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Momentum
Introduction
As was pointed out in the previous chapter, some of the most powerful tools in physics
are based on conservation principles. The idea behind a conservation principle is
that there are some properties of systems that don’t change, even though other things
about the system may. In this chapter we will be introducing the concept of
momentum which, like energy, is a conserved property of any closed system. The
Conservation of Momentum principle is a very powerful tool in physics and is, as will
be shown later, a necessary consequence of Newton’s Second and Third Laws.
Momentum of a Single Object
The momentum of a single object is simply equal to the product of its mass and its
velocity. The symbol for momentum is “p”. Since mass is a scalar and velocity is a
vector, momentum is also a vector. The direction of momentum is always the same as
that of the object’s velocity.
p = mv

Momentum is a vector so it has a magnitude and a velocity.

Its magnitude is the product of its mass and velocity, p = mv.

Its direction is the same as the direction of its velocity.
Units of Momentum
The unit of momentum can be derived from the above equation.
p = mv
The SI units of mass are kilograms (kg) and of velocity are meters / second (m/s).
Therefore, the units of momentum are kg∙ m/s.
There is no special name for the unit of momentum.
Example 1: A 20 kg object has a velocity of 4.5 m/s in the positive x direction. What is
its momentum?
p = mv
p = (20kg) (4.5 m/s)
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p = 90 kg∙ m/s
p = 90 kg∙ m/s in the positive x direction
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Example 2: A 60 kg object has a velocity of 1.5 m/s in the negative x direction. What is
its momentum?
p = mv
p = (60kg) (-1.5 m/s)
= -90 kg∙ m/s
p = 90 kg∙ m/s in the negative x direction
Please notice two things in the above examples. First, each answer required a
magnitude and a direction. Second, since momentum is the product of mass and
velocity, objects of different mass can have equal amounts of momentum. That would
just require the less massive object to have greater velocity.
Momentum of a Closed System of Objects
To determine the momentum of a system that has more than one object you have to
add together the momenta of all the individual objects.
psystem = p1 + p2 + p3 +…
psystem = m1v1 + m2v2 + m3v3 + …
Or
psystem = ∑p
psystem = ∑mv
So if a system were comprised of the objects in Example 1 and 2, above, then the
momentum of that total system would simply be the sum of those two momenta.
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Example 3: Determine the momentum of a system that consists of the two objects from
Example 1 and 2, above.
psystem = ∑p
psystem = p1 + p2
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psystem = (90 kg∙ m/s) + (-90 kg∙ m/s)
psystem = 0
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Example 4: Determine the momentum of a system that consists of the two objects.
One object, m1, has a mass of 2 kg and a velocity of +5 m/s and the second object, m2,
has a mass of 20 kg and a velocity of +3 m/s.
psystem = ∑p
psystem = p1 + p2
psystem = m1v1 + m2v2
psystem = (2 kg) (+5 m/s) + (20 kg)(+3 m/s)
psystem = (10 kg∙ m/s) + (60 kg∙ m/s)
psystem = 70 kg∙ m/s
psystem = 70 kg∙ m/s in the positive x direction
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Example 5: Determine the momentum of a system that consists of the two objects.
One object, m1, has a mass of 4 kg and a velocity of 17 m/s towards the east and the
second object, m2, has a mass of 70 kg and a velocity of 4 m/s towards the east.
First, we need to decide how to orient our positive and negative axes. The simplest
choice would be motion towards the east positive. Then, both velocities will be positive.
psystem = ∑p
psystem = p1 + p2
psystem = m1v1 + m2v2
psystem = (4 kg) (17 m/s) + (70 kg)(4 m/s)
psystem = (68 kg∙ m/s) + (280 kg∙ m/s)
psystem = 348 kg∙ m/s
psystem = 348 kg∙ m/s towards the east
Please note that in the last step we translated back to the directions that were given in
the problem. The person who wrote the problem doesn’t know which way we decided
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was positive, so we have to give our answer based on what we were give, in this case
east not positive.
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Example 6: Determine the momentum of a system that consists of the two objects.
One object, m1, has a mass of 4 kg and a velocity of 17 m/s towards the east and the
second object, m2, has a mass of 70 kg and a velocity of 14 m/s towards the west.
