8.3 Elastic and Inelastic Collisions

Section Key Terms

Elastic and Inelastic Collisions

When objects collide, they can either stick together or bounce off one another, remaining separate. In this section, we’ll cover these two different types of collisions, first in one dimension and then in two dimensions.

In an elastic collision, the objects separate after impact and don’t lose any of their kinetic energy. Kinetic energy is the energy of motion and is covered in detail elsewhere. The law of conservation of momentum is very useful here, and it can be used whenever the net external force on a system is zero. Figure 8.6 shows an elastic collision where momentum is conserved.

Figure 8.6 The diagram shows a one-dimensional elastic collision between two objects.

An animation of an elastic collision between balls can be seen by watching this video. It replicates the elastic collisions between balls of varying masses.

Perfectly elastic collisions can happen only with subatomic particles. Everyday observable examples of perfectly elastic collisions don’t exist—some kinetic energy is always lost, as it is converted into heat transfer due to friction. However, collisions between everyday objects are almost perfectly elastic when they occur with objects and surfaces that are nearly frictionless, such as with two steel blocks on ice.

Now, to solve problems involving one-dimensional elastic collisions between two objects, we can use the equation for conservation of momentum. First, the equation for conservation of momentum for two objects in a one-dimensional collision is

Substituting the definition of momentum p = mv for each initial and final momentum, we get

where the primes (') indicate values after the collision; In some texts, you may see i for initial (before collision) and f for final (after collision). The equation assumes that the mass of each object does not change during the collision.

Now, let us turn to the second type of collision. An inelastic collision is one in which kinetic energy is not conserved. A perfectly inelastic collision (also sometimes called completely or maximally inelastic) is one in which objects stick together after impact, and the maximum amount of kinetic energy is lost. This lack of conservation means that the forces between colliding objects may convert kinetic energy to other forms of energy, such as potential energy or thermal energy. The concepts of energy are discussed more thoroughly elsewhere. For inelastic collisions, kinetic energy may be lost in the form of heat. Figure 8.7 shows an example of an inelastic collision. Two objects that have equal masses head toward each other at equal speeds and then stick together. The two objects come to rest after sticking together, conserving momentum but not kinetic energy after they collide. Some of the energy of motion gets converted to thermal energy, or heat.

Figure 8.7 A one-dimensional inelastic collision between two objects. Momentum is conserved, but kinetic energy is not conserved. (a) Two objects of equal mass initially head directly toward each other at the same speed. (b) The objects stick together, creating a perfectly inelastic collision. In the case shown in this figure, the combined objects stop; This is not true for all inelastic collisions.

Since the two objects stick together after colliding, they move together at the same speed. This lets us simplify the conservation of momentum equation from

to

for inelastic collisions, where v′ is the final velocity for both objects as they are stuck together, either in motion or at rest.

Solving Collision Problems

The Khan Academy videos referenced in this section show examples of elastic and inelastic collisions in one dimension. In one-dimensional collisions, the incoming and outgoing velocities are all along the same line. But what about collisions, such as those between billiard balls, in which objects scatter to the side? These are two-dimensional collisions, and just as we did with two-dimensional forces, we will solve these problems by first choosing a coordinate system and separating the motion into its x and y components.

One complication with two-dimensional collisions is that the objects might rotate before or after their collision. For example, if two ice skaters hook arms as they pass each other, they will spin in circles. We will not consider such rotation until later, and so for now, we arrange things so that no rotation is possible. To avoid rotation, we consider only the scattering of point masses—that is, structureless particles that cannot rotate or spin.

We start by assuming that F_net = 0, so that momentum p is conserved. The simplest collision is one in which one of the particles is initially at rest. The best choice for a coordinate system is one with an axis parallel to the velocity of the incoming particle, as shown in Figure 8.8. Because momentum is conserved, the components of momentum along the x- and y-axes, displayed as p_x and p_y, will also be conserved. With the chosen coordinate system, p_y is initially zero and p_x is the momentum of the incoming particle.

Figure 8.8 A two-dimensional collision with the coordinate system chosen so that m₂ is initially at rest and v₁ is parallel to the x-axis.

Now, we will take the conservation of momentum equation, p₁ + p₂ = p′₁ + p′₂ and break it into its x and y components.

Along the x-axis, the equation for conservation of momentum is

In terms of masses and velocities, this equation is

But because particle 2 is initially at rest, this equation becomes

The components of the velocities along the x-axis have the form v cos θ . Because particle 1 initially moves along the x-axis, we find v_1x = v₁. Conservation of momentum along the x-axis gives the equation

where ${\theta }_1$ and ${\theta }_2$ are as shown in Figure 8.8.

