falls vertically into an open cart rolling
along a straight horizontal track with negligible
friction. As a result of the accumulating water, the
speed of the cart
1. increases.
2. does not change.
3. decreases.
4. not sure.
Concept Question: Losing Mass But
Not Momentum
Consider an ice skater gliding on ice holding a bag of
sand that is leaking straight down with respect to the
moving skater. As a result of the leaking sand, the speed
of the skater
1.  increases.
2.  does not change.
3.  decreases.
4.  not sure.
Continuous Recoil
The material continually is ejected from the object,
resulting in a recoil of the object. For example when fuel
is ejected from the back of a rocket, the rocket recoils
forward.
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Mini-Experiment:
Recoiling Balloons
Rocket Equation:
Consider the propulsion as a
series of recoils
Rocket Motion in Empty Space
A rocket at time t = 0 is initially at rest in empty space. Fuel is
ejected continually backward with speed u relative to the rocket
during the interval [0, t] . We shall divide up this time interval
into a sequence of N time intervals in which the amount of fuel
Δm f is ejected in each interval:
[0, t1 = Δt1 ]
[t1 , t2 = t1 + Δt2 ]

[t N −1 , t N = t N −1 + Δt N ]
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Rocket Motion in Empty Space
For each interval we will calculate the change in speed of the
rocket due to recoil
For interval [0, t1 = Δt1 ] calculate Δvr ,1
For interval [t1 , t2 = t1 + Δt2 ] calculate Δvr , 2

For interval [t N −1 , t N = t N −1 + Δt N = t] calculate Δvr , N
Rocket Motion in Empty Space
To find the final speed of the rocket, we then sum up the
changes in speed for each interval
j= N
vr (t) = ∑ Δvr , j
j=1
Now we take the limit that Δvr , j → 0, N → ∞
Our sum then goes to an integral
vr (t) =
vr (t )
∫
dvr
vr (0)=0
Recoil During Interval [0, t1]
In the first interval, recoil
of rocket can be calculated
as follows:
px, i = px, f ⇒ 0 = −Δm f (u − Δvr ,1 ) + mr ,1Δvr ,1 ⇒
Δm f u − Δm f Δvr ,1 = mr ,1Δvr ,1
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Recoil During Interval [0, t1]
The speed of the ejected fuel is much greater than the small recoil speed of the
rocket and therefore
Δm f u >> Δm f Δvr ,1
So our momentum principle enables us to calculate the recoil speed for the first
interval
Δm f u = mr ,1Δvr ,1 ⇒ Δvr ,1 =
(In the limit that
Δm f
mr ,1
u
Δt1 → 0 , Δvr ,1 → 0 , and even for small u, Δm f u >> Δm f Δvr ,1 .)
Recoil During Interval [t1 , t2]
Δvr ,2 =
Δm f
mr ,2
u
In the second interval, we choose
a new reference frame S1 that is
moving with speed Δvr ,1 with
respect to our first reference S0 .
Then the rocket is again at rest at
t1 in this new reference frame and
so our calculation for the second
recoil is identical to our previous
case
Recoil During Interval [tj-1 , tj]
For an arbitrary interval the same
argument applies where S j−1 is the
reference frame which is moving
with speed
vr , j−1 = Δvr ,1 + Δvr ,1 ++ Δvr , j−1
Δvr , j =
Δm f
mr , j
u
with respect to S0 . The rocket is at
t j−1
rest in this frame at time
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Recoil During Interval [tj-1 , tj]
For an arbitrary interval same
argument applies where S j−1is the
reference frame which is moving
with speed
vr , j−1 = Δvr ,1 + Δvr ,2 ++ Δvr , j−1
Δvr , j =
Δm f
mr , j
with respect to S0 . The rocket is
at rest in this frame at time t j−1
u
Recoil Contributions to Speed of Rocket
The speed at time t j is the
sum of contributions
vr , j = Δvr ,1 + Δvr ,2 ++ Δvr , j
vr , j =
Δm f
mr ,1
u+
Δm f
mr ,2
u ++
Δm f
mr , j
u
Because the mass of the rocket is decreasing
mr ,1 > mr ,2 >  > mr , j
The recoil contributions are increasing
Δvr ,1 < Δvr , 2 <  < Δvr , j
Rocket Equation
As we shrink the jth interval to zero, denote
mr , j → mr
Δm f → dm f
Δvr , j → dvr
Δt j → dt
Then the recoil equation for the interval [t j ,t j + Δt j ] becomes
Δvr , j =
dvr =
Δm f
mr , j
dm f
mr
u→
u
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Rocket Equation: Conservation of
Mass
Conservation of mass: rate of decrease of mass of rocket
equals rate of ejection of fuel
dm f = −dmr
The recoil equation is then
−dmr u = mr dvr
Thus the differential change in the speed of the rocket is
dvr = −
dmr
u
mr
Rocket Motion in Empty Space
dvr = −
Rocket Equation:
Integrate both sides
vr ( mr )
∫
vr ,i
dmr
u
mr
mr
dvr = −u ∫
mr ,i
dmr′
mr′
Find speed as a function of mass of the rocket
⎛ m ⎞
⎛ mr, i ⎞
vr (mr ) − vr (mr, i ) = −u(ln mr − ln mr, i ) = −u ln ⎜ r ⎟ = u ln ⎜
⎝ mr ⎟⎠
⎝ mr, i ⎠
In our example, the rocket started from rest so
vr (mr ) = u ln
mr, i
mr
Table Problem: Rocket Sled
A rocket sled ejects gas backwards at a speed u relative to the rocket sled.
The mass of the fuel in the rocket sled is equal to one half the initial total mass mr,0
(including fuel) of the sled. The rocket sled starts from rest on a frictionless track.
You may ignore air resistance.
a)  Derive a relation between the differential of the speed of the rocket sled, dvr , and
the differential of the total mass of the rocket, dmr
b)  Integrate the above relation to find the speed of the rocket sled as a function of
mass, vr(m), as the rocket sled speeds up.
c)  What is the final speed of the rocket sled after all the fuel has been burned?
Express your answers in terms of the quantities u, and mr,0 as needed.
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