8.1 Linear Impulse and Linear Momentum
Fig. 8 – 1: Different types of hammers
produce different types of impulsive
forces. DK2
In engineering, linear impulse is defined as the integral over time of force.
When the force acts over a small amount of time it’s called an impulsive force. The
hammer, for instance, is designed to create impulsive forces, as shown in Fig. 8 – 1.
Whether the force is impulsive or not, the linear impulse that is created by the force is
responsible for changing the object’s speed. The relationship between linear impulse
and speed is governed by the linear impulse-linear momentum principle.
This section deals first with the linear impulse-momentum principle for a
single particle. Then, the impulsive force is examined more closely, revealing general
properties that displacements, velocities and accelerations satisfy. Next, the linear
impulse-momentum principle is extended to a system of particles. Finally, this section
gives special attention to collision problems. Direct collisions problems are first treated
followed by oblique collision problems.
Linear Momentum of a Single Particle
Linear Momentum
Consider a single particle of mass m moving at velocity v. The linear
momentum vector of the particle is defined as
Linear Impulse
p  mv.
(8 – 1)
Assume that the particle is subjected to a resultant force R. Integrating
Newton’s Second Law with respect to time yields
(8 – 2)
t2
t1
t
t1
Rdt   2 madt  mv 2  mv 1
where v1 = v(t1) and v2 = v(t2). Equation (8 – 2) can be rewritten as
(8 – 3)
G12  p 2  p1 ,
where
(8 - 4)
Linear Impulse-Momentum
Principle for a Particle
t
t1
G1-2 =  2 Rdt
is called the linear impulse of the resultant force R acting on the particle from time t1
to time t2. Equation (8 – 3) states that the linear impulse of the resultant force acting
on a particle from time t1 to time t2 is equal to the change in the linear momentum
of the particle. Equation (8 – 3) is called the linear impulse-momentum principle.
As a simple illustration, picture a 40 kg block resting on a smooth horizontal
surface. A constant 10 N force then acts on the block in the horizontal direction. After
5 seconds, according to the linear impulse-momentum principle, the block has been
subjected to a 10(5) = 50 lb-s impulse increasing the block’s linear momentum by 50
kg-m/s. Since the block was originally at rest, its velocity after 5 seconds becomes
50/40 = 1.25 m/s.
When no resultant force is acting on the particle, the linear impulsemomentum equation states that the linear momentum of the particle is conserved (mv1
= mv2), from which it follows that the velocity of the particle is constant (v1 = v2). This
is actually Newton’s First Law.
The Impulsive Force
Fig. 8 – 2: It is difficult to measure an
impulsive force because it acts so
quickly.
A very large force that is applied and removed very quickly is called an
impulsive force. Impulsive forces occur during explosions and collisions. Say that you
strike a nail with a hammer, or that you hit a baseball with a bat, or that a piston hits
the bottom of a cylinder. In all of these situations very large forces occur over very
small amounts of time (See Fig. 8 – 2). Theoretically, the impulsive force can be taken
to be infinitely large and to act over an infinitesimally small interval of time, although
this is clearly an idealization. In contrast, the non-impulsive force, such as the weight
force and the spring force shown in Fig. 8 – 3, is a bounded function, meaning that it
can’t become very large over a very small amount of time. The linear impulse that a
non-impulsive force creates over an infinitesimal amount of time is therefore
infinitesimal, whereas the linear impulse created over an infinitesimal amount of time
by an impulsive force is finite.
Whether a force is impulsive or not, the linear impulse it creates is never
infinite. For otherwise, Eq. (8 – 2) would imply that an object of finite mass could
appear in one place at one instant only to reappear somewhere else in the span of an
instant. Mathematically, the force is said to be an integrable function.
In practice, it is difficult to accurately measure and predict impulsive forces,
given that they occur so fast. Also, during the very short amount of time that they do
act, the impulsive force is not constant. It changes depending on the material properties
at the point of application of the force. Fortunately, though, the motion of objects
subject to impulsive forces can be predicted without knowing the impulsive forces
themselves, or how long they act. As Eq. (8 – 3) indicates, the linear impulse produced
by an impulsive force is the quantity of interest, not explicitly the force or the duration
of time.
Fig. 8 – 3: The weight force and the
spring force are non-impulsive forces.
The average force is defined as
Fave 
(8 – 5)
1
t 2  t1
t2
t1
Fdt
Equation (8 – 5) can be used to determine the average force from a known linear
impulse if the duration of time t2 – t1 is known (See Fig. 8 – 2).
The Nature of Acceleration, Velocity, and Position
Fig. 8 – 4: A hammer strikes a block.
The impulsive force affects the motion of a particle in a very specific way.
The way in which the motion is affected depends on the general nature of acceleration,
velocity, and position. To appreciate what is meant by this statement, consider a block
resting on a smooth surface when all of a sudden it’s subjected to a resultant impulsive
force R, as shown in Fig. 8 – 4. Assume that the impulsive force is constant over the
contact time – between the times t1 = 0 and t2 = t. From Eq. (8 – 4), the linear
impulse created by the force is
2
 Rt , 0  t  t
t
G1-2 =  2 Rdt  
,
t1
t  t
 Rt ,
(8 – 6)
Equation (8 – 6) tells us that the impulse is growing during the contact time t until it
reaches Rt, after which the impulse remains constant. From Eqs. (8 – 3) and (8 – 6)
the velocity of the block is
R
 m t , 0  t  t
1 t
v=
Rdt

