Consider Again the One Dimensional Elastic Collision Discused in Section R8.2.

Learning Objectives

By the end of this section, y'all will exist able to:

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

Let united states of america 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 tin can be used whenever the net external force on a system is nix.

We start with the elastic collision of ii objects moving forth the same line—a one-dimensional problem. An elastic standoff is 1 that also conserves internal kinetic energy. Internal kinetic energy is the sum of the kinetic energies of the objects in the arrangement. Effigy 1 illustrates an elastic collision in which internal kinetic energy and momentum are conserved.

Truly elastic collisions tin but be achieved with subatomic particles, such as electrons striking nuclei. Macroscopic collisions tin can exist very virtually, but not quite, elastic—some kinetic energy is always converted into other forms of energy such as rut transfer due to friction and audio. One macroscopic collision that is nearly elastic is that of 2 steel blocks on ice. Another nearly elastic collision is that betwixt 2 carts with bound bumpers on an air track. Icy surfaces and air tracks are virtually frictionless, more readily allowing almost rubberband collisions on them.

Elastic Standoff

An rubberband collision is one that conserves internal kinetic free energy.

Internal Kinetic Energy

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

Figure 1. An elastic one-dimensional two-object collision. Momentum and internal kinetic energy are conserved.

Now, to solve problems involving one-dimensional rubberband collisions between 2 objects we can utilise the equations for conservation of momentum and conservation of internal kinetic energy. First, the equation for conservation of momentum for two objects in a i-dimensional collision is

p ₁ +p ₂ =p′_one +p′_two (F _net = 0)

m ₁ v ₁ +grand _two v ₂ =yard _i v′₁ +chiliad ₂ v′₂ (F _net = 0),

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

[latex]\frac{ane}{2}m_1{v_1}^2+\frac{1}{2}m_2{v_2}^2=\frac{ane}{2}m_1{v′_1}^2+\frac{1}{2}m_2{5′_2}^2\left(\text{ii-object elastic collision}\right)\\[/latex]

expresses the equation for conservation of internal kinetic energy in a ane-dimensional standoff.

Instance 1. Calculating Velocities Following an Elastic Collision

Calculate the velocities of ii objects following an rubberband standoff, given thatthou _one = 0.500 kg,one thousand _two = 3.50 kg,v ₁ = 4.00 m/due south, andfive ₂ = 0.

Strategy and Concept

First, visualize what the initial conditions mean—a small-scale object strikes a larger object that is initially at rest. This situation is slightly simpler than the situation shown in Figure 1 where both objects are initially moving. We are asked to detect 2 unknowns (the concluding velocities v′_ane and v′₂). To find two unknowns, we must use two independent equations. Considering this standoff is elastic, we can use the higher up two equations. Both can exist simplified by the fact that object two is initially at rest, and thus v ₂=0. Once nosotros simplify these equations, we combine them algebraically to solve for the unknowns.

Solution

For this problem, note that five _ii=0 and utilise conservation of momentum. Thus,

p _one =p′ ₁ +p′ ₂ org _ane v _i=m ₁ v′₁+m ₂ 5′_two.

Using conservation of internal kinetic energy and that five ₂=0,

[latex]\frac{1}{two}m_1{v_1}^2=\frac{ane}{2}m_1{5′_1}^2+\frac{1}{ii}m_2{v′_2}^ii\\[/latex]

Solving the starting time equation (momentum equation) for five′_ii, nosotros obtain

[latex]v′_2=\frac{m_1}{m_2}\left(v_1-v′_1\correct)\\[/latex].

Substituting this expression into the second equation (internal kinetic energy equation) eliminates the variable five′₂, leaving only v′₁ every bit an unknown (the algebra is left equally an exercise for the reader). There are two solutions to any quadratic equation; in this instance, they are

v′₁ = 4 . 00 thousand/southward andv′₁=−3.00 m/s.

Every bit noted when quadratic equations were encountered in earlier chapters, both solutions may or may not be meaningful. In this instance, the first solution is the same equally the initial condition. The offset solution thus represents the situation before the collision and is discarded. The second solution (v′_ane=−3.00 m/s) is negative, meaning that the commencement object bounces backward. When this negative value of five′₁ is used to find the velocity of the 2d object afterward the collision, we get

[latex]v′_2=\frac{m_1}{m_2}\left(v_1-v′_1\right)=\frac{0.500\text{ kg}}{three.50\text{ kg}}\left[4.00-\left(-3.00\correct)\right]\text{m/s}\\[/latex]

v′_ii=1.00 m/southward.

Discussion

The effect of this instance is intuitively reasonable. A small object strikes a larger one at rest and bounces backward. The larger one is knocked forward, simply with a low speed. (This is like a compact auto bouncing astern off a full-size SUV that is initially at rest.) As a check, attempt computing the internal kinetic energy before and afterward the collision. Yous volition encounter that the internal kinetic energy is unchanged at four.00 J. Likewise check the full momentum before and after the collision; you will discover it, besides, is unchanged.

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

Making Connections: Accept-Habitation Investigation—Ice Cubes and Elastic Collision

Observe a few ice cubes which are about the aforementioned size and a shine kitchen tabletop or a table with a glass top. Place the water ice cubes on the surface several centimeters abroad from each other. Picture one ice cube toward a stationary ice cube and notice the path and velocities of the ice cubes after the standoff. Try to avoid edge-on collisions and collisions with rotating ice cubes. Have you created approximately rubberband collisions? Explain the speeds and directions of the water ice cubes using momentum.

PhET Explorations: Collision Lab

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

Collision Lab screenshot.

Click to run the simulation.

Department Summary

An elastic standoff is one that conserves internal kinetic energy.
Conservation of kinetic energy and momentum together allow the last velocities to be calculated in terms of initial velocities and masses in i dimensional two-body collisions.

Conceptual Questions

What is an elastic collision?

Problems & Exercises

Two identical objects (such as billiard balls) have a one-dimensional collision in which i is initially motionless. Afterwards the collision, the moving object is stationary and the other moves with the same speed as the other originally had. Show that both momentum and kinetic free energy are conserved.
Professional Awarding.Ii manned satellites approach one another at a relative speed of 0.250 yard/south, intending to dock. The first has a mass of 4.00 × 10³ kg, and the 2d a mass of seven.50 × 10 kg. If the two satellites collide elastically rather than dock, what is their final relative velocity?
A 70.0-kg water ice hockey goalie, originally at rest, catches a 0.150-kg hockey puck slapped at him at a velocity of 35.0 m/due south. Suppose the goalie and the ice puck have an rubberband collision and the puck is reflected back in the direction from which it came. What would their concluding velocities exist in this case?

Glossary

elastic standoff: a collision that too conserves internal kinetic energy

internal kinetic energy: the sum of the kinetic energies of the objects in a system

Selected Solutions to Issues & Exercises

2. 0.250 m/southward

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Source: https://courses.lumenlearning.com/physics/chapter/8-4-elastic-collisions-in-one-dimension/