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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. [link] 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.

The system of interest contains a smaller mass m sub1 and a larger mass m sub2 moving on a frictionless surface. M sub 2 moves with velocity V sub 2 and momentum p sub 2 and m sub 1 moves behind m sub 2, with velocity V sub 1 and momentum p sub 1 toward the right direction. P 1 plus P 2 equals p total. The net force is zero. After collision m sub 1 moves toward the left with velocity V sub 1 while m sub 2 moves toward the right with velocity V sub 2 on the same frictionless surface. The momentum of m sub 1 becomes p 1 prime and m 2 becomes p 2 prime now. P 1 prime plus p 2 prime equals p total.
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 size 12{ left (F rSub { size 8{"net"} } =0 right )} {}

or

m 1 v 1 + m 2 v 2 = m 1 v 1 + m 2 v 2 F net = 0 , size 12{ left (F rSub { size 8{"net"} } =0 right )} {}

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,

1 2 m 1 v 1 2 + 1 2 m 2 v 2 2 = 1 2 m 1 v 1 2 + 1 2 m 2 v 2 2 (two-object elastic collision)

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

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 c m = m A v A + m B v B
( 3.0 ) v c m = ( 1 ) ( 12 ) + ( 2 ) ( 12 )
v c m = 4.0  m/s

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

( m A + m B ) v c m = ( m A + m B ) v f i n a l
( 3.0 ) v c m = ( 3 ) ( 4.0 )
v c m = 4.0  m/s

The total momentum of the system before the collision is:

m A v A + m B v B = ( 1 ) ( 12 ) + ( 2 ) ( 12 ) = 12  kg m/s

The total momentum of the system after the collision is:

( m A + m B ) v f i n a l = ( 3 ) ( 4 ) = 12  kg m/s

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.

Questions & Answers

Propose a force standard different from the example of a stretched spring discussed in the text. Your standard must be capable of producing the same force repeatedly.
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Practice Key Terms 2

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Source:  OpenStax, College physics for ap® courses. OpenStax CNX. Nov 04, 2016 Download for free at https://legacy.cnx.org/content/col11844/1.14
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