9.4 Two-Dimensional Collisions

1
9.4 Two-Dimensional Collisions
2
Two-Dimensional Collisions

• The momentum is conserved in all directions
• Use subscripts for
• identifying the object
• indicating initial or final values
• the velocity components
• If the collision is elastic, use conservation of
  kinetic energy as a second equation
• Remember, the simpler equation can only be used
  for one-dimensional situations

3
Two-Dimensional Collision, 2

• Qualitative Analysis
• Physical Principles The same as in One-Dimension
• 1. Conservation of VECTOR momentum
• P1x P2x P1x? P2x? P1y
  P2y P1y? P2y?
• 2. Conservation of Kinetic Energy
• ½m1v12 ½m2v22 ½m1v12 ½m2v22

4
Two-Dimensional Collision, 3

• For a collision of two particles in two
  dimensions implies that the momentum in each
  direction x and y is conserved
• The game of billiards is an example for such two
  dimensional collisions
• The equations for conservation of momentum are
• m1v1ix m2v2ix m1v1fx m2v2fx
• m1v1iy m2v2iy m1v1fy m2v2fy
• Subscripts represent
• (1,2) Objects
• (i,f) Initial and final values
• (x,y) Component direction

5
Two-Dimensional Collision, 4

• Particle 1 is moving at velocity v1i and particle
  2 is at rest
• In the x-direction, the initial momentum is m1v1i
• In the y-direction, the initial momentum is 0

6
Two-Dimensional Collision, final

• After the glancing collision, the conservation of
  momentum in the x-direction is
• m1v1i m1v1f cosq m2v2f cosf (9.24)
• After the collision, the conservation of momentum
  in the y-direction is
• 0 m1v1f sinq m2v2f sinf (9.25)

7
Active Figure 9.13
8
Example 9.8 Collision at an Intersection (Example
9.10 Text Book)

• Mass of the car mc 1500kg
• Mass of the van mv 2500kg
• Find vf if this is a perfectly inelastic
  collision (they stick together).
• Before collision
• The cars momentum is
• Spxi mcvc ?
• Spxi (1500)(25) 3.75x104 kgm/s
• The vans momentum is
• Spyi mvvv ?
• Spyi (2500)(20) 5.00x104 kgm/s

9
Example 9.8 Collision at an Intersection, 2

• After collision, both have the same x- and
  y-components
• Spxf (mc mv )vf cos?
• Spyf (mc mv )vf sin?
• Because the total momentum is both directions is
  conserved
• Spxf Spxi ?
• 3.75x104 kgm/s (mc mv )vf cos? (1)
• Spyf Spyi ?
• 5.00x104 kgm/s (mc mv )vf sin? (2)

10
Example 9.8 Collision at an Intersection, final

• Since (mc mv ) 400kg. ?
• 3.75x104 kgm/s 4000 vf cos? (1)
• 5.00x104 kgm/s 4000vf sin? (2)
• Dividing Eqn (2) by (1)
• 5.00/3.75 1.33 tan? ?
• ? 53.1
• Substituting ? in Eqn (2) or (1) ?
• 5.00x104 kgm/s 4000vf sin53.1 ?
• vf 5.00x104/(4000sin53.1 ) ?
• vf 15.6m/s

11
9.5 The Center of Mass

• There is a special point in a system or object,
  called the center of mass (CM), that moves as if
  all of the mass of the system is concentrated at
  that point
• The system will move as if an external force were
  applied to a single particle of mass M located at
  the CM
• M Smi is the total mass of the system
• The coordinates of the center of mass are
• (9.28) (9.29)

12
Center of Mass, position

• The center of mass can be located by its position
  vector, rCM
• (9.30)
• ri is the position of the i th particle, defined
  by

13
Active Figure 9.16
14
Center of Mass, Example

• Both masses are on the x-axis
• The center of mass (CM) is on the x-axis
• One dimension, x-axis
• xCM (m1x1 m2x2)/M
• M m1m2
• xCM (m1x1 m2x2)/(m1m2)

15
Center of Mass, final

• The center of mass is closer to the particle with
  the larger mass
• xCM (m1x1 m2x2)/(m1m2)
• If x1 0, x2 d m2 2m1
• xCM (0 2m1d)/(m12m1) ?
• xCM 2m1d/3m1 ?
• xCM 2d/3

16
Active Figure 9.17
17
Center of Mass, Extended Object

• Up to now, weve been mainly concerned with the
  motion of single (point) particles.
• To treat extended bodies, weve approximated the
  body as a point particle treated it as if it
  had all of its mass at a point!
• How is this possible?
• Real, extended bodies have complex motion,
  including translation, rotation, vibration!
• Think of the extended object as a system
  containing a large number of particles

18
Center of Mass, Extended Object, Coordinates

• The particle separation is very small, so the
  mass can be considered a continuous mass
  distribution
• The coordinates of theCM of the object are
• (9.31)
• (9.32)

19
Center of Mass, Extended Object, Position

• The position of CM can also be found by
• (9.33)
• The CM of any symmetrical object lies on an axis
  of symmetry and on any plane of symmetry
• An extended object can be considered a
  distribution of small mass elements, Dm
• The CM is located at position rCM

20
Example 9.9 Three Guys on a Raft

• A group of extended bodies, each with a known CM
  and
• equivalent mass m.
• Find the CM of the group.
• xCM (Smixi)/Smi
• xCM (mx1 mx2 mx3)/(mmm) ?
• xCM m(x1 x2 x3)/3m (x1 x2 x3)/3 ?
• xCM (1.00m 5.00m 6.00m)/3 4.00m

21
Example 9.10 Center of Mass of a Rod (Example
9.14 Text Book)

• Find the CM position of a rod of mass M and
  length L
• The location is on the x-axis
• (yCM zCM 0)
• (A). Assuming the road has a uniform mass per
  unit length
• ? M/L (Linear mass density)
• From Eqn 9.31

22
Example 9.10 Center of Mass of a Rod, 2

• But ? M/L ?
• (B). Assuming now that the linear mass density of
  the road is no uniform ? ?x
• The CM will be

23
Example 9.10 Center of Mass of a Rod, final

• But mass of the rod and ? are related by
• ? The CM will be

24
Material for the Final

• Examples to Read!!!
• Example 9.12 (page 269)
• Example 9.13 (page 272)
• Example 9.16 (page 276)
• Homework to be solved in Class!!!
• Problems 43