R
m 0
 t ,
t  t
m
(8 – 7)
Fig. 8 – 5: These graphs reveal the
general nature of acceleration, velocity,
and position.
By integrating Eq. (8 – 7) the position of the block is found to be
(8 – 8)
 Rt 2
,
0  t  t

t
x =  vdt   2m
2
0
 R(t )  Rt (t  t ),
t  t

2m
m
Equation (8 – 7) tells us that the velocity of the block is growing during the contact
time, after which the velocity is constant. Equation (8 – 8) tells us that the position of
the block is growing with the square of time until it reaches the end of the contact time
after which the position grows linearly.
For example, assume that the mass of the block is m = 5 slug block, that the
constant impulsive force is R = 500 lb, and that the contact time of t = 0.01 s. As Fig.
8 – 5 shows, just after the impulse, the block’s speed is 1 ft/s and it has displaced 0.005
ft.
It was stated earlier that the force acting on any particle is an integrable
function. Since the resultant force and acceleration are related by R = ma, it follows
that acceleration is an integrable function, too. This means that the velocity at the end
of the contact time, Rt, is finite. So, over the infinitesimal amount of time t in Eq. (8
– 7), the block’s velocity jumps to a finite amount from 0 to Rt/m. The velocity of the
block is said to be continuous except when it’s subjected to an impulsive force, during
which time its velocity jumps a finite amount. The velocity is a piecewise continuous
function with jump discontinuities occurring each time the block is subjected to an
impulsive force.
From Eq. (8 – 8), the displacement at the contact time t is Rt2/2 =
(Rt)t/2. But Rt is finite, so (Rt)t/2 is infinitesimal. In other words, the position at
time t is infinitesimal. Thus, position is a continuous function.
To summarize, acceleration, velocity, and position are very different types of functions,
by their nature. Acceleration is an integrable function. Velocity is a piecewise
continuous function, having a jump discontinuity when the particle is subjected to
an impulsive force. Position is a continuous function.
3
Linear Momentum of a System of Particles
Fig. 8 – 6: A General System of Particles
In the case of a system of particles acted on by a resultant external force R,
Newton’s Second Law for a system of particles is R = maC. Integrating with respect to
time yields the same equation as Eq. (8 – 3),
G12  p 2  p1 ,
(8 – 9)
but now G1-2 represents the linear impulse produced by the external forces acting on
the system and p represents the linear momentum of the system, given by
Linear Impulse-Momentum
Principle for a System
Conservation
of Linear
Impulsive
Force Momentum
n
p  mv C   mi v i .
(8 – 10)
i 1
Equation (8 – 10) states that the linear impulse of the resultant external force acting
on a system from time t1 to time t2 is equal to the change in the linear momentum
of the system. This is the linear impulse-momentum principle for a system of particles.
Notice that the linear impulse-momentum principle for a system of particles is the same
as the linear impulse-momentum principle for a single particle. When no external force
is acting on the system, the linear momentum of the system is constant. This statement
is called conservation of linear momentum for a system of particles.
For example, say that a 50 lb object is at rest when it explodes into two pieces.
The impulsive forces acting on the object are internal. No impulsive external forces
would be acting on the object so the linear momentum of the object (both pieces) is
conserved over the short time of the explosion. If one piece is 25% of the weight of the
other piece, then it follows that a) just after the explosion the two pieces fly apart in
opposite directions, and b) just after the explosion the heavier piece moves at 25% of
the speed of the lighter piece.
As this example illustrates, the impulse-momentum principle, Eq. (8 – 9), is
particularly easy to use when no external forces are acting on a system. It is good
practice, when looking at the free-body diagrams of a system, to look at the external
forces acting on the system and check whether they are zero in any given direction.
When this happens, linear momentum is conserved in the directions in which the
resultant external force is zero. Also, notice that the linear impulse-momentum
equations, Eqs. (8 – 9), represent three equations. They can be used instead of the
equations of motion determined directly from Newton’s Second Law in any or all of
the three directions.
Collisions
Fig. 8 – 7: When two objects collide,
first identify the line of force and then
the plane of impact.
Line of Force
Plane of Impact
Figure 8 – 7 shows two objects that are about to collide. A coordinate system
has been set up with the x-axis directed along the line of the internal forces acting
between the objects during the collision. The x-axis is called the line of force. The yaxis and the z-axis make up the plane of impact, which is perpendicular to the line of
force. Notice that the x-axis is directed from object A to object B. With this convention,
the objects are sure to collide if vAx > vBx where vAx is the x-component of the velocity
of object A just before the collision and vBx is the x-component of the velocity of object
B just before the collision. After the collision, it will necessarily be the case that
vBx  vAx where v Bx is the x-component of the velocity of object B just after the
4
collision and v Ax is the x-component of the velocity of object A just after the collision.
If this were not the case, object A would pass through object B, which is impossible.
Direct Collisions
Oblique Collisions
Collisions in which the motions of the objects are along the direction of the
line of force are called direct collisions. Collisions in which the motions are not along
the line of force are called oblique collisions. In direct collision problems,
v Ay , v Az , v By , v Bz , v Ay , v Az , v By , and v Bz are all zero. In other words, there is no
motion in the plane of impact in direct collision problems.
The Coefficient of Restitution
Fig. 8 – 8: Two objects deform a very
small amount during a collision.
Let’s first consider direct collisions. Our interest lies in determining the
velocities of objects A and B just after a collision, given the velocities of the objects
just before the collision. The two objects are regarded as a system of two particles.
Let’s look at the linear impulse-momentum principle for this system of particles over
the collision time from t = 0 to t = t. Assume that no external impulsive forces act on
the particles during the collision. Linear momentum is then conserved over the
collision time, so
m A v A  mB v B  m A v A  mB v B .
(8 – 11)
Fig. 8 – 9: When the deformations are
small the normal force is approximated
by a tangent plane. Its general form is
given by Eq. (8 – 12).
Fig. 8 - 10: The normal force between
colliding objects can be modeled as a
spring and a damper.
The subscript x has been dropped temporarily because it’s the only direction we’re
considering right now. Equation (8 – 11) is one equation expressed in terms of the two
unknown velocities v’A and v’B at the end of the collision. An additional equation is
needed in order to be able to find v’A and v’B. We know that the additional equation
must depend on the material properties of the objects that are colliding. After all, two
objects that are colliding can respond in different ways depending on how they deform,
even though the deformations are very small (See Fig. 8 – 8). The deformation is