Along the y-axis, the equation for conservation of momentum is

or

But v₁y is zero, because particle 1 initially moves along the x-axis. Because particle 2 is initially at rest, v₂y is also zero. The equation for conservation of momentum along the y-axis becomes

The components of the velocities along the y-axis have the form v sin $\theta$ . Therefore, conservation of momentum along the y-axis gives the following equation:

Worked Example

Calculating Velocity: Inelastic Collision of a Puck and a Goalie

Find the recoil velocity of a 70 kg ice hockey goalie who catches a 0.150-kg hockey puck slapped at him at a velocity of 35 m/s. Assume that the goalie is at rest before catching the puck, and friction between the ice and the puck-goalie system is negligible (see Figure 8.9).

Figure 8.9 An ice hockey goalie catches a hockey puck and recoils backward in an inelastic collision.

Strategy

Momentum is conserved because the net external force on the puck-goalie system is zero. Therefore, we can use conservation of momentum to find the final velocity of the puck and goalie system. Note that the initial velocity of the goalie is zero and that the final velocity of the puck and goalie are the same.

Solution

For an inelastic collision, conservation of momentum is

$$m_1{v}_1+m_2{v}_2=\left(m_1+m_2\right)v^{\prime }\text{,}$$

8.8

where v′ is the velocity of both the goalie and the puck after impact. Because the goalie is initially at rest, we know v₂ = 0. This simplifies the equation to

$$m_1{v}_1=\left(m_1+m_2\right)v^{\prime }\text{.}$$

8.9

Solving for v′ yields

$${\text{v}}^{\prime }=\left(\frac{m_1}{m_1+m_2}\right){\text{v}}_1\text{.}$$

8.10

Entering known values in this equation, we get

$$\begin{array}{ccc}\hfill {\text{v}}^{\prime }& =& \left(\frac{0.150\text{kg}}{70.0\text{kg}+0.150\text{kg}}\right)(35\text{m/s})\hfill \\ & =& 7.48\times {10}^{-2}\text{m/s.}\hfill \end{array}$$

8.11

Discussion

This recoil velocity is small and in the same direction as the puck’s original velocity.

Worked Example

Calculating Final Velocity: Elastic Collision of Two Carts

Two hard, steel carts collide head-on and then ricochet off each other in opposite directions on a frictionless surface (see Figure 8.10). Cart 1 has a mass of 0.350 kg and an initial velocity of 2 m/s. Cart 2 has a mass of 0.500 kg and an initial velocity of −0.500 m/s. After the collision, cart 1 recoils with a velocity of −4 m/s. What is the final velocity of cart 2?

Figure 8.10 Two carts collide with each other in an elastic collision.

Strategy

Since the track is frictionless, F_net = 0 and we can use conservation of momentum to find the final velocity of cart 2.

Solution

As before, the equation for conservation of momentum for a one-dimensional elastic collision in a two-object system is

$$m_1{v}_1+m_2{v}_2=m_1{v}^{\prime }{}_1+m_2{v^{\prime }}_2\text{.}$$

8.12

The only unknown in this equation is v′₂. Solving for v′₂ and substituting known values into the previous equation yields

$$\begin{array}{ccc}\hfill {{\text{v}}^{\prime }}_{\text{2}}^{}& \text{=}& \frac{m_{\text{1}}{\text{v}}_{\text{1}}\text{+}{m}_{\text{2}}{\text{v}}_{\text{2}}-m_{\text{1}}{{\text{v}}^{\prime }}_{\text{1}}^{}}{m_{\text{2}}}\hfill \\ & \text{=}& \frac{\left(\text{0}\text{.350}\text{kg}\right)\left(\text{2}\text{.00}\text{m/s}\right)\text{+}\left(\text{0}\text{.500 kg}\right)\left(-\text{0}\text{.500 m/s}\right)-\left(\text{0}\text{.350}\text{kg}\right)\left(-\text{4}\text{.00}\text{m/s}\right)}{\text{0}\text{.500}\text{kg}}\hfill \\ & \text{=}& \text{3}\text{.70}\text{m/s.}\hfill \end{array}$$

8.13

Discussion

The final velocity of cart 2 is large and positive, meaning that it is moving to the right after the collision.

