7.3

Collisions in One Dimension

As discussed in the last section, the total linear momentum is conserved when two objects collide, provided they constitute an isolated system. When the objects are atoms or subatomic particles, the total kinetic energy of the system is often conserved also. In other words, the total kinetic energy of the particles before the collision equals the total kinetic energy of the particles after the collision, so that kinetic energy gained by one particle is lost by another.
In contrast, when two macroscopic objects collide, such as two cars, the total kinetic energy after the collision is generally less than that before the collision. Kinetic energy is lost mainly in two ways: (1) it can be converted into heat because of friction, and (2) it is spent in creating permanent distortion or damage, as in an automobile collision. With very hard objects, such as a solid steel ball and a marble floor, the permanent distortion suffered upon collision is much smaller than with softer objects and, consequently, less kinetic energy is lost.
Collisions are often classified according to whether the total kinetic energy changes during the collision:

1. | Elastic collision—One in which the total kinetic energy of the system after the collision is equal to the total kinetic energy before the collision.
2. | Inelastic collision—One in which the total kinetic energy of the system is not the same before and after the collision; if the objects stick together after colliding, the collision is said to be completely inelastic.

The boxcars coupling together in Figure 7-10 are an example of a completely inelastic collision. When a collision is completely inelastic, the greatest amount of kinetic energy is lost. Example 7 shows how one particular elastic collision is described using the conservation of linear momentum and the fact that no kinetic energy is lost.

Problem solving insight As long as the net external force is zero, the conservation of linear momentum applies to any type of collision. This is true whether the collision is elastic or inelastic.

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| Example 7 | | | A Head-On Collision

Figure 7-12 illustrates an elastic head-on collision between two balls. One ball has a mass of and an initial velocity of +5.00 m/s. The other has a mass of and is initially at rest. No external forces act on the balls. What are the velocities of the balls after the collision?

Figure 7-12 A 0.250-kg ball, traveling with an initial velocity of , undergoes an elastic collision with a 0.800-kg ball that is initially at rest.

Reasoning Three facts will guide our solution. The .rst is that the collision is elastic, so that the kinetic energy of the two-ball system is the same before and after the balls collide. The second fact is that the collision occurs head-on. This means that the velocities before and after the balls collide all point along the same line. In other words, the collision occurs in one dimension. Lastly, no external forces act on the balls, with the result that the two-ball system is isolated and its total linear momentum is conserved. We expect that ball 1, having the smaller mass, will rebound to the left after striking ball 2, which is more massive. Ball 2 will be driven to the right in the process.
Knowns and Unknowns The data for this problem are given in the following table:

Description | Symbol | Value | Comment
Explicit Data
Mass of ball 1 | m1 | 0.250 kg |
Initial velocity of ball 1 | v01 | +5.00 m/s | Before collision occurs.
Mass of ball 2 | m2 | 0.800 kg |
Implicit Data
Initial velocity of ball 2 | v02 | 0 m/s | Before collision occurs, ball initially at rest.
Unknown Variables
Final velocity of ball 1 | vf 1 | ? | After collision occurs.
Final velocity of ball 2 | vf 2 | ? | After collision occurs.

Modeling the Problem

| Step 1 Elastic Collision
| Since the collision is elastic, the kinetic energy of the two-ball system is the same before and after the balls collide:

Here we have utilized the fact that ball 2 is at rest before the collision. Thus, its initial velocity v02 is zero and so is its initial kinetic energy. Solving this equation for , the square of the velocity of ball 1 after the collision, gives Equation 1 at the right. To use this result, we need a value for vf2, which we will obtain in Step 2.

| (1)

| Step 2 Conservation of Linear Momentum
| The total linear momentum of the two-ball system is conserved, because no external forces act on the system; momentum conservation does not depend on whether or not the collision is elastic. Conservation of linear momentum indicates that

We have again utilized the fact that ball 2 is at rest before the collision; since its initial velocity v02 is zero, so is its initial momentum. Solving the expression above for vf2 gives

which can be substituted into Equation 1, as shown at the right.

Solution Algebraically combining the results of each step, we find that
Solving for vf1 shows that

| (7.8a)

Substituting the expression for vf1 into Equation 2 gives

| (7.8b)

The negative value for vf1 indicates that ball 1 rebounds to the left after the collision in Figure 7-12, while the positive value for vf2 indicates that ball 2 moves to the right, as expected.
Related Homework: Conceptual Question 17, Problems 26, 32, 40

We can get a feel for an elastic collision by dropping a steel ball onto a hard surface, such as a marble floor. If the collision is elastic, the ball will rebound to its original height, as Figure 7-13a illustrates. In contrast, a partially deflated basketball exhibits little rebound from a relatively soft asphalt surface, as in part b, indicating that a fraction of the ball’s kinetic energy is dissipated during the inelastic collision. The very deflated basketball in part c has no bounce at all, and a maximum amount of kinetic energy is lost during the completely inelastic collision.