described in terms of the elasticity and the plasticity of the objects. Elasticity refers to
the spring-like force that acts between the objects during the collision and plasticity
refers to the damping-like force that acts between the objects during the collision.
As Fig. 8 – 9 shows, the normal force N between two colliding bodies is some
function of the deformation d between the bodies and the deformation rate d . When
the deformation between the bodies is small, the normal force approximately follows
the tangent plane. The normal force is essentially a linear function of the elastic
deformation d and the plastic deformation rate d , written
N  kd  cd
(8 – 12)
Linear Elastic-Plastic Collision
where the constants k and c are constants that depend on the material. The normal
force given in Eq. (8 – 12) can be represented by a spring and a damper, as shown in
Fig. 8 – 10. The material constant k represents a spring constant and c represents a
damping constant. Equation (8 – 12) is called the linear elastic-plastic model of the
impulsive force, and the collision is called a linear elastic-plastic collision.
In Chapter 10 Eq. (8 – 12) will be used to show that the relative velocity
between the objects just after the collision is linearly proportional to the relative
velocity of the objects just before the collision, written
5
e
(8 – 13)
Coefficient of Restitution
v B  v A
v A  vB
where the material constant e is called the coefficient of restitution1. We won’t need
to further use Eq. (8 – 12) in this section, but Eq. (8 – 13) will be very helpful.
The values of the coefficient of restitution in Eq. (8 – 13) are between 0 and 1. When e
= 0, the collision is called a purely plastic collision. This case arises when no rebound
takes place and the two objects stay together after the collision. The largest amount of
energy is lost in a purely plastic collision. At the other extreme, when e = 1, the
collision is called a purely elastic collision. The relative velocities between the objects
after the collision is largest in a purely elastic collision and no energy is lost in a purely
elastic collision. The purely plastic collision (e = 0) and the purely elastic collision (e
= 1) are idealizations. In reality, collisions are elastic-plastic, in which case 0 < e < 1.
Equation (8 – 13) is the second equation that we were looking for. The
unknown velocities v’A and v’B. at the end of the collision can now be determined from
the two linear algebraic equations, Eq. (8 – 11) and (8 – 13). The solution is
Fig. 8 – 11: Remember to direct the line
of force from object A to object B.
v A 
1
(m A  emB )v A  (1  e)mB v B ,
m A  mB
v B 
1
(1  e)m Av A  (mB  em A )v B .
m A  mB
(8 – 14)
Remember that the line of force in Eq. (8 – 14) is assumed to be directed from object A
to object B, as shown again in Fig. 8 – 11.
A special case of interest occurs when object A collides with a stationary wall.
In this case, the mass mB of the wall is very large compared to the mass mA of the
object and the velocity of vB of the wall is 0. Substituting mA/mB = 0 and vB = 0 into
Eq. (8 – 14) yields,
vA  ev A ,
(8 – 15)
vB  0
Notice that object A changes direction and that the wall doesn’t move. Also, notice
when the collision is purely elastic (e = 1) that the ball rebounds from the wall at the
same speed it approaches the wall. A few popular coefficients of restitution are given
in Table 7 – 3.
Notice that the numerator in Eq. (8 – 13) subtracts the velocity of object B from
object A while the denominator subtracts the velocity of object A from the velocity of
object B. As a result, the coefficient of restitution given by Eq. (8 – 13) is never
negative.
1
6
Objects in sports
basketball and court
golf ball and club
tennis ball and racket
Baseball and bat
Volleyball and court
Coefficient of restitution e
0.85
0.80
0.73
0.55
0.75
Table 7 – 3: The coefficient of restitution in several popular sports
Oblique Collisions
The components of the velocities of the
objects in the plane of impact do not
change over the small time of the
collision.
Equation (8 - 14) predicts the velocities of two colliding objects just after they
collide. The velocities in Eqs. (8 – 14) are actually components in the x-direction,
where the x-direction has been taken along the line of force. In oblique collisions, there
are velocity components in the y and z directions, too. However, there are no impulsive
forces acting on either of the objects in the y and z directions during the collisions.
Therefore, over the short duration of the collision, the linear impulses acting on each of
the objects in both the y and z directions are zero. It follows from Eq. (8 – 9) that the
components of the velocities of the objects in the y and z directions do not change over
the time of the collision. The velocity components of the two objects in the y and z
directions are
(8 – 16)
v Ay  v Ay ,
v By  v By ,
v Az  v Az ,
v Bz  v Bz .
New Terms
Average Force, Coefficient of Restitution, Collision, Conservation of Linear Momentum, Contact Time, Continuous
Function, Direct Collision, Elasticity, Impulse, Impulsive Force, Integrable Function, Jump Discontinuity, Linear
Momentum, Linear Impulse, Linear Impulse-Momentum Principle, Non-impulsive Force, Oblique Collision, Piecewise
Continuous Function, Plasticity, Purely Elastic Collision, Purely Plastic Collision
Review Questions
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
Define linear momentum.
State the linear impulse-momentum principle for a single particle.
What is an impulsive force? Give examples.
What is a non-impulsive force? Give examples.
Are there any restrictions to the types of functions that can represent a force?
Describe the nature of acceleration, velocity, and position.
State the linear impulse-momentum principle for a system of particles.
When is linear momentum conserved?
What should you look for in your free body diagrams to determine whether linear momentum is conserved?
What are the line of force and the plane of impact?
What’s the difference between a direct collision and an oblique collision?
What is elasticity? What is plasticity?
How is the coefficient of restitution defined?
How do the components in the plane of impact of the velocities of two objects change as soon as the two objects collide with each
other?
15. What is a purely elastic collision? What is a purely plastic collision?
16. What is the equation that describes what happens to the velocity of an object when it hits a wall?
7
Examples
Fig. 1a
8 – 1 A 20 kg block lies on a horizontal surface when it’s subjected to a horizontally
applied force of F = at2 N where a = 100 N/s2. The kinetic coefficient of friction
between the block and the surface is  = 0.2. If the block starts at rest at t1 = 0,
determine its speed after t2 = 2 seconds. What is the average applied force acting on the
block over this time? What would have been the speed of the block had the block been
subjected to the average applied force instead of the actual applied force?
Solution: From the free-body diagram, the resultant force acting on the block in the xdirection is made up of the applied force and the friction force, both of which change
the block’s linear momentum. From the impulse-momentum principle in the xdirection
G  mv 2  mv1 ,
Set-up – The block is a particle
moving in the x-direction.
Transition – Draw the free body
diagram of the block.
t
t
t
a
2
G   2 R x dt   2 (at 2  mg )dt   t 3  umgt 
0
0
3
0
(1)
a