First, we need to decide how to orient our axes. The simplest choice would be to make
motion towards the east positive. That means that motion in the opposite direction, to
the west, will be negative. Then,
psystem = ∑p
psystem = p1 + p2
psystem = m1v1 + m2v2
psystem = (4 kg) (17 m/s) + (70 kg)(-4 m/s)
psystem = (68 kg∙ m/s) + (-280 kg∙ m/s)
psystem = -212 kg∙ m/s
psystem = 212 kg∙ m/s towards the west
Once again, note that in our last step we interpret the negative direction to mean
towards the west.
Conservation of Momentum in a closed system
We learned in the previous chapter that the energy of a closed system can only be
changed by an outside force doing Work on it. The change in the energy of the system
is equal to the work done by that outside force.
Similarly, the momentum of a closed system can only be changed by an outside force
providing an impulse to it. The change in the momentum of the system is equal to the
impulse provided by that outside force. This can be stated symbolically as:
p0 + I = pf
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where I is the symbol for impulse, p0, represents the initial momentum of the system
and pf represents the final momentum of that same system.
In many cases, there will be no outside forces acting on a closed system. In those
cases, the momentum will not change regardless of what goes on within the system.
Let’s first look at those cases, where the impulse provided a system is zero.
p0 + I = pf
but I = 0 so
p0 = pf
This means that if we measure the total momentum of a system at any point in time, its
momentum will not change if it is not affected by something outside the system. The
objects can collide, explode, break apart, stick together, etc. Nothing that happens
within the system will change its momentum.
When we’re doing the following example we’re going to change how we indicate initial
and final. That’s because, as you’ll see, the subscripts can get pretty confusing. There
are just too many of them. So, the first time, we’ll first write out vo1 for the initial velocity
of the first object and vf1 for its final velocity. Then we’re going to change to easier
notation that we’ll use from now on in doing momentum problems. We’ll indicate the
initial velocity of the first object with v1 and its final velocity with v´1. We think you’ll find
it’s easier to keep track of everything that way. Especially as you write out your own
work.
Example 7: A closed system consists of two objects. One object, m1, has a mass of 15
kg and a velocity of 6 m/s towards the east and the second object, m2, has a mass of 20
kg and a velocity of 3 m/s towards the west. These two objects collide and stick
together. What is the velocity of the combined object?
Once again, let’s choose east to be positive and west to be negative. Then,
p0 + I = pf
Since there are no external forces, I = 0 so
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p0 = pf
m1vo1 + m2vo2 = m1vf1 + m2vf2
Simplifying the subscripts gives us
m1v1 + m2v2 = m1 v´1 + m2 v´2
Since the objects stick together they must have
the same velocity afterwards, v´1 = v´2 = v´
m1v1 + m2v2 = m1 v´ + m2 v´
m1v1 + m2v2 = (m1+ m2) v´
v´ = (m1v1 + m2v2) / (m1+ m2)
v´ = ((15 kg) (6 m/s) + (20 kg)(-3 m/s)) / (15 kg + 20 kg)
v´ = ((90 kg∙ m/s) + (-60 kg∙ m/s)) / (35 kg)
v´ = (30kg∙ m/s) / (35 kg)
v´ = 0.86 m/s
v´ = 0.86 m/s towards the east
Example 8: A closed system consists of a stationary object that explodes into two
pieces. After the explosion, one piece, m1, has a mass of 25 kg and a velocity of 8 m/s
towards the north and the second object, m2, has a mass of 40 kg. Determine the
velocity of the second piece.
Let’s choose north to be positive and south to be negative. Then,
p0 + I = pf
Since there are no external forces, I = 0 so
p0 = pf
Since the object had no initial velocity, p0 = 0
0 = m1 v´1 + m2 v´2
Then we solve for v´2
m2 v´2 = - m1 v´1
v´2 = (- m1 v´1) / m2
And then substitute in numbers
v´ = (- (25 kg)( 8 m/s) / (4 kg)
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v´ = (- (25 kg)( 8 m/s)) / (4 kg)
v´ = (- 200 kg∙ m/s) / (4 kg)
v´ = (- 50 m/s)
Then we interpret our negative velocity in
terms of direction.
v´ = 50 m/s
towards the south
So this is how you solve all momentum problems if no outside forces are acting on the
system. There are a whole class of problems, called Collision problems where this is
the case.
Collisions
The collisions described in this section are all between objects within a closed system.
No external forces are affecting the system, so there is no impulse acting on the
system, I = 0.
