### Problem Set 4 Momentum and Continuous Mass Flow Solutions

```MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
Physics 8.01
Fall 2012
Problem Set 4 Momentum and Continuous Mass Flow Solutions
Problem 1:
a) Explain why the total force on a system of particles is only due to the sum of
external forces.
Answer: All internal forces are part of interaction pairs and by Newton’s Third Law the
sum of the forces of each interaction pair is zero hence all internal forces all cancel in
pairs. Therefore only the external forces sum to give the force on a system of particles.
b) Explain under what conditions does the center of mass of a system of particles
move with constant velocity?
Answer: Using Newton’s Second and Third Laws, the external force on a system of a
particles is equal to the product of mass of the system with the acceleration of the center
of mass of the he system. If the external force is zero, the acceleration of the center of
mass is zero and hence the center of mass of a system of particles move with constant
velocity.
c) Explain what is meant by the expression “the thrust on a rocket” in terms of the
relative speed of the ejected fuel and the rate that fuel is burned. How is this thrust
related to the conservation of momentum when there are no external forces acting
on the system consisting of rocket and fuel?
Answer: The magnitude of the thrust on a rocket is the product of the time rate of change
that mass of the fuel that is ejected from the rocket with the relative speed of that fuel
with respect to the rocket. The direction of the thrust is in the direction of the acceleration
of the rocket. The thrust term is actually the time rate of change in the momentum of the
burned fuel. The direction of this change in momentum of the fuel is opposite the
direction of the acceleration of the rocket. If there are no external forces acting on the
rocket, the change in the momentum of the rocket is equal in magnitude to the change in
momentum of the fuel but opposite in direction as the rocket recoils forward.
!
d) The same force F is applied separately to each of the three objects shown on the
right with masses satisfying m1 > m2 > m3 . Which of the following statements is
true? (Note: more than one statement may be true.)
Figure 1
Figure 2
Figure 3
1. In all cases the center of mass of the system has the same acceleration.
2. In all cases the magnitudes of the acceleration of the center of mass of the
system are different.
3. The magnitude of the acceleration of the center of mass of the system is
largest in figure (1).
4. The magnitude of the acceleration of the center of mass of the system is
largest in figure (2).
5. The magnitude of the acceleration of the center of mass of the system is
largest in figure (3).
Answer: 1. In each case the external force on the system is the same, so the center of
mass has the same acceleration.
Problem 2: Jumping off the Ground
A person, of mass m , jumps off the ground. Suppose the person pushes off the ground
with a constant force of magnitude F for a time interval !t . What was the magnitude of
the maximum displacement of the center of mass of the person?
Solution: We divide the jump into two parts, pushing off the ground, and then in the air.
For the first part, we begin by computing how much the center of mass displaces during
the jumping. Let’s set t = 0 at he beginning of the jump. During the time interval
[0 < t < !t] , the person is pushing against the ground. We can first compute the
acceleration of the center of mass using
!
!
!
!
Fext t = p cm (t) ! p cm (t = 0) = mv cm (t) .
(1)
Let’s choose positive y -axis pointing up then the external force is given by
!
!
!
Fext = Fground , pertson + Fgravity, person = (F ! mg)ˆj .
(2)
Therefore substituting Eq. (2) into Eq. (1) and taking the y -components yields
(F ! mg)t = mvcm, y (t) .
(3)
Thus the y -component of the velocity of the center of mass is given by
"F
%
vcm, y,1 (t) = \$ ! g ' t; 0 ( t ( )t .
#m
&
(4)
We can integrate this to find the displacement of the center of mass
!ycm,1 = ycm,1 (!t) " ycm,1 (0) =
t # = !t
\$
vcm, y,1 (t # )dt #
t # =0
t # = !t
%F
(
(
1% F
!ycm,1 = \$ ' " g * t # )dt # = ' " g * (!t)2
2& m
&m
)
)
t # =0
.
(5)
The y -component of the velocity of the center of mass when the jump is finished at
t = !t is then
#F
&
vcm, y,1 (!t) = % " g ( !t .