  t 22  umg t 2 ,
3

mv1  20  0  0,
Equation – Write out the impulsemomentum equation.
which is one equation in terms of the unknown speed v2. Solving for v2
Answer – Solve for v2.
.
(2)
 at 3

 100  2 3

v 2   2  g t 2  
 0.2  9.81   2  22 .7 m/s.
 3  20

 3m





Next, let’s calculate the average applied force and the speed resulting from the
average applied force. From Eq. (8 – 5)
(3)
Fave 
1
t 2  t1
t2
t
Fdt 
1
1
t2
t2
0
at 2 dt 
a 2
t 2  133 N.
3
Notice that at time t2 the actual applied force F  at 22 is three times greater than the
average applied force given in Eq. (3). From the impulse-momentum principle
(4)
G  mv 2  mv1 ,
2
Knowledge – The average applied
force is three times smaller than the
peak value of the actual applied
force. The resulting speed of the
block is the same whether subjected
to the actual applied force or the
average applied force.
2
2
2 t
a

G   R x dt   (a 2  mg )dt   t 22 t  umgt  ,
0
0
3
3

0
mv1  20  0  0,
so
 at 3

v 2   2  g t 2  22.7 m/s,
 3m



(2)
8
which is the same speed that was obtained with the actual applied force. Both the
applied force and the average force, even though they’re quite different, create
the same impulse and hence impart the same resulting speed.
Fig. 2a
(use Fig. 8.1 – 2)
8 – 2 A block is acted on by an impulsive force over the small period T. The impulsive
t
t
force is the parabolic function F  4F0 1   where t is time and F0 is a constant.
T T
Determine the impulse associated with the force, the maximum value of the force, and
the average force.
Solution: This example deals with the relationship between force and impulse. From
Eq. (8 – 3)
Equation – Write out the equation
of an impulse and the equation of an
average force.
Answer – Solve for the impulse and
the average force from the given
equations. Determine the maximum
force either from calculus or
recognizing that it occurs at T/2.
T
(1)
T
0
T
4 F0
0
G   Fdt 
 t2
2F T
t 
t
t3 

1  dt 4 F0 
  0
2
T T
3
 2T 3T  0
The maximum force occurs at T/2 so
Fmax  4F0
(2)
(T / 2)  (T / 2) 
1 
  F0
T 
T 
From Eq. (8 – 5), the average force is
(3)
Knowledge – The average force is
two-thirds of the maximum value of
the force.
Fig. 3a
Fav 
1
t 2  t1
T
0
Fdt 
1  2 F0T  2 F0
.