All collisions fall into one of three categories, Perfectly Inelastic, Inelastic or Perfectly
Elastic. In the case of perfectly inelastic collisions, the colliding objects stick together.
While momentum is conserved, as it is in all these cases, mechanical energy is not
conserved. Some of the mechanical energy goes into heat, sound, deforming the
material, etc. While the total energy is conserved, it is now in forms with which we
haven’t yet worked. That is, the total of the gravitational potential, kinetic and elastic
potential energy is not conserved in these types of collisions.
That is true of any inelastic collision. While in the case of perfectly inelastic collisions
the two colliding objects stick together, in the case of inelastic collisions, while they don’t
stick together, some mechanical energy is still lost to other forms.
The only case where mechanical energy is conserved is during a perfectly elastic
collision. In this instance, the two objects bounce off each other with no loss of
mechanical energy. (A good example of this is when billiard balls collided.) In that
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case, and in that case alone, we can combine our equations for conservation of energy
with our equations for conservation of momentum.
Problem Solving Strategies for Collisions
The first step in solving a collision problem is to determine the type of collision. You
can’t assume that; the problem must tell you either directly or indirectly. Obviously if
they simply tell you the type of collision, that’s fine. On the other hand, they could tell
you indirectly through hints. For instance, if you are told that the objects stick together,
then you know it was a perfectly inelastic collision without being told that outright.
From that information you know that mechanical energy was not conserved and, most
importantly, that v´1 = v´2 = v´. That last fact is critical to solving the problem. An
example of this type can be found in Example 7.
If you are not told the type of collision, that the objects stick together or that mechanical
energy is conserved, then you can only assume this was an elastic collision. In that
case, they must tell you some more information in order to solve the problem, for
instance, the velocity and mass of one of the objects after the collision. Without that
added information these problems can’t be solved.
Example 9: A bullet if fired at a piece of wood which is at rest on a frictionless surface.
The bullet has a mass of 0.025 kg and a velocity of 300 m/s and the mass of the wood
is 2.0 kg. After passing through the wood the velocity of the bullet is 200 m/s. What is
the velocity of the wood?
Since they don’t tell us that this is a perfectly elastic or perfectly inelastic collision, that
the objects stick together or that energy is conserved, we have to assume it is a
inelastic collision. But they did give us the added information of the velocity of the bullet
after the collision.
p0 + I = pf
Since there are no external forces, I = 0 so
p0 = pf
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mbvb + mwvw = mbv´b + mwv´w
But the wood was originally at rest so vw = 0
mbvb = mbvb´ + mw vw´
mbvb - mbvb´ = mw vw´
mw vw = mbvb - mbvb´
vw = mb (vb - vb) / mw
vw = (.025kg)(300 m/s - 200m/s) / (2 kg)
vw = (2.5 kg∙ m/s ) / (2 kg)
vw = (1.25 m/s ) in the same direction as the bullet was traveling
If you are told that mechanical energy is conserved, then you know it was a perfectly
elastic collision. That allows you to use what we learned about conservation of energy
in solving the problem. It turns out in this case that a very important general result can
be obtained for perfectly elastic collisions. That is the difference of velocities prior to the
collision is equal to the opposite of that difference after the collision. (This is derived at
the end of this chapter.)
v1 - v2 = v2´ - v1´
Or another way of saying that is that the total of the velocities before and after the
collision are the same for both objects.
v1 + v1´= v2 + v2´
Example 10: Two objects collide and bounce off of each other such that mechanical
energy is conserved. One object, m1, has a mass of 2.0 kg and a velocity of 8 m/s
towards the east and the second object, m2, has a mass of 4.0 kg and a velocity of 3
m/s towards the west. What are the velocities of the two objects after the collision?
