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MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
Physics 8.01
Fall 2007
Problem Set 6 Momentum and Impulse
Solutions
Problem 1 (10 points): Estimation and concept Questions
Part 1) (2 points) Estimate the magnitude of the momentum of the following moving
objects: a) a person walking; b) a car moving along a highway; c) a truck moving along a
highway; d) a passenger jet at cruising speed; e) a freight train transporting coal.
Indicate your choice of reference frame for each case.
Part 2) (4 points): A force of magnitude F is applied to a dumbbell first as in (a) and
then as in (b). In which case does the dumbbell acquire the greater center-of-mass
acceleration? Explain your reasoning.
Part 3) (4 points): Skater A of mass 75 kg and skater B of mass 50 kg are initially at rest
some distance apart. Each skater holds tightly onto a rope of negligible mass. Skater A
pulls on the rope with constant force so that the skaters approach each other and meet.
The ice is completely frictionless. Which statements below are true and which are false?
Explain your reasoning.
1. Only skater B moves relative to the ice.
2. The magnitude of the acceleration of skater A is less than the magnitude of the
acceleration of skater B.
3. Just before they meet, the speed of skater A is less than the speed of B.
4. While the skaters are moving, their momentum vectors have equal magnitudes.
Answers for Part 1):
These must be “answers” instead of “solutions,” as a wide range of values for the
estimated quantities could be used. Note that for common experience or everyday
examples, it’s sometimes convenient to make estimates in English units and convert to
metric. For these answers, scientific notation is not always helpful.
All of the following estimates are with respect to a stationary observer on the ground.
Using a frame moving with the walker or the respective vehicle is not really in the spirit
of the problem.
a) Time yourself while walking. For some of us, 1m ⋅ s −1 is more of a stroll, while
2 m ⋅ s −1 is a fairly fast clip. So, using a mass of 100 kg and a walking speed of
1.5m ⋅ s −1 , the magnitude of the momentum is 150 kg ⋅ m ⋅ s −1 . Larger walkers
(such as the writer of these answers, who could stand to lose a few pounds) have
in general a larger mass and a longer stride and a faster walking pace.
b) For a small car, use a mass of 1000 kg . For highway speeds, the exact conversion
in English units (still used in this country) is 60 mph = 88 ft / sec , or about 25 30 m ⋅ s −1 , for an approximate momentum magnitude of ∼ 2500 kg ⋅ m ⋅ s −1 .
c) Trucks vary greatly in mass, from lightweight pickups to the tandem eighteenwheelers. I’m used to a ton-and-a-half flatbed (that’s the capacity, not the
weight), so five tons (the weight limit of a small bridge) would mean a mass of
roughly 5000 kg . Using the same speed from part b) of 25m ⋅ s −1 gives an order
of magnitude momentum magnitude of 105 kg ⋅ m ⋅ s −1 .
d) For the mass of the plane, let’s use the value of 5 ×104 kg cited in Problem Set 3
and a cruising speed of 200 m ⋅ s −1 (roughly two-thirds the speed of sound) for a
momentum magnitude of 106 kg ⋅ m ⋅ s −1 .
e) Lots of crude estimates are needed here. We’re not given the number of cars,
whether the coal is anthracite or bituminous, and some of us would have to guess
at the size of a car and the density of coal. Suppose instead that each car has the
same magnitude of momentum as the truck in part c) and that we have a train of
ten cars (seems low, but this train may be dealing with mountainous terrain). In
this case, the magnitude of the momentum is 106 kg ⋅ m ⋅ s −1 . If anything, this
estimate is low, in that it’s a short train at slow speeds, and we haven’t included
the engine.
Part 2) answer:
It doesn’t matter. For practical purposes, it might be hard to maintain a force in a
constant direction for case (b), as the dumbbell would tend to rotate (we’ll deal with this
when we get to rotational motion, of course), but as long as the forces are applied in the
same direction, the acceleration of the center of mass will be the same. You might
anticipate (or you might have seen in some other physics class) that in case (b) the upper
end of the dumbbell, where the force is applied, will initially move twice as fast as the
center of mass. Still, the center of mass will accelerate the same in both cases.
Part 3) answer:
Before giving the answers, consider that saying that “Skater A only pulls on the rope” is
perhaps misleading (not intentionally, of course). The force that skater A exerts on the
rope and the force that the rope exerts on skater A are an action-reaction pair, as are the
forces that skater B and the rope exert on each other. Thus, to hold onto the rope,
skater B must be pulling on the rope as well; as the problem states, skater B “holds
tightly” to the rope. If we assume a massless rope, as given in the problem statement,
then the magnitude of the net force on each skater is the same.
