Elastic and Inelastic Collisions, Momentum | Physics

Learning objectives

Section Key Terms

elastic collision

inelastic collision

point masses

recoil

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_08_03_Elastic shows an elastic collision where momentum is conserved.

View original figure (external; image not reproduced in this text edition)

Original figure description: An illustration shows before and after diagrams of two boxes moving toward each other on a frictionless surface. The box on the left is labeled m one and the box on the right is labeled m two. Both diagrams are labeled System of Interest. In the before diagram, a velocity vector, v one, points from m one toward m two. A second, shorter velocity vector, v two, points from the left side of m two toward m one. Two equations are shown: p one plus p one equals p total and net F equals zero. In the after diagram, both velocity vectors point away from each mass, m one and m two. The vector on the left is shorter than the one on the right. The equation p one prime plus p one prime equals p total is shown.

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

p_{1} + p_{2} = p^{'}_{1} + p^{'}_{2} (F_{net} = 0) .

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

m_{1} v_{1} + m_{2} v_{2} = m_{1} {v^{'}}_{1} + m_{2} {v^{'}}_{2},

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_08_03_Inelastic 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.

View original figure (external; image not reproduced in this text edition)

Original figure description: An illustration shows before and after diagrams of two boxes moving toward each other on a frictionless surface. Both boxes are labeled m. Both diagrams are labeled System of Interest. In the before diagram, a velocity vector, v, points from the left box to the right box. A second vector of equal magnitude, negative v, points from right box toward the left box. Two equations are shown: p one plus p two equals zero and net F equals zero. In the after diagram, the boxes are touching. The equations p one prime plus p two prime equals zero and v equals zero are shown.

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

m_{1} v_{1} + m_{2} v_{2} = m_{1} {v^{'}}_{1} + m_{2} {v^{'}}_{2}

m_{1} v_{1} + m_{2} v_{2} = (m_{1} + m_{2}) v^{'}

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

Introduction to Momentum

This video reviews the definitions of momentum and impulse. It also covers an example of using conservation of momentum to solve a problem involving an inelastic collision between a car with constant velocity and a stationary truck. Note that Sal accidentally gives the unit for impulse as Joules; it is actually N $\cdot$ s or kg $\cdot$ m/s.

Introduction to momentum | Impacts and linear momentum | Physics | Khan Academy

How would the final velocity of the car-plus-truck system change if the truck had some initial velocity moving in the same direction as the car? What if the truck were moving in the opposite direction of the car initially? Why?

If the truck was initially moving in the same direction as the car, the final velocity would be greater. If the truck was initially moving in the opposite direction of the car, the final velocity would be smaller.
If the truck was initially moving in the same direction as the car, the final velocity would be smaller. If the truck was initially moving in the opposite direction of the car, the final velocity would be greater.
The direction in which the truck was initially moving would not matter. If the truck was initially moving in either direction, the final velocity would be smaller.
The direction in which the truck was initially moving would not matter. If the truck was initially moving in either direction, the final velocity would be greater.

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_08_03_dimension. 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.

View original figure (external; image not reproduced in this text edition)

Original figure description: An illustration shows before and after diagrams of two balls, one moving toward the other. The first ball is labeled m one and the second ball is labeled m two. Both diagrams are labeled System of Interest, but the diagram on the left specifies before collision and the diagram on the right specifies after collision. In the before diagram, a velocity vector, v one, points from m one to m two. The following are also shown: v two equals zero, net F equals zero, and p one equals p total. In the after diagram, m one is moving in an upward direction and m two is moving in a downward direction. Vectors labeled v one prime and v two prime show their directional movements. M one and m two are shown forming angles with the x-axis. The equation p one prime plus p one prime equals p total is shown.

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

p_{1x} + p_{2x} = p^{'}_{1x} + p^{'}_{2x} .

In terms of masses and velocities, this equation is

m_{1} v_{1 x} + m_{2} v_{2 x} = m_{1} {v^{'}}_{1 x} + m_{2} {v^{'}}_{2 x} .

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

m_{1} v_{1 x} = m_{1} {v^{'}}_{1 x} + m_{2} {v^{'}}_{2 x} .

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 v₁_x = v₁. Conservation of momentum along the x-axis gives the equation

m_{1} v_{1} = m_{1} {v^{'}}_{1} \cos θ_{1} + m_{2} {v^{'}}_{2} \cos θ_{2},

where $θ_{1}$ and $θ_{2}$ are as shown in Figure_08_03_dimension.

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

p_{1 y} + p_{2 y} = {p^{'}}_{1 y} + {p^{'}}_{2 y},

m_{1} v_{1 y} + m_{2} v_{2 y} = m_{1} {v^{'}}_{1 y} + m_{2} {v^{'}}_{2 y} .