Let’s choose east to be positive. Then,
p0 + I = pf
Since there are no external forces, I = 0 so
p0 = pf
m1v1 + m2v2 = m1 v´1 + m2 v´2
Since the collision is elastic, v1 - v2 = v2´ - v1´
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Solving for v2´ yields, v2´ = v1 - v2 + v1´
Then substituting that for v2´ yields
m1v1 + m2v2 = m1 v´1 + m2 (v1 - v2 + v1´)
Distributing m2 yields
m1v1 + m2v2 = m1 v´1 + m2 v1 - m2v2 + m2v1´
Leaving all the terms with v´1 on the right but
moving the rest to the left
m1v1 - m2 v1 + m2v2 + m2v2 = m1 v´1 + m2v1´
Factoring out and combining common terms
(m1- m2) v1 + 2m2v2 = v´1 (m1 + m2)
Switching v´1 (m1 + m2) to the right
v´1 (m1 + m2)= (m1- m2) v1 + 2m2v2
Solving form v´1 by dividing by (m1 + m2)
v´1 = ((m1- m2) v1 + 2m2v2) / (m1 + m2)
Substituting numbers
v´1 = ((2kg- 4kg)(8m/s) + 2(4kg)(-3m/s) / (2 kg + 4 kg)
v´1 = ((-16 kg∙ m/s) + -24 kg∙ m/s) / (6 kg)
v´1 = -6.7 m/s
Then substituting this back into the equation we got from conservation of energy
v2´ = v1 - v2 + v1´
v2´ = 8 m/s – (-3 m/s) + (-6.7 m/s)
v2´ = 8 m/s + 3 m/s -6.7 m/s
v2´ = 4.3 m/s
v´1 = 6.7 m/s towards the west
v2´ = 4.3 m/s towards the east
Change of Momentum and Impulse
If no external force acts on a closed system, its momentum will not change. However, if
an external force does act on a system, the systems momentum will be altered changed
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by the Impulse applied by that external force. So the initial amount of momentum of a
system can be altered by an amount equal to the Impulse applied. The result is that the
final momentum of the system has changed.
p0 + I = pf
I = pf - p0
Now let’s use Newton’s Second Law to get an equation for Impulse (I).
We know that p = mv, so let’s substitute that in for p
I = mvf - mv0
Now factor out m
I = m (vf - v0)
I = m Δv
But acceleration is defined as a = Δv / Δt
So we can replace Δv by a Δt
I = ma Δt
But if there is one force acting on a system then
F = ma so we can replace ma by F
I = F Δt
So the impulse on a system, or an object, due to an external force is just equal to the
force that’s applied to it times by the amount of time that that force is applied. Since this
is also equal to the change in the momentum of the system, or object, we can also write
this as
I = Δp = F Δt
It can be seen from this equation that the units of impulse must also be equal to he units
of momentum, kg∙ m/s.
Example 11: A stationary ball whose mass is 4.5 kg is subject to an impulse of 36 kg∙
m/s towards the north. What will its velocity be after this? What will the change in its
momentum be?
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p0 + I = pf
pf = p0 + I
mv´= mv + I
But the ball was initially at rest so v = 0
mv´= I
v´= I/m
v´= 36 kg∙ m/s / 4.5 kg
v´= 36 kg∙ m/s / 4.5 kg
v´= 8 m/s
v´= 8 m/s towards the north
I = Δp
So Δp = 36 kg∙ m/s towards the north
Example 12: If the force on the ball in example 11 had been applied for 1.8s, what was
its average magnitude and direction?
I = F Δt
F = I / Δt
F = (36 kg∙ m/s( / (1.8s)
F = 20 kg∙ m/s2 towards the north
F = 20 N towards the north
The force and impulse are always in the same direction since Δt is a scalar.
Example 13: A 5.0 kg ball is traveling at a velocity of 25 m/s to the east when it
bounces off of a wall and rebounds with a velocity of 20 m/s to the west. (a) What was
its change in momentum? (b)What impulse did the wall deliver to the ball? (c)If the ball
was in contact with the wall for 0.25s, what was the average force acting on the ball?
(d)What was the average force acting on the wall? (e)What was the impulse delivered
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by the wall by ball? (Remember to answer all questions about vectors with both
magnitudes and directions.)
Let’s take east to be the positive direction.
(a)
Δp = pf - po
Δp = pf - po
Δp = mvf - mvo
Δp = m(vf - vo)
Δp = 5.0 kg(-20 m/s -25 m/s)
Δp = 5.0 kg(-45 m/s)
Δp = 220 kg∙ m/s to the west
(b)
I = Δp = 220 kg∙ m/s to the west
(c)
I = F Δt
F = I / Δt
F = 220 kg∙ m/s / .25s
F = 880 N towards the west
(d)
By Newton’s third law we know that the force acting on the wall from the ball must be
equal and opposite to the force acting on the ball due to the wall. Therefore,
The force on the wall must be 880 N towards the east.