Thus, statement (1) is false. Statements (2) and (3) are true, and of course are sort of the
same statement, in slightly different forms, given that both skaters are initially at rest.
Since the mass of the rope is neglected, the magnitude of the net force is the same, as
mentioned above, and as there is no net force on the skaters/rope system, the momenta of
the skaters must be equal in magnitude (but opposite in direction).
Problem 2 (10 points): Change of Momentum and Instantaneous Collisions:
Plane landing on runway
A light plane with mass 1100 kg makes an emergency landing on a short runway. With
its engine off, it lands on the runway at a speed of 35 m ⋅ s −1 . A hook on the plane snags a
cable attached to a 100-kg sandbag and drags the sandbag along. You may assume that
the collision with the sandbag is instantaneous. If the coefficient of friction between the
sandbag and the runway is μk = 0.4 , and if the plane’s brakes give an additional retarding
force of magnitude 1300 N , how far does the plane go before it comes to a stop?
Solution:
Given that the collision of the plane with the sandbag-cable combination is instantaneous
(which might be dangerous, but that’s not part of the problem), the speed v f of the planesandbag combination after the collision is obtained from the constancy of the momentum
in the direction of motion of the plane,
mplanevplane, 0 + msand vsand, 0 = mplanevplane, f + msand vsand, f
(1100 kg ) ( 35 m ⋅ s −1 ) + (100 kg )( 0 ) = (1200 kg ) v f
⎛ 1100 ⎞
−1
−1
vf = ⎜
⎟ 35 m ⋅ s = 32 m ⋅ s
⎝ 1200 ⎠
(2.1)
to two significant figures.
To find the distance traveled after the sandbag is hooked, we find the magnitude of the
deceleration from the known forces and Newton’s Second Law and hence the time the
plane takes to come to a stop and hence the distance traveled. Symbolically, the
magnitude of the net force is Fnet = Fbrake + μk msand g and the magnitude of the deceleration
is a = Fnet / mtotal . The time to come to a stop is tstop = v f / a and in this time the plane
travels a distance
v 2f
1 2
Δx = a tstop =
2
2a
v 2f
(2.2)
mtotal
.
=
2 Fbrake + μ k msand g
The above numerical calculation was done keeping arbitrary precision in all intermediate
calculations, including v f found above and 9.81m ⋅ s −2 . The numerical result is
Δx = 360 m
(2.3)
to two significant figures. (Using g = 9.8m ⋅ s −2 gives Δx = 370 m to significant figures.)
We will see in Chapter 9 that using the work-energy theorem allows a more direct
calculation of the distance.
Problem 3 (10 points): Impulse and Center of Mass Velocity:
Elevator Ride
A person is riding the Building 16 elevator, while standing on the force platform that was
used in the lecture demonstration on Monday/Tuesday Oct 15/16. The force platform
measures the force of the person’s feet on the platform. The data in Figure 3.1a shows a
plot of the force of platform on the person (normal force) vs. time for the person in the
elevator.
Figure 3.1a
Figure 3.1b
Figure 3.1b highlights the area for the time interval [t1 = 2.75 s, t 2 = 6.35 s ] . The impulse
for this interval is:
I [t1 , t2 ] =
t2 =6.35 s
∫
F total (t ) dt = ( −174 N ⋅ s ) ˆj
(3.1)
t1 = 2.75 s
Figure 3.2 shows a plot of the vertical component of the center of mass velocity of the
person as a function of time.
Figure 3.2: Velocity as a function of time
a) What does the negative sign signify in Equation (3.1) in terms of the choice of positive
direction for a relevant coordinate axis? Is the choice of positive direction up or down?
Based on the plot in Figure 3.1a, describe the motion of the elevator during the interval
⎡⎣ t1 = 2.75 s, t f = 12.0 s ⎤⎦ . In particular, identify any distinct stages. For each distinct stage
describe whether the elevator is moving up or down, whether or not the acceleration is
constant (or zero), or non-constant. Explain your reasoning.
b) Explain why the shaded area in Figure 3.1b is with respect to a baseline force of
700 N and not 0 N .
c) Explain how you can use the data in Figure 3.1a to calculate the change in the vertical
component of the velocity of the center of mass of the person. In particular, what is the
vertical component of the velocity of the center of mass of the person at (i) t1 = 2.75 s
and (ii) t 2 = 6.35 s ? How does your calculation compare with the results for the vertical
component of the velocity of the center of mass of the person shown in the graph in
Figure 3.2?
d) What do you expect the impulse to equal for the time interval ⎡⎣ t1 = 2.75 s, t f = 12.0 s ⎤⎦ ?
