## 10.1. Introduction

**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

## 10.2. Impulse

**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

_{ }

_{ }

**collision impulse J**:

_{ }

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

_{ }

_{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 ?

_{ }

_{ }

_{ }

## 10.3. Collisions in One-Dimension

**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

_{ }

_{ }

_{1f}and v

_{2f}) which can be solved. The first equation can be rewritten as

_{ }

_{ }

_{1}can now be calculated by dividing the last two expressions

_{ }

Figure 10.1. Collision in One-Dimension.

This gives_{ }

_{1}can now be obtained

_{ }

_{2}can also be obtained

_{ }

_{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

_{ }

**Some Special Cases**

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

v_{1f}= 0 m/s

v_{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

_{ }

_{ }

**Massive Projectile**: m_{1}>> m_{2}. Using the previously derived expressions for v_{1f }and v_{2f}we obtain

_{ }

_{ }

## 10.5. Motion of the Center of Mass

**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).**

_{ }

**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:

_{ }

_{ }

_{ }

_{ }

_{ }

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

_{ }

_{ }

_{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

_{ }

_{ }

_{ }

Figure 10.2. Example Problem 10-1.

Substituting the expressions for v_{1f}and v

_{2f}in the expression for x, we obtain

_{ }

_{1}= m

_{2}. In this case, x = d. Note: v

_{1f}= 0 m/s and v

_{2f}= v

_{1i}.

## 10.6. Collisions in One-Dimension: Inelastic

**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

The final momentum of the system is_{i}= m_{1}v_{i}
p

Conservation of linear momentum implies_{f}= (m_{1}+ m_{2}) v_{f}
m

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

The initial 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

The final velocity of the block plus bullet is v_{i}= m_{1}v_{i}_{f}. The final momentum is

p

Conservation of linear momentum implies_{f}= (m_{1}+ m_{2}) v_{f}
m

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

_{ }

_{ }

_{ }

_{ }

_{ }

## 10.7. Collisions in Two-Dimensions

**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):

_{ }

_{ }

_{ }

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

_{ }

_{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

_{ }

_{ }

Figure 10.5. Example Problem 10-3.

Substituting this in the first equation we obtain_{ }

_{ }

_{ }

_{4}

_{ }

_{4}

_{ }

_{ }

**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

_{ }

_{ }

_{ }

_{ }

_{ }

_{3}= 1014 m/s. The energy supplied by the explosion is now easy to calculate:

_{ }

Figure 10.6. Example Problem 10-4.

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