(e)
I = F Δt
Since the force acting on the wall is equal and opposite to that acting on the ball, and Δt
is the same, that means that the impulse delivered to the wall by the ball must be equal
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and opposite the impulse given the ball by the wall. So the impulse is 220 kg∙ m/s to the
east.
It is always true that if two objects collide that the impulse one delivers to the other is
equal and opposite to the impulse it receives.
Derivation of relationship of velocities before and after an elastic collision.
p0 + I = pf
There are no outside forces so I = 0
p0 = pf
m1v1 + m2v2 = m1 v1´ + m2 v2´
Reorganize this to put all the m1 terms on
one side and all the m2 terms on the other
m1v1 - m1 v1´ = m2 v2´ - m2v2
Factor out the masses
m1 (v1 - v1´) = m2 (v2´ - v2)
Leave that result for later. Now let’s use
mechanical energy conservation
for perfectly elastic collisions to get a second
equation
E0 + W = Ef
But there are no outside forces so W = 0
E0 = E f
There are no height changes or springs
involved so all the energy is kinetic
KE0 = KEf
½ m1v12 + ½ m2v22 = ½ m1v1´2 + ½ m2v2´2
Multiplying by 2 simplifies this
m1v12 + m2v22 = m1v1´2 + m2v2´2
Now reorganize this to put all the m1 terms on
one side and all the m2 terms on the other
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m1v12 – m1 v1´2 = m2 v2´2 - m2v22
Factor out the masses
m1(v12 - v1´2)= m2(v2´2 - v22)
Now factor using difference of squares
m1 (v1 - v1´) (v1 + v1´) = m2 (v2´- v2) (v2´+ v2)
Now divide this equation, from energy
conservation by the one we got from
momentum conservation.
m1(v1 - v1´) (v1 + v1´) = m2 (v2´- v2) (v2´+ v2)
m1 (v1 - v1´)
=
m2 (v2´ - v2)
v1 + v1´ = v2´+ v2
Which is fine, or you can say that
v1 - v2 = v2´ - v1´
Which indicates that the difference of velocities
is reversed during a collision
Most importantly, notice that this result is true regardless of the masses. They can be
the same or different, it has no affect on this result.
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Chapter Questions
1. What is formula for momentum? What is the unit of momentum?
2. What is the difference between momentum and mass?
3. Which has greater mass, a bowling ball at rest or a rolling basketball? Which has
greater momentum?
4. Is linear momentum conserved, when two objects collide and stick together? Is
kinetic energy conserved?
5. Why, when you release an inflated but untied balloon, does it fly across the
room?
6. How can a rocket change direction when it is far out of space and is essentially in
a vacuum?
7. A boy stands in the middle of a perfectly smooth, frictionless, frozen lake. How
can he set himself in motion?
8. When rain falls from the sky, what happens to its momentum as it hits the
ground?
9. What is impulse? What is the unit of impulse?
10. What is the difference between impulse and momentum? How does impulse
relate to momentum?
11. When a glass falls, will the impulse be greater if it lands on a plush carpet than if
it lands on a hard floor?
12. When a small car meshes with a large truck in a head-on collision, which of these
two experiences greater change in momentum?
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Chapter Problems
Momentum of a single object
1. What is the momentum of a 3000 kg truck traveling at 25 m/s?
2. A 1500 kg ferryboat has a momentum of 25000 kg∙m/s. What is the speed of the
ferryboat?
3. A car travels at a constant speed of 24 m/s and has a momentum of 28800
kg∙m/s. What is the mass of the car?
4. An 8 kg bowling ball rolls at a constant speed of 3 m/s. What is the momentum of
the ball?
5. A bicyclist travels at a constant speed of 7 m/s. What is the total mass of the
bicycle and the boy, when the total momentum is 490 kg∙m/s?
6. When a 45 kg cannon ball leaves a barrel it has a momentum of 14000 kg∙m/s.
What is the speed of the ball at the end of the barrel?
Momentum of a closed system of objects
7. Determine the momentum of a system that consists of two objects. One object,
m1, has a mass of 6 kg and a velocity of 13 m/s towards the east and a second
object, m2, has a mass of 14 kg and a velocity of 7 m/s in that same direction.
8. Determine the momentum of a system of the two objects. One object, m 1, has a
mass of 35 kg and a velocity of 3.7 m/s towards the east and the second object,
m2, has a mass of 57 kg and a velocity of 4.3 m/s towards the west.