Solutions:
a) Note: Due to an editing error, the unit vector ˆj in Equation (3.1) was not included in
earlier versions of this problem. We regret this oversight, and apologize for any
confusion.
The downward force that the person exerts on the scale has decreased in magnitude, and
hence the magnitude of the upward force that the scale exerts on the person has
decreased. The force that the scale exerts on the person is less than the person’s weight,
and hence the minus sign in Equation (3.1). The positive direction must be upward;
scales as a rule do not measure forces tending to move the top part of the scale, where the
person is standing, upward. The plot in Figure 3.1a is for the elevator (and the scale, and
the person) descending by accelerating downward, moving at more or less constant speed
(zero acceleration) for about three seconds and then accelerating upward while still
moving down until the elevator comes to a stop.
b) There are two forces on the person, the downward gravitational force (the person’s
weight) and the upward force that the scale exerts on the person’s feet. The net impulse
is found from the difference between these two forces, and hence the person’s weight,
approximately 700 N as seen from the graph before the elevator began its descent and
after it stopped, is subtracted from the scale reading. To restate one of the answers to
part a), for the period when the elevator is accelerating downward, the net impulse is
negative if the positive direction is upward.
c) The product of the person’s mass and velocity is the integral of the net force with
respect to time, which is the area between the horizontal line representing the person’s
weight and the curve representing the force on the scale. For calculation purposes, this
area can be calculated numerically by a computer. From an eyeball examination of the
graph in Figure 3.2, the elevator is not moving at t1 = 2.75 s . We are given (via a
numerical integration using LabView) that the vertical component of the net impulse in
the interval [ 2.75 s, 6.35 s ] was −174 N ⋅ s . The person’s mass is 700 N / g , giving a
vertical component of velocity of −2.44 m ⋅ s −1 (which is likely more significant figures
than we need or want).
d) Between these two times, the elevator has descended and ideally has come to a stop.
The impulse, the net change in momentum, should be zero. From Figure 3.2, this is
almost the case, and if the elevator and person are not at rest at the end of the trip, either
the elevator is dangerous or there was some error in the computer’s numerical integration
process.
Problem 4 (10 points): Recoil
A person of mass m1 is standing on a cart of mass m2 that is on ice. Assume that the
contact between the cart and the ice is frictionless. The person throws a ball of mass m3
at an angle of θ with respect to the horizontal as measured by the person in the cart. The
ball is thrown with a speed v0 with respect to the cart.
a) What is the final velocity of the ball as seen by an observer fixed to the ground?
b) What is the final velocity of the cart as seen by an observer fixed to the ground?
c) What angle, with respect to the horizontal, does the fixed observer see the ball
leave the cart?
Solution:
a) Our reference frame is fixed to the ground. We shall take as our initial state to be
before the ball is thrown and our final state to be after the ball is thrown. As stated in the
problem, we are assuming that the contact between the wheels and ice is frictionless so
there are no external forces acting in the horizontal direction.
The initial x -component of the total momentum is zero,
pxtotal
,0 = 0 .
(4.1)
After the ball is thrown, the cart and person have a final momentum
p f ,cart = − ( m2 + m1 ) v f ,cart ˆi
(4.2)
as measured by an observer on the ground, where v f ,cart is the speed of the person and
cart. The ball is thrown with a speed v0 and at an angle θ with respect to the horizontal
as measured by the person in the cart. Therefore the person in the cart throws the ball
with velocity
v′f ,ball = v0 cosθ ˆi + v0 sin θ ˆj .
(4.3)
Since the cart is moving in the negative x -direction with speed v f ,cart just as the ball
leaves the person’s hand, the x -component of the velocity of the ball as measured by the
observer on the ground is given by
(v
) = ( v cosθ − v
f , ball x
0
f ,cart
).
(4.4)
The ball appears to have a smaller x -component of the velocity according to the observer
on the ground. The velocity of the ball as measured by the observer on the ground is
v f , ball = ( v0 cosθ − v f , cart ) ˆi + v0 sin θ ˆj .
(4.5)
b) The final momentum of the ball according to the observer on the ground is
p f , ball = m3 ( v0 cosθ − v f , cart ) ˆi + m3v0 sin θ ˆj .