\$m
'
(6)
For the second part, once the person’s feet leaves the ground, the pushing force F = 0 , so
the acceleration is now
acm, y,2 = !g .
(7)
We now reset our clock to zero. Integrating Eq. (7) yields the change in the y component of the velocity of the center of mass
vcm, y,2 (t) ! vcm, y,2 (0) =
t " =t
#
!g dt " = !gt .
(8)
t " =0
We note that vcm, y,1 (!t) = vcm, y,2 (0) and so we can solve Eq. (8) for the y -component of
the velocity of the center of mass.
vcm, y,2 (t) = vcm, y,1 (!t) = "gt .
(9)
The person reaches the highest point of the jump when
0 = vcm, y,2 (ttop ) = vcm, y,1 (!t) " gttop
(10)
ttop = vcm, y,1 (!t) / g
(11)
Therefore
The additional displacement is given by integrating Eq. (8)
!ycm,2 =
t " = !t
#
vcm, y,2 (t " )dt " =
t " = !t
t " =0
#
(vcm, y,1 (!t) \$ gt " )dt "
(12)
t " =0
= vcm, y,1 (!t)ttop \$ (1 / 2)gt
2
top
Substituting Eq. (11) and Eq. (6) into Eq. (12) yields
2
!ycm,2
&
1 #F
= (1 / 2)vcm, y,1 (!t) / g =
" g ( !t 2
%
2g \$ m
'
2
(13)
So the maximum displacement of the center of mass of the person through both parts is
determined by adding Eqs. (5) and (13),
2
!ycm,max = !ycm,1 + !ycm,2
&
&
1# F
1 #F
= % " g ( (!t)2 +
" g ( !t 2
%
2\$ m
2g \$ m
'
'
(14)
A little algebra then yields
!ycm,max =
&
F(!t)2 # F
" g( .
%
2mg \$ m
'
(15)
Problem 3: Center of Mass of a Rod
A thin rod has length L and total mass M .
a) Suppose the rod is uniform. Find the position of the center of mass with respect to
the left end of the rod.
b) Now suppose the rod is not uniform but has a linear mass density that varies with
the distance x from the left end according to
!=
!0 2
x
L2
(1)
where !0 is a constant and has SI units [kg ! m -1 ] . Find !0 and the position of the
center of mass with respect to the left end of the rod.
Solution: (a) Choose a coordinate system with the rod aligned along the x-axis and origin
located at the left end of the rod. The center of mass of the rod can be found using the
definition
!
! r dm
!
!
1
body
R cm =
! r dm =
M body
! dm
(2)
body
!
In this expression dm is an infinitesimal mass element and r is the vector from the
origin to the mass element dm .
Choose an infinitesimal mass element dm located a distance x ! from the origin. In this
problem . x ! will be the integration variable. Let the length of the mass element be dx ! .
Then
dm =
M
dx !
L
(3)
!
The vector r = x! ˆi . The center of mass is found by integration
!
!
1
1 x
1 2
R cm =
x!
# r dm =
# x!dx! ˆi =
M body
L x!=0
2L
x!= L
x!= 0
ˆi = 1 ( L2 " 0) ˆi = L ˆi
2L
2
(4)
(b) For a non-uniform rod, the mass element is found using Eq. (1)
dm = ! ( x")dx" = ! =
!0 2
x" dx" .
L2
(5)
!
The vector r = x! ˆi . The total mass is found by integrating the mass element over the
length of the rod
x=L
M = \$ dm = \$ ! ( x")dx" =
body
x" = 0
x"= L
!0 x = L 2
!
!
!
x" dx" = 02 x"3 "= = 02 ( L3 # 0) = 0 L
2 \$
x 0
L x" = 0
3L
3L
3
(6)
Therefore
!0 =
3M
L
(7)
The center of mass is again found by integration
!
1
3 x
3 x 3
!
R cm =
! r dm =
! " ( x # ) x #dx # ˆi = 3 ! x # dx # ˆi
M body
"0 L x # =0
L x # =0
.
!
