In a collision, strong mutual forces act between a few particles for a short
time. These internal forces are significantly larger than any external forces
during the time of the collision. The laws of conservation of linear momentum
and energy, applied to the "before" and "after" situations, often allows us to
predict the outcome of a collision. **A great deal can be learned about the
interactions between the colliding particles from the observed collision
products.**

** Note**:

- external forces are small (and are ignored) during the
collision
- particles before and after the collision can be different (for example:
nuclear reactions)
- the collision force does not need to be a contact force

Suppose a force F acts during a collision. The result of the collision force
will be a change in the momentum of the particles involved. The amount of
change depends not only on the average value of the force, but also on the time
period during which it acts. The change in momentum **dp** is related to
the collision force **F** as

_{
}

The total change in the of momentum during the collision is given by

_{
}

The right hand side of this equation is a measure of both the strength
and the duration of the collision force. It is called the
**collision impulse J**:

_{
}

The unit of the impulse is N ^{.} s. From the definition of
the impulse J we see that the relation between the impulse and the change of
momentum is given by

_{
}

If the average collision force F_{av} acts during a time period
equal to [Delta]t, the impulse J is equal to

F_{av} [Delta]t = J

**Sample Problem 10-1**

A baseball of mass m in horizontal flight with speed v_{i} is
struck by a batter. After leaving the bat, the ball travels in the opposite
direction with a speed v_{f}. What impulse J acts on the ball while it
was in contact with the bat ?

_{
}

If the impact time is [Delta]t, what is the average force that acts on the baseball ?

_{
}

What is the average acceleration of the baseball during this period ?

_{
}

Consider the collision shown in Figure 10.1. If there are no
external forces acting on this system (consisting of the two masses) the total
momentum of the system is conserved. The first class of collisions we will
discuss are the **elastic collisions**. Collisions are
called elastic collisions if the total kinetic energy of the system is
conserved. Applying conservation of linear momentum to the collision shown in
Figure 10.1 gives

_{
}

Conservation of the total kinetic energy gives

_{
}

We now have two equations with two unknown (v_{1f} and
v_{2f}) which can be solved. The first equation can be rewritten as

_{
}

The second equation can be rewritten as

_{
}

The final velocity of mass m_{1} can now be calculated by
dividing the last two expressions

_{
}

Figure 10.1. Collision in One-Dimension.

This gives

_{
}

The final velocity of m_{1} can now be obtained

_{
}

The final velocity of mass m_{2} can also be obtained

_{
}

It is clear that the velocity of m_{2} is always positive. The
velocity of m_{1} can be either positive or negative, depending on the
masses of the two objects: v_{1f} is negative if m_{2} >
m_{1}, and positive if m_{2} < m_{1}.

In Chapter 9 we have shown that the motion of the center of mass is unaffected by the collision. The velocity of the center of masscenter of mass velocity can be calculated easily

_{
}

It can be easily verified that the velocity of the center of mass after the collision is the same as it was before the collision (as it should be of course since there are no external forces acting on the system).

** Some Special Cases**

**Equal Mass**: m_{1}= m_{2}.The previously derived equations show that in this case

v

_{1f}= 0 m/sv

_{2f}= v_{1i}In head-on collisions, particles of equal mass simply exchange velocities.

**Massive Target**: m_{2}>> m_{1}.Using the previously derived expressions for v

_{1f }and v_{2f}we obtain_{ }_{ }The projectile simply bounces back and the final velocity of the target will be a very small fraction of the initial velocity of the projectile.

**Massive Projectile**: m_{1}>> m_{2}._{}_{}Using the previously derived expressions for v_{1f }and v_{2f}we obtain_{ }_{ }The velocity of the projectile is almost unchanged while the target will move with twice the initial velocity of the projectile.

The collision force acting between the target and the projectile is an internal force of the system under consideration consists of these two objects. The motion of the center of mass of a number of objects is solely determined by the external forces acting on the system (see Chapter 9).

_{
}

This equation shows that if no external forces act on the system, the velocity of its center of mass is constant.

**Sample Problem 10-4**

In a nuclear reactor, newly-produced fast neutrons must be slowed down before
they can participate effectively in the chain-reaction process. By what
fraction is the kinetic energy of a neutron (mass m_{1}) reduced in a
head-on collision with a nucleus of mass m_{2} (initially at rest) ?

Suppose v_{1i} is the initial velocity of the neutron. Its final
velocity, v_{1f}, can be obtained using one of the previously derived
equations:

_{
}

The initial kinetic energy of the neutron is given by

_{
}

The final kinetic energy of the neutron is given by

_{
}

The fraction of the kinetic energy of the neutron lost in the collision is given by

_{
}

This is equal to

_{
}

Since the mass units in the equation for f cancel, we can use atomic mass units:

mass Pb: 206 amu => f = 0.019 = 1.9 %

mass C: 12 amu => f = 0.28 = 28 %

mass H: 1 amu => f = 1.00 = 100 %

We conclude that any compound that contains large amounts of H will be a good moderator.

**Example Problem 10-1**

A glider whose mass is m_{2} rests on an air track. A
second glider, whose mass is m_{1}, approaches the target glider with a
velocity v_{1i} and collides elastically with it. The target glider
rebounds elastically from the end of the track and meets the projectile glider
a second time (see Figure 10.2). At what distance from the end of the air
track will the second collision occur ?

The velocity of m_{1} and m_{2} after the first collision are
given by

_{
}

_{
}

Suppose the second collision occurs a distance x from the end of the
track (see Figure 10.2). At that point, mass m_{1} has traveled a
distance (d - x) after the first collision and mass m_{2} has traveled
a distance (d + x). Both masses must cover these distances of course in the
same time

_{
}

This can be rewritten as

_{
}

or

_{
}

Figure 10.2. Example Problem 10-1.

