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MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
8.01
Problem Solving Session 3: Momentum and Continuous Mass Transfer
Solution
ICW05D3-1 Group Problem 1: Rocket Sled Slowing Down
A rocket sled can eject gas backwards or forwards at a speed u relative to the sled. The
mass of the fuel in the sled is equal to dry mass of the sled, m0 . At t  0 the rocket sled
has speed v0 and starts to eject fuel in the forward direction in order to slow down. You
may ignore air resistance. You may treat u as a given constant in the following
questions.
a) Derive a relation between the differential of the speed of the rocket sled, dvr , and
the differential of the mass of the rocket sled, dmr .
b) Integrate the above relation to find the speed of the rocket sled as a function of
mass, vr (mr ) , as the rocket sled slows down.
c) What was the initial speed v0 of the rocket sled if the sled came to rest just as all
the fuel was burned?
Solution:
Take as the system the rocket sled, fuel, and the small amount of fuel of mass m f that is
ejected during the interval [t ,t  t ] . Let the positive x -direction be to the right in the
figures below. We first show the momentum diagram at time t :
The smaller square represents the small amount of fuel of mass m f that is
ejected during the interval [t ,t  t ] . In the above figure mr (t) is the combined
mass of the rocket sled, the dry mass of the rocket sled plus the mass of fuel
inside the rocket sled at time t that does not leave the rocket sled during the
interval. The x -component of the velocity of the rocket sled is denoted by vr (t) .
In analyzing our momentum diagrams we shall drop the explicit reference to time
dependence for the mass of the rocket and the x -component of the velocity of
the rocket sled, denoting them respectively by mr and vr Returning to our
momentum diagram we see that at time t , the x -component of the momentum of
the system pxsys (t) is given by
pxsys (t)  (mr  m f )vr
(1)
By the end of the interval the rocket has recoiled backward with x -component of the
velocity vr  vr , here vr  0 . (Note we did not put a minus sign in but let the
differential be negative. As you will see later on the sign takes care of itself.) The fuel is
ejected forwards relative to the rocket sled with speed u , hence with a speed vr  vr  u
relative to the fixed reference frame depicted in the figures. After the interval has ended
the momentum diagram of the system is shown below.
At time t  t , the x -component of the momentum of the system pxsys (t  t) is given by
pxsys (t  t) = mr (vr  vr )  m f (vr  vr  u)
(2)
and so the change in x -component of the momentum of the system during the interval
[t ,t  t ] is
pxsys  pxsys (t  t)  pxsys (t)
 mr (vr  vr )  m f (vr  vr  u)  (mr  m f )vr .
(3)
 mr vr  m f vr  m f u
There are no external forces acting in the x -direction, therefore the momentum
principle
pxsys
Fxext  lim
(4)
t
t0
becomes
0  lim
mr vr  m f vr  m f u
t
t0
.
(5)
We note that
lim
m f vr
t
t0
 0.
(6)
Then using the definition of the derivative, we find that the differential equation
describing the motion of the system is given by
dvr
0  mr
dt

d mf
dt
u
(7)
Because the fuel is being ejected at a positive rate and the mass of the rocket sled is the
rate of decrease of the mass of the rocket sled is negative we have that
dmr
dt

dm f
(8)
dt
We can now substitute Eq. (8) into the Eq. (7) and find that
0  mr
dvr
dt

d mr
dt
u.
(9)
We can solve this equation by the technique of separation of variables. First rewrite the
equation as
dv
d mr
mr r 
u.
(10)
dt
dt
Multiple each side by dt and divide by through by mr . Thus
dvr 
d mr
mr
u.
(11)
b) Integrate the above relation to find the speed of the rocket sled as a function of
mass, vr (mr ) , as the rocket sled speeds up.
From the information given in the statement of the problem, the dry mass m0 of the
rocket sled is equal to the initial mass of the fuel so the initial mass of the rocket is
mr,0  2m0 .
(12)
We can integrate both sides of Eq. (11), (making sure that our choice of limits for the
integrals are consistent),
vr  vr (mr )