3 4 x#= L ˆ
3 4
3
R cm = 3 x #
i = 3 (L \$ 0) ˆi = L ˆi
x # =0
4
4L
4L
.
(8)
Problem 4 Walking on a Cart
A person of mass mp is standing at the right end of a cart of mass mc and length s .
The cart and the person are initially at rest. The person then walks to the left end and
stops. You may assume that there is zero rolling resistance between the cart and the
ground.
a) What is the speed of the cart when the person has finished walking?
b) When the person is finished walking in what direction and how far has the cart
moved?
c) If the person takes a time interval !t1 to go from rest to a speed u relative to the
cart, what is the average force the person exerts on the cart during this time
interval? Express your answer only in terms of m p , mc , !t1 , and u as needed.
Solution:
a)
There are no external horizontal forces acting on the system consisting of the
person and the cart. Because the momentum of the cart and person is is zero before the
person started walking, it also must be zero after the person has stopped walking.
Therefore the final speed of the cart must be zero.
b)
Because the center of mass was initially at rest and there are no external forces in
the horizontal direction, the center of mass does not accelerate and hence remains at rest
and in the same place. So we will calculate the position of the center of mass before the
person started walking and after the person finished walking, and set them equal to find
out how far the cart has moved to the right. Chose a coordinate system with the origin at
the left end of the cart before the person has started to walk.
Then the center of mass is located at
xcm =
mc xcm,cart + m p x p,o
mc + m p
We assume the cart is uniform so xcm,cart = s / 2 and the person is initially standing at the
end of the cart so x p,o = s . Therefore
xcm =
mc s / 2 + m p s
mc + m p
.
(1)
We the person reaches the left end of the cart and stops, the cart has moved a distance
x p, f . The center of mass of the cart is now located at xcm,cart = x p, f + s / 2 . Therefore
xcm =
mc (x p, f + s / 2) + m p x p, f
mc + m p
.
(2)
Now we equate Eqs. (1) and (2)
mc s / 2 + m p s
mc + m p
=
mc (x p, f + s / 2) + m p x p, f
mc + m p
and find that the final position of the person and also the distance that cart has moved is
x p, f =
mps
mc + m p
.
(3)
c) If the person takes a time interval !t1 to go from rest to a speed u relative to the cart,
what is the average force the person exerts on the cart during this time interval? Express
your answer only in terms of m p , mc , !t1 , and u as needed.
The average force that the person exerts on the cart has equal magnitude and opposite
direction of the average force that the car exerts on the person. The impulse that the car
exerts on the person is equal to the change in momentum of the person. Therefore
I x,ave = (Fx,ave ) person,cart !t1 = !px, person .
(4)
When the person is walking to the left with speed u relative to the cart, and the cart is
moving to the right with speed vc,g relative to the ground. Then the person is moving to
the left with speed v p,g = u ! vc,g relative to ground. The momentum of the person and the
cart is zero relative to the ground and so
0 = !px, person = "mp v p,g + mc vc,g = "mp (u " vc,g ) + mc vc,g
Therefore we can solve for the speed of the cart relative to the ground
vc,g =
mp
mc + mp
u.
Hence the speed of the person relative to the ground is
v p,g = u ! vc,g = u !
mp
mc + mp
u=
mc
u.
mc + mp
The change in momentum of the person is
!px, person = "mp v p,g = "
mp mc
mp + mc
u.
(5)
Substitute Eq. (5) into Eq. (4) and solve for the average force of the person by the cart
(Fx,ave ) person,cart = !
mp mc
u
.
mp + mc "t1
(6)
By Newton’s Third Law, the force on the cart by the person is then
(Fx,ave )cart , person = !(Fx,ave ) person,cart =
mp mc
u
.
mp + mc "t1
(7)
Problem 5 Stopping a Bullet
A bullet of mass m1 traveling horizontally with speed u is stopped in a block of mass
m2 that is originally at rest. Assume that the collision is nearly instantaneous. The block
then slides horizontally a distance d on a surface with kinetic friction coefficient µ k , and
then falls off the surface at a height h as shown. Neglect air resistance. Gravity with
constant gravitational acceleration g is acting on the system. Assume that all distances
are large compared to the size of the block, and ignore small corrections that arise from
the finite size of the block.
a) Determine a condition on the initial bullet speed u so that the block falls off the
surface.
b) Assume that the initial speed of the bullet u is large enough for the block to fall
off the surface (as shown). How far x f from the bottom edge of the cliff does the
block hit the ground?