Substituting the expressions for v_{1f} and v_{2f} in
the expression for x, we obtain

_{
}

A special case occurs when m_{1} = m_{2}. In this
case, x = d. Note: v_{1f} = 0 m/s and v_{2f} =
v_{1i}.

If no external forces act on a system, its momentum is conserved. However, kinetic energy is not always conserved. An example of an inelastic collision is a collision in which the particles stick together (after the collisions). This type of a collision is a completely inelastic collision (of course, even in a completely inelastic collision, the total energy is conserved. The lost kinetic energy is converted into another form of energy: for example, thermal energy, energy of deformation etc.)

**Example Problem 10-2**

Suppose a mass m_{1} is moving with an initial velocity v_{i}
and collides with a mass m_{2}, which is initially at rest (see Figure
10.3). The two masses stick together. What is the final velocity of the
system, and what is the change in the kinetic energy of the system ?

Figure 10.3. Completely Inelastic Collision.

The initial momentum of the system is

p_{i} = m_{1} v_{i}

The final momentum of the system is

p_{f} = (m_{1} + m_{2}) v_{f}

Conservation of linear momentum implies

m_{1} v_{i} = (m_{1} + m_{2})
v_{f}

or

_{
}

The final velocity of the combined system will always be less than that of the incoming object.

The initial kinetic energy of the system is

_{
}

The final kinetic energy of the system is

_{
}

**Note**: not all the kinetic energy can be lost, even in a
completely inelastic collision, since the motion of the center of mass must
still be present. Only if our reference frame is chosen such that the
center-of-mass velocity is zero, will the final kinetic energy in a completely
inelastic collision be zero.

**Sample Problem 10-5**: The ballistic pendulum.

Suppose a bullet of mass m_{1} hits a large block of wood of mass
m_{2}. As a result, the block plus bullet swings upwards (maximum
height is h). What is the velocity of the bullet ?

Suppose the velocity of the bullet is v_{i}. The initial momentum of
the system is

p_{i} = m_{1} v_{i}

The final velocity of the block plus bullet is v_{f}. The
final momentum is

p_{f} = (m_{1} + m_{2}) v_{f}

Conservation of linear momentum implies

m_{1} ^{.} v_{i} = (m_{1} +
m_{2}) v_{f}

or

_{
}

The initial kinetic energy of the block plus bullet is

_{
}

Since it is assumed that the block plus bullet swing frictionless, the total mechanical energy of the block plus bullet must be conserved. At its highest point, the mechanical energy of the block plus bullet is equal to

_{
}

Conservation of mechanical energy implies that

_{
}

or

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}

The velocity of the bullet can now be calculated

_{
}

Suppose a mass m_{1}, with initial velocity v_{1i},
undergoes a collision with a mass m_{2} (which_{ }is initially
at rest). The particles fly of at angles [theta]_{1} and
[theta]_{2}, as shown in Figure 10.4. Since no external forces act on
the collision system, linear momentum is conserved (in both x and y
direction):

_{
}

and

_{
}

If the collision is elastic, kinetic energy also needs to be conserved:

_{
}

Figure 10.4. A Collision in Two Dimensions.

The variables are:

- mass: m
_{1}and m_{2} - velocity: v
_{1i}, v_{1f}and v_{2f} - angle: [theta]
_{1}and [theta]_{2}

**Example Problem 10-3**

A beam of nuclei with mass m_{1} and velocity v_{1} is
incident on a target nucleus with mass m_{2} which us initially at
rest. The velocity and scattering angles of both reaction products is
measured. Determine the masses of the reaction products and the change in
kinetic energy.

The collision is schematically shown in Figure 10.5. Conservation of linear momentum along the x-axis requires

_{
}

where p_{1}, p_{3} and p_{4} are the momenta of
particle 1, particle 3 and particle 4, respectively. Conservation of linear
momentum along the y-axis requires

_{
}

The last equation can be rewritten as

_{
}

Figure 10.5. Example Problem 10-3.

Substituting this in the first equation we obtain

_{
}

This immediately shows that

_{
}

or

_{
}

This relation can be used to find p_{4}

_{
}

and m_{4}

_{
}

The change in the kinetic energy of the system can now be calculated

_{
}

**Example Problem 10-4**

A 20 kg body is moving in the direction of the positive x-axis with a speed of 200 m/s when, owing to an internal explosion, it breaks into three parts. One part, whose mass is 10 kg, moves away from the point of explosion with a speed of 100 m/s along the positive y-axis. A second fragment, with a mass of 4 kg, moves along the negative x-axis with a speed of 500 m/s.

a). What is the speed of the third (6 kg) fragment ?

b) How much energy was released in the explosion (ignore gravity) ?

The problem is schematically illustrated in Figure 10.6. Since the force of the explosion is an internal force, the total momentum of the system will be conserved. We therefore can apply this conservation law both along the x-axis and along the y-axis. Conservation of linear momentum along the x-axis requires

_{
}

Conservation of linear momentum along the y-axis requires

_{
}

These equations can be rewritten as

_{
}

_{
}

Squaring both equations and adding them gives

_{
}

Solving this equation gives v_{3} = 1014 m/s. The energy
supplied by the explosion is now easy to calculate:

_{
}

Figure 10.6. Example Problem 10-4.

Send comments, questions and/or suggestions via email to **wolfs@pas.rochester.edu** and/or visit the home page of Frank Wolfs.