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Momentum and Impulse
Challenge Problem Solutions
Problem 1: Estimation
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.
Problem 1 Solution:
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 eighteen-
wheelers. 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
5! 1
of magnitude momentum magnitude of 10 kg " m s".
4
d) For the mass of the plane, let’s use the value of 5! 10 kg cited in Problem Set 3
and a cruising speed of 200 m s
! 1
" (roughly two-thirds the speed of sound) for a
6! 1
momentum magnitude of 10 kg " m s".
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
6! 1
this case, the magnitude of the momentum is 10 kg " m " s. If anything, this
estimate is low, in that it’s a short train at slow speeds, and we haven’t included
the engine.
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 17.1b is with respect to a baseline force of
700 N and not^ 0 N.
c) Explain how you can use the data in Figure 17.1a 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) t 1
= 2.75 s ,
(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 17.2?
d) What do you expect the impulse to equal for the time interval (^) [ t
1
= 2.75 s, t f
= 12.0 s]?
Problem 2 Solutions:
a) 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 expression for impulse. 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 (^) t 1
= 2.75 s. We are given ( via^ a
numerical integration) 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.
Solving Eq.( 3. 3 ) at t =! t col
yields
v
! t col
y , o
( 3. 4 )
a cm
Substitute Eq. ( 3. 4 ) into Eq. ( 3. 2 ) and solve for the acceleration giving
( v )
2
a =!
y , o
( 3. 5 ) cm
2 " y
Substitute Eq. ( 3. 5 ) into Eq. ( 3. 4 ) and solve for the collision time
2! y
! t = ( 3. 6 ) col
v y , o
The initial y-velocity component from Eq. ( 3. 1 ) is v =! 2 gh , the displacement is y , o 0
! y = "! d , so with these values the collision time is
3
! t s ( 3. 7 ) col
2! d
2 gh 0
2
m)
2(9.8 m $ s
)(2.0 m)
b) Find the average force of the ground on the person during the collision.
If we treat the person as the system then there are two external forces acting on the
person, the gravitational force, F grav
=! mg
j , and a normal force between the ground and
the person, F = N
j. This normal force varies with time but we shall consider the ground
time average, ( F ground
ave
= N
ave
j.
The impulse is equal to the change in momentum so
total
F! t = " mv ( 3. 8 ) y col y ,
Using our expressions for the total force, and the collision time and, we have
2 " y
( N^!^ mg^ )% ( =^! mv^ y ,
ave
v y , o '
Solving for the average normal force
m ( v y ,
2
N
ave
= mg! ( 3. 10 )
2 " y
Finally substituting in the initial y – component of the velocity and the displacement. the
average normal force is
h (^) (2.0 m)
N
ave
= mg (1 +
0
) = (60 kg)(9.8 m " s
)(1 + ) = 1.2 # 10
5
N ( 3. 11 )
! d (^) (1 # 10
$ 2
m)
c) What is the ratio of the average (normal) force of the ground on the person to the
gravitational force on the person? Can we effectively ignore the gravitational force during
the collision?
N h $^ (2.0 m) ' ave 0 = (1 + ) = 1 +
2
N ( 3. 12 )
mg! d %
2
m)
h 0
h 0
Notice that the factor (1 + )! = 200 so we can effectively ignore the external
! d! d
gravitational force during the collision.
d) Will the person break his ankle?
We approximate the compressional force per area by the average normal force divided by
the smallest cross sectional area of the tibia
N
ave
mg h 0
P = = (1 + ) = (1.2 " 10
5
N)/(3.2 " 10
4
m
2
) = 3.7 " 10
8
N $ m
( 3. 13 )
A A! d
The compressive force per area necessary to break the tibia in the lower leg is about
F / A = 1.6! 10
8
N " m
, so this fall is enough to break the tibia.
Remark: In order to find the minimum displacement to avoid breaking the bone, we
could set
mg h 0
mg h 0 -
(1 + )! ( ) = 1.6 " 10
8
N # m ( 3. 14 )
A! d A min
! d min
Thus
Problem 4: Momentum and Impulse
A superball of m 1
, starting at rest, is dropped from a height h 0
above the ground and
bounces back up to a height of h f
. The collision with the ground occurs over a time
interval! t. c
a) What is the momentum of the ball immediately before the collision?
b) What is the momentum of the ball immediately after the collision?
c) What is the average force of the ground on the ball?
d) What impulse is imparted to the ball?
e) What is the change in the kinetic energy during the collision?