Solution: We first analyze the collision in order to determine the speed va of the bullet
and block after the collision. Then we will use Newton’s Second Law to analyze the
sliding motion along the surface to determine the speed of the bullet and block when it
just leaves the surface. We can finally use projectile motion kinematics to determine
where the block and bullet hit the ground.
Let’s choose positive x -direction to the right (direction of motion of bullet). During the
collision,
Fext ,x !tcoll = !psys,x .
(1)
Because we are assuming that the collision is nearly instantaneous, we have that
0 = (m1 + m2 )va ! m1u .
(2)
Therefore the speed of the bullet and block after the collision is
va =
m1
u.
m1 + m2
(3)
While the bullet and block are sliding on the surface, the force diagram is shown in the
figure below.
There is a kinetic friction force in the horizontal direction so Newton’s Second Law in
the x -direction is
!uk (m1 + m2 )g = (m1 + m2 )ax .
(4)
ax = !uk g .
(5)
The acceleration is therefore
Let’s set our clock t = 0 immediately after the collision, and choose our origin at the
collision point as shown in the figure below.
Then integrating Eq. (5) yields
vx (t) ! vx (0) =
t " =t
#
t " =t
#
ax dt " =
t " =0
!uk g dt " = !uk gt .
(6)
t " =0
Note that vx (0) = va and so we can solve Eq. (6) for the x -component of the velocity.
vx (t) = va ! µ k gt .
(7)
We can integrate vx (t) to find the displacement
x(t) ! x(0) =
t " =t
#
vx (t " ) dt " =
t " =0
t " =t
# (v
a
t " =0
1
! uk gt " ) dt " = vat ! uk gt 2 .
2
(8)
Let t1 denote the time that the bullet and block reach the edge of the level surface, and
using x(t1 ) = d and x(0) = 0 ), Eq. (8) becomes
1
d = vat1 ! uk gt12 .
2
(9)
Setting t = t1 in Eq. (7) and solve for t1 :
t1 =
va ! vx (t1 )
.
µk g
(10)
Substituting Eq. (10) into Eq. (9) yields
2
2
2
" va ! vx (t1 ) % 1
" va ! vx (t1 ) %
1 va
1 vx (t1 )
d = va \$
' ! uk g \$ µ g ' = 2 µ g ! 2 µ g .
# µk g & 2
#
&
k
k
k
(11)
We can now solve Eq. (11) for the x -component of the velocity of the bullet and the
block at time t1
vx (t1 ) = va2 ! 2d µ k g .
(12)
Substitute Eq. (3) into Eq. (12) yielding
2
! m1
\$
vx (t1 ) = #
u & ' 2d µ k g .
" m1 + m2 %
(13)
We now apply our kinematic equations for parabolic motion when the bullet and block
leave the edge of the surface. Reset our origin at the base of the vertical wall and set
t = 0 when the bullet and block just leave the edge.
Then when the bullet and block hit the ground at t = t f
1 2
gt
2 f
x(t f ) = vx (t1 )t f
0 = y(t f ) = h !
(14)
(15)
We can solve Eq. (14) for t f
t f = 2h / g
(16)
Substitute Eq. (16) into Eq. (15) in order to find out where the bullet and block hit the
ground
x(t f ) = vx (t1 )t f = vx (t1 ) 2h / g
(17)
Now substitute Eq. (13) into Eq. (17) yielding our result that the bullet and block hit the
ground at
2
x(t f ) =
\$
2h ! m1
u & ' 4hd µ k .