Figure 7-13 (a) A hard steel ball would rebound to its original height after striking a hard marble surface if the collision were elastic. (b) A partially deflated basketball has little bounce on a soft asphalt surface. (c) A very deflated basketball has no bounce at all.

The next example illustrates a completely inelastic collision in a device called a “ballistic pendulum.” This device can be used to measure the speed of a bullet.

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| Example 8 | | | A Ballistic Pendulum

A ballistic pendulum can be used to measure the speed of a projectile, such as a bullet. The ballistic pendulum shown in Figure 7-14a consists of a stationary 2.50-kg block of wood suspended by a wire of negligible mass. A 0.0100-kg bullet is fired into the block, and the block (with the bullet in it) swings to a maximum height of 0.650 m above the initial position (see part b of the drawing). Find the speed with which the bullet is fired, assuming that air resistance is negligible.

Figure 7-14 (a) A bullet approaches a ballistic pendulum. (b) The block and bullet swing upward after the collision.

The physics of measuring the speed of a bullet.

Reasoning The physics of the ballistic pendulum can be divided into two parts. The first is the completely inelastic collision between the bullet and the block. The total linear momentum of the system (block plus bullet) is conserved during the collision, because the suspension wire supports the system’s weight, which means that the sum of the external forces acting on the system is nearly zero. The second part of the physics is the resulting motion of the block and bullet as they swing upward after the collision. As the system swings upward, the principle of conservation of mechanical energy applies, since nonconservative forces do no work (see Section 6.5). The tension force in the wire does no work because it acts perpendicular to the motion. Since air resistance is negligible, we can ignore the work it does. The conservation principles for linear momentum and mechanical energy provide the basis for our solution.
Knowns and Unknowns The data for this problem are as follows:

Description | Symbol | Value | Comment
Explicit Data
Mass of bullet | m1 | 0.0100 kg |
Mass of block | m2 | 2.50 kg |
Height to which block plus
bullet swings | hf | 0.650 m | Maximum height of swing.
Implicit Data
Initial velocity of block | v02 | 0 m/s | Before collision, block is stationary.
Unknown Variable
Speed with which bullet is fired | v01 | ? | Before collision with block.

Modeling the Problem

| Step 1 Completely Inelastic Collision
| Just after the bullet collides with it, the block (with the bullet in it) has a speed . Since linear momentum is conserved, the total momentum of the block–bullet system after the collision is the same as it is before the collision:

Note that the block is at rest before the collision (), so its initial momentum is zero. Solving this equation for v01 gives Equation 1 at the right. To find a value for the speed vf in this equation, we turn to Step 2.

| (1)

| Step 2 Conservation of Mechanical Energy
| The speed vf immediately after the collision can be obtained from the maximum height hf to which the system swings, by using the principle of conservation of mechanical energy:

This result can be solved for vf to show that

which can be substituted into Equation 1, as shown at the right. In applying the energy-conservation principle, it is tempting say that the total potential energy at the top of the swing is equal to the total kinetic energy of the bullet before it strikes the block and solve directly for v01. This is incorrect, however, because the collision between the bullet and the block is inelastic, so that some of the bullet’s initial kinetic energy is dissipated during the collision (due to friction and structural damage to the block and bullet).

Solution Algebraically combining the results of each step, we find that
Related Homework: Problem 29

Check Your Understanding 3

Two balls collide in a one-dimensional, elastic collision. The two balls constitute a system, and the net external force acting on them is zero. The table shows four possible sets of values for the initial and final momenta of the two balls, as well as their initial and final kinetic energies. Only one set of values could be correct. Which is it? (The answer is given at the end of the book.)

|  | Initial (Before Collision) | Final (After Collision)
|  | Momentum | Kinetic Energy | Momentum | Kinetic Energy
a. | Ball 1: | +4 kg·m/s | 12 J | −5 kg·m/s | 10 J
| Ball 2: | −3 kg·m/s | 5 J | −1 kg·m/s | 7 J
b. | Ball 1: | +7 kg·m/s | 22 J | +5 kg·m/s | 18 J
| Ball 2: | +2 kg·m/s | 8 J | +4 kg·m/s | 15 J
c. | Ball 1: | −5 kg·m/s | 12 J | −6 kg·m/s | 15 J
| Ball 2: | −8 kg·m/s | 31 J | −9 kg·m/s | 25 J
d. | Ball 1: | +9 kg·m/s | 25 J | +6 kg·m/s | 18 J
| Ball 2: | +4 kg·m/s | 15 J | +7 kg·m/s | 22 J

Background: Momentum, kinetic energy, the principle of conservation of linear momentum, and the notion of an elastic collision are involved in this question. Remember that momentum is a vector quantity, and kinetic energy is a scalar quantity.
For similar questions (including conceptual counterparts) consult Self-Assessment Test 7.2. This test is described at the end of Section 7.5.

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