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v 2 = m 1 m 2 v 1 sin θ 1 sin θ 2

Entering known values into this equation gives

v 2 = 0 . 250 kg 0 . 400 kg 1 . 50 m/s 0 . 7071 0 . 7485 . size 12{ { {v}} sup { ' } rSub { size 8{2} } = - left ( { {0 "." "250"" kg"} over {0 "." "400"" kg"} } right ) left (1 "." "50"" m/s" right ) left ( { {0 "." "7071"} over { - 0 "." "7485"} } right ) "." } {}

Thus,

v 2 = 0 . 886 m/s . size 12{ { {v}} sup { ' } rSub { size 8{2} } =0 "." "886"" m/s"} {}

Discussion

It is instructive to calculate the internal kinetic energy of this two-object system before and after the collision. (This calculation is left as an end-of-chapter problem.) If you do this calculation, you will find that the internal kinetic energy is less after the collision, and so the collision is inelastic. This type of result makes a physicist want to explore the system further.

A purple ball of mass m1 and velocity v one moves in the right direction into a dark room. It collides with an object of mass m two of value zero point four zero milligrams which was initially at rest and then leaves the dark room from the top right hand side making an angle of forty-five degrees with the horizontal and at velocity v one prime. The net external force on the system is zero. The momentum before and after collision remains the same. The velocity v two prime of the mass m two and the angle theta two it would make with the horizontal after collision not given.
A collision taking place in a dark room is explored in [link] . The incoming object m 1 size 12{m rSub { size 8{1} } } {} is scattered by an initially stationary object. Only the stationary object’s mass m 2 size 12{m rSub { size 8{2} } } {} is known. By measuring the angle and speed at which m 1 size 12{m rSub { size 8{1} } } {} emerges from the room, it is possible to calculate the magnitude and direction of the initially stationary object’s velocity after the collision.

Elastic collisions of two objects with equal mass

Some interesting situations arise when the two colliding objects have equal mass and the collision is elastic. This situation is nearly the case with colliding billiard balls, and precisely the case with some subatomic particle collisions. We can thus get a mental image of a collision of subatomic particles by thinking about billiards (or pool). (Refer to [link] for masses and angles.) First, an elastic collision conserves internal kinetic energy. Again, let us assume object 2 m 2 size 12{ left (m rSub { size 8{2} } right )} {} is initially at rest. Then, the internal kinetic energy 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 .

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.

Questions & Answers

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Source:  OpenStax, Introduction to applied math and physics. OpenStax CNX. Oct 04, 2012 Download for free at http://cnx.org/content/col11426/1.3
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