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

0 = m_{1} {v^{'}}_{1} y + m_{2} {v^{'}}_{2} y .

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

0 = m_{1} {v^{'}}_{1} \sin θ_{1} + m_{2} {v^{'}}_{2} \sin θ_{2}

Collision Lab

In this simulation, you will investigate collisions on an air hockey table. Place checkmarks next to the momentum vectors and momenta diagram options. Experiment with changing the masses of the balls and the initial speed of ball 1. How does this affect the momentum of each ball? What about the total momentum? Next, experiment with changing the elasticity of the collision. You will notice that collisions have varying degrees of elasticity, ranging from perfectly elastic to perfectly inelastic.

If you wanted to maximize the velocity of ball 2 after impact, how would you change the settings for the masses of the balls, the initial speed of ball 1, and the elasticity setting? Why? HintPlacing a checkmark next to the velocity vectors and removing the momentum vectors will help you visualize the velocity of ball 2, and pressing the More Data button will let you take readings.

Maximize the mass of ball 1 and initial speed of ball 1; minimize the mass of ball 2; and set elasticity to 50 percent.
Maximize the mass of ball 2 and initial speed of ball 1; minimize the mass of ball 1; and set elasticity to 100 percent.
Maximize the mass of ball 1 and initial speed of ball 1; minimize the mass of ball 2; and set elasticity to 100 percent.
Maximize the mass of ball 2 and initial speed of ball 1; minimize the mass of ball 1; and set elasticity to 50 percent.

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_08_03_Goalie).

View original figure (external; image not reproduced in this text edition)

Original figure description: An illustration shows before and after diagrams of a hockey puck and an ice hockey player on a frictionless ice surface. The hockey puck is labeled m one and the hockey player is labeled m two. Both diagrams are labeled System of Interest. In the before diagram, a velocity vector, v one, points from the puck to the player. The following are shown: p one equals p total, v two equals zero, and net F equals zero. In the after diagram, the hockey puck has made contact with the player. A velocity vector, v prime, points from the hockey player to the right. The equation p one prime plus p two prime equals p total is shown. The equation v one prime plus v two prime equals v prime is also shown.

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

Solution

For an inelastic collision, conservation of momentum is

m_{1} v_{1} + m_{2} v_{2} = (m_{1} + m_{2}) v^{'},

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} = (m_{1} + m_{2}) v^{'} .

Solving for v′ yields

v^{'} = (\frac{m_{1}}{m_{1} + m_{2}}) v_{1} .

Entering known values in this equation, we get

\begin{matrix} v^{'} & = & (\frac{0.150 kg}{70.0 kg + 0.150 kg}) (35 m/s) \\ = & 7.48 \times 10^{- 2} m/s. \end{matrix}

Discussion

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

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_08_03_Carts). 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 −0.94 m/s. What is the final velocity of cart 2?

View original figure (external; image not reproduced in this text edition)

Original figure description: An illustration shows before and after diagrams of two carts moving toward each other on a frictionless surface. The cart on the left is labeled m one and the car on the right is labeled m two. Both diagrams are labeled System of Interest. In the before diagram, a velocity vector, v one, points from the cart on the left to the cart on the right. A second, shorter velocity vector, v two, points from the cart on the right to the cart on the left. Two equations are shown: p one plus p one equals p total and net F equals zero. In the after diagram, the velocity vectors are equal in magnitude and point away from the carts. They are labeled v one prime and v two prime. The equation p one prime plus p one prime equals p total is shown.

Two carts collide with each other in an elastic collision.

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^{'}_{1} + m_{2} {v^{'}}_{2} .

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

\begin{matrix} {v^{'}}_{2}^{} & = & \frac{m_{1} v_{1} + m_{2} v_{2} - m_{1} {v^{'}}_{1}^{}}{m_{2}} \\ = & \frac{(0 .350 kg) (2 .00 m/s) + (0 .500 kg) (- 0 .500 m/s) - (0 .350 kg) (- 0 .94 m/s)}{0 .500 kg} \\ = & 1 .56 m/s. \end{matrix}

Discussion

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

Calculating Final Velocity in a Two-Dimensional Collision

Suppose the following experiment is performed (Figure_08_03_Scattered). 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 $θ_{2}$ ) of the 0.400 kg object after the collision.

