College Physics for AP® Courses

# 8.4Elastic Collisions in One Dimension

College Physics for AP® Courses8.4 Elastic Collisions in One Dimension

### Learning Objectives

By the end of this section, you will be able to:

• Describe an elastic collision of two objects in one dimension.
• Define internal kinetic energy.
• Derive an expression for conservation of internal kinetic energy in a one-dimensional collision.
• Determine the final velocities in an elastic collision given masses and initial velocities.

The information presented in this section supports the following AP® learning objectives and science practices:

• 4.B.1.1 The student is able to calculate the change in linear momentum of a two-object system with constant mass in linear motion from a representation of the system (data, graphs, etc.). (S.P. 1.4, 2.2)
• 4.B.1.2 The student is able to analyze data to find the change in linear momentum for a constant-mass system using the product of the mass and the change in velocity of the center of mass. (S.P. 5.1)
• 5.A.2.1 The student is able to define open and closed systems for everyday situations and apply conservation concepts for energy, charge, and linear momentum to those situations. (S.P. 6.4, 7.2)
• 5.D.1.1 The student is able to make qualitative predictions about natural phenomena based on conservation of linear momentum and restoration of kinetic energy in elastic collisions. (S.P. 6.4, 7.2)
• 5.D.1.2 The student is able to apply the principles of conservation of momentum and restoration of kinetic energy to reconcile a situation that appears to be isolated and elastic, but in which data indicate that linear momentum and kinetic energy are not the same after the interaction, by refining a scientific question to identify interactions that have not been considered. Students will be expected to solve qualitatively and/or quantitatively for one-dimensional situations and only qualitatively in two-dimensional situations. (S.P. 2.2, 3.2, 5.1, 5.3)
• 5.D.1.5 The student is able to classify a given collision situation as elastic or inelastic, justify the selection of conservation of linear momentum and restoration of kinetic energy as the appropriate principles for analyzing an elastic collision, solve for missing variables, and calculate their values. (S.P. 2.1, 2.2)
• 5.D.1.6 The student is able to make predictions of the dynamical properties of a system undergoing a collision by application of the principle of linear momentum conservation and the principle of the conservation of energy in situations in which an elastic collision may also be assumed. (S.P. 6.4)
• 5.D.1.7 The student is able to classify a given collision situation as elastic or inelastic, justify the selection of conservation of linear momentum and restoration of kinetic energy as the appropriate principles for analyzing an elastic collision, solve for missing variables, and calculate their values. (S.P. 2.1, 2.2)
• 5.D.2.1 The student is able to qualitatively predict, in terms of linear momentum and kinetic energy, how the outcome of a collision between two objects changes depending on whether the collision is elastic or inelastic. (S.P. 6.4, 7.2)
• 5.D.2.2 The student is able to plan data collection strategies to test the law of conservation of momentum in a two-object collision that is elastic or inelastic and analyze the resulting data graphically. (S.P.4.1, 4.2, 5.1)
• 5.D.3.2 The student is able to make predictions about the velocity of the center of mass for interactions within a defined one-dimensional system. (S.P. 6.4)

Let us consider various types of two-object collisions. These collisions are the easiest to analyze, and they illustrate many of the physical principles involved in collisions. The conservation of momentum principle is very useful here, and it can be used whenever the net external force on a system is zero.

We start with the elastic collision of two objects moving along the same line—a one-dimensional problem. An elastic collision is one that also conserves internal kinetic energy. Internal kinetic energy is the sum of the kinetic energies of the objects in the system. Figure 8.9 illustrates an elastic collision in which internal kinetic energy and momentum are conserved.

Truly elastic collisions can only be achieved with subatomic particles, such as electrons striking nuclei. Macroscopic collisions can be very nearly, but not quite, elastic—some kinetic energy is always converted into other forms of energy such as heat transfer due to friction and sound. One macroscopic collision that is nearly elastic is that of two steel blocks on ice. Another nearly elastic collision is that between two carts with spring bumpers on an air track. Icy surfaces and air tracks are nearly frictionless, more readily allowing nearly elastic collisions on them.

### Elastic Collision

An elastic collision is one that conserves internal kinetic energy.

### Internal Kinetic Energy

Internal kinetic energy is the sum of the kinetic energies of the objects in the system.

Figure 8.9 An elastic one-dimensional two-object collision. Momentum and internal kinetic energy are conserved.