T 3 
3
Notice that the average force is two-thirds of the maximum value of the parabolic
force. The impulse created by the actual force and the average force are the same.
8 – 3 A system of particles is subjected to applied forces, as shown. The initial
velocities of the particles are also given. Determine the velocity and displacement of
the mass center after 2 seconds.
Solution: The figure showing the particles is really a free-body diagram of the
particles. From the linear impulse-momentum principle
2
G  p 2  p1   (FA  FB  FC )dt
0
Set-up – The example is set up from
the beginning as a system of
particles. The motion is planar.
Transition – The figure of the
particles is the free-body diagram of
the system.
(1a, b)
 [(10  50 sin 25)t  3t 2 ]i  [(50 cos 25)t  4t 2 ]j,
p1  m A v A1  m B v B1  mC v C1  5(10 i  10 j)  10 (20 i )  20 (15 j)
 250 i  350 j kg - m/s.
The linear impulse was left as a function of time because it will later need to be
integrated. Equation (1) is a two-dimensional vector equation that represents 2 scalar
Equation – Write out the impulsemomentum equation for a system.
9
Answer – Solve for vC. Integrate
with respect to time and solve for rC.
Knowledge – Notice that particle A
has a very small effect on the
answer because particle A’s mass
and initial velocity are both
relatively small.
equations in terms of the vector unknown p2 = mv2, which is two scalar unknowns.
Solving for the final velocity of the mass center v2
v2 
(2)
1
p1  G  5.02i  6.95 j m/s.
m
Integrating Eq. (1a) with respect to time yields the displacement vector of the mass
center after 2 seconds.
2
r2  r1   vdt
(3)
0
 16 .3i  599 j m.
Notice that we did not integrate Eq. (2) to determine the displacement of the mass
center because it was already evaluated at the final time.
Fig. 4a
Set-up – Two particles are moving
in the x-direction.
Transition – The free-body diagram
of the system is drawn. Notice that
no external forces are acting in the
x-direction.
Equation – Write out the
conservation of linear momentum
equation for the x-direction.
Answer – Solve for vC.. Integrate
and solve for rC..
Knowledge – When docking a small
boat, stand up at the end of the boat
closest to the dock.
8 – 4 A mF = 140 lb fisherman is returning from a long day out on the water. When he
docks, the left end of his mB = 100 lb, L = 10 ft boat is just touching the dock and the
fisherman is seated at the right end of the boat. The fisherman gets up and walks over
to the left end of the boat, only to discover that the boat has drifted to the right. The
weary fisherman attempts to jump onto the dock. Assuming that he is capable of
jumping a maximum of 4 feet from the boat to the dock, will he make it to the dock or
will the fisherman fall into the water?
Solution: The boat and the fisherman are a system of two particles. From the free-body
diagram, no external forces act on the system, so the linear momentum of the system is
conserved. Since the linear momentum of the system is initially zero, it remains zero. It
follows that the velocity of the mass center of the system is zero and hence the position
of the mass center of the system is a constant.
Measuring from the dock, the initial position of the mass center and the final
position of the mass center are
x C1 
1
L
(m F L  m B ),
mF  mB
2
xC 2 
1
L
(m F d  m B (d  )),
mF  mB
2
(1)
where d is the final distance between the left end of the boat and the dock. From xC1 =
xC2,
d
(2)
mF
L  5.83 ft,
mF  mB
which is greater than 4 ft. The fisherman does not make it to the dock and falls into the
water.
Fig. 5a
8 – 5 Two trains are locked to each other by allowing one train to gently ram into the
other, while both trains’ brakes are disengaged. Let train A, which weighs 10,000 lb
coast at 2 mph while train B, which weighs 15,000 lb, is stationary. Find the average
locking force if the locking is observed to take place in 0.1 s.
10
Solution: This is a direct collision problem for which e = 0. From Eq. (8 – 14)
v B2 
Fig. 5b
(1)

mA
mA g
v A1 
v A1
m A  mB
m A g  mB g
10,000
 5280 
2
  1.17 ft/s  0.8 mph.
10,000  15,000  3600 
From the impulse-momentum equation, Eq. (8 - 9), applied to train B,
Set-up – This is a system of two
particles that move in the xdirection.
Transition – The free-body
diagram of the system is drawn.
No external forces are acting on
the system.
Equation – Write out the solution
to the collision problem and the
impulse-momentum equation.
Answer –
immediate.
The
answers
are
Knowledge – Disengage the
brakes on both trains when
locking the two trains.
1
1
G  m B v B 2  m B v B1 
t
t
1 15,000
1.17  0  5470 lbs.