The momentum flow diagram is then:
(4.6)
Since the total x -component of the momentum of the system is constant, we have
0 = ( p f , cart ) + ( p f , ball ) = − ( m2 + m1 ) v f , cart + m3 ( v0 cosθ − v f , cart ) .
x
x
(4.7)
We can solve Equation (4.7) for the final velocity of the cart as measured by the observer
on the ground,
v f , cart =
m3v0 cosθ
.
m2 + m1 + m3
(4.8)
Note that the y -component of the momentum is not constant because as the person is
throwing the ball they are pushing off the cart, and the normal force that the ground
exerts on the cart exceeds the gravitational force, and so the net external force in the y direction is non-zero.
c) The angle (we’ll call it φ ) that the ball is thrown as seen by the observer on the ground
is given by
⎡ ( v f , ball ) ⎤
⎡
⎤
v0 sin θ
y
⎥ = tan −1 ⎢
φ = tan ⎢
⎥
v0 cosθ − v f , cart ⎦⎥
⎢ ( v f , ball ) ⎥
⎢
⎣
x ⎦
⎣
⎡
⎤
v0 sin θ
⎥
= tan −1 ⎢
T
⎢⎣ v0 cosθ − ( m3v0 cosθ / m ) ⎥⎦
−1
(4.9)
where mT = m1 + m2 + m3 is the total mass. After a little algebra we get for the angle
⎡
⎤
⎡
⎛
mT sin θ
m3 ⎞ ⎤
−1
⎥ = tan ⎢ tan θ ⎜ 1 +
⎟⎥ .
+
+
m
m
cos
θ
m
m
(
)
1
2
1
2
⎝
⎠⎦
⎣
⎣
⎦
φ = tan −1 ⎢
(4.10)
From Equation (4.10), φ > θ ; the observer on the ground sees the ball moving closer to
the vertical.
Problem 5 (10 points): Continuous Mass Transport:
Rocket Take-off
A rocket ascends from rest in a uniform gravitational field by ejecting exhaust with
constant speed u relative to the rocket. Assume that the rate at which mass is expelled is
given by dm / dt = γ m , where m is the instantaneous mass of the rocket and γ is a
positive constant. The rocket is retarded by air resistance with a force of magnitude
F = bm v that is proportional to the instantaneous momentum of the rocket, where b is a
positive constant with dimensions of inverse time, ⎡⎣ time −1 ⎤⎦ . Find the velocity of the
rocket as a function of time. (You may need to first derive a differential equation using
the idea that external force changes the momentum of a system. Make sure you use the
sign conventions given in the problem.)
Solution:
To find the rocket’s velocity as a function of time, we first need to find how the velocity
changes with respect to time, the differential equation mentioned in the problem
statement. Three methods will be presented here, all using momentum considerations.
The relation dm / dt = γ m , γ > 0 may seem inconsistent with other usages, but the
problem statement is correct; the fuel is being exhausted at a rate γ m > 0 . A more
dm
= γ m . In any event, the expulsion of the
mathematically precise statement would be
dt
exhausted fuel must provide an upward thrust to the rocket.
For all methods, take the positive direction to be upward and assume that the rocket
velocity is upward as well. For purposes of space and clarity, the positive upward
direction is shown as being to the right in the figures. The gravitational force and air
resistance force are then in the negative direction, to the left in the figures.
Method I:
As suggested by the figure, take the system to be the rocket and fuel combination. The
small square (red, if viewed in color) represents the differential mass dm of fuel ejected
in the differential time dt . The initial momentum in the forward (upward, but to the right
in the figure) direction is
pinitial = m v
(4.11)
where m is the combined mass of the rocket and fuel. The final momentum is
pfinal = ( m − dm )( v + dv ) + dm ( v + dv − u )
= ( m v + m dv − v dm ) + ( v dm − u dm )
(4.12)
= m v + m dv − u dm
and so the change in momentum is
dp = pfinal − pinitial = m dv − u dm .
(4.13)
Note that in going from the second line in (4.12) to the third, the second-order differential
terms dm dv cancelled nicely; if they had not cancelled, we would have discarded them
anyway.
Method II:
Again, the system is the rocket-fuel combination, but with m denoting the mass of the
rocket after the fuel has been ejected. For this model, we have
pinitial = ( m + dm ) v
pfinal = m ( v + dv ) + dm ( v + dv − u )
pfinal − pinitial
= m v + m dv + v dm − u dm + dm dv
= m dv − u dm + dm dv
(4.14)
= m dv − u dm.
Note that we did have to discard the dm dv term “by hand.”
Both Methods I and II, which lead to the same expression for the change in momentum
of the fuel-rocket combination, next use that the change in momentum divided by the
differential time dt is the net external force,
m dv − u dm
= −m g − b m v
dt
1 dm
dv
=u
− g −bv
dt
m dt
= γ u − g − b v.