#
g " m1 + m2 %
(18)
Problem 6
A rocket accelerates upward in a uniform gravitational field pointing downwards with
constant g = 9.8 m ! s "2 . The rocket has a dry mass (empty of fuel) mr ,0 = 3.6 ! 107 kg ,
and initially carries fuel with mass m f ,0 = 4.2 ! 107 kg . The fuel is ejected at a speed
u = 2.2 ! 103 m " s-1 relative to the rocket. The fuel burn time is !tburn = 160 s . What is the
final speed of the rocket after all the fuel has burned?
Solution
Choose the positive x -direction for the direction of motion of the rocket. The rocket
!
burns fuel that is then ejected backward with velocity u = !uˆi relative to the rocket,
where u > 0 is the relative speed of the ejected fuel. This exhaust velocity is independent
of the velocity of the rocket. The rocket must exert a force to accelerate the ejected fuel
backwards and therefore by Newton’s Third law the fuel exerts a force that is equal in
magnitude but opposite in direction resulting in propelling the rocket forward. The rocket
!
velocity is a function of time, v r (t) = vr ,x (t)ˆi , and the x -component increases at a rate
dvr ,x / dt . Because fuel is leaving the rocket, the mass of the rocket is also a function of
time, mr (t) , and is decreasing at a rate dmr / dt .
Let t = ti denote the instant the rocket begins to burn fuel and let t = t f denote the instant
the rocket has finished burning fuel. At some arbitrary time t during this process, the
!
rocket has velocity v r (t) = vr ,x (t)ˆi where and the mass of the rocket is mr (t) ! mr .
During the time interval [t,t + !t] , where !t is taken to be small interval, a small
amount of fuel of mass !m f is ejected backwards with speed u relative to the rocket.
The fuel was initially traveling at the speed of the rocket and so undergoes a change in
momentum. Also the rocket recoils forward also undergoing a change in momentum. In
order to keep track of all momentum changes, we define our system to be the rocket and
the small amount of fuel that is ejected during the interval !t . At time t , the fuel has not
yet been ejected so it is still inside the rocket. The figure below represents a momentum
diagram at time for our system relative to a fixed inertial reference frame.
The x - component of the momentum of the system at time t is therefore
psys, x (t) = (mr (t) + !m f )vr, x (t) .
(1)
During the interval [t,t + !t ] the fuel is ejected backwards relative to the rocket with
speed u . The rocket recoils forward with an increased x - component of the velocity
vr, x (t + !t) = vr, x (t) + !vr, x , where !vr, x represents the increase the rocket’s x component of the velocity. As usual let’s assume that all the fuel element with mass !m f
has completely left the rocket at the end of the time interval, so the x - component of the
velocity of the rocket is v f , x = vr.x + !vr, x " u . The momentum diagram of the system at
time t + !t is shown in the figure below.
The x - component of the momentum of the system at time t + !t is therefore
psys, x (t + !t) = mr (t)(vr, x (t) + !vr, x ) + !m f (vr, x (t) + !vr, x " u) .
(2)
In the figure below we show the diagram depicting the change in the x - component of
the momentum of the ejected fuel and rocket.
Therefore the change in the x - component of the momentum of the system is given by
!psys,x = !pr ,x + !p f ,x = mr (t)!vr ,x + !m f (!vr ,x " u) .
(3)
We again note that !p f ,x = !m f (!vr ,x " u) ! "!m f u , and we show the modified diagram
for the change in the x - component of the momentum of the system in the figure below
We can now apply Newton’s Second Law in the form of the momentum, for the system
consisting of the rocket and exhaust fuel:
Fext ,x = lim
psys,x (t + !t) # psys,x (t)
!t"0
!t
= lim
!psys,x
!t
!t"0
= lim
!t"0
!pr ,x
!t
+ lim
!t"0
!p f ,x
!t
.
(4)
From our diagram depicting the change in the x - component of the momentum of the
system, we have that
!m f (!vr ,x # u)
m (t)!vr ,x
Fext ,x = lim r
+ lim
.
(5)
!t"0
!t"0
!t
!t
We note that !m f !vr, x is a second order differential, therefore
lim
!m f !vr ,x
!t"0
We also note that
dvr ,x
dt
!t
! lim
"vr ,x
"t
"t#0
and
dm f
dt
! lim
"m f
"t#0
Therefore Eq. (5) becomes
Fext ,x = mr (t)
dvr ,x
dt
= 0.