View original figure (external; image not reproduced in this text edition)

Original figure description: An illustration shows before and after diagrams of two balls, one moving toward the other. The first ball is labeled m one and the second ball is labeled m two and point four kilograms. In the before diagram, a velocity vector, v one, points from m one to m two. In the after diagram, m one is moving in an upward direction with a velocity vector, v one prime, indicating its direction. It forms an angle of forty-five degrees with the x-axis. The following are also shown: before, p one equal p total; after, prime plus p two prime equals p total; net F equals zero; v two equals zero; v two prime is unknown; angle two is unknown.

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.

Solution

First, we’ll solve both conservation of momentum equations ( $m_{1} v_{1} = m_{1} v^{'}_{1} \cos θ_{1} + m_{2} v^{'}_{2} \cos θ_{2}$ and $0 = m_{1} v^{'}_{1} \sin θ_{1} + m_{2} v^{'}_{2} \sin θ_{2}$ ) for v′₂ sin $θ_{2}$ .

For conservation of momentum along x-axis, let’s substitute sin $θ_{2}$ /tan $θ_{2}$ for cos $θ_{2}$ so that terms may cancel out later on. This comes from rearranging the definition of the trigonometric identity tan $θ$ = sin $θ$ /cos $θ$ . This gives us

m_{1} v_{1} = m_{1} v^{'}_{1} \cos θ_{1} + m_{2} v^{'}_{2} \frac{\sin θ_{2}}{\tan θ_{2}} .

Solving for v′₂ sin $θ_{2}$ yields

v^{'}_{2} \sin θ_{2} = \frac{(m_{1} v_{1} - m_{1} v^{'}_{1} \cos θ_{1}) (\tan θ_{2})}{m_{2}} .

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

v^{'}_{2} \sin θ_{2} = \frac{- (m_{1} v^{'}_{1} \sin θ_{1})}{m_{2}} .

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

\frac{(m_{1} v_{1} - m_{1} v^{'}_{1} \cos θ_{1}) (\tan θ_{2})}{m_{2}} = \frac{- (m_{1} v^{'}_{1} \sin θ_{1})}{m_{2}} .

Solving this equation for tan $θ_{2}$ , we get

\tan θ_{2} = \frac{v^{'}_{1} \sin θ_{1}}{v^{'}_{1} \cos θ_{1} - v_{1}} .

Entering known values into the previous equation gives

\tan θ_{2} = \frac{(1.50) (0.707)}{(1.50) (0.707) - 2.00} = - 1.129 .

Therefore,

θ_{2} = \tan^{- 1} (- 1.129) = 312° .

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^{'}_{2} = - \frac{m_{1}}{m_{2}} v^{'}_{1} \frac{\sin θ_{1}}{\sin θ_{2}}

Entering known values into this equation gives

{v^{'}}_{2} = - \frac{(0.250)}{(0.400)} (1.50) (\frac{0.7071}{- 0.7485}) .

Therefore,

{v^{'}}_{2} = 0.886 m/s .

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.

Practice Problems

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In an elastic collision, an object with momentum 25 kg ⋅ m/s collides with another that has a momentum 35 kg ⋅ m/s. The first object’s momentum changes to 10 kg ⋅ m/s. What is the final momentum of the second object?

10 kg ⋅ m/s
20 kg ⋅ m/s
35 kg ⋅ m/s
50 kg ⋅ m/s

Check Your Understanding

What is an elastic collision?

An elastic collision is one in which the objects after impact are deformed permanently.
An elastic collision is one in which the objects after impact lose some of their internal kinetic energy.
An elastic collision is one in which the objects after impact do not lose any of their internal kinetic energy.
An elastic collision is one in which the objects after impact become stuck together and move with a common velocity.

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What is the equation for conservation of momentum for two objects in a one-dimensional collision?

p₁ + p₁′ = p₂ + p₂′
p₁ + p₂ = p₁′ + p₂′
p₁ − p₂ = p₁′ − p₂′
p₁ + p₂ + p₁′ + p₂′ = 0

Section Summary

If kinetic energy is conserved, the collision is elastic. If they stick together, the collision is perfectly inelastic.
Kinetic energy is conserved in an elastic collision, but not in an inelastic collision.
The approach to two-dimensional collisions is to choose a convenient coordinate system and break the motion into components along perpendicular axes. Choose a coordinate system with the x-axis parallel to the velocity of the incoming particle.
Two-dimensional collisions of point masses, where mass 2 is initially at rest, conserve momentum along the initial direction of mass 1, or the x-axis, and along the direction perpendicular to the initial direction, or the y-axis.
Point masses are structureless particles that cannot spin.