vr  v0
mr  mr
dvr  u
dmr
.
mr
mr  2 m0

(13)
Integration yields
 m 
vr (mr )  v0  u ln  r 
 2m0 
(14)
The speed of the rocket sled vr (mr ) as a function of mass, is thus
 m 
vr (mr )  v0  u ln  r  .
 2m0 
(15)
Note that mr  2m0 , so that ln(mr / 2m0 )  0 and hence vr (mr )  v0 . The signs did in fact
take care of themselves!
c) What was the initial speed v0 of the rocket sled if the sled came to rest just as all
the fuel was burned?
The final mass of the rocket is then mr, f  m0 . After substituting vr (mr , f )  0 into Eq.
(15) we can solve for the initial speed v0 of the rocket sled
v0  u ln(2) .
(0.16)
ICW05D3-2 Group Problem 2: Car on Barge
QuickTime™ and a
decompressor
are needed to see this picture.
A workman is unloading the last car (mass mc ) from a barge (mass mb ) tied to the pier
by a short cable. He drives with a constant acceleration from a standing start and attains a
speed v f in the time t f it takes to reach the end of the barge.
a) Does this put tension in the cable or does it push the barge against the dock?
b) Depending on your answer to a) determine an expression for either the tension in the
cable T (t) or the magnitude of the horizontal force of the pier on the barge F(t) during
the time t  0 to t  t f . Express your answer in terms of some or all of the quantities mc ,
mb , v f and t f .
Solution:
a) Consider the car and the barge as the system. The momentum of the system is
increasing toward the right. Thus there must be an external force on the system pointing
right. That implies tension in the cable.
b) We begin by noting that the car moves with constant acceleration, so ac  v f / t f and
by integration
vc (t)  (v f / t f )t
using the fact that the car starts from rest, vc (0)  0 .
There are several possible ways to approach this problem.
r
r
First Approach: F  dp / dt applied to system consisting of the car and the barge.
r
r
We shall apply F  dp / dt to this system that is under the influence of an external force,
the tension in the rope:
d
T (t)  (mbvb  mc vc )
dt
Thus
dv
T (t)  mc c  mc v f / t f .
dt
r
r
Second Approach: We choose the car as our system and apply F  mca c to the car.
Since the car is accelerating, it is experiencing a force by the barge on the car
r
r
Fb,c  mca c . Because the car and barge from an interaction pair , By Newton's Third Law,
r
r
r
the force on the barge due to the car is Fc,b  Fb,c  mca c  mc ac î . But the barge, is
stationary, so the horizontal forces must sum to zero. The barge experiences two forces:
r
r
the force from the car Fc,b and the tension force of the rope on the barge Tr ,b (t)  T (t) î .
r
r
r
Thus Tr ,b (t)  Fc,b  0 implies that after taking the x -component mc ac  T (t)  0 . So
the tension in the rope is
T (t)  mc ac  mc v f / t f
in agreement with our result above.
Third Approach: Motion of the Center of Mass:
We can relate the acceleration of the center of mass to the external force, the tension,
r
Tr ,b (t)  T (t) î . Let L be the length of the boat. Choose an origin at the left side of the
boat. Treat the barge as a point-like object located a distance L / 2 from the origin. Then
the x -component of the center of mass is
X cm (t) 
mc xc  mb xb
mc  mb

mc xc  (L / 2)xb
mc  mb
.
Differentiating twice gives the x -component of the acceleration of the center of mass:
Acm (t) 
mc ac  mb ab
mc  mb