"t
!
(6)
,
(7)
.
(8)
dm f
dt
u.
(9)
The rate of decrease of the mass of the rocket, dmr / dt , is equal to the negative of the
rate of increase of the exhaust fuel
dm f
dmr
=!
.
(10)
dt
dt
Therefore substituting Eq. (10) into Eq. (9), we find that the differential equation
describing the motion of the rocket and exhaust fuel is given by
Fext ,x !
dv
dmr
u = mr (t) r ,x .
dt
dt
(11)
Eq. (11) is called the rocket equation. For a rocket launch at the surface of the earth
Fext ,x (t) = !mr (t)g .
(12)
Substitute Eq. (12) into Eq. (11) and multiply each side by dt :
!mr (t)gdt = mr (t)dvr + dmr u
(13)
Now divide through by mr (t) and after rearranging terms we have
dvr = !gdt !
dmr
u
mr (t)
(14)
We now integrate both sides
vr, f
!
vr,i =0
mr , f
dvr = "
!
mr ,i
t
f
dmr
u " ! g dt .
mr (t)
0
(15)
where mr ,i is the initial mass of the rocket and the fuel. Integration yields
" mr , f %
vr , f (t f ) = !u ln \$
' ! gt f .
# mr ,i &
(16)
After all the fuel is burned at t = t f , the mass of the rocket is equal to the dry mass
mr , f = mr ,d and so
vr , f (t f ) = u ln R ! gt f .
(17)
The initial mass of the rocket included the fuel is
mr ,i = mr ,0 + m f ,0 = 3.6 ! 107 kg + 4.2 ! 107 kg = 7.8 ! 107 kg
(18)
The ratio of the initial mass of the rocket (including the mass of the fuel) to the final dry
mass of the rocket (empty of fuel) is
R=
mr ,i
mr ,d
=
7.8 ! 107 kg
= 2.2
3.6 ! 107 kg
(19)
The final speed of the rocket is then
vr , f = u ln R ! gt f = (2.2 " 103 m # s-1 ) ln(2.2) ! (9.8 m # s !2 )(160 s) = 1.33 " 102 m # s-1 .(20)
Problem 7 Filling a Freight Car
A freight car of mass mc,0 , open at the top, is coasting along a level track with negligible
friction at a speed vc,0 , when it begins to rain hard. The raindrops fall vertically with
respect to the ground.
a) In a time interval [t , t + !t ] , an amount of water !mw enters the freight car.
Choose a system. Is a component of the momentum of your system constant?
Write down a differential equation that results from the analysis of the momentum
changes inside your system, in terms of the mass mc of the freight car and rain at
time t the horizontal component vc of the velocity of the freight car at time t , the
infinitesimal change dmc in the mass of the freight car due to the added rain, and
the infinitesimal change dvc in the horizontal component of the velocity of the
freight car.
b) What is the speed of the freight car when it has collected a mass mw of rain? You
may solve this part by integrating your differential equation or by using an
alternate method. If you use an alternate method, please state the physical
principles that are involved.
Solution.
Choose positive x -direction to the right in the figure below. Define the system at
time t to be the car with whatever rain is in it and the rain that falls into it during the
time interval [t,t + !t] . Denote the mass of the car and rain at time t by mc (t) and
let !mw denote the rain that falls into the car during the time interval [t,t + !t] . The
rain has no x -component of velocity. At time t , the car is moving with x -component
of the velocity vc (t) . At time t + !t the car is moving with x -component of the
velocity vc (t) + !vc . The momentum diagrams for time t and t + !t are shown
below.
The diagram representing the change in the change in the x -component of the
momentum is shown below.
There are no external forces in the x -direction acting on the system, Fext ,x = 0 , so the
momentum principle becomes
0 = !psys,x = !pc,x + !pw,x
(1)
From our diagram showing the change in the x -component of the momentum of the
elements of the system, Eq. (1) becomes
0 = mc !vc + !mwvc (t)
(2)
where we can ignore the contribution form the second order term !m!vc . Because the
cart’s mass is increasing due to the material entering we have that
!mw = !mc .