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

$p 1 + p 2 = p′ 1 + p′ 2 F net = 0 p 1 + p 2 = p′ 1 + p′ 2 F net = 0 size 12{ left (F rSub { size 8{"net"} } =0 right )} {}$
8.43

or

$m1v1+m2v2=m1v′1+m2v′2Fnet=0,m1v1+m2v2=m1v′1+m2v′2Fnet=0, size 12{ left (F rSub { size 8{"net"} } =0 right )} {}$
8.44

where the primes (') indicate values after the collision. By definition, an elastic collision conserves internal kinetic energy, and so the sum of kinetic energies before the collision equals the sum after the collision. Thus,

$12m1v12+ 12m2 v22 = 12 m1 v′12 + 12 m2 v′22 (two-object elastic collision)12m1v12+ 12m2 v22 = 12 m1 v′12 + 12 m2 v′22 (two-object elastic collision)$
8.45

expresses the equation for conservation of internal kinetic energy in a one-dimensional collision.

### Making Connections: Collisions

Suppose data are collected on a collision between two masses sliding across a frictionless surface. Mass A (1.0 kg) moves with a velocity of +12 m/s, and mass B (2.0 kg) moves with a velocity of −12 m/s. The two masses collide and stick together after the collision. The table below shows the measured velocities of each mass at times before and after the collision:

Time (s) Velocity A (m/s) Velocity B (m/s)
0 +12 −12
1.0 s +12 −12
2.0 s −4.0 −4.0
3.0 s −4.0 −4.0
Table 8.1

The total mass of the system is 3.0 kg. The velocity of the center of mass of this system can be determined from the conservation of momentum. Consider the system before the collision:

$( m A + m B ) v cm = m A v A + m B v B ( m A + m B ) v cm = m A v A + m B v B$
8.46
$(3.0) v cm =(1)(12)+(2)(−12) (3.0) v cm =(1)(12)+(2)(−12)$
8.47
8.48

After the collision, the center-of-mass velocity is the same:

$( m A + m B ) v cm =( m A + m B ) v final ( m A + m B ) v cm =( m A + m B ) v final$
8.49
$(3.0) v cm =(3)(−4.0) (3.0) v cm =(3)(−4.0)$
8.50
8.51

The total momentum of the system before the collision is:

8.52

The total momentum of the system after the collision is:

8.53

Thus, the change in momentum of the system is zero when measured this way. We get a similar result when we calculate the momentum using the center-of-mass velocity. Since the center-of-mass velocity is the same both before and after the collision, we calculate the same momentum for the system using this method both before and after the collision.

### Example 8.4

#### Calculating Velocities Following an Elastic Collision

Calculate the velocities of two objects following an elastic collision, given that

$m 1 = 0 . 500 kg, m 2 = 3 . 50 kg, v 1 = 4 . 00 m/s, and v 2 = 0 . m 1 = 0 . 500 kg, m 2 = 3 . 50 kg, v 1 = 4 . 00 m/s, and v 2 = 0 . size 12{m rSub { size 8{1} } =0 "." "500"" kg, "m rSub { size 8{2} } =3 "." "50"" kg, "v rSub { size 8{1} } =4 "." "00"" m/s, and "v rSub { size 8{2} } =0 "." } {}$
8.54

Strategy and Concept

First, visualize what the initial conditions mean—a small object strikes a larger object that is initially at rest. This situation is slightly simpler than the situation shown in Figure 8.9 where both objects are initially moving. We are asked to find two unknowns (the final velocities $v′1v′1$ and $v′2v′2 size 12{v rSub { size 8{2} } '} {}$). To find two unknowns, we must use two independent equations. Because this collision is elastic, we can use the above two equations. Both can be simplified by the fact that object 2 is initially at rest, and thus $v2=0v2=0 size 12{v rSub { size 8{2} } =0} {}$. Once we simplify these equations, we combine them algebraically to solve for the unknowns.

#### Solution

For this problem, note that $v2=0 v2=0 size 12{v rSub { size 8{2} } =0} {}$ and use conservation of momentum. Thus,

$p 1 = p′ 1 + p′ 2 p 1 = p′ 1 + p′ 2 size 12{p rSub { size 8{1} } =p' rSub { size 8{1} } +p' rSub { size 8{2} } } {}$
8.55

or

$m1v1=m1v′1+m2v′2.m1v1=m1v′1+m2v′2. size 12{m rSub { size 8{1} } v rSub { size 8{1} } =m rSub { size 8{1} } { {v}} sup { ' } rSub { size 8{1} } +m rSub { size 8{2} } { {v}} sup { ' } rSub { size 8{2} } } {}$
8.56

Using conservation of internal kinetic energy and that $v2=0v2=0 size 12{v rSub { size 8{2} } =0} {}$,