mc ac
mc  mb

(mc v f / t f )
mc  mb
.
r
r
The tension in the rope is then found by using Fext  msystem A cm , where msystem  mc  mb
r
r
and Fext  Tr ,b (t)  T (t) î . Thus
T (t)  (mc  mb ) Acm (t)  (mc  mb )
(mc v f / t f )
mc  mb
 mc v f / t f .
Remark: The x -component of momentum is not constant due to the presence of an
external force. Therefore mc vc (t)  mbvb (t) . It is incorrect to draw the following
conclusions that vb (t)  mc vc (t) / mb hence ab (t)  mc ac (t) / mb  (mc / mb )(v f / t f ) ,
and finally T (t)  mb ab (t)  mc v f / t f .Two things to note. First, the barge is stationary
so ab (t)  0 . Second, this incorrect result for the tension differs only by a minus sign
from the correct result but it is a very important minus sign because the direction of the
tension force on the barge must be in the positive x -direction.
ICW05D3-3 Group Problem 3: Extra Table Problem (if there is time): Acrobat and
Clown
An acrobat of mass m A jumps upwards off a trampoline with an initial y-component of
the velocity v y,0  v0 . At a height h0 , the acrobat grabs a clown of mass m B . Assume that
the time the acrobat takes to grab the clown is negligibly small.
How high do the acrobat and clown rise? How high would the acrobat go if the acrobat
and the clown have the same mass?
Solution: Choose the acrobat and the clown as the system. The first important
observation to make is that there is a collision between the acrobat and the clown. This
collision is completely inelastic in that the two bodies collide and “stick’ together after
the collision. The details of the collision are determined by the internal forces in the
system. Since this is a one- dimensional motion, let’s choose an origin at the trampoline
and the positive y-axis upwards.
There are two important states to identify in this problem.
State 1: Immediately before the collision. Acrobat A just arrives at platform located at
y1, A  y2, B  h0 with velocity v1, A  v1, Aˆj , immediately before grabbing Clown B. Denote
this time by t1 .
The collision lasts a time tcol . During this time interval, acrobat A grabs Clown B.
State 2: Immediately after the collision. At the end of the interval, the two acrobats rise
together with velocity v 2  v2 ˆj . Denote the time at the end of this interval by
t2  t1  tcol . The key assumption is that the collision time is instantaneous tcol  0 .
The impulse delivered by the external gravitational force is zero during the collision
because the collision was assumed to be instantaneous. Therefore during the collision, the
total momentum of the system is constant. If the collision lasts a significant length of
time, there would be some slowing down of the acrobat A during the collision. then we
need to calculate this effect. However by assuming the collision is instantaneous, we can
ignore this slowing down, and therefore the change in the system momentum is zero.
From one-dimensional kinematics, the y-component of the velocity of the acrobat A at
the t1 is given by
(v1, A ) y  (v02  2gh0 )1/ 2 .
(17)
State 1 to State 2:
Momentum in state 1 (immediately before collision) is only due to acrobat A
p1, A  mA (v1, A ) y j  mA (v02  2 gh0 )1/ 2 j .
(18)
The momentum in state 2 (immediately after the collision) is
p2  (mA  mB )v2, y j .
(19)
mA (v02  2 gh0 )1/ 2 j  (mA  mB )v2, y j .
(20)
Since momentum is unchanged,
The y-component of the velocity of the acrobat A at the t 2 is given by
. v2, y 
mA
mA  mB
(v02  2gh0 )1/ 2
(21)
Again from one-dimensional kinematics, the final height of the acrobat and the clown is
given by
hf 
1
(v2, y )2  h0 .
2g
(22)
We can use our above result for the y-component of the velocity immediately after the
collision to find the final height in terms of the initial y-component of the velocity of
acrobat A and the initial height of clown B,
2
hf 
mA
1
(v02  2gh0 )  h0 .
2
2g (mA  mB )
(23)
When the mass of the acrobat is equal to the mass of the clown mA  mB , the mass ratio
becomes
mA2
(mA  mB )2

1
and so the height becomes
4
2
mA2
1
1 2
1 v0
2
hf 
(v  2gh0 )  h0 
(v  2gh0 )  h0  (  3h0 ) . (24)
2g (mA  mB )2 0
8g 0
4 2g