0.1 32 .2
Fave 
(2)
The average locking force could have also been calculated by looking at train
A instead of train B in Eqs. (1) and (2).
The average locking force is quite large, and the peak locking force could be
considerably greater. Moreover, if the second train’s brakes were inadvertently
engaged during the collision, preventing it from moving during the collision, then the
locking force would become still greater. When the second train’s brakes are engaged,
the average locking force becomes
1
1
m Av A2  m Av A1 
G
t
t
1 10,000  5280 


2
  9110 lbs,
0.1 32 .2  3600 
Fave  
(3)
which roughly doubles the average locking force.
Fig. 6a
Fig. 6b
8 – 6 Blocks A and B are at rest on a smooth table when a hammer strikes block A. The
impact of the hammer imparts an initial velocity of vA1 to block A. Determine the
speeds of the blocks when the spring later undergoes its largest compression. Also
determine the speeds of the blocks when the spring later undergoes no compression.
The masses of the blocks are mA and mB. The answers should be in terms of mA, mB,
and vA1.
Solution: The two blocks form a system of two particles. We set up a coordinate
system to describe the motion of each particle in the horizontal direction. During the
infinitesimal time of the impact, the only impulsive force acting on block A is produced
by the hammer. The spring force is non-impulsive. Therefore, immediately after the
impact, the velocity of block A can change from 0 to vA1 which is not equal to zero, but
block B must remain at rest at that instant. Let point 1 be just after the impact, point 2
is when the spring compresses the most, and point 3 is when the spring is not
compressed.
11
After point 1, no resultant external force acts on the system in the x direction.
Therefore, linear momentum is conserved. Since the work done by the nonconservative forces acting on the system is zero, too, energy of the system is also
conserved.
First, let’s determine the speeds of the blocks at point 2. From conservation of
linear momentum of the system at points 1 and 2
Set-up – This is a system of two
particles that move in the xdirection.
Transition – The free-body diagram
of the system is drawn. Notice that
the hammer exerts the only
impulsive force. After the impact no
external forces are acting on the
system.
Equation – Write out conservation
of linear momentum of the system
in the x-direction and conservation
of energy of the system. When the
compression is the largest vA2 = vB2
and when there is no compression s
= 0.
Answer – To determine the speeds
when the compression is the largest,
solve a linear algebraic equation. To
determine the speeds when the
compression is zero, solve 2 nonlinear algebraic equations. You can
factor out one of the solutions
(which you know), to reduce it to
linear equations.
m A v A1  m A v A2  m B v B 2 .
(1)
At point 2, the relative velocity between the blocks is zero, so vA2 = vB2. From Eq. (1)
v A2  v B 2 
(2)
mA
v A1 .
m A  mB
Next, let’s determine the speeds of the blocks at point 3 when the spring
undergoes no compression. Equation (1) still applies, replacing point 2 with point 3.
However, at point 3 the relative velocity of the blocks is not zero, so Eq. (1) represents
one equation in terms of the two unknowns vA3 and vB3. The additional equation that is
needed expresses conservation of energy, E1 = E3, so
m Av A1  m Av A3  mB v B3
(3a)
(3b)
1
1
1
m A v A21  m A v A2 3  m B v B2 3
2
2
2
Equations (3a,b) are two non-linear algebraic equations expressed in terms of the
unknowns vA3 and vB3. By substituting Eq. (3a) into (3b), we’ll get a quadratic equation.
The problem will actually be relatively easy to solve because we already know one of
the two solutions. Recall that the spring is not compressed at point 1, so one of the
solutions is the speed of a block at point 1.
Solving for vB3 in Eq. (3a) and substituting the result into Eq. (3b), yields
(4)
 m (v  v A3 ) 
1
1
1

m A v A21  m A v A2 3  m B  A A1
2
2
2
mB


2
which is a quadratic equation in terms of the unknown vA3. We know that one of the
solutions is vA3 = vA1, which is now factored out. Multiplying Eq. (4) by 2 and
rearranging terms,
2
(5)
 m (v  v A3 ) 
  m A (v A21  v A2 3 )  m A (v A1  v A3 )(v A1  v A3 )
m B  A A1
m
B


Dividing by mA( v A1  v A3 ) and solving for vA3 yields
(6)
v A3 
m A  mB
2m A
v A1 , v B3 
v A1
m A  mB
m A  mB
where vB3 was found by substituting vA3 back into Eq. (3a).
12
Knowledge – The hammer only
imparts an initial speed to block A.
Some
interesting
trends
were
observed. For example, when the
masses of the blocks are equal, notice
from Eq. (6) that the block A becomes
momentarily stationary at the instant
the spring is not compressed.
Fig. 7a
Note that Eq. (6) can be verified by checking that the velocity of the mass center
of the system at point 3 is equal to the velocity of each of the blocks at point 2. Indeed,
from Eq. (6) mA vA3 + mBvB3 = (mA/(mA+mB))vA1.
Among the observations that can be made, notice from Eqs. (2) and (3) when
mA >> mB and when the spring undergoes its largest compression that the velocities of
the blocks are equal to the initial velocity that was imparted to block A by the hammer.
This is also the same as the velocity of block A when the spring undergoes no
compression. The speed of block B, however, is double.
8 – 7 In the game of croquet, a wooden mallet strikes a wooden ball. The coefficient of
restitution of the collision is about e = 0.7 and the mallet is about 6 times heavier than
the ball. Determine the relationship between the speed of the ball vB2 just after the
collision and the speed of the mallet vM1 just before the collision. How does the speed
of the mallet change as a result of the collision?
Solution: This is a direct collision problem whose solution was already worked out in
Eq. (8 – 14). Let particle A represent the mallet M and let particle B represent the ball.
Also, let’s define the ratio  = mM /mB = 1/6 and substitute it into Eq. (8 – 14) by
dividing both the numerator and the denominator in Eq. (8 – 14) by mM, to get
1  e
v M 1  0.757 v M 1
1 
1 e