Problem 4 Solutions:
For all parts, take the upward direction to be the positive
j direction, and use the ground
to be the zero point of gravitational potential energy.
a) The initial mechanical energy is E = m gh ; all potential, no kinetic energy. The 0 1 0
energy when the ball first hits the ground is E = m v
2
/ 2 , all kinetic, no potential energy. 1 1 1
Equating these energies (with the assumption that there are no nonconservative forces)
and solving for the speed v 1
gives v 1
= 2 gh 0
, for a momentum
p
= m 2 gh! j. ( 4. 1 ) ( )
1 1 0
b) In order to find the momentum, we need to know the speed. The speed is found, again,
through conservation of mechanical energy. Denote the upward speed after the collision
as v 2
, with corresponding mechanical energy E = m v
2
/ 2. This energy must be the 2 1 2
same as the potential energy at the top of the bounce, E = m g h. Solving for the 3 1 f
rebound speed v 2
gives v 2
= 2 g h f
, for a momentum
p
= m 2 gh ( )
j. ( 4. 2 ) 2 1 f
c) While the ball is in contact with the ground, the average net force is the change in
momentum divided by the time interval,
F =
! p p " p (^) f 0
! t
h + h
2 1
= m 1
2 g j. ( 4. 3 ) net, ave
! t! t c c
However, there are two forces on the ball, the contact force that the ground exerts on the
ball and the gravitational force. The net force is the sum of these forces, and so
F = F + F = F! m g
j. ( 4. 4 ) net ball, ground grav ball, ground 1
The gravitational force must be presumed to be constant, and hence equal to its average
value. Combining the terms (watch the subscripts),
f 0
% t
h + h
F = F + m g
j = m #
2 g + g $
j. ( 4. 5 ) ground, ball, ave net, ave 1 1
c & '
d) The impulse is the change in momentum,
)
! p = p " p = m j. ( 4. 6 ) 2 1 1
e) The change in kinetic energy during the collision is the difference between the energies
found from the heights before and after the collision,
! K = E " E = m gh " m gh = m g h (
" h )
0 2 1 0 1 f 1 0 f
( (^0)
f
g h + h
b) Let us first calculate the kinetic energy of the ball just before it hits the ground.
Consider the states B and C. The mechanical energy of the system is conserved between
these states, as expressed in the equations
( U
C
! U
B
) + ( K
C
! K
B
K
C
= mg y max
Next determine the kinetic energy of the ball just after it bounces up from the ground.
Consider the states C and D. The mechanical energy of the system is conserved between
these states, so that
K y ( 5. 3 ) D
= mg max
%
The change in the kinetic energy of the ball due to its collision with the ground is then
mg y max
K. ( 5. 4 )
D
! K
C
= mg y
'
! mg y =! max max
c) Choose the ball alone as the system. The impulse imparted to the system from the
ground is the change in momentum of the system (ball);
I =! p
= m! v
= m v D
j " v C
" j_. ( 5. 5 )_
Determine the velocity at points C and D from the kinetic energies.
From Equation ( 5. 2 ),
mv C
2
= mg y , ( 5. 6 )
max
which gives
v = 2 g y. ( 5. 7 ) C max
From Equation ( 5. 3 ),
2
mv ( 5. 8 ) D
= mg y max
%
which gives
v C
= g y. ( 5. 9 )
max
Using Equations ( 5. 7 ) and ( 5. 9 ) in Equation ( 5. 5 ) yields the expression for the velocities
in Equation ( 5. 5 ),
I = (^) ( D
)
j = m
m v g y + 2 g y j = m g y max
j. ( 5. 10 ) max max
d) The impulse exerted by the ground on the ball is also I = F ave
! t collision
where
! t is the time of collision. Using the expression for the impulse from collision
Equation ( 5. 10 ) to determine the average force,
! t
m g y max ! I ˆ F = = j. ( 5. 11 ) ave
! t
