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Solution

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

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

or

m 1 v 1 = m 1 v 1 + m 2 v 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} } } {}

Using conservation of internal kinetic energy and that v 2 = 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 . 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} } } {}

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

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 )} {}

Substituting this expression into the second equation (internal kinetic energy equation) eliminates the variable v 2 size 12{ { {v}} sup { ' } rSub { size 8{2} } } {} , leaving only v 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 size 12{ { {v}} sup { ' } rSub { size 8{1} } =4 "." "00"`"m/s"} {}

and

v 1 = 3 . 00 m/s . size 12{ { {v}} sup { ' } rSub { size 8{1} } = - 3 "." "00"" m/s"} {}

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 ) 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 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 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"} {}

or

v 2 = 1 . 00 m/s . size 12{ { {v}} sup { ' } rSub { size 8{2} } =1 "." "00"`"m/s"} {}

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.

Collision Lab

Section summary

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

Conceptual questions

What is an elastic collision?

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Problems&Exercises

Two identical objects (such as billiard balls) have a one-dimensional collision in which one is initially motionless. After 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 energy are conserved.

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Professional Application

Two manned satellites approach one another at a relative speed of 0.250 m/s, intending to dock. The first has a mass of 4 . 00 × 10 3 kg size 12{4 "." "00" times "10" rSup { size 8{3} } " kg"} {} , and the second a mass of 7 . 50 × 10 3 kg size 12{7 "." "50" times "10" rSup { size 8{3} } " kg"} {} . If the two satellites collide elastically rather than dock, what is their final relative velocity?

0.250 m/s

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A 70.0-kg ice hockey goalie, originally at rest, catches a 0.150-kg hockey puck slapped at him at a velocity of 35.0 m/s. Suppose the goalie and the ice puck have an elastic collision and the puck is reflected back in the direction from which it came. What would their final velocities be in this case?

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Questions & Answers

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Practice Key Terms 2

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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