v M 1  1.48v M 1
1 
vM 2 
(1)
vB2
Knowledge – A stationary object,
when struck by a moving object,
gains a speed of less than twice the
initial speed of the moving object,
depending on the energy lost in the
collision and the relative masses of
the objects.
Fig. 8a
The speed of the ball is about 50% greater than the initial speed of the mallet. The
speed of the mallet, after the collision, decreases by about 25%. Looking at Eq. (1), had
this been a purely elastic collision, the speed of the ball would have been about 86%
larger than the speed of the mallet. In fact, had the collision been purely elastic and had
the mallet been infinitely heavier than the ball, the theoretically highest speed of the
ball becomes twice the speed of the mallet.
Collision problems like this one, in which a large moving object hits a small
stationary object, are quite common in sports. This same situation arises when serving
in tennis and racquetball, and when hitting fly balls in baseball. In each of these
situations the speed of the ball is slightly less than double the speed of the object
hitting it.
8 – 8 In billiards, a ball approaches a cushion at ○. At what angle  does it leave
the cushion? Assume a coefficient of restitution of e = 0.75. Also, how different is 
than when the approaching angle becomes either larger or smaller than 45 ○?
Solution: This is an oblique collision problem in which one object is much larger than
the other – the cushion being the very large object. The line of force is perpendicular to
the cushion. This problem has already been solved; the solutions are given in Eqs. (8 –
15) and (8 – 16). First, consider the x-direction. From Eq. (8 – 15)
(1)
v Ax  v sin  ,
13
vAx  ev Ax  ev sin 
Set-up – This is an oblique
collision problem of two particles
that move in the x- and ydirections.
In the y direction, from Eq. (8 – 16)
Transition – The figure shows
the direction of the line of force.
So, from the figure,
Equation – Determine tan from
the figure.
(3)
v Ay  v cos ,
(2)
tan  
v Ay  v Ay  v cos
vAx ev sin 

 e tan  ,
vAy
v cos
and so
Knowledge – The angle changes the
most when the ball approaches the
cushion at 45○.
  tan 1 (e tan  )
(4)
From Eq. (4), the angles corresponding to different initial angles are:
Fig. 8b
deg)
  tan 1 (0.75 tan  ) 

0
20
45
70
90
0
15.3
36.9
64.1
90
0
4.7
8.1
5.9
0
They’re also shown in Fig. 8b. The angle that the ball leaves the cushion differs most
from the approaching angle when the approaching angle is 45○. With a little calculus,
the precise angle  that maximizes can be determined. Is the precise angle 45○?
Fig. 9a
Set-up – This is an oblique
collision problem but the direction
of the line of force is not known.
Transition – Notice that linear
momentum is conserved in both
directions.
Equation – Write out the
conservation of linear momentum
in the x- and y-directions.
8 – 9 A police officer arrives at the scene of an accident. Driver A claims to have been
driving at 55 mph (the speed limit) when driver B went through the intersection; failing
to come to a complete stop at the stop sign. Driver B claims that driver A was traveling
at what appeared to him to be 100 mph. He claimed to have come to a complete stop
but accelerated at the intersection to about 10 or 20 mph when driver A barreled into
him. The physical evidence shows that the cars locked and skidded together at an angle
of 20○, as shown. The masses of the cars are about the same. Which driver is likely
telling the truth and why?
Solution: This is an oblique collision problem, although the direction of the line of
force is not obvious. Nevertheless, linear momentum of the system of two cars is
conserved during the collision in both directions. Let’s solve this problem two ways.
By the first way, assume that driver A’s information is accurate and by the second way
assume that driver B’s information is accurate. From conservation of linear momentum
in the x- and y-directions,
(1)
m A v Ax  (m A  m B )v x  (m A  m B )v  cos ,
m B v By  (m A  m B )v y  (m A  m B )v  sin  .
From Eq. (1)
tan  
(2)
Answer – Solve the linear
algebraic equations for each of the
two hypothetical cases.
14
m B v By
m A v Ax
.
If driver A’s information is accurate and he was traveling at 55 mph, then from (2)
(3)
v By  tan 
mA
v Ax  tan 20  (1)55  20 .0 mph.
mB
On the other hand, if driver B’s information is accurate and he was traveling at 20 mph,
then from (2)
Knowledge – Driver B appears
guilty of failing to come to a
complete stop and lying to a police
officer.
1 mB
1
v By 
(1)20  55 .0 mph.
tan  m A
tan 20 
Driver A’s statement is consistent with these calculations and with driver B’s statement
that he was going 20 mph when the cars collided. However, driver B’s statement that
driver A was traveling at about 100 mph is not consistent with the facts. It appears that
driver B not only failed to come to a complete stop, but also lied to the officer.
(4)
v Ax 
Problems
8 -1(L)
A block on a smooth horizontal surface is subjected to a horizontally applied constant force of 100 lb. If the block was
originally at rest, use the linear impulse-momentum principle to determine its speed after 5 seconds.
Answer: v = 166 ft/s.
8 – 2(L)
A 175 lb man runs into a wall at 10 mph and does not rebound from the collision. If the collision took place in about 0.1
seconds, what is the average force exerted on the man by the wall?
Answer: Fave = 797 lbs.
8 – 3(L)
A 150 lb engineering student jumps 12 inches in the air. If the landing on the ground takes 0.01 seconds, what is the
average force exerted on the student by the ground?
Answer: Fave = 3740 lbs.
8 – 4 (L)
Develop a formula for the average of the triangular force over the time T.
Answer: Fave = 0.5F0.
8 – 5(L)
Develop a formula for the average of the sinusoidal force over the time T.
Answer: Fave = 0.637F0 .
8 – 6(L)
In a manufacturing line, three chutes are lined up above a cart. A 10-kg part comes out of each chute at 5 m/s, one after the
other, into a 100 kg free-rolling cart. If the speed of the cart is 2 m/s before any of the parts were dropped into it, what is
the speed of the cart after the first part is dropped into it?, after the second part is dropped into it?, and after the third part is
dropped into it?
Answer: v1 = 2.21 m/s, v2 = 2.39 m/s, v1 = 2.54 m/s.
15
8 – 7(L)
A 350 gram bullet shot from a colt 45 is traveling at 20 m/s when it hits a 2 kg can sitting on a wood post. If the bullet gets
caught in the can, at what speed does the can fly off the post?
Answer: v = 2.98 m/s.
8 – 8(L)
A 50 kg stone is exploded into two large pieces. One piece takes off at 2 m/s and the other at 3 m/s. Determine the mass of
each piece.
Answer: m1 = 30 kg, m2 = 20 kg.
8 – 9(L)
A 100 kg stone is exploded into three large pieces that shoot out in a plane. Let their speeds be vA = 3 m/s, vB = 5 m/s, and
vC = 2 m/s. The angles are 1 = 30○,2 = 60○, and3 = 60○. Determine the mass of each piece.
Answer: mA = 32.5 kg, mB = 11.3 kg, mC = 56.3 kg.
8 – 10(L)
A 275 lb skater dives for a puck on the ice at 10 mph, which is in the direction of a 125 lb skater who is diving for the same
puck at 15 mph. The two skaters collide directly into each other, grabbing on to each other during the collision. How fast
do they slide on the ice after the collision?
Answer: v  2.19 mph.
8 – 11(M)
  2t  
A general form of an impulsive force is F  F0 1  1    for n = 1, 2, … As n increases, the force F rises to F0 and
  T  
falls to zero more rapidly, making the function look more like the constant F0. Develop a general formula for the average
force over the time T.
1 

Answer: Fave  F0 1 
 for n  1.
 2n  1 
2n
8 – 12(M)
A 10,000 lb stone floating out in deep space and initially located at point O, is exploded into three large pieces. A camera
records their positions after 0.5 seconds. They are rA = (1.5 2 0) ft, rB = (0.5 0 2) ft, and rC = (-4 -1 -13) ft. Neglecting the
effect of gravity, determine the weights of each of the pieces.
Answer: WA = 625 lb, WB = 8130 lb, WC = 1250 lb.
8 – 13(M)
A shuffleboard puck is released at v1 = 2.5 m/s, slides d1 = 4 m, and then directly hits a second puck. How far does the
second puck go? Assume that the kinetic friction coefficient between the pucks and the surface is  = 0.04 and that the
coefficient of restitution of the collision is e = 0.8.
Answer: d2 = 3.21 m.
8 – 14(M)
Billiard ball A is moving with no spin at a speed of vA when it strikes billiard ball B obliquely. What is the speed vB’ and
direction  of billiard ball B after it was struck? Assume a coefficient of restitution of e = 0.9.
Answer: vB’ = 0.823vA,  = 30○.
8 – 15(M)
Particle A is moving at 10 m/s and particle BC is at rest when particle A strikes particle BC, causing it to break into two
pieces B and C. Particle BC weighed 5 times as much as particle A. A high speed camera recorded the direction of the
motion of each particle just after the collision. Determine the speeds of the three particles after the collision. Assume that
the ground does not apply an appreciable impulse to the particles during the collision.
Answer: v A  5.36 m/s, v B  0.583 m/s, vC  4.04 m/s.
16
8 – 16(H)
Ball A of mass mA is held slightly above ball B of mass mB and then both are released from rest about h1 feet from the
ground. Ball B will hit the ground first, ball A will then hit ball B, and then ball B will shoot up h2 feet into the air. Assume
that h1 is large compared to the radius of the balls and the initial gap between them, and that all of the collisions are purely
elastic. Develop a formula for h2 in terms of mA, mB, and h1. What is h2 when mB/mA = 7?
3 mB m A 1
Answer: h2 
h1 , h2  2.5h1 .
mB m A  1
8 – 17(H)
Determine a general expression for the energy loss that objects A and B undergo as a result of a linear elastic-plastic
collision. Also show that the energy loss is zero in a purely elastic collision (  = 1) and that the energy loss is the most in a
purely plastic collision ( = 0).
m A mB
(1   2 )( v Ax  v Bx ) 2 .
Answer: E 
2(m A  m B )
More problems are under preparation.
In the published book, the answers are given in the back of the text instead of directly following the problems.
17