9. Determine the momentum of a system that consists of two objects. One object,
m1, has a mass of 6 kg and a velocity of 13 m/s in the direction of the positive xaxis and a second object, m2, has a mass of 14 kg and a velocity 7 m/s in the
direction of the negative x-axis.
10. Determine the momentum of a system of the two objects. One object, m 1, has a
mass of 35 kg and a velocity of 3.7 m/s towards the north and the second object,
m2, has a mass of 57 kg and a velocity of 4.3 m/s towards the south.
11. Determine the momentum of a system that consists of three objects. One object,
m1, has a mass of 7 kg and a velocity 23 m/s towards the north, a second object,
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m2, has a mass of 9 kg and a velocity 7 m/s towards the north and the third
object, m3 has a mass of 5 kg and a velocity of 42 m/s towards the south .
12. Determine the momentum of a system that consists of three objects. One object,
m1, has a mass of 12 kg and a velocity 120 m/s towards the east, a second
object, m2, has a mass of 25 kg and a velocity 18 m/s towards the west and the
third object, m3 has a mass of 1 kg and a velocity of 350 m/s towards the east .
Conservation of Momentum and Inelastic Collisions
13. A 13,500 kg railroad freight car travels on a level track at a speed of 4.5 m/s. It
collides and couples with a 25,000 kg second car, initially at rest and with brakes
released. What is the speed of the two cars after collision?
14. A 0.01 kg bullet has a speed of 700 m/s before it strikes a 0.95 kg wooden block
that is stationary on a horizontal frictionless surface and remains inside of it.
What is the speed of the block after the bullet becomes embedded in it?
15. A cannon ball with a mass of 100 kg flies in horizontal direction with a speed of
600 m/s and strikes a railroad freight car filled with sand and initially at rest. The
total mass of the car and send is 25,600 kg. Find the speed of the car after the
ball becomes embedded it the send.
16. A 40 kg boy skates at 3.5 m/s on ice toward his 65 kg friend who is standing still,
with open arms. As they collide and hold each other, what is the speed of the two
boys?
17. A hunter in a boat throws a 15.6 kg rifle in horizontal direction with a velocity of
7.5 m/s. Calculate the velocity of the boat immediately after, assuming that the
initial velocity of the boat was 1.8 m/s and it was moving in the same direction as
the rifle. The mass of the hunter is 75 kg and the boat’s mass is 90 kg.
18. A 45 kg boy jumps off a 150 kg moving boat. Find the boat velocity immediately
after the boy jumps, assuming that the boat’s initial velocity is 2.2 m/s and the
boy’s velocity after jumping is 5.2 m/s (opposite to the boat’s initial velocity) with
respect to the stationary water.
19. A 55 kg boy jumps off a 100 kg moving boat. Find the boat’s velocity immediately
after the boy jumps, assuming that the boat’s initial velocity is 0.2 m/s and the
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boy’s velocity is 2.5 m/s (in the same direction as the boat’s initial velocity) with
respect to the stationary water.
20. A 0.01 bullet is fired at a 0.5 kg block initially at rest. The bullet, moving with an
initial speed of 400 m/s, emerges the block with a speed of 300 m/s. What is the
speed of the block after the collision?
21. A 0.015 bullet is fired at a 1.5 kg block initially at rest. The bullet, moving with an
initial speed of 500 m/s, emerges the block with a speed of 400 m/s. What is the
speed of the block after the collision?
22. A 55 kg skater at rest on a frictionless rink throws a 3 kg ball, giving the ball a
velocity of 8 m/s. What is the velocity of the skater immediately after?
23. Two football players with mass 85 kg and 110 kg run directly toward each other
with speeds 4 m/s and 7 m/s respectively. If they grab each other as they collide,
what is the combined speed of the players just after the collision?
24. An air track car with a mass of 0.55 kg and velocity of 5.8 m/s to the right collides
and couples with a 0.45 kg car moving to the left with a velocity of 3.9 m/s. What
is the combined velocity of the cars just after the collision?
25. An air track car with a mass of 0.25 kg and velocity of 3.4 m/s to the right collides
and couples with a 0.45 kg car moving to the left with a velocity of 3.9 m/s. What
is the combined velocity of the cars just after the collision?
26. A 15000 kg railroad car travels on a horizontal track with a constant speed of 12
m/s. A 6000 kg load is dropped onto the car. What will be the car’s speed?