(4.15)
Method III:
Consider just the rocket, of mass m , as the system. Then, the force that the ejected fuel
exerts on the rocket is an external force. Denote the impulse of this force, delivered over
a time dt , as dI rocket, fuel , the order of the subscripts being a reminder that this is the
impulse on the rocket. The change in momentum of the rocket is then the sum of the
external impulses,
dp = m dv = dI rocket, fuel + dI rocket, gravity + dI rocket, air
(4.16)
These three differential impulses are known to be:
dI rocket, fuel = −dI fuel, rocket = −dm ( −u )
dI rocket, gravity = −m g dt
(4.17)
dI rocket, air = −b m v dt.
Inserting the expressions in (4.17) into Equation (4.16) and dividing by dt yields
m
dv
dm
=u
− m g −bmv ,
dt
dt
which is equivalent to the expression in (4.15).
(4.18)
As a check for either method, note that the signs of the terms come out correctly; the term
representing the ejected fuel is a positive upward force (a good idea for rockets), while
gravity and air resistance both tend to bring the rocket back to earth. The quicker the fuel
is burned and the faster the fuel is ejected, the more thrust on the rocket.
On to the Differential Equation:
We’ve done the physics, but there are several interesting aspects to the calculation of the
velocity as a function of time. The mathematical problem was considered in Section 5.10
of the Course Notes, “Linear Friction with Gravity,” reproduced here in
abbreviated form. Be sure to realize that the symbol “ γ ” is used differently in the notes,
and that in the notes, which address a falling object, the positive direction is downward.
Rewrite the last expression in (4.15) or Equation (4.18) as
dv
= γ u − g −bv
dt
dv / dt
1d
=−
ln (γ u − g − b v ) = 1
b dt
γ u − g −bv
(4.19)
⎡γ u − g − b v ⎤
ln ⎢
⎥ = −bt
⎣ γ u−g ⎦
where the term in the denominator of the argument of the logarithm corresponds to a
constant of integration, set so that v ( t = 0 ) = 0 .
Exponentiating both sides of the last expression in (4.19) and solving for the
velocity gives
v (t ) =
γ u−g
b
= vterminal
(1 − e )
(1 − e )
− bt
(4.20)
− bt
where vterminal = (γ u − g ) / b .
There are several aspects of the result in (4.20) that are worth considering. First, unless
γ u > g , the rocket doesn’t get off the ground. Next, the rocket continues to accelerate,
only reaching terminal velocity in the limit t → ∞ ; the air resistance force seems to have
wimped out. Also, what happens in free space, where g → 0 and b → 0 ?
Explaining the last question first might make things clearer. Consider that as b → 0 ,
2
1 − e −bt 1 − (1 − bt + bt / 2 +
=
b
b
) ≅t;
(4.21)
in the first expression in Equation (4.20), v ( t ) → (γ u − g ) t , constant acceleration!
Indeed, if we had set b = 0 in the first expression in Equation (4.19), we would have
recognized the equation for constant acceleration.
For those not familiar with the Taylor series for e − bt used in Equation (4.21), consider the
substitution s = bt and the limit
⎡ 1 − e− s ⎤
1 − e − bt
d
= lim ⎢t
= t (1 − e − s ) = t .
⎥
s →0
s
→
0
s =0
b
s ⎦
ds
⎣
lim
(4.22)
The key point in considering the result of (4.20) is in the model for the rate at which the
fuel is exhausted,
dm
=γ m ,
dt
(4.23)
where on the left side of (4.23) m is the total mass of all the ejected fuel and on the right
m is the sum of the mass of the rocket and the remaining fuel. Equation (4.23) would
tell us that the mass of the rocket and remaining fuel is m = M 0 e −γ t . The mass being
accelerated decreases exponentially, the gravitation force and the air resistance force,
being proportional to the mass, decrease as well, and so the rocket can continue to
accelerate indefinitely.
The product u ( dm / dt ) is sometimes called the “thrust” (ask your Course XVI friends);
check to see that the thrust has dimensions of force. In the model for this problem, we
would have that the “thrust force” is
Fthrust = u
dm
=γ um,
dt
(4.24)
an unlikely feature of realistic rocket design. At best, even if Equation (4.24) were an
approximate model, the thrust would have to vanish when the fuel runs out.
The bottom line is that the model for the fuel burn rate was given in the form it was in
order to make solving for the velocity as a function of time possible. Maybe a neat math
problem, but the physics was done once we found the rate of change of momentum.