Key Equations

conservation of momentum in any collision, including elastic collisionselastic collision	$m_{1} v_{1} + m_{2} v_{2} = m_{1} {v^{'}}_{1} + m_{2} {v^{'}}_{2},$
conservation of momentum in a perfectly inelastic collision	$m_{1} v_{1} + m_{2} v_{2} = (m_{1} + m_{2}) v^{'}$
conservation of momentum along x-axis for 2D collisions	$m_{1} v_{1} = m_{1} {v^{'}}_{1} \cos θ_{1} + m_{2} {v^{'}}_{2} \cos θ_{2}$
conservation of momentum along y-axis for 2D collisions	$0 = m_{1} v^{'}_{1} \sin θ_{1} + m_{2} v^{'}_{2} \sin θ_{2}$

Concept Items

In the equation p₁ + p₂ = p'₁ + p'₂ for the collision of two objects, what is the assumption made regarding the friction acting on the objects?

Friction is zero.
Friction is nearly zero.
Friction acts constantly.
Friction before and after the impact remains the same.

What is an inelastic collision?

when objects stick together after impact, and their internal energy is not conserved
when objects stick together after impact, and their internal energy is conserved
when objects stick together after impact, and always come to rest instantaneously after collision
when objects stick together after impact, and their internal energy increases

Critical Thinking

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Problems

In an elastic collision, an object with momentum 25 kg⋅m/s (moving to the left) collides with another that has a momentum 35 kg⋅m/s, moving toward it to the right. The first object’s momentum changes to 15 kg⋅m/s, and it is now moving to the right. What is the magnitude of the momentum of the second object after the collision?

5 kg⋅m/s
10 kg⋅m/s
25 kg⋅m/s
75 kg⋅m/s

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Performance Task

You will need the following:

balls of different weights
a ruler or wooden strip
some books
a paper cup

Make an inclined plane by resting one end of a ruler on a stack of books. Place a paper cup on the other end. Roll a ball from the top of the ruler so that it hits the paper cup. Measure the displacement of the paper cup due to the collision. Now use increasingly heavier balls for this activity and see how that affects the displacement of the cup. Plot a graph of mass vs. displacement. Now repeat the same activity, but this time, instead of using different balls, change the incline of the ruler by varying the height of the stack of books. This will give you different velocities of the ball. See how this affects the displacement of the paper cup.

Test Prep Multiple Choice

In which type of collision is kinetic energy conserved?

Elastic
Inelastic

In physics, what are structureless particles that cannot rotate or spin called?

Elastic particles
Point masses
Rigid masses

Two objects having equal masses and speeds collide with each other and come to rest. What type of a collision is this and why?

Elastic collision, because internal kinetic energy is conserved
Inelastic collision, because internal kinetic energy is not conserved
Elastic collision, because internal kinetic energy is not conserved
Inelastic collision, because internal kinetic energy is conserved

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Test Prep Short Answer

Two objects collide with each other and come to a rest. How can you use the equation of conservation of momentum to describe this situation?

m₁v₁ + m₂v₂ = 0
m₁v₁ − m₂v₂ = 0
m₁v₁ + m₂v₂ = m₁v₁′
m₁v₁ + m₂v₂ = m₁v₂

Source link

What is the equation for conservation of momentum along the x-axis for 2D collisions in terms of mass and velocity, where one of the particles is initially at rest?

m₁v₁ = m₁v₁′cos⁡θ₁
m₁v₁ = m₁v₁′cos⁡θ₁ + m₂v₂′cos⁡θ₂
m₁v₁ = m₁v₁′cos⁡θ₁ − m₂v₂′cos⁡θ₂
m₁v₁ = m₁v₁′sin⁡θ₁ + m₂v₂′sin⁡θ₂

What is the equation for conservation of momentum along the y-axis for 2D collisions in terms of mass and velocity, where one of the particles is initially at rest?

0 = m₁v₁′sin⁡θ₁
0 = m₁v₁′sin⁡θ₁ + m₂v₂′sin⁡θ₂
0 = m₁v₁′sin⁡θ₁ − m₂v₂′sin⁡θ₂
0 = m₁v₁′cos⁡θ₁ + m₂v₂′cos⁡θ₂

Test Prep Extended Response

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What approach would you use to solve problems involving 2D collisions?

Break the momenta into components and then choose a coordinate system.
Choose a coordinate system and then break the momenta into components.
Find the total momenta in the x and y directions, and then equate them to solve for the unknown.
Find the sum of the momenta in the x and y directions, and then equate it to zero to solve for the unknown.

Glossary

elastic collision: collision in which objects separate after impact and kinetic energy is conserved
inelastic collision: collision in which kinetic energy is not conserved
perfectly inelastic collision: collision in which objects stick together after impact and kinetic energy is not conserved
point masses: structureless particles that cannot rotate or spin
recoil: backward movement of an object caused by the transfer of momentum from another object in a collision

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