Worked Example

Calculating Final Velocity in a Two-Dimensional Collision

Suppose the following experiment is performed (Figure 8.11). An object of mass 0.250 kg (m₁) is slid on a frictionless surface into a dark room, where it strikes an initially stationary object of mass 0.400 kg (m₂). The 0.250 kg object emerges from the room at an angle of 45º with its incoming direction. The speed of the 0.250 kg object is originally 2 m/s and is 1.50 m/s after the collision. Calculate the magnitude and direction of the velocity (v′₂ and ${\theta }_2$ ) of the 0.400 kg object after the collision.

Figure 8.11 The incoming object of mass m₁ is scattered by an initially stationary object. Only the stationary object’s mass m₂ is known. By measuring the angle and speed at which the object of mass m₁ emerges from the room, it is possible to calculate the magnitude and direction of the initially stationary object’s velocity after the collision.

Strategy

Momentum is conserved because the surface is frictionless. We chose the coordinate system so that the initial velocity is parallel to the x-axis, and conservation of momentum along the x- and y-axes applies.

Everything is known in these equations except v′₂ and θ₂, which we need to find. We can find two unknowns because we have two independent equations—the equations describing the conservation of momentum in the x and y directions.

Solution

First, we’ll solve both conservation of momentum equations ( $m_1{\text{v}}_1=m_1{\text{v}}^{\prime }{}_1\cos {\theta }_1+m_2{\text{v}}^{\prime }{}_2\cos {\theta }_2$ and $0=m_1{\text{v}}^{\prime }{}_1\sin {\theta }_1+m_2{\text{v}}^{\prime }{}_2\sin {\theta }_2$ ) for v′₂ sin ${\theta }_2$ .

For conservation of momentum along x-axis, let’s substitute sin ${\theta }_2$ /tan ${\theta }_2$ for cos ${\theta }_2$ so that terms may cancel out later on. This comes from rearranging the definition of the trigonometric identity tan $\theta$ = sin $\theta$ /cos $\theta$ . This gives us

$$m_1{v}_1=m_1{v}^{\prime }{}_1\cos {\theta }_1+m_2{v}^{\prime }{}_2\frac{\sin {\theta }_2}{\tan {\theta }_2}\text{.}$$

8.14

Solving for v′₂ sin ${\theta }_2$ yields

$$v^{\prime }{}_2\sin {\theta }_2=\frac{(m_1{v}_1-m_1{v}^{\prime }{}_1\cos {\theta }_1)(\tan {\theta }_2)}{m_2}\text{.}$$

8.15

For conservation of momentum along y-axis, solving for v′₂ sin ${\theta }_2$ yields

$$v^{\prime }{}_2\sin {\theta }_2=\frac{-(m_1{v}^{\prime }{}_1\sin {\theta }_1)}{m_2}\text{.}$$

8.16

Since both equations equal v′₂ sin ${\theta }_2$, we can set them equal to one another, yielding

$$\frac{(m_1{v}_1-m_1{v}^{\prime }{}_1\cos {\theta }_1)(\tan {\theta }_2)}{m_2}=\frac{-(m_1{v}^{\prime }{}_1\sin {\theta }_1)}{m_2}\text{.}$$

8.17

Solving this equation for tan ${\theta }_2$, we get

$$\tan {\theta }_2=\frac{v^{\prime }{}_1\sin {\theta }_1}{v^{\prime }{}_1\cos {\theta }_1-v_1}\text{.}$$

8.18

Entering known values into the previous equation gives

$$\tan {\theta }_2=\frac{(1.50)(0.707)}{(1.50)(0.707)-2.00}=-1.129\text{.}$$

8.19

Therefore,

$${\theta }_2={\tan }^{-1}(-1.129)={312}^0\text{.}$$

8.20

Since angles are defined as positive in the counterclockwise direction, m₂ is scattered to the right.

We’ll use the conservation of momentum along the y-axis equation to solve for v′₂.

$$v^{\prime }{}_2=-\frac{m_1}{m_2}{v}^{\prime }{}_1\frac{\sin {\theta }_1}{\sin {\theta }_2}$$

8.21

Entering known values into this equation gives

$${{\text{v}}^{\prime }}_2=-\frac{\left(0.250\right)}{\left(0.400\right)}\left(1.50\right)\left(\frac{0.7071}{-0.7485}\right)\text{.}$$

8.22

Therefore,

$${v^{\prime }}_2=0.886\text{ m/s}.$$

8.23

Discussion

Either equation for the x- or y-axis could have been used to solve for v′₂, but the equation for the y-axis is easier because it has fewer terms.