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1 2 mv 1 2 = 1 2 mv 1 2 + 1 2 mv 2 2 .

Because the masses are equal, m 1 = m 2 = m size 12{m rSub { size 8{1} } =m rSub { size 8{2} } =m} {} . Algebraic manipulation (left to the reader) of conservation of momentum in the x size 12{x} {} - and y size 12{y} {} -directions can show that

1 2 mv 1 2 = 1 2 mv 1 2 + 1 2 mv 2 2 + mv 1 v 2 cos θ 1 θ 2 .

(Remember that θ 2 size 12{θ rSub { size 8{2} } } {} is negative here.) The two preceding equations can both be true only if

m v 1 v 2 cos θ 1 θ 2 = 0 .

There are three ways that this term can be zero. They are

  • v 1 = 0 : head-on collision; incoming ball stops
  • v 2 = 0 : no collision; incoming ball continues unaffected
  • cos ( θ 1 θ 2 ) = 0 : angle of separation ( θ 1 θ 2 ) is 90º after the collision

All three of these ways are familiar occurrences in billiards and pool, although most of us try to avoid the second. If you play enough pool, you will notice that the angle between the balls is very close to 90º size 12{"90"°} {} after the collision, although it will vary from this value if a great deal of spin is placed on the ball. (Large spin carries in extra energy and a quantity called angular momentum , which must also be conserved.) The assumption that the scattering of billiard balls is elastic is reasonable based on the correctness of the three results it produces. This assumption also implies that, to a good approximation, momentum is conserved for the two-ball system in billiards and pool. The problems below explore these and other characteristics of two-dimensional collisions.

Connections to nuclear and particle physics

Two-dimensional collision experiments have revealed much of what we know about subatomic particles, as we shall see in Medical Applications of Nuclear Physics and Particle Physics . Ernest Rutherford, for example, discovered the nature of the atomic nucleus from such experiments.

Test prep for ap courses

Two cars of equal mass approach an intersection. Car A is moving east at a speed of 45 m/s. Car B is moving south at a speed of 35 m/s. They collide inelastically and stick together after the collision, moving as one object. Which of the following statements is true about the center-of-mass velocity of this system?

  1. The center-of-mass velocity will decrease after the collision as a result of lost energy (but not drop to zero).
  2. The center-of-mass velocity will remain the same after the collision since momentum is conserved.
  3. The center-of-mass velocity will drop to zero since the two objects stick together.
  4. The magnitude of the center-of-mass velocity will remain the same, but the direction of the velocity will change.

(b)

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Car A has a mass of 2000 kg and approaches an intersection with a velocity of 38 m/s directed to the east. Car B has a mass of 3500 kg and approaches the intersection with a velocity of 53 m/s directed 63° north of east. The two cars collide and stick together after the collision. Will the center-of-mass velocity change as a result of the collision? Explain why or why not. Calculate the center-of-mass velocity before and after the collision.

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Section summary

  • 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 (the x -axis), stated by m 1 v 1 = m 1 v 1 cos θ 1 + m 2 v 2 cos θ 2 and along the direction perpendicular to the initial direction (the y -axis) stated by 0 = m 1 v 1 y + m 2 v 2 y .
  • The internal kinetic before and after the collision of two objects that have equal masses is
    1 2 mv 1 2 = 1 2 mv 1 2 + 1 2 mv 2 2 + mv 1 v 2 cos θ 1 θ 2 .
  • Point masses are structureless particles that cannot spin.

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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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