$1 2 m 1 v 1 2 = 1 2 m 1 v′ 1 2 + 1 2 m 2 v′ 2 2 . 1 2 m 1 v 1 2 = 1 2 m 1 v′ 1 2 + 1 2 m 2 v′ 2 2 . size 12{ { {1} over {2} } m rSub { size 8{1} } v rSub { size 8{1} rSup { size 8{2} } } = { {1} over {2} } m rSub { size 8{1} } v"" lSub { size 8{1} } ' rSup { size 8{2} } + { {1} over {2} } m rSub { size 8{2} } v rSub { size 8{2} } ' rSup { size 8{2} } } {}$
8.57

Solving the first equation (momentum equation) for $v′2v′2 size 12{ { {v}} sup { ' } rSub { size 8{2} } } {}$, we obtain

$v′2 = m 1 m 2 v 1 − v′1 . v′2 = m 1 m 2 v 1 − v′1 . size 12{ { {v}} sup { ' } rSub { size 8{2} } = { {m rSub { size 8{1} } } over {m rSub { size 8{2} } } } left (v rSub { size 8{1} } - { {v}} sup { ' } rSub { size 8{1} } right )} {}$
8.58

Substituting this expression into the second equation (internal kinetic energy equation) eliminates the variable $v′2v′2 size 12{ { {v}} sup { ' } rSub { size 8{2} } } {}$, leaving only $v′1v′1 size 12{ { {v}} sup { ' } rSub { size 8{1} } } {}$ as an unknown (the algebra is left as an exercise for the reader). There are two solutions to any quadratic equation; in this example, they are

$v′1 = 4 . 00 m/s v′1 = 4 . 00 m/s size 12{ { {v}} sup { ' } rSub { size 8{1} } =4 "." "00""m/s"} {}$
8.59

and

$v′1=−3.00 m/s.v′1=−3.00 m/s. size 12{ { {v}} sup { ' } rSub { size 8{1} } = - 3 "." "00"" m/s"} {}$
8.60

As noted when quadratic equations were encountered in earlier chapters, both solutions may or may not be meaningful. In this case, the first solution is the same as the initial condition. The first solution thus represents the situation before the collision and is discarded. The second solution $(v′1=−3.00 m/s)(v′1=−3.00 m/s) size 12{ $${ {v}} sup { ' } rSub { size 8{1} } = - 3 "." "00""m/s"$$ } {}$ is negative, meaning that the first object bounces backward. When this negative value of $v′1v′1 size 12{ { {v}} sup { ' } rSub { size 8{1} } } {}$ is used to find the velocity of the second object after the collision, we get

$v′2 = m 1 m 2 v 1 − v′1 = 0 . 500 kg 3 . 50 kg 4 . 00 − − 3 . 00 m/s v′2 = m 1 m 2 v 1 − v′1 = 0 . 500 kg 3 . 50 kg 4 . 00 − − 3 . 00 m/s size 12{ { {v}} sup { ' } rSub { size 8{2} } = { {m rSub { size 8{1} } } over {m rSub { size 8{2} } } } left (v rSub { size 8{1} } - { {v}} sup { ' } rSub { size 8{1} } right )= { {0 "." "500""kg"} over {3 "." "50""kg"} } left [4 "." "00" - left ( - 3 "." "00" right ) right ]"m/s"} {}$
8.61

or

$v′2=1.00 m/s.v′2=1.00 m/s. size 12{ { {v}} sup { ' } rSub { size 8{2} } =1 "." "00""m/s"} {}$
8.62

#### Discussion

The result of this example is intuitively reasonable. A small object strikes a larger one at rest and bounces backward. The larger one is knocked forward, but with a low speed. (This is like a compact car bouncing backward off a full-size SUV that is initially at rest.) As a check, try calculating the internal kinetic energy before and after the collision. You will see that the internal kinetic energy is unchanged at 4.00 J. Also check the total momentum before and after the collision; you will find it, too, is unchanged.

The equations for conservation of momentum and internal kinetic energy as written above can be used to describe any one-dimensional elastic collision of two objects. These equations can be extended to more objects if needed.

### Making Connections: Take-Home Investigation—Ice Cubes and Elastic Collision

Find a few ice cubes which are about the same size and a smooth kitchen tabletop or a table with a glass top. Place the ice cubes on the surface several centimeters away from each other. Flick one ice cube toward a stationary ice cube and observe the path and velocities of the ice cubes after the collision. Try to avoid edge-on collisions and collisions with rotating ice cubes. Have you created approximately elastic collisions? Explain the speeds and directions of the ice cubes using momentum.

### PhET Explorations

#### Collision Lab

Investigate collisions on an air hockey table. Set up your own experiments: vary the number of discs, masses and initial conditions. Is momentum conserved? Is kinetic energy conserved? Vary the elasticity and see what happens.

Figure 8.10
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