(3)
and so Eq. (2) can now be written after taking limits as !t " 0
mc (t)
dvc
dm
= ! c vc (t) .
dt
dt
(4)
We can multiply both sides of Eq. (4) by dt yielding
mc (t)dvc = !dmc vc (t) .
This is now separable and collecting terms we have
(5)
dvc
dmc
.
=!
vc (t)
mc (t)
(6)
We can integrate both sides paying careful attention to make a consistent choice for the
limits of the integrals
vc! = vc (t )
m ! = m (0)
c
c
dvc!
dmc!
=
#
" v!
" m! .
v ! = v (0)
m ! = m (0)
c
c
(7)
! v (t) \$
! m (t) \$
! m (0) \$
ln # c & = ' ln # c & = ln # c & .
" vc (0) %
" mc (0) %
" mc (t) %
(8)
vc (t) mc (0)
.
=
vc (0) mc (t)
(9)
c
c
c
c
Integrating yields
Exponentiating yields
The x -component of the velocity of the cart at time t is then
vc (t) =
mc (0)
v (0) .
mc (t) c
(10)
Very Simple Alternative Method:
Because the rain does not add any horizontal momentum and there are no external forces,
the x -component of the momentum of the system must be constant. Therefore
mc (t)vc (t) = mc (0)vc (0) .
(11)
We immediately solve this equation for x -component of the velocity of the cart at time t
vc (t) =
mc (0)
v (0) .
mc (t) c
which is in agreement with our first approach.
(12)
Problem 8 Bicycling in the Rain
A bicycle rider is caught in a fierce rainstorm that begins at t = 0 when the rider is
traveling at speed v0 . At ground level the rain is moving horizontally at the rider with
speed u relative to the ground. The rider stops pedaling as the rain sticks to the rider. The
initial mass of the bicyclist is m0 (including mass of bicycle). You may ignore all
resistance.
a) Derive a relation between the differential of the speed of the bicyclist, dv , and the
differential of the total mass of the bicyclist, dm .
b) Integrate the above relation to find the speed of the bicyclist as a function of mass,
v(m) .
c) Assume that the rate that the mass is added to bicyclist is constant,
dm / dt = b > 0 , (this is a simplifying assumption that is only an approximation).
How long does it take the bicyclist to come to a stop?
Solution:
Define the mass in our system as the mass mb (t) of the bicyclist at time t and small mass
of the rain that strikes and sticks to the bicyclist during the time interval [t, t + !t] . We
show the momentum diagrams for time t and t + !t below.
The change in momentum of the system consisting of the bicyclist and the added rain is
shown in the figure below.
There are no external forces in the x -direction so the momentum principle becomes
0 = mb (t)
dvb dmr
+
(v (t) + u) .
dt
dt b
(13)
dmb dmr
=
.
dt
dt
(14)
We know that
So Eq. (13) becomes
mb (t)
dvb
dm
= ! b (vb (t) + u) .
dt
dt
(15)
After some rearrangement we have that
dvb
dmb
.
=!
(vb (t) + u)
mb (t)
(16)
We can integrate both sides
vb! = vb (t )
"
vb! = v0
dvb!
=#
( vb! + u)
mb! = mb (t )
"
mb! = m0
dmb!
.
mb!
(17)
yielding
! v (t) + u \$
! m \$
ln # b
= ln # 0 & .
&
" v0 + u %
" mb (t) %
(18)
vb (t) + u
m0
.
=
v0 + u
mb (t)
(19)
Exponentiation yields
We can solve Eq. (19) for the x -component of the velocity of the bicyclist at time t
vb (t) =
m0
m
m
(v0 + u) ! u = u( 0 ! 1) + v0 0 .
mb (t)
mb (t)
mb (t)
(20)
Note that in the limit that t ! " , mb (t) ! " , and vb (t) ! "u which is what we expect.
The bicyclist moves backwards with the rain.
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