Perfectly Elastic Collisions
27. A ball of mass 0.34 kg moving with a speed of 2.7 m/s to the right collides headon with a 0.24 kg ball at rest. If the collision is perfectly elastic, what is the speed
and direction of each ball after the collision?
28. An ice puck of mass 0.54 kg moving with a speed of 5m/s to the right collides
with a 0.28 kg piece of ice moving with a speed of 4.2 m/s to the right. If the
collision is perfectly elastic, what is the speed and direction of each mass after
the collision?
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29. An air track car with a mass of 0.75 kg and velocity of 8.5 m/s to the right
collides elastically with a 0.65 kg car moving to the left with a velocity of 7.2 m/s.
What is the combined velocity of the cars just after the collision?
30. An air track car with a mass of 0.85 kg and velocity of 3.4 m/s to the right collides
elastically with a 0.95 kg car moving to the left with a velocity of 4.9 m/s. What is
the combined velocity of the cars just after the collision?
31. A ball of mass 6.5 kg moving with a speed of 15 m/s to the right collides head-on
with a 3.5 kg ball which is at rest. If the collision is perfectly elastic, what is the
speed and direction of each ball after the collision?
32. An ice puck of mass 7.5 kg moving with a speed of 18 m/s to the right collides
with a 2.5 kg piece of ice moving with a speed of 4.2 m/s to the right. If the
collision is perfectly elastic, what is the speed and direction of each mass after
the collision?
Change of Momentum and Impulse
33. A 1200 kg car accelerates from 13 m/s to 17 m/s. Find the change in momentum
of the car?
34. A 0.17 kg hokey puck slows down from 54 m/s to 35 m/s when it slides on
horizontal ice surface. Find the change in momentum of the puck?
35. A 15000 kg air jet accelerates from rest to 45 m/s before it takes off. What is the
change in momentum of the jet?
36. A 0.01 kg bullet is fired at 250 m/s into a wooden block that is fixed. The bullet
emerges from the block with a speed of 120 m/s. What is the change in
momentum of the bullet?
37. A 0.025 kg piece of clay is thrown into a wall and has a speed of 9 m/s before it
strikes the wall. Find the change in momentum of the clay and the impulse
exerted on the clay if it does not bounce from the wall.
38. A small object with a momentum of 6 kg∙m/s to the west approaches head-on a
large object at rest. The small object bounces straight back with a momentum of
5 kg∙m/s. What is the change in the momentum of the small object? What is the
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impulse exerted on the small ball? What is the impulse exerted on the large
object?
39. A 0.05 kg tennis ball moves at a speed of 10 m/s and is struck by a racket
causing it to rebound in the opposite direction at a speed 16 m/s. What is the
change in momentum of the ball? What is the impulse exerted on the ball? What
is the impulse exerted on the racket?
40. A 0.25 kg beach ball rolling at a speed of 7 m/s collides with a heavy exercise
ball at rest. The beach ball bounces straight back with a speed of 4 m/s. That is
the change in momentum of the beach ball? What is the impulse exerted on the
beach ball? What is the impulse exerted on the exercise ball?
41. A 0.15 kg apple falls from a height of 2.5 m. What is the change in momentum of
the apple just before it strikes the ground?
42. A 0.07 kg tennis ball leaves a racket with a speed of 56 m/s. If the ball is in
contact with the racket for 0.04 s, what is the average force on the ball by the
racket?
43. A 0.03 kg golf ball is hit off the tee at a speed of 34 m/s. The golf club was in
contact with the ball for 0.003 s. What is the average force on ball by the golf
club?
44. A 0.16 kg hokey puck is moving on an icy horizontal surface with a speed of 5
m/s. A player strikes the puck by a hokey stick, after the impact the puck moves
in opposite direction with a speed of 9 m/s. If the puck was in contact with the
stick for 0.005 s, what is the average force on the puck by the stick?
45. A 0.145 kg baseball reaches a speed of 36 m/s when a bat strikes. If the average
force of 500 N was applied on the ball by the bat, what is the impact time?
46. A toy rocket, of mass 0.3 kg, achieves a velocity of 55 m/s after 3 s, when fired
straight up. What average force does the rocket engine exert?
47. A constant force of 12 N acts for 5 s on a 5 kg object. What is the change in
object’s velocity?
48. A small object with a mass of 1 kg moves in a circular path with a constant speed
of 5 m/s. What is the change in momentum during